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4 Phys460.nb

2 Time-Independent Perturbation

2.1. Overview

2.1.1. General question Assuming that we have a Hamiltonian, H=H 0 +λH 1 (2.1)

where λ is a very small real number. The eigenstates of the Hamiltonian should not be very different from the eigenstates of H 0. If we already know all eigenstates of H0, can we get eigenstates of H 1 approximately? Bottom line: we are studying an approximate method.

2.1.2. Why perturbation theory? Why we need to study this approximation methods? (considering the fact that numerical methods can compute the eigenstates very efficiently and accurately for any Hamiltonian that we consider in this course) Reason number I: It is part of the history (QM was born before electronic computer becomes a powerful tool in scientific research). Reason number II: It reveals to us universal principles, which are very important and cannot be obtained from just numerical simulations Reason number III: The idea of perturbation theory has very deep and broad impact in many branches of . Perturbation is in many cases the only theoretical technique that we have to handle various complex systems (quantum and classical). Examples: in quantum theory (which is in fact a nonlinear generalization of QM), most of the efforts is to develop new ways to do perturbation theory (Loop expansions, 1/N expansions, 4-ϵ expansions).

2.1.3. Assumptions Assumption #1: we know all eigenstates of H 0, as well as their corresponding eigenenergies

0 0 0 0 H ψn =E n ψn  (2.2)

Assumption #2: we know the perturbation H '. What do we mean by knowing H '? Here, we mean that we can write down H ' using the 0 0 0 complete basis of ψn , i.e., we know the value of ψn H' ψm  for any m and n.

Assumption #3: we only consider quantum states with discrete eigenenergies In general, the spectrum of a quantum system (i.e. all eigenvalues of the Hamiltonian) falls into one of the following three general possibilities ◼ A discrete spectrum: eigenenergies can only take certain discredited values (example: infinite deep potential wells, e.g. harmonic potential

En = (n+1/2) ℏω) Phys460.nb 5

◼ A continuous spectrum: eigenenergies can take any (real) values in certain allowed range (example: a constant potential. Here, any E≥V is an eigenenergy) ◼ A mixed spectrum: some parts of the spectrum are continuous, while other parts has discrete eigenenergies. (example: a finite potential well. Here, we may have some discrete states inside the well. But for E above the top of the potential well, we have a continuous spectrum). Q: Consider the energy spectrum of an attractive Coulomb (1/r) potential. Is it discrete, continuous or mixed? A: It is mixed. When we consider the an attractive Coulomb potential, we mostly focus on the negative energy states (E<0). This part of the spectrum is discrete, as we all know very well from the study of a . But if we look at states with positive , there is a continuous spectrum for E> 0. For E> 0, the system is NOT a bound state, i.e. the proton and the electron doesn’t form an atom. In other words, we have a high probability found the proton and the electron to be separated far from each other. There, the attractive potential is very small and negligible, so we have two free particles and only need to consider their kinetic energies. For free particles, we know that any positive energy is an allowed eigenenergy (i.e. we have a continuum spectrum for E> 0). Bottom line: in this chapter, our perturbation theory only consider discrete spectrum or the discrete part of a mixed spectrum. Another version of assumption #3: we only consider confined states. (In QM, in most cases, confined states=discrete energy and unconfined states=continuous energy). Comment: In QM, we only study discrete states in a perturbation theory. But this is NOT true for other branches of physics. For example, in , perturbation theory is applied to continuous spectral.

2.2. Non-degenerate Perturbation Theory

2.2.1. Assumptions

0 0 0 1 Key assumption: we consider a specific state ψn . Here, we assume that En -E m is much larger than λH for any other 0 eigenstate ψm 

2.2.2. Preparation #1 wavefunctions

0 Since the eigenstates of H0 form a compete basis, we can write down any quantum state as a linear superposition of ψm 

0 ψ〉= am ψm  m (2.3)  Now, if we consider an eigenstate of H, ψn, it can also be written in a similar fashion

 0 ψn= am ψm  m (2.4)

 0 As discussion above, if λ is small, an eigenstate of H would be similar to an eigenstate of H0. Here, we assume that ψn is very close to ψn . 0 This means that an ≈ 1 and for other values of m≠n, a m ~0, To highlight this, we separate the term for ψn  out from the sum

 0 0 ψn=a n ψn + am ψm  m≠n (2.5)

It turns out that it is usually more convenient to use unnormalized eigenstates. Now, let us define unnormalized eigenstates of H

1 am  0 0 ψn〉= ψn= ψn + ψm  m≠n (2.6) an an

For simplicity, we will now call am /a n =c m

0 0 ψn〉= ψn + cn ψm  m≠n (2.7)

Because am ~ 0 and an ~ 1, we know that cm ~0 for small λ. Comment #1: This state is NOT normalized

2 〈ψn ψn〉=1+ cm ≥1 m≠n (2.8) 6 Phys460.nb

But we can easily normalized it, if we want to  1 ψ = ψn n (2.9) 2 1+ ∑m≠n cm

0 Comment #2: (almost) any quantum states can be written in the form of Eq. (2.3). This is because ψm  forms a complete basis. Q: what does the word “almost” mean here?

0 A: If a state is orthogonal to ψn , we cannot write the state the form of Eq. (2.3). But we don’t need to worry about it here, because we are 0 doing perturbation theory and we know that the eigenstates of H is close to eigenstates of H 0. So it can not be orthogonal to ψn . Bottom line: we are not making any assumptions or approximations here. It is just a new way to write down eigenstates of H.

Comment #3: cms are functions of λ, i.e. cm(λ). For small λ, we can use the Taylor series:

(1) (2) 2 (3) 3 cm =c m λ+c m λ +c m λ + … (2.10)

0 Here, the Taylor series doesn’t contain the 0th order term of λ (i.e. the constant term). This is because when λ= 0, ψn〉= ψn , and thus

cm(λ) = 0 at λ= 0. As a result,

0 0 0 ∞ (k) k 0 0 ∞ k (k) 0 ψn〉= ψn + cm(λ) ψm = ψn + cm λ ψm = ψn + λ  cm ψm  m≠n m≠n k=1 k=1 m≠n (2.11)

If we define

k (k) 0 ψn = cm ψm  m≠n (2.12)

we get

0 1 2 2 ψn〉= ψn +λ ψn +λ ψn +… (2.13)

This is Eq. [6.5] in the textbook.

1 2 0 0 Important: ψn , ψn  … doesn’t contain ψn . In other words, all corrections are orthogonal to ψn .

2.2.3. Preparation #2 eigenenergies Eigenenergies of H are also functions of λ, and for small λ, we can use the Taylor series:

0 1 2 2 En(λ) =E n +λE n +λ En + … (2.14)

This is Eq. [6.6] in the textbook.

2.2.4. Schrodinger Equation in the perturbation theory

H ψn〉=E n ψn〉 (2.15)

0 1 2 2 0 1 2 2 0 1 2 2 (H0 +λH') ψn +λ ψn +λ ψn +…=E n +λE n +λ En + …  ψn +λ ψn +λ ψn +… (2.16)

0 1 0 2 2 1 0 0 0 1 1 0 2 0 2 1 H0 ψn + λH 0 ψn +H' ψn +λ H0 ψn +H' ψn +…=E n ψn + λE n ψn +E n ψn +λ En ψn +E n (2.17) 1 2 0 ψn +E n ψn +…

1 2 In the perturbation theory, we need to compute two sets of quantities (1) energy corrections at each order En , En ,... and (2) wavefunc- 1 2 3 tion corrections at each order, ψn , ψn , ψn . It turns out that these two set of quantities are entangled together and we need to compute both of them. At each order, we will first compute energy corrections, and then the wavefunction corrections.

2.2.5. Zeroth order The leading order terms in the equation is λ0 = constant

0 0 0 H0 ψn =E n ψn  (2.18) Phys460.nb 7

This is identical to the case of λ= 0, i.e. the unperturbed system.

2.2.6. First order To the order of λ, we have

1 0 0 1 1 0 H0 ψn +H' ψn =E n ψn +E n ψn  (2.19)

1 0 Here, we first compute the energy correction En . This is done by multiplying on both sides ψn

0 1 0 0 0 0 1 0 1 0 ψn H0 ψn + ψ n H' ψn = ψ n En ψn + ψ n En ψn  (2.20)

For the first term on the l.h.s., we use the fact that

0 0 0 ψn H0 = ψn En (2.21)

1 0 1 0 1 0 0 1 For the last term on the r.h.s., we use the fact that En is a number (not a quantum operator), and thus ψn En ψn =E n ψn ψn =E n

0 0 1 0 0 0 0 1 1 ψn En ψn + ψ n H' ψn = ψ n En ψn +E n (2.22)

0 0 1 ψn H' ψn =E n (2.23)

The first order correction in energy is the expectation value of H '.

0 0 0 2 En =E n + λ ψn H' ψn +Oλ = (2.24) 0 0 0 0 2 0 0 2 0 0 2 ψn H0 ψn + ψ n λH' ψn +Oλ = ψ n H0 +λH' ψn +Oλ = ψ n H ψn +Oλ 

Bottom line: to the first order (or say up to corrections at the order of λ2), we can use the old wavefunction (the zeroth order wavefunction).

1 0 Then we compute the first order correction for the wavefunction ψn . To do that, we multiply both sides of the equation with ψm where m≠n

0 1 0 0 0 0 1 0 1 0 ψm H0 ψn + ψ m H' ψn = ψ m En ψn + ψ m En ψn  (2.25)

For the first term on the l.h.s., we use the fact that

0 0 0 ψm H0 = ψm Em (2.26)

0 1 0 0 1 0 0 1 For the two terms on the r.h.s., we use the fact that En and En are both numbers (not quantum operators), so ψm En ψn =E n ψm ψn  0 1 0 1 0 0 and ψm En ψn =E n ψm ψn = 0. Here, we used the fact that when m≠n, the two quantum states are orthogonal and thus 0 0 ψm ψn =0.

0 0 1 0 0 0 0 1 Em ψm ψn + ψ m H' ψn =E n ψm ψn  (2.27)

So 0 0 ψm H' ψn  0 1 ψm ψn = (2.28) 0 0 En -E m

1 According to the definition of ψn 

1 (1) 0 ψn = cm ψm  m≠n (2.29) we have 0 0 ψm H' ψn  (1) 0 1 cm = ψm ψn = (2.30) 0 0 En -E m

And therefore, 8 Phys460.nb

0 0 ψm H' ψn  1 0 ψn =  ψm  (2.31) m≠n 0 0 En -E m

So 0 0 ψm H' ψn  0 0 ψn〉= ψn +λ  ψm  + … (2.32) m≠n 0 0 En -E m

2.2.7. Second order

2 1 0 2 1 1 2 0 H0 ψn +H' ψn =E n ψn +E n ψn +E n ψn  (2.33)

2 0 Here, we first compute the energy correction En . This is done by multiplying on both sides ψn

0 2 0 1 0 0 2 0 1 1 0 2 0 ψn H0 ψn + ψ n H' ψn = ψ n En ψn + ψ n En ψn + ψ n En ψn  (2.34)

0 0 2 0 1 0 0 2 1 0 1 2 ψn En ψn + ψ n H' ψn = ψ n En ψn +E n ψn ψn +E n (2.35)

0 1 The second term on the r.h.s. is zero, because we required ψn ψn = 0 at the beginning.

2 0 1 En = ψn H' ψn  (2.36)

Bottom line: to compute the second order perturbation, we need to know wavefunction at the first order. This conclusion is in fact generically true. We need wavefunction at lower order to compute energy correction at one order higher. 0 0 0 0 0 0 ψm H' ψn  ψn H' ψm  ψm H' ψn  2 0 1 0 0 En = ψn H' ψn =  ψn H' ψm  =  (2.37) m≠n 0 0 m≠n 0 0 En -E m En -E m

0 0 0 0 ψn H' ψm  ψm H' ψn  0 0 0 2 3 En =E n + λ ψn H' ψn +λ  +Oλ  (2.38) m≠n 0 0 En -E m

2 0 The we compute the second order correction for the wavefunction ψn . To do that, we multiply both sides of the equation with ψm where m≠n

0 2 0 1 0 0 2 0 1 1 0 2 0 ψm H0 ψn + ψ m H' ψn = ψ m En ψn + ψ m En ψn + ψ m En ψn  (2.39)

0 0 2 0 1 0 0 2 1 0 1 2 0 0 Em ψm ψn + ψ m H' ψn =E n ψm ψn +E n ψm ψn +E n ψm ψn  (2.40)

0 0 0 2 0 1 1 0 1 En -E m  ψm ψn = ψ m H' ψn -E n ψm ψn  (2.41)

0 1 1 ψm H' ψn  En 0 2 0 1 ψm ψn = - ψm ψn = 0 0 0 0 En -E m En -E m 0 0 0 0 0 0 0 0 ψm H' ψm'  ψm' H' ψn  ψn H' ψn  ψm' H' ψn  0 0  -  ψm ψm'  = m'≠n 0 0 0 0 m'≠n 0 0 0 0 En -E m En -E m' En -E m En -E m' 0 0 0 0 0 0 0 0 (2.42) ψm H' ψm'  ψm' H' ψn  ψn H' ψn  ψm' H' ψn   -  δm,m' = m'≠n 0 0 0 0 m'≠n 0 0 0 0 En -E m En -E m' En -E m En -E m' 0 0 0 0 0 0 0 0 ψm H' ψm'  ψm' H' ψn  ψn H' ψn  ψm H' ψn  - m'≠n m'≠n 0 0 0 0 0 0 2 En -E m En -E m' En -E m 

2 According to the definition of ψn 

2 (2) 0 ψn = cm ψm  m≠n (2.43)

We have

(2) 0 2 cm = ψm ψn  (2.44) Phys460.nb 9

and thus

0 0 0 0 0 0 0 0 ψm H' ψm'  ψm' H' ψn  ψn H' ψn  ψm H' ψn  2 0 ψn = - ψm  m≠n m'≠n (2.45) 0 0 0 0 0 0 2 En -E m En -E m' En -E m 

2.2.8. Third order Same as the second order, we can use the same method to show that

3 0 2 En = ψn H' ψn  (2.46)

So,

0 0 0 0 0 0 0 0 0 0 0 0 ψn H' ψm  ψm H' ψm'  ψm' H' ψn  ψn H' ψn  ψm H' ψn  ψn H' ψm  3 En = - m≠n m'≠n (2.47) 0 0 0 0 0 0 2 En -E m  En -E m'  En -E m 

And one can keep doing this for higher and higher order

2.2.9. Summary For a Hamiltonian H=H 0 +λH 1 (2.48)

0 0 0 0 assuming that we know all the eigenstates of H  ψn ), and we know the expectation values ψm1 H' ψm2  for any two eigenstates of H0,

0 0 ψm1  and ψm2 ), then we can write down eigenstates of H as a power series expansions of λ

0 0 0 0 ψn H' ψm  ψm H' ψn  0 1 2 2 0 0 0 2 En =E n +λE n +λ En + … =E n + λ ψn H' ψn +λ  + … (2.49) m≠n 0 0 En -E m and 0 0 ψm H' ψn  0 1 2 2 0 0 ψn〉= ψn +λ ψn +λ ψn +…= ψn +λ  ψm  + … (2.50) m≠n 0 0 En -E m

2.2.10. Second order perturbation always reduces the energy of the One key conclusion from the perturbation theory is that the second order correction always makes the energy of the ground state lower (in 2 comparison to the unperturbed one). This can be seen by looking at E0 0 0 0 0 0 0 2 ψn H' ψm  ψm H' ψn  ψn H' ψm  2 En =  =  (2.51) m≠n 0 0 m≠n 0 0 En -E m En -E m

0 0 0 0 0 0 2 In the numerator, ψn H' ψm  is the complex conjugate of ψm H' ψn , so it is ψn H' ψm  , which is non-negative

0 0 2 ψn H' ψm  ≥0 (2.52)

0 0 The denominator En -E m < 0, if n is the ground state for the unperturbed Hamiltonian (if it is the ground state, then its eigenenergy must be smaller than eigenenergy of any other states). And therefore 0 0 2 ψn H' ψm  ≤0 (2.53) 0 0 En -E m

So 0 0 2 ψn H' ψm  2 En =  ≤0 (2.54) m≠n 0 0 En -E m

0 0 The equal sign only arise when ψn H' ψm = 0 for ALL m≠n. (if this is the case, we don’t need to do perturbation theory. The first order 2 perturbation become exact). As long as we ignore this very special case, we find that En < 0 for the ground state, regardless of details. 10 Phys460.nb

This conclusion is very important in , because in many systems, the first order perturbation of the ground state happens to 1 be zero. En = 0. There,

0 2 2 En =E n +λ En + … (2.55)

The energy correction is dominated by the second order term, which must be negative for the ground state. Without any calculation, we know immediately that

0 En

In the first homework, we will see that this relation implies that the speed of light in a (linear) medium can only be slower than the vacuum. 0 (i.e., if En >E n , we will violate the special relativity).

2.3. Brillouin-Wigner Perturbation Theory

2.3.1. Negative sides of Rayleigh–Schrödinger perturbation theory The perturbation theory discussed above is known as Rayleigh–Schrödinger perturbation theory. It is presented for most of the textbooks. However, this approach has some limitations and is not sufficient enough for some cases. 1. Too complicated to go to higher order (e.g. third order or fourth order correction) 2. The physical meaning is less clear (Why do we need to sum over all other quantum state? How should we think about the sum.) 3. One needs to compute energy and wavefunctions at the same time (if we only want to know the eigenenergy, can we compute only energy without bothering to do wavefunction?) One way to resolve these problems: Brillouin-Wigner Perturbation Theory

2.3.2. Brillouin-Wigner Perturbation Theory Brillouin-Wigner Perturbation Theory considers the same setup and the final conclusions are exactly the same. However it has a couple of advantages 1. It offers a nice and simple physical interpretation (a baby version of Feynman diagrams used in quantum field theory) 2. It is easier to compute higher order corrections (If we want to compute the eigenenergy using a computer, this perturbation theory just needs one very simple iteration) 3. One can compute energy along, without worry about wavefunctions. Let’s start from the same setup

(H0 +λH') ψn〉=E n ψn〉 (2.57)

where ψn〉 represents the same unnormalized eigenstate of H

0 0 ψn〉= ψn + cn ψm  m≠n (2.58)

We can rewrite the equation above as

(E-H 0) ψn〉=λH' ψn〉 (2.59)

and

-1 ψn〉=λ(E-H 0) H' ψn (2.60)

-1 Note: here (E-H 0) is the matrix inverse, instead of a number inverse, because H0 is an operator, instead of a number.  Q: What is a function of operator? e.g, f(Q )? A: First, we write down the same function as a number function and do a power-law expansion

2 f(x) =a 0 +a 1 x+a 2 x + … (2.61) Phys460.nb 11

  then, f(Q ) represents the same power series, but with number x substitute by operator Q    2 f(Q ) =a 0 +a 1 Q +a 2 Q + … (2.62) 2   where Q =Q Q, etc.

-1 Here, the inverse function of operator (E-H 0) shall be understood the same way Because 1 1 x x2 = + + + … (2.63) E-x E E2 E3 we know that   2  -1 1 H0 H0 E-H 0 = + + + … (2.64) E E2 E3

Now, back to the derivation above: λ 0 0 -1 0 ψm ψn= ψ m λ(E-H 0) H' ψn= ψm H' ψn 0 (2.65) E-E m

Remember that from the definition, of ψn〉

0 0 ψn〉= ψn + cm ψm  m≠n (2.66) we have

0 ψm ψn=c m (2.67) for any m≠n. And thus, we get 1 0 0 0 ψn〉= ψn +λ  ψm  ψm H' ψn m≠n 0 (2.68) E-E m  If we define a quantum operator X as 1  0 0 R =  ψm  ψm m≠n 0 (2.69) E-E m we get

0   ψn〉= ψn +λR H ' ψn (2.70)

Thus,

   0 I -λR H ' ψn= ψn  (2.71)  where I is the identity operator So,

   -1 0 ψn〉=I -λR H ' ψn  (2.72)    Again, we emphasize that here, I -λR H '-1 represent matrix inverse. For matrix inverse, we can use Taylor expansions to write it out. We know that (1-a) -1 =1+a+a 2 +a 3 + … (2.73)

So similarly, we have   -1             I-λR H ' =I+λR H '+λ 2 R H 'R H '+λ 3 R H 'R H 'R H '+… (2.74)

So we have

  -1 0 0   0 2     0 3       0 ψn〉=I-λR H ' ψn = ψn +λR H ' ψn +λ R H 'R H ' ψn +λ R H 'R H 'R H ' ψn + (2.75) 12 Phys460.nb

So we find that

k   k 0 ψn =R H ' ψn  (2.76)

Previous, we found that

1 0 0 En = ψn H' ψn  (2.77)

2 0 1 En = ψn H' ψn  (2.78)

3 0 2 En = ψn H' ψn  (2.79)

In fact, we can use the same procedure to show that for kth order,

k 0 k-1 En = ψn H' ψn  (2.80)

k-1   k-1 0 Because we have found that ψn =R H' ψn 

k 0 k-1 0   k-1 0 En = ψn H' ψn = ψ n H'R H ' ψn  (2.81)

So, we have

1 0 0 En = ψn H' ψn  (2.82) 1 2 0 0 0 0 0 0 En = ψn H'RH' ψn =  ψn H' ψm  ψm H' ψn  m≠n 0 (2.83) En -E m 1 1 3 0 0 0 0 0 0 0 0 En = ψn H'RH'RH' ψn =   ψn H' ψm  ψm H' ψm'  ψm' H' ψn  m≠n m'≠n 0 0 (2.84) En -E m En -E m'

... (2.85)

From these formula we see a pattern.

k 0 1. For any En , if we look at the formula from right to left, one always start from unperturbed state ψn  and eventually goes back to the 0 same state ψn .

0 0 0 0 2. In the path from ψn  to ψ n , we go through several intermediate states ψm , ψm' …. For kth order perturbation, we have k- 1 intermediate states.

3 3. To turn from a state to another along the path (e.g. from n to m’ or from m’ to m in En ), we use the perturbation H' 4. For each intermediate state,we have an denominator 1 En-Em0

2.3.3. Diagrammatic representation

k We can represent the En using diagrams. 1. For each intermediate state, we represent 1 as a solid line with integer m labeling the state. En-Em0

0 0 0 0 2. For each ψm H' ψm' , we represent it as a dot. And we use Vm m' to represent ψm H' ψm' 

k 3. Connect everything together in the same order as in En

0 4. At the two ends of the line, we use two short line to present that we start from and end at the same state ψn  First order: Phys460.nb 13

Second order:

Third order

By making the line longer, we can write down easily perturbation terms to any order. Relations to QFT: In QFT, we use very similar diagrams, known as the Feynman diagrams. There, solid lines are propagator of a particle 1 where ω is ω-ϵ0 frequency, pretty much the same as energy En and ϵ0 is the unperturbed energy of the particle (energy ignore interactions between particles). In faction, the diagrams we show here are baby versions of the diagrams of Feynman. Physics meaning discussed in class: (example: two electrons exchange to get E&M interactions).

2.3.4. How to compute the energy using Brillouin-Wigner Perturbation Theory? First, let’s define some abbreviation to make the formula shorter,

0 0 Vij = ψi H' ψ j  (2.86) and thus

Vn m Vm n Vn m V V Vn m V V V 0 2 3 m m' m'n 4 m m' m'm'' m''n En =E n +λV nn +λ +λ +λ + ... 0 0 0 0 0 0 (2.87) En -E m En -E m  En -E m'  En -E m  En -E m'  En -E m'' 

Here all the ms are summed over but they cannot be the same as n. It may looks like that we can find En using this formula, but it is not quite the case yet. This is because on the r.h.s., the denominator contains also En, i.e. IT is a equation for En and En arises on both sides. This equation can be solved easily using iterative method (e.g. using a computer code). One start from zeroth order, and then go to first, second, third order …, every time we need En in the kth order calculation, we just use the (k-1)th order En on the r.h.s.. Here is how it is done First run

(1) 0 En =E n +λV nn (2.88)

Second run

Vn m Vm n (2) 0 2 En =E n +λV nn +λ (1) 0 (2.89) En -E m

Third run

Vn m Vm n Vn m V V (3) 0 2 3 m m' m'n En =E n +λV nn +λ +λ (2) 0 (2) 0 (2) 0 (2.90) En -E m En -E m  En -E m' 

Fourth run 14 Phys460.nb

Vn m Vm n Vn m V V Vn m V V V (4) 0 2 3 m m' m'n 4 m m' m'm'' m''n En =E n +λV nn +λ +λ +λ (3) 0 (3) 0 (3) 0 (3) 0 (3) 0 (3) 0 (2.91) En -E m En -E m  En -E m'  En -E m  En -E m'  En -E m'' 

... Another option is using analytic methods, as will be discussed below.

2.3.5. Preparation Consider the following function

f(x) =x 2 g(x) =x 21+ax+bx 2 +cx 3 + ... (2.92)

If we want to keep f(x) to O(x n), we only need to keep g(x) to Oxn-2. Similarly, for the following function x2 x2 f(x) = = (2.93) g(x) 1+ax+bx 2 +cx 3 + ...

If we want to keep f(x) to O(x n), we only need to keep g(x) to Oxn-2. This will be something that useful for us latter

2.3.6. iterative method

Vn m Vm n Vn m V V Vn m V V V 0 2 3 m m' m'n 4 m m' m'm'' m''n En =E n +λV nn +λ +λ +λ + ... 0 0 0 0 0 0 (2.94) En -E m En -E m  En -E m'  En -E m  En -E m'  En -E m'' 

Zeroth order (no En on the l.h.s., so job done)

0 En =E n +O(λ) (2.95)

First order (no En on the l.h.s., so job done)

0 2 En =E n +λV nn +Oλ  (2.96)

2 Second order, we use En obtained at zeroth order for the λ term

Vn m Vm n Vn m Vm n 0 2 3 0 2 3 En =E n +λV nn +λ +Oλ =E n +λV nn +λ +Oλ  0 0 0 (2.97) En -E m En -E m

This is because the third term on the r.h.s. already has a λ2 prefactor. Thus to keep to Oλ2, we only need to keep the denominator to Oλ0.

Third order,

Vn m Vm n Vn m V V 0 2 3 m m' m'n 4 En =E n +λV nn +λ +λ +Oλ = 0 0 0 En -E m En -E m  En -E m'  (2.98) Vn m Vm n Vn m V V 0 2 3 m m' m'n 4 En +λV nn +λ +λ +Oλ  0 0 0 0 0 0 En +λV nn-E m En -E m  En -E m' 

2 3 In the λ term, we now need to keep to O(λ). In the λ term, we just need to keep En to the zeroth order. For the λ2 term, we can expand it for small λ

Vn m Vm n Vn m Vm n Vn m Vm n Vnn λ2 =λ 2 -λ 3 + … 0 0 0 0 0 0 0 0 (2.99) En +λV nn-E m En -E m En -E m En -E m

So

Vn m Vm n Vn m V V Vn m Vm n Vn n 0 2 3 m m' m'n 4 En =E n +λV nn +λ +λ  - +Oλ  0 0 0 0 0 0 0 0 2 (2.100) En -E m En -E m  En -E m'  En -E m 

Fourth order, Phys460.nb 15

Vn m Vm n 0 2 En =E n +λV nn +λ + 0 2 Vn m' Vm'n 0 En +λV nn +λ 0 0 -E m En -Em' (2.101) Vn m Vm m' Vm'n Vn m Vm m' Vm'm'' Vm''n λ3 +λ 4 +Oλ 5 0 0 0 0 0 0 0 0 0 0 En +λV nn -E m  En +λV nn -E m'  En -E m  En -E m'  En -E m'' 

Vn m Vm n Vn m V V Vn m Vm n Vn n 0 2 3 m m' m'n En =E n +λV nn +λ +λ  - + 0 0 0 0 0 0 0 0 2 En -E m En -E m  En -E m'  En -E m 

Vn m V V V Vn m Vm n V V Vn m Vm n Vn m V V Vn n 4 m m' m'm'' m''n n m' m'n 2 m m' m'n λ  - + Vnn -2 + (2.102) 0 0 0 0 0 0 0 0 2 0 0 0 0 3 0 0 2 0 0 En -E m  En -E m'  En -E m''  En -E m  En -E m' En -E m  En -E m  En -E m'  Oλ5

2.4. Degenerate Perturbation Theory

0 In the previous section, we studied the effect of a small perturbation λH ' on an eigenstate of H 0, ψn . The key assumption there is that 0 before we turn on the perturbation (i.e. at λ= 0), the eigenenergies of all other eigenstates of H 0 are very far away from En

0 0 0 0 En -E m |> > ψi λH' ψ j  (2.103)

This section, we will consider the opposite situation, where there is at least one other eigenstate of H0 which has the same eigenenergy 0 as ψn . Two states having the same eigenenergy is known as “degeneracy”. So this perturbation theory is known as the degenerate perturbation theory.

2.4.1. Why non-degenerate perturbation theory fails in the presence of degeneracy? In the presence of degeneracy, the perturbation theory that we learned before will fail. To see this, we just need to look at the second order perturbation of the eigenenergy 0 0 0 0 ψn H' ψm  ψm H' ψn  0 1 2 2 0 0 0 2 En =E n +λE n +λ En + … =E n + λ ψn H' ψn +λ  + … (2.104) m≠n 0 0 En -E m

Here, we focus on the second order correction: 0 0 0 0 ψn H' ψm  ψm H' ψn  λ2  (2.105) m≠n 0 0 En -E m

0 If H0 has another eigenstate ψn'  with the same eigenenergy, at least one term in this sum will have zero in the denominator and thus will 0 0 1 diverge, i.e., when En =E n' , 0 0 → ∞, and thus the theory becomes ill-defined. En -En'

1 NOTE: the same divergence will arise also in higher order corrections. But there is no divergence in the first order correction En . In power-law expansions, infinite coefficient doesn’t always mean singularity. It means that we missed something in the lower order correction. Here is a simple example: Let’s consider a function f(x), which can be written as the following Taylor expansion at small x

2 f(x) =a 0 +a 1 x+a 2 x + … (2.106)

Now, assume that I made a mistake in the Taylor expansion for the coefficient a1. Instead of the correction value, a1, I used a wrong coefficient for the linear term, say b1.

2 f(x) =a 0 +b 1 x+(a 1 -b 1)x+a 2 x + … (2.107)

In other words, here I missed part of the linear term, (a1 -b 1)x. And thus coefficients of the higher order terms will also need to be adjusted to absorb this mistake. Let’s try to use the x2 term to correct this error, i.e. a -b 1 1 2 f(x) =a 0 +b 1 x+ +a 2 x + … (2.108) x 16 Phys460.nb

Let me define b =a + a1-b1 2 2 x

2 f(x) =a 0 +b 1 x+b 2 x + … (2.109)

Now, once again, I wrote my function as a power-law expansion. Because I used a wrong coefficient for the linear term, b1, my second order

term needs to use this new coefficient. This new coefficient b2 is infinite at small x. This is transparent if we notice that when x→0

a1 -b 1 b2 =a 2 + → ∞ (2.110) x

Bottom line: infinite coefficient in the second order term (and higher order term) means that the first order result is incorrect and needs to be revised.

2.4.2. What to do? Here, let’s first take another look at the second order correction 0 0 0 0 ψn H' ψm  ψm H' ψn  2 En =  (2.111) m≠n 0 0 En -E m

As we know, the problem arises because E 0 =E 0 for certain m, and thus we get 1 =∞. To avoid this singularity, the only thing that we need n m 0 0 0 to do is to request that the numerator also vanish whenever the denominator is zero. i.e., if En =E m , we must make sure that 0 0 ψm H' ψn = 0.

0 0 0 0 * NOTE: the two factors in the numerator are complex conjugate to each other: ψn H' ψm = ψ m H' ψn  , and thus if one of them is zero, the other is also zero. Bottom line: For degenerate states, before we start the procedure described in the non-degenerate perturbation theory, we need to first 0 0 make sure that for any degenerate states, ψm H' ψn =0

2.4.3. Whenever there is an degeneracy, we have an option to choose the basis A good example, a free particle. Consider a free particle with mass m. p2 ℏ2 ⅆ2 H0 = = - (2.112) 2m 2m ⅆx 2

The eigenstates of H0 arises in pairs (i.e. there is a degeneracy for any excited states). The static Schrodinger equation here is ℏ2 ⅆ2 - ψ(x) =Eψ(x) (2.113) 2m ⅆx 2

It is a second order and we know the solution are just plane waves ψ=Aⅇ ⅈkx +Bⅇ -ⅈkx (2.114)

The eigenenergy for this state is E=p 2 2m=(ℏk) 2 2m, i.e. the kinetic energy. Here, A and B are two arbitrary coefficients.

For each fixed k, we have one eigenenergy E=(ℏk) 2 2m, but infinite number of eigenstates ψ=Aⅇ ⅈkx +Bⅇ -ⅈkx , i.e. a degeneracy. This example is known as two-fold degeneracy, or we say that two states have the same energy. The reason we say “two states” here is because not all the eigenstates are linear independent. In fact, we just need two states, ⅇⅈkx and ⅇ-ⅈkx , all other eigenstates can be written as linear superposi-

tion of these two. Bottom line: two-fold degeneracy means that any linear combination of these two states is an eigenstate of H0 with the same eigenenergy. Now, let’s look at the same second order differential equation again. ℏ2 ⅆ2 - ψ(x) =Eψ(x) (2.115) 2m ⅆx 2

we know that we can also write the solution for this equation as ψ=C coskx+D sinky (2.116) Phys460.nb 17

i.e., instead of using exponentials, we can use sin or cos functions to represent plane waves. Here, once again we find infinite number of eigenstates with the same eigenenergy, and once again, they are not all independent. We just need two states coskx or sinkx. And all other eigenstates are just linear superpositions of they two. So, again, we reach the same conclusion, the system have a two-fold degeneracy. But early on, we said that the two states are ⅇⅈkx and ⅇ-ⅈkx , but now for the two degenerate states, we use coskx or sinkx. These different choices are just different basis to represent all the eigenstates. We can choose to use ⅇ ⅈkx and ⅇ-ⅈkx or coskx or sinkx. There is no difference between them. In fact, we can choose any two linear independent states

ⅈkx -ⅈkx ψ1 =A 1 ⅇ +B 2 ⅇ (2.117)

ⅈkx -ⅈkx ψ2 =A 2 ⅇ +B 2 ⅇ (2.118)

And then, we can say that we have two degenerate states ψ1(x) and ψ 2(x). and then, we can represent any other eigenstates (with the same eigenenergy) as

ψ(x) =Xψ 1(x) +Yψ 2(x) (2.119)

Q: Why do we usually use ⅇⅈkx and ⅇ-ⅈkx or coskx or sinkx? Why not use ⅇ ⅈkx and cos k x. A: There is no problem (mathematically) if we choose to use ⅇⅈkx and cos k x. However, for convenience, it is usually better using orthonormal bases

ⅈk 1 x * ⅈk 2 x  ⅆx ⅇ  ⅇ =2πδ(k 1 -k 2) (2.120)

*  ⅆx(coskx) sinkx=0 (2.121)

Q: Why do we use ⅇⅈkx and ⅇ-ⅈkx more often than coskx or sinkx in quantum mechanics?

ⅈkx -ⅈkx A: For H0, there is little difference between the two choices. However, if we consider other quantum operators, like momentum, ⅇ and ⅇ is a better choice. This is because ⅇⅈkx and ⅇ-ⅈkx are eigenstates of the momentum operator too! So they have not only well-defined energy, but also well defined momenta (ℏ k and -ℏ k respectively). coskx or sinkx don’t have well defined momentum. They have 50% chance having momentum ℏk and another 50% chance having momentum -ℏk. Bottom line: when you cannot decide which choice of basis is better, look at another quantum operator.

These conclusions are true generically. Let’s start from two-fold degenerate. Assuming that for H0, there are two degenerate eigenstates:

0 0 0 H0 ψa =E ψa  (2.122) and

0 0 0 H0 ψb =E ψb  (2.123)

We assume that these two states are orthogonal to each other (otherwise, we make them orthogonal, using Gram-Schmidt procedure). We 0 0 0 0 assume that ψa H' ψb  ≠ 0. Here we first prove a fact: if ψa  and ψb  are both eigenstates of H 0 and they have the same eigen- 0 0 0 value E , then any linear superposition of ψa  and ψb  is also an eigenstate of H0 with the same eigenenergy. Let's define 0 0 0 ψ =α ψa +β ψb 

0 0 0 0 0 0 0 0 0 H0 ψ =H 0α ψa +β ψb =αH 0 ψa +βH 0 ψb =αE 0 ψa +βE 0 ψb =E 0αH 0 ψa +βH 0 ψb = (2.124) 0 E0 ψ 

0 0 Bottom line: if H0 has two degenerate eigenstates ψa  and ψb , we have infinite eigenstates with the same 0 0 0 eigenvalue ψ =α ψa +β ψb 

0 0 To represent these infinite eigenstates, we need to choose two states as basis, e.g. ψa  and ψb . Then, any eigenstates with eigenenergy E0 can be written as a superposition of them. When we have one set of basis, we know that we can choose another set of basis (i.e. we can 0 0 change to a different set of basis): for example, we can define ψ1  and ψ2 , where

0 0 0 ψ1 =α 1 ψa +β 1 ψb  (2.125) 18 Phys460.nb

0 0 0 ψ2 =α 2 ψa +β 2 ψb  (2.126)

0 0 Here, we request ψ1  and ψ2  to be orthogonal to each other (otherwise, we make them orthogonal, using Gram-Schmidt procedure)

0 0 0 * 0 * 0 0 * * ψ1 ψ2 = ψ a α1 + ψb β1  α2 ψa +β 2 ψb =α 1 α2 +β 1 β2 =0 (2.127)

0 0 and we assume that ψ1  and ψ2  are normalized

0 0 0 * 0 * 0 0 * * 2 2 ψ1 ψ1 = ψ a α1 + ψb β1  α1 ψa +β 1 ψb =α 1 α1 +β 1 β1 = α1 + β1 =1 (2.128)

0 0 0 * 0 * 0 0 * * 2 2 ψ2 ψ2 = ψ a α2 + ψb β2  α2 ψa +β 2 ψb =α 2 α2 +β 2 β2 = α2 + β2 =1 (2.129)

0 0 0 0 Bottom line, instead of our old states ψa  and ψb , we can use ψ1  and ψ2  instead as our basis.

2.4.4. Which basis shall we use?

0 0 As mentioned above, in general, ψa H' ψb  ≠ 0. If so, the native perturbation theory will have singularity at the second order. Now, we 0 0 0 0 learned that we can choose to use a different set of unperturbed eigenstates ψ1  and ψ2 , so can we make ψ1 H' ψ2 = 0? If so, it will save the day and get ride of the singularity. The way to do it is very simple. As we know early on, if we cannot decide which basis to use, we shall look at another quantum operator. Here we do have one more quantum operator, which is H'. We first write down a 2×2 matrix 0 0 0 0 ψa H' ψa  ψa H' ψ  W= b 0 0 0 0 (2.130) ψb H' ψa  ψb H' ψb 

To make the formula shorter, we define

0 0 Wij = ψi H' ψ j  (2.131)

So

Waa Wab W=  (2.132) Wba Wbb

1. W is a Hermitian matrix (W† = W), so its eigenvalues are real  *  †  This is pretty straightforward to prove, because ψ1 X ψ2 = ψ2 X ψ1. In particular, if X is an Hermitian operator (the quantum operator   †  *  of any physics observable is Hermitian), we have X =X and thus ψ1 X ψ2 = ψ2 X ψ1. Thus it is easy to notice that

* * * Waa =W aa andW bb =W bb andW ba =W ab (2.133) * * †  * Waa Wba Waa Wab W =W  = * * = =W (2.134) Wab Wbb Wab Wbb

α1 α2 2. W has two eigenvalues E+ and E-, and each of them has a vector,   for E+ and   for E- β1 β2

Waa Wab α1 α1    =E +  (2.135) Wba Wbb β1 β1

and

Waa Wab α2 α2    =E -  (2.136) Wba Wbb β2 β2

where E+ and E- are the two eigenvalues.

0 0 3. We can use these two eigenvectors to define our ψ1  and ψ2  as

0 0 0 ψ1 =α 1 ψa +β 1 ψb  (2.137)

0 0 0 ψ2 =α 2 ψa +β 2 ψb  (2.138)

As will be shown below, these two states are precisely what we should use for the perturbation theory. Phys460.nb 19

4. The two eigenvalues are the first order corrections to the eigenenergy

0 2 E1 =E +λE + +Oλ  (2.139)

0 2 E2 =E +λE - +Oλ  (2.140)

Q: Why we have two eigenenergies here?

0 0 A: Because we started from two degenerate eigenstates. At λ= 0, the two states ψ1  and ψ2  have the same energy. Now, if we turn on a 0 2 small perturbation λH ', we find that these two states (in general) have different eigenenergies. One of them is E 1 =E +λE + +Oλ  and the 0 2 other E2 =E +λE - +Oλ . NOTE #1: We say that the perturbation “lifted the degeneracy”.

0 0 NOTE #2: After we lift the degeneracy, ψ1  and ψ2  no longer have the same energy. If the perturbation is small enough, we can now do non-degenerate perturbation theory, i.e. problem solved.

2.4.5. A key conclusion: in quantum mechanics, perturbations will in general lift all degeneracy, unless there is a reason saying that the degeneracy shall not be lifted. In general, in the study of qua tum physics, we can never include all terms in the Hamiltonian in our theoretical calculation. We always need some approximations (i.e. drop some small/unimportant part of the Hamiltonian). For example. in the study of a Hydrogen atom, we ignored relativistic effects. We also ignored the magnetic interactions between the electron and the nucleon (remember that both particles have spin. Whenever a charged particle starts to spin, there is a magnetic dipole. Because both the electron and the proton have magnetic dipoles, there should be an dipole-dipole interaction between them, which was ignored). In addition,we also ignored the earth magnetic field, which is always in presence when we do an experiment (unless we screen it out using some special devices.)

Let’s use Hreal to represent the full Hamiltonian of a real system and Hmodel to represent the Hamiltonian that we used to theoretically analyze the system. We know that these two are not the same, because we always need some approximation to simplify a real problem, i.e.

Hreal =H model +δH (2.141) we can treat δH as a perturbation.

Now here comes the questions, if we found that two (or more) states have the same eigenenergy (degeneracy) using Hmodel, are these states really degenerate in a real system? The general answer is no (unless there is a reason), because our first order degenerate perturbation theory told as that any small perturbation will in general lift the degeneracy.

The only except is: if there is a reason (usually based on symmetry) to tell us that Waa is exactly the same as Wbb and Wab =W ba = 0 precisely. (in a real physics system, in most of the case, we cannot say that the value of a quantity is precisely this number. What we really mean is that there is an argument to show that the difference between Waa and Wbb is unmeasurably small and Wab and Wba is unmeasurably small).

0 0 2.4.6. Prove: 〈ψ1 H' ψ2 〉=0 Here the proof contains two steps: first

0 0 * * Waa Wab α2 ψ1 H' ψ2 =(α 1 β1 )    (2.142) Wba Wbb β2 and then we will show

* * Waa Wab α2 (α 1 β1 )   =0 (2.143) Wba Wbb β2

0 0 The first step is very straightforward (it is from the definition of ψ1  and ψ2  )

0 0 0 ψ1 =α 1 ψa +β 1 ψb  (2.144)

0 0 0 ψ2 =α 2 ψa +β 2 ψb  (2.145) so 20 Phys460.nb

0 0 0 * 0 * 0 0 0 * 0 * 0 0 ψ1 H' ψ2 = ψ a α1 + ψb β1 H' α 2 ψa +β 2 ψb = ψ a α1 + ψb β1  α2 H' ψa +β 2 H' ψb = (2.146) * 0 0 * 0 0 * 0 0 * 0 0 α1 α2 ψa H' ψa +α 1 β2 ψa H' ψb +β 1 α2 ψb H' ψa +β 1 β2 ψb H' ψb 

0 0 The r.h.s. of the equation is in fact exactly the same, if we remember the definition of the W matrix Wij = ψi H' ψ j 

* * Waa Wab α2 * * * * (α 1 β1 )   =α 1 α2 Waa +α 1 β2 Wab +β 1 α2 Wba +β 1 β2 Wbb (2.147) Wba Wbb β2

0 0 * * Waa Wab α2 So, we proved that ψ1 H' ψ2 =(α 1 β1 )    Wba Wbb β2

For the second step, we first consider the situation that E+ ≠E -. According to the eigenequations, we have

Waa Wab α2 α2    =E -  (2.148) Wba Wbb β2 β2

* * If we multiply on both sides (α1 ,β 1 ), we get

* * Waa Wab α2 * * α2 * * α2 (α1 ,β 1 )   =(α 1 ,β 1 )E - =E -(α1 ,β 1 )  (2.149) Wba Wbb β2 β2 β2

Similarly, if we start from the other eigenequation

Waa Wab α1 α1    =E +  (2.150) Wba Wbb β1 β1

we get

* * Waa Wab α1 * * α2 * * α1 (α2 ,β 2 )   =(α 2 ,β 2 )E + =E +(α2 ,β 2 )  (2.151) Wba Wbb β1 β2 β1

If we take a conjugate on both sides

* * Waa Wab α2 * * α2 (α1 ,β 1 )   =E +(α1 ,β 1 )  (2.152) Wba Wbb β2 β2

here we used the fact that W is Hermitian and thus W† =W. Notice that we have shown

* * Waa Wab α2 * * α2 (α1 ,β 1 )   =E +(α1 ,β 1 )  (2.153) Wba Wbb β2 β2

and

* * Waa Wab α2 * * α2 (α1 ,β 1 )   =E -(α1 ,β 1 )  (2.154) Wba Wbb β2 β2

If E+ ≠E - the only way that these two equations can both be valid is that

* * Waa Wab α2 * * α2 * * α2 (α1 ,β 1 )   =E +(α1 ,β 1 ) =E -(α1 ,β 1 ) =0 (2.155) Wba Wbb β2 β2 β2

0 0 So we proved that ψ1 H' ψ2 = 0.

Q: what will happen is E+ =E - ?

0 0 A: Turns out that this is the simple case. If E+ =E -, as will be shown below, Wab = ψa H' ψb = 0. So, there is no divergence from the beginning. We can start the perturbation theory without worrying about these divergence.

2.4.7. First order perturbation

0 0 0 The calculation described above provides to us the zeroth order wavefunctions (i.e., we should use ψ1  or ψ2 ) , instead of ψa  or 0 ψb  as our unperturbed wavefunction). As we learned early on (in non-degenerate perturbation theory), the first order correction of energy is just

1 0 0 En = ψn H' ψn  (2.156)

i.e., we use the zeroth order wavefunction and compute the expectation value for H '. Here, for the zeroth order wavefunctions, we have two of 0 0 them, ψ1  and ψ2 , so we need to compute the first order energy correction for each of them. And we will prove in this section Phys460.nb 21

1 0 0 E1 = ψ1 H' ψ1 =E + (2.157) and

1 0 0 E2 = ψ2 H' ψ2 =E - (2.158)

0 0 i.e., the first order energy corrections for ψ1  and ψ2  are precisely the two eigenvalues of the W matrix.

0 0 0 * 0 * 0 0 0 * 0 * 0 0 ψ1 H' ψ1 = ψ a α1 + ψb β1 H' α 1 ψa +β 1 ψb = ψ a α1 + ψb β1  α1 H' ψa +β 1 H' ψb = (2.159) * 0 0 * 0 0 * 0 0 * 0 0 α1 α1 ψa H' ψa +α 1 β1 ψa H' ψb +β 1 α1 ψb H' ψa +β 1 β1 ψb H' ψb 

0 0 If we remember the definition of the W matrix Wij = ψi H' ψ j , we realized immediately that this formula is exactly the same as

* * Waa Wab α1 * * * * (α 1 β1 )   =α 1 α1 Waa +α 1 β1 Wab +β 1 α1 Wba +β 1 β1 Wbb (2.160) Wba Wbb β1

So we found

1 0 0 * * Waa Wab α1 E1 = ψ1 H' ψ1 =(α 1 β1 )    (2.161) Wba Wbb β1 α Because  1  is an eigenvector of B β1

Waa Wab α1 α1    =E +  (2.162) Wba Wbb β1 β1

1 0 0 * * Waa Wab α1 * * α1 E1 = ψ1 H' ψ1 =(α 1 β1 )   =E +(α 1 β1 )  (2.163) Wba Wbb β1 β1

α1 * * α1 * * Because we have required   to be normalized, (α 1 β1 ) =α 1 α1 +β 1 β1 =1 β1 β1

1 0 0 * * Waa Wab α1 * * α1 E1 = ψ1 H' ψ1 =(α 1 β1 )   =E +(α 1 β1 ) =E + (2.164) Wba Wbb β1 β1

Similarly, we can show that

1 0 0 * * Waa Wab α2 * * α2 E2 = ψ2 H' ψ2 =(α 2 β2 )   =E -(α 2 β2 ) =E - (2.165) Wba Wbb β2 β2

2.4.8. Eigenvalues of the matrix W In this part, we review basic ideas of eigenvalues and eigenvectors. We starts from the eigenequation defined in the previous section

Waa Wab α α    =E  (2.166) Wba Wbb β β

This means that

Waa α+W ab β=Eα (2.167)

Wba α+W bb β=Eβ (2.168)

These two equations have an obvious and trivial solution α=β= 0. This solution is NOT what we want and we will not consider this trivial solution. To get a nontrivial solution, the eigenvalue E cannot be an arbitrary value. It can only be one of two values, as will be seeing below. Using the first equation, we get

Wab α= β (2.169) E-W aa

Using the second equation, we get

E-W bb α= β (2.170) Wba

The first relation means 22 Phys460.nb

α Wab = (2.171) β E-W aa

but the second relation requires

α E-W bb = (2.172) β Wba

So we have

α Wab E-W bb = = (2.173) β E-W aa Wba

In general, Wab ≠ E-Wbb , so we find an contradiction. This contradiction means that for a general value of E, we will only have the trivial E-Waa Wba solution α=β= 0. To get a nontrivial solution, we have to request Wab = E-Wbb . This equation is often written in a different form E-Waa Wba

Wab E-W bb = (2.174) E-W aa Wba

(E-W aa)(E-W bb) =W ab Wba (2.175)

(E-W aa)(E-W bb)-W ab Wba =0 (2.176)

E-W aa -Wab det =0 (2.177) -Wba E-W bb

or equivalently

E0 Waa Wab det  - =0 (2.178) 0E Wba Wbb

det(E-W)=0 (2.179)

Here, W is the matrix that we define above

Waa Wab W=  (2.180) Wba Wbb

and number E here means E times the identity matrix E* 1 0 = E0  0 1 0E (2.181)

det(E-W)= 0 means

(Waa -E)(W bb -E)-W ab Wba =0 (2.182)

And thus

2 E -(Waa +W bb)E+(W aa Wbb -W ab Wba) =0 (2.183)

By definition, trW=W aa +W bb and detW=W aa Wbb -W ab Wba. Therefore, we can write the same equation as E2 - trWE+ detW=0 (2.184)

This equation has two solutions

trW± (trW) 2 - 4 detW E± = (2.185) 2

As shown above, these two solutions, E± are the first order correction to the eigenenergy. In the perturbation theory, the eigenenergies of these two quantum states are

0 2 E=E +E + λ+Oλ  (2.186)

and Phys460.nb 23

0 2 E=E +E - λ+Oλ  (2.187) at small λ.

Comment #1. tr W and detW are both real. This is straightforward to prove, if we notice that W is Hermitian. Because * * * 2 Waa =W aa andW bb =W bb andW ba =W ab , Waa and Wbb are real. And Wab Wba = Wab is also real, so trW=W aa +W bb and detW=W aa Wbb -W ab Wba are both real.

2 Comment #2. (trW) - 4 detW≥ 0. Therefore, E ± are both real.

2 2 2 2 (trW) - 4 detW=(W aa +W bb) -4W aa Wbb +4W ab Wba = (Waa -W bb) +4 Wab ≥0 (2.188)

* here we used the fact that Wab =W ba.

2 Comment #3. There are in general two possible situations (a) If (trW) - 4 detW> 0, E+ >E -. i.e. the two eigenvalues are NOT the same. (b) 2 If Waa =W bb and Wab =W ba = 0, (trW) - 4 detW= 0, and thus E+ =E - = trW2.

0 0 The situation (b) is the easy case, because Wab =W ba = 0 means ψa H' ψb = 0. Remember that the problem we had from the beginning is 0 0 0 0 〈ψa H' ψb 〉 〈ψb H' ψa 〉 0 0 that the second order correction will diverge, 0 0 , because Ea =E b . For situation (b), the numerator is zero, so there is no Ea -Eb divergence. And thus we can just do non-degenerate perturbation theory. The situation (a) is the more generic case. There, as we have shown early on, the perturbation H ' lift the degeneracy.

2.4.9. Eigenvectors of the matrix W In this section, we will assume that (trW) 2 - 4 detW> 0, i.e. situation (a) discussed in the previous section. We have two eigenvalues. For each

2 eigenvalue, we can solve for the corresponding eigenvector. For the eigenvalue E = trW+ (trW) -4 detW , we have + 2

Waa Wab α1 α1    =E +  (2.189) Wba Wbb β1 β1

2 and for the other eigenvalue E = trW- (trW) -4 detW , we have - 2

Waa Wab α2 α2    =E -  (2.190) Wba Wbb β2 β2

We will use the first one as example (equation for E+). There, the matrix equation can be written as two separate equations

Waa α1 +W ab β1 =E + α1 (2.191)

Wba α1 +W bb β1 =E + β1 (2.192)

Using the first equation, we get

Wab α1 = β1 (2.193) E+ -W aa

Using the second equation, we get

E+ -W bb α1 = β1 (2.194) Wba

These two relations are actually identical, because for any eigenvalue E, we have Wab = E-Wbb as we proved early on. E-Waa Wba

* * In addition, we know that α1 α1 +β 1 β1 = 1, i.e. the normalization condition. So we have

Wab α = 1 (2.195) 2 2 Wab +(E+ -W aa)

E+ -W aa β = 1 (2.196) 2 2 Wab +(E+ -W aa)

Similarly, we have 24 Phys460.nb

Wab α = 2 (2.197) 2 2 Wab +(E- -W aa)

E- -W aa β = 2 (2.198) 2 2 Wab +(E- -W aa)

In conclusion, we found that

W E+ -W aa 0 0 0 ab 0 0 ψ =α ψa +β ψ = ψa + ψ  1 1 1 b b (2.199) 2 2 2 2 Wab +(E+ -W aa) Wab +(E+ -W aa)

W E- -W aa 0 0 0 ab 0 0 ψ =α ψa +β ψ = ψa + ψ  2 2 2 b b (2.200) 2 2 2 2 Wab +(E- -W aa) Wab +(E- -W aa)

2.4.10. the very special case

In general, we expect E+ ≠E -, i.e. the generacy is lifted. What will happen if E+ =E -. From the equation

trW± (trW) 2 - 4 detW E± = (2.201) 2

2 2 we know that E+ =E - can only arise when (trW) - 4 detW =0, i.e. (trW) - 4 detW=0. As shown above

2 2 2 (trW) - 4 detW=(W aa -W bb) +4 Wab (2.202)

Both the two terms on the r.h.s. are non-negative, and thus if we want the whole thing to be zero, we must have

2 (Waa -W bb) =0 (2.203)

and

2 4 Wab =0 (2.204)

i.e., Waa =W bb and Wab = 0.

With Waa =W bb and Wab = 0, W is actually proptional to an identity matrix. Waa 0 1 0 W= =W aa   (2.205) 0W aa 0 1

This situation is highly unlikely to arise (unless there is a reason) because, in general, the W matrix has four free values to pick W aa, Wbb, the

real part of Wab and the imaginary part of Wab (note 1: Waa and Wbb are real, so they don’t have imaginary part. note 2: Wba is the complex

conjugate of Wab, so we don’t need to consider it here as a seperate degree of freedom). If you have four real values, what is the probability for

these four real values to satisify that Waa matches exactly Wbb without any error bar, and both the real and imarginaly parts of Wab vanishes exactly without any error bar? Without a reason, the chance is zero. So this is a sitation that we don’t need to worry much, unless there is a reason. In most cases, such a specal case arises due to symmetry. For example, time reversal symmetry tells us that there should be two degenerate

states (a state and its time reversal state). Then, for H0 these two states degenerates and for H, they should still be degenerate, so E+ =E -. For that situation, it turns out that one can directly start from non-degenrate pertubation theory (no singularities will arise), although the states are degenerate. We will dicuss a more generic situation later, which covers this case.

2.4.11. Review: quantum states and quantum operators as matrices Once we choose a set of basis, any quantum state can be written as a vector (i.e., a N-by-1 matrix).

For a complete set of basis, { ψi〉}, we can write any quantum states as

ψ〉= ci ψi i (2.206) Phys460.nb 25

where ci are complex numbers. Here, we find that if we want to describe a state, we just need to know all the coefficient ci. We can write these ci as a vector

c1 c2 (2.207) c3 ⋮

These coefficients are

ci = 〈ψi ψ〉 (2.208)

To see this, we multiply 〈ψ j for both sides of ψ〉= ∑i ci ψi〉

〈ψ j ψ〉= ci 〈ψ j ψi〉= ci δij =c j i i (2.209)

Bottom line: a quantum state is a column vector

c1 〈ψ1 ψ〉 c2 〈ψ2 ψ〉 ψ〉 → = (2.210) c3 〈ψ3 ψ〉 ⋮ ⋮

Conjugate states is the represented by the conjugate vector. By definition, we know that

* ψ = 〈ψi ci i (2.211)

* so, we can write all these ci as a row vector

* * * (c 1 c2 c3 … ) = ( 〈ψ ψ1〉 〈ψ ψ2〉 〈ψ ψ3〉…) (2.212)

Here, we used the fact that 〈ψ ψi〉 is the complex conjugate of 〈ψi ψ〉 Inner produce of two states are product of a row vector and a column vector If we know two quantum states

c1 c2 ψ〉 → (2.213) c3 ⋮ and

d1 d2 ϕ〉 → (2.214) d3 ⋮ then, we know

* * * 〈ϕ →(d 1 d2 d3 …) (2.215) so

c1 * * * c2 * * * 〈ϕ ψ〉 →(d 1 d2 d3 …) =d 1 c1 +d 2 c2 +d 3 c3 + … (2.216) c3 ⋮

Q: How about a quantum operator? A: Once we choose a set of basis, a quantum operator is a matrix. To understand this, we just need to realize that a quantum operator transforms a quantum state into a different state  X ψ= ϕ (2.217)

As we have known, |ψ〉 is a column vector, and |ϕ〉 is another column vector. Which object transfers a column vector to a different column 26 Phys460.nb

vector? We know that a matrix can do such a job

x11 x12 x13 … c1 x11 c1 +x 12 c2 +x 13 c3 + … d1 x21 x22 x23 ... c2 x21 c1 +x 22 c2 +x 23 c3 + … d2 = = (2.218) x31 x32 x33 … c3 x31 c1 +x 32 c2 +x 33 c3 + … d3 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮

So a quantum operator is really similar to a matrix. In fact, the matrix elements xijs are very easy to compute  xij = ψi X ψ j (2.219)

Q: How about eigenvalues and eigenstates? A: Matrices also have eigenvalues and eigenstates

x11 x12 x13 … c1 c1 x21 x22 x23 ... c2 c2 =W (2.220) x31 x32 x33 … c3 c3 ⋮ ⋮ ⋮ ⋱ ⋮ ⋮

1. The matrix of a Hermitian operator is a Hermitian matrix 2. An N×N Hermitian matrix has N eigenvalues,each of which has an eigenvector 3. Eigenvalues of the matrix is the same as the eigenvalues of the corresponding quantum operator

c1 c2  4. Each eigenvector corresponds to a eigenstate, i.e. If is an eigenvector with eigenvalue W, then ψ〉= ∑i ci ψi〉 is an eigenstate of X c3 ⋮ with eigenvalue W. Final conclusion: for a quantum system, we just needs to play with matrices Only one problem: these matrices are huge (∞×∞) It is extremely hard to handle big matrices (say 100 million by 100 million). So this approach doesn’t make our life easier.

2.4.12. Degenerate perturbation theory

H=H 0 +λH' (2.221)

Using eigenstates of H0 as basis, then H0 corresponds to a diagonal matrix

0 H0 ψi〉=E i ψi (2.222)

where i= 1, 2, 3,… and we request this is an orthonormal basis

〈ψi ψ j〉=δ ij (2.223)

a matrix element of the matrix is

0 0 0 〈ψi H0 ψ j〉= ψ i E j ψ j=E j 〈ψi ψ j〉=E j δi,j (2.224)

So, 0 E1 0 0… 0 0E 2 0 ... H0 → 0 (2.225) 0 0E 3 … ⋮ ⋮ ⋮ ⋱

This conclusion is generically true. If we use the eigenstates of an operator as our basis, then this operator is a diagonal matrix (i.e. off-diagonal terms are all zero). And along the diagonal line, we just have all the eigenvalues of this quantum operator.

〈ψ1 H' ψ1〉 〈ψ1 H' ψ2〉 〈ψ1 H' ψ3〉… 〈ψ2 H' ψ1〉 〈ψ2 H' ψ2〉 〈ψ2 H' ψ3〉 ... λH'→λ (2.226) 〈ψ3 H' ψ1〉 〈ψ3 H' ψ2〉 〈ψ3 H' ψ3〉… ⋮ ⋮ ⋮ ⋱ Phys460.nb 27

In general, H ' is NOT an diagonal matrix 0 E1 0 0… 〈ψ1 H' ψ1〉 〈ψ1 H' ψ2〉 〈ψ1 H' ψ3〉… 0 0E 2 0 ... 〈ψ2 H' ψ1〉 〈ψ2 H' ψ2〉 〈ψ2 H' ψ3〉 ... H=H 0 +λH'→ +λ = 0 0 0E 3 … 〈ψ3 H' ψ1〉 〈ψ3 H' ψ2〉 〈ψ3 H' ψ3〉… ⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋱ (2.227) 0 E1 + λ 〈ψ1 H' ψ1〉 λ 〈ψ1 H' ψ2〉 λ 〈ψ1 H' ψ3〉… 0 λ 〈ψ2 H' ψ1〉E 2 + λ 〈ψ2 H' ψ2〉 λ 〈ψ2 H' ψ3〉 ... 0 〈ψ3 H' ψ1〉 λ 〈ψ3 H' ψ2〉E 3 + λ 〈ψ3 H' ψ3〉… ⋮ ⋮ ⋮ ⋱

This is a very large matrix and is very hard to handle in general. However, if there are 2 degenerate states, ⋱ ⋮ ⋮… ...E 0 0 ... H = 0 ... 0E 0 … (2.228) ⋮ ⋮ ⋮ ⋱ i.e., two of the eigenvalues of H0 coincides, or say two of two numbers along the diagonal line of the matrix of H0 happens to be the same, the we don’t need to handle the whole big matrix, if λ is small. Here, we can do degenerate perturbation theory, and to the first order, we can forget all other quantum states and only look at the two degenerate ones. What does this mean? Remember, that in general, a set of complete basis contains infinite number of states ψi〉 with i= 1, 2,…∞. As a result, the matrix of a quantum operator has dimension ∞×∞,  0 0 ψi X ψ j with i= 1, 2,…∞ and j = 1, 2,…∞. Now, if we only limit ourselves to the two degenerate states ψa  and ψb , then the matrix of our quantum operator only has dimensions 2×2, because my i and j here can only be a or b

0  0 0  0  ψa X ψa  ψa X ψ  X → b 0  0 0  0 (2.229) ψb X ψa  ψb X ψb 

For H0, its matrix is

E0 0 H0 →   (2.230) 0E 0 and for H ', the matrix is 0 0 0 0 ψa H' ψa  ψa H' ψ  H'→ b 0 0 0 0 (2.231) ψb H' ψa  ψb H' ψb 

So our H is 0 0 0 0 E0 + λ ψa H' ψa  λ ψa H' ψb  H=H 0 +λH'→ 0 0 0 0 (2.232) λ ψb H' ψa E 0 + λ ψb H' ψb 

The eigenvalues of this matrix are E0 +λE + and E0 +λE -. And the eigenvector is the same as we computed in previous section Bottom line: for degenerate perturbation, we can drop all other states (with different eigenenergies), and consider a much smaller Hilbert space (only the degenerate states are considered here). Then, our Hamiltonian becomes a very small matrix, and we can diagonalize this small matrix. The eigenvalues are the eigenenergies to the first order. And the eigenvectors give us eigenwavefunctions to zeroth order.

2.4.13. n-fold degeneracy

If H0 as n-fold degeneracy, and we want to do perturbation theory for these n degenerate states, we just ignore all other states and only keep these n states.

E0 0 0… 0E 0 0 ... H0 → (2.233) 0 0E 0 … ⋮ ⋮ ⋮ ⋱ n×n 28 Phys460.nb

0 0 and H ' is a n×n matrix with matrix elements ψi H' ψj , where i= 1,…,n and j= 1,…,n Then, there are two (equivalent) ways to do the calculation

1 1 Option #1: compute eigenvalues for the n×n matrix of H ', as E1 …En . Then the eigenenergy to the first order correction is

1 2 Ei =E 0 +λE i +Oλ  (2.234)

where i= 1, 2,…,n.

Option #2: direction compute eigenvalues of the n×n matrix of H =H 0 +λH'. You will find n eigenvalue, they are

1 E0 +λE i (2.235)

where i= 1, 2,…,n.

2.4.14. Nearly-degenerate perturbation theory

0 0 What if we have two states that are not totally degenerate, but nearly degenerate, i.e. two eigenstates of H0 ψa  and ψb  has very similar 0 0 energies Ea ~E b , but not exactly the same.

0 0 Case 1. if λ H ' << Ea -E b , we can do non-degenerate perturbation theory

0 0 0 0 0 0 0 Case 2. if λ H' >> Ea -E b , but λ H ' << Ea -E m and λ H ' << Eb -E m for any other eigenstates of H0, where Em represent 0 0 eigenenergy of another eigenstate of H0 (beyond ψa  and ψb ), we can do nearly-degenerate perturbation theory.

0 0 Here, the procedure is similar to the degenerate perturbation theory, we ignore all other states and only consider ψa  and ψb . Now, every quantum operator becomes a 2×2 matrix. E 0 0 H → a 0 0 (2.236) 0E b

and for H ', the matrix is 0 0 0 0 ψa H' ψa  ψa H' ψ  V V H'→ b = aa ab  0 0 0 0 (2.237) ψb H' ψa  ψb H' ψb  Vba Vaa

0 0 Notice that H0 has two different diagonal components Ea ≠E b . This is the difference between degenerate and nearly-degenerate perturbation theory. Now we consider H , which is E 0 +λV λV H=H +λH'→ a aa ab 0 0 (2.238) λV ba Eb +λV bb

Then, we can get eigenvalues of this matrix, and these are the eigenvalues of H up to first order in perturbation theory.

0 0 0 0 Case 3. if λH ' is too large, even larger than | Ea -E m | and |Eb -E m |, then λH ' is too large and thus cannot be considered as a perturbation and as a result, we cannot do perturbation theory anymore. This result can be easily generalized to cases where we have more than 2 nearly-degenerate states.

2.4.15. Philosophy behind degenerate and nearly-degenerate perturbation theory

0 0 Assuming that H0 have n eigenstates who have very similar (or exactly the same) eigenenergies  ψa , ψb … and they all have energy near 0 E0), but all other eigenstates of H0 have energies very different from these states ( ψm〉 has eigenenergy Em and m is not one of the nearly 0 degenerate states. For any m, we have Em very different from E0). Then when we start our system from one (or some superposition) of these n

states, and then perturb the Hamiltonian by a small amount H=H 0 +λH ' with λ being very small. Then, because any state ψm〉 has an

energy much different from E0, when the perturbation is small, it is (almost) impossible for the system to reach a state ψm〉 from one state

with energy E0. Note: for a classical system, this would be totally impossible due to energy conservation. In a classical system, if we start from a state with

energy E0 and then add a small amount of energy δE to the system, the final states must have energy E0 +δE, which would be very close to E 0,

if δE is small. So, it is absolutely impossible to have a final states with energy very different from E0. But for a quantum system, anything is Phys460.nb 29

possible (think about quantum tunneling, classically it is impossible, but for a quantum system it become possible). However, we know that in quantum mechanics, the probability for us to reach such a final states is small (although not exactly zero). Since the probability is small, to the 0 leading order, we can ignore that probability. This is the key reason why we can ignore all those states ψm .

Since it is highly unlikely to reach ψm〉, we can ignore them to the leading order approximation. After we ignore all of them, our Hilbert space becomes very small, only n quantum states now. And thus our quantum operators becomes n×n matrices. If n is 2, we can easily find the eigenvalue. If n=3 or 4, we can get the eigenvalue (with analytic form) with a little bit of help (e.g. software like Mathematica). If n> 4 but not extremely huge), say a couple of hundreds or smaller, we can easily get the eigenvalue numerically using available software packages. If n is 10 or 100 million, it is not easy to get all the eigenvalues for the system, but we can easily get the smallest several or the largest several numeri- cally using techniques like Lanczos .

0 After taking care of the n×n matrices, we may be able to take ψm  back into consideration by going to higher orders in the perturbation theory.

2.4.16. Example: (textbook page 262) Consider a 3D infinite cubical potential well V(x,y,z) = 0 if 0

The Hamiltonian for this system is P2 ℏ2 2 H0 = +V(x,y,z) = - ∇ +V(x,y,z) (2.240) 2m 2m

The eigenstates of H0 are sin waves

2 3/2 nx π ny π nz π 0 ψ nx ny nz (x,y,z) = sin x sin y sin z (2.241) a a a a where nx, ny and nz are positive integers. The eigenenergy for such a state is π2 ℏ2 0 2 2 2 E nx ny nz = nx +n y +n z  (2.242) 2ma 2

2 2 2 2 The ground state is obviously n =n =n = 1, which has energy E0 = 3π ℏ . For simplicity, we will call this energy E 0 = 3π ℏ , where the x y z 111 2ma 2 0 2ma 2 subscript 0 represents the ground states.

There are three (degenerate) first excited states with nx,n y andn z being (1, 1, 2) or (1,2,1) or (2,1,1). π2 ℏ2 π2 ℏ2 0 0 0 2 E 112 =E 121 =E 211 = 1+1+2 =3 (2.243) 2ma 2 m a2

0 For simplicity, we will call the energy of this first excited states E1 where the subscript 1 means that this is for the first excited states. In addition, for simplicity, we define

ψa =ψ 112 andψ b =ψ 121 andψ c =ψ 211 (2.244)

Now, consider a perturbation

H=H 0 +λH' (2.245) where V if 0

For the ground state, we shall do non-degenerate perturbation theory, because there is no degeneracy, and the first order correction to the energy is 30 Phys460.nb

a/2 a/2 a 1 0 * 0 0 2 E 111 = 〈ψ111 H' ψ111〉=  ⅆxⅆyⅆz ψ 111(x,y,z) H'ψ 111(x,y,z) =V 0  ⅆx  ⅆy  ⅆz ψ 111(x,y,z) 0 0 0 2 3 a/2 π a/2 π a π 2 3 1 a 1 a 1 V (2.247) 2 2 2 0 =V 0  ⅆx sin x  ⅆy sin y  ⅆz sin z =V 0 × × × a= a 0 a 0 a 0 a a 2 2 2 2 2 4

So, the energy of the ground states now becomes 3π 2 ℏ2 V 0 1 0 E111 =E 111 +λE 111 + … = +λ + … (2.248) 2ma 2 4

For the first excited states, there is a three-fold degeneracy, so we need to define a 3×3 matrix

Waa Wab Wac 〈ψa H' ψa〉 〈ψa H' ψb〉 〈ψa H' ψc〉 W= Wba Wbb Wbc = 〈ψb H' ψa〉 〈ψb H' ψb〉 〈ψb H' ψc〉 (2.249) Wca Wcb Wcc 〈ψc H' ψa〉 〈ψc H' ψb〉 〈ψc H' ψc〉

For diagonal terms a/2 a/2 a 0 * 0 0 2 Waa = 〈ψa H' ψa〉=  ⅆxⅆyⅆz ψ 112(x,y,z) H'ψ 112(x,y,z) =V 0  ⅆx  ⅆy  ⅆz ψ 112(x,y,z) 0 0 0 2 3 a/2 π a/2 π a 2π 2 3 1 a 1 a 1 V (2.250) 2 2 2 0 =V 0  ⅆx sin x  ⅆy sin y  ⅆz sin z =V 0 × × × a= a 0 a 0 a 0 a a 2 2 2 2 2 4

Similarly, we can show that

Waa =W bb =W cc =V 0 /4 (2.251)

For Wab, we have

0 * 0 Wab = 〈ψa H' ψb〉=  ⅆxⅆyⅆz ψ 112(x,y,z) H'ψ 121(x,y,z)

2 3 a/2 π a/2 π 2π a π 2π (2.252) 2 =V 0  ⅆx sin x  ⅆy sin y sin y  ⅆz sin z sin z a 0 a 0 a a 0 a a

One can show that the last integral aⅆz sin π z sin 2π z is zero. So W = 0. Similarly, W =W = 0. ∫0 a a aa ab ac

Finally, for Wbc

0 * 0 Wab = 〈ψb H' ψc〉=  ⅆxⅆyⅆz ψ 121(x,y,z) H'ψ 211(x,y,z)

2 3 a/2 π 2π a/2 π 2π a π 16 (2.253) 2 =V 0  ⅆx sin x sin x  ⅆy sin y sin y  ⅆz sin z = V0 a 0 a a 0 a a 0 a 9π 2

Thus, we have V0 0 0 4 W= 0 V0 16 V 4 9π 2 0 (2.254) 0 16 V V0 9π 2 0 4

For this matrix, it has three eigenvalues (all are real numbers). The equation for eigenvalues is det(EI-W)=0 (2.255)

E- V0 0 0 4 det 0E- V0 - 16 V =0 4 9π 2 0 (2.256) 0- 16 V E- V0 9π 2 0 4

And thus 2 2 V0 V0 16 E-  E- - V0 =0 (2.257) 4 4 9π 2 Phys460.nb 31

V0 V0 16 V0 16 E- E- + V0 E- - V0 =0 (2.258) 4 4 9π 2 4 9π 2

So the solutions are V 16 V 8 2 1 0 0 E1 = - V0 = 1-  (2.259) 4 9π 2 4 3π V 1 0 E2 = (2.260) 4 V 16 V 8 2 1 0 0 E3 = + V0 = 1+  (2.261) 4 9π 2 4 3π

So the energies of the first excited states are

2 2 2 E 0 +λE 1 + … =3 π ℏ +λ V0 1- 8   state#1 1 1 m a2 4 3π 2 2 E = E 0 +λE 1 + … =3 π ℏ +λ V0 state#2 1 1 2 m a2 4 (2.262) 2 2 2 E 0 +λE 1 + … =3 π ℏ +λ V0 1+ 8   state#3 1 3 m a2 4 3π

Now, for the eigenstates of the W matrix V0 0 0 4 α1 α1 2 α1 V0 8 0 V0 16 V β =E 1 β = 1-  β 4 9π 2 0 1 1 1 1 (2.263) γ γ 4 3π γ 0 16 V V0 1 1 1 9π 2 0 4 1 0 0 2 α1 2 α1 V0 0 1 8  V0 8 3π β1 = 1-  β1 (2.264) 4 2 γ 4 3π γ 0 8  1 1 1 3π

By canceling V0 /4 on both sides, we get 1 0 0 2 α α 0 1 8  1 8 2 1 3π β1 =1-  β1 (2.265) 2 γ 3π γ 0 8  1 1 1 3π

It means that 8 2 α1 =1-  α1 (2.266) 3π 8 2 8 2 β1 + γ1 =1-  β1 (2.267) 3π 3π 8 2 β1 +γ 1 =β 1 (2.268) 3π

The first equation means α1 = 0 and the last two means β1 = -γ1. In addition, we know that normalization condition requires α 2 + β 2 + γ 2 = 1, so 0 1 α1 β1 = 2 (2.269) 1 γ1 - 2

So, 32 Phys460.nb

π 2π 2π π 3/2 sin x sin y- sin x sin y ψb -ψ c ψ121 -ψ 211 2 a a a a π ψ1(x,y,z) =α 1 ψa +β 1 ψb +γ 1 ψc = = = sin z (2.270) 2 2 a 2 a

Using the same approach, we find that

α2 1 β2 = 0 (2.271) γ2 0

and thus 2 3/2 2π π π ψ2(x,y,z) =α 2 ψa +β 2 ψb +γ 2 ψc =ψ a =ψ 211 = sin x sin y sin z (2.272) a a a a

Finally, for the third eigenstate, we can use the same method to show that 0 1 α3 β3 = 2 (2.273) 1 γ3 2

and thus

π 2π 2π π 3/2 sin x sin y+ sin x sin y ψb +ψ c ψ121 +ψ 211 2 a a a a π ψ3(x,y,z) =α 3 ψa +β 3 ψb +γ 3 ψc = = = sin z (2.274) 2 2 a 2 a

2.4.17. A small trick for finding ψ1 and ψ2 We use two-fold degenerate here as an example, but the conclusion here can be easily generalized.

0 0 As we have shown above, the key in degenerate perturbation theory is to find a good set of basis, such that ψ1 H' ψ2 = 0. In the most 0 0 0 0 general situation, we state from a set of states ψa  and ψb . If it is a good set already (i.e., ψa H' ψb = 0) , we don't need to find another basis. We can just use them and

0 0 0 2 E=E + λ ψa H' ψa +Oλ  (2.275)

and

0 2 E=E + λ 〈ψb H' ψb〉+Oλ  (2.276)

0 0 In general, we would be so lucky. i.e., if we just randomly choose a a set of states ψa  and ψb , the chance for this basis to be a good set of 0 0 basis is extremely low (we will in general have 〈ψa H' ψb〉 ≠ 0). Is there a way to help us pick ψa  and ψb ? The answer is yes, for some cases.   If there is another quantum operator A, which compute with both H0 and H ', then we can use the common eigenstates of A and H0 as the basis

for H0

0 0 H0 ψa =E 0 ψa  (2.277)

0 0 H0 ψb =E 0 ψb  (2.278)

 0 0 A ψa =A a ψa  (2.279)

 0 0 A ψb =A b ψb  (2.280)

0 0 In particular, if Aa ≠A b, then ψa  and ψb  are already a good set of basis.

0 0 To see this, we just need to prove that ψa H' ψb =0    Because we have assumed that A,H'=A H'-H'A = 0, Phys460.nb 33

0   0 0= ψ a A H'-H'A ψb = (2.281) 0  0 0  0 0 0 0 0 0 0 ψa A H' ψb - ψ a H'A ψb =A a ψa H' ψb -A b ψa H' ψb =(A a -A b) ψa H' ψb 

0 0 If Aa ≠A b, this equation means that ψa H' ψb = 0.

0 0 For this situation, although we have a degeneracy, one can just do non-degenerate perturbation for ψa  and ψb  (separately) and there will be no singularities at all.

2.5. the fine structure of a hydrogen atom

2.5.1. Relativistic correction In QMI, we solved an ideal model for a hydrogen atom (i.e. a particle in 1/r potential). In a real hydrogen atom, that model missed some of the physics, and one of them is relativistic effects.

Q: what is the energy of a particle, if the particle is moving at speed v and the rest mass m. m E=Mc 2 = c2

2 (2.282) 1- v c2

Q: what is the momentum of a particle, if the particle is moving at speed v and the rest mass m. m p =Mv= v

2 (2.283) 1- v c2

As a result,

E= p2 c2 +m 2 c4 (2.284)

To prove this relation, we start from the r.h.s.,

p2 c2 +m 2 c4 =

v2 2 2 2 2 2 m2 c41-  2 2 2 2 4 2 2 2 2 4 m v m v c c2 m v c +m c -m c v m c m (2.285) c2 +m 2 c4 = + = = = c2 =E v2 v2 v2 v2 v2 1- 2 1- 2 1- 2 1- 2 1- 2 2 c c c c c 1- v c2

This relation between E and p is an very important relation for relativistic physics! Q: what is kinetic energy?

A: First, find the energy of a particle when it is not moving p = 0. Then we measure the energy again when it is moving (with momentum p). The energy difference between them is the kinetic energy for this particle.

p 2 T=E-mc 2 = p2 c2 +m 2 c4 -mc 2 =mc 2 1+ -1 (2.286) m c

When particle is moving at low velocity (v <

So, 1 p 2 1 p 4 p2 p4 T=mc 2 - + … = - + … (2.289) 2 m c 8 m c 2m 8m 3 c2

The first term here is the kinetic energy in classical mechanics. In relativistic physics, the kinetic energy is NOT just p2 2m. Instead, we have a lot of corrections. These corrections are small if a particle is moving at low speed. There, we can treat them as perturbation p4 H'=- (2.290) 8m 3 c2

NOTE: this treatment is NOT the rigorous way to combine special relativity with quantum mechanics, because this treatment has one major flaw. At larger p or small m (e.g. consider a very light particle), the series will diverge. This problem comes from the fact that we used square root in the definite of the Hamiltonian. Square root is NOT an analytic function near small x, and thus will cause trouble

(to see this, think about f (x) = x , one can easily show that for the first order derivative, x=0 is infinite limx→0 f'(x)→∞). The correct way m c2 p c to do it is to use a matrix. Notice that for a matrix, square root arises naturally (e.g., the eigenvalue of are p c-m c 2 ± p2 c2 +m 2 c4 . We get square root without having any square root in the matrix). The person who figured this out is Dirac and this is Dirac’s theory for relativistic fermions. If we ignore higher order terms, our hydrogen atom should follow this Hamiltonian

H=H 0 +H' (2.291)

2 where H = p +V(r). With the perturbation H ', the energy of a hydrogen atom will be different from what we computed early on. How large 0 2m is the difference? This question can be answered by the perturbation theory.

We have already known the energy spectrum of H0, 13.6 eV 0 En = - withn= 1, 2, 3,… (2.292) n2

More precisely, 1 m e2 2 0 En = - (2.293) 2 2 n 2ℏ 4πϵ 0

We often define Bohr radius a as 2 ℏ 4πϵ 0 a= (2.294) m e2

And then, 1 1 e2 0 En = - (2.295) 2 n 2 4πϵ 0 a

2 For En, there are n degenerate quantum states (ignore spin at this moment) ψn l m where l is the angular momentum quantum number

l= 0, 1, 2…n- 1 and m is the quantum number for Lz and m=-l,-l+ 1,…0,…,l- 1,l

2 2 L ψn l m =ℏ l(l+1)ψ n l m (2.296)

Lz ψn l m =ℏmψ n l m (2.297)

For, n= 1 there is no degeneracy, and we can do non-degenerate perturbation theory. For any n> 1, there are n degenerate states, and thus we should do degenerate perturbation theory. However, we are very lucky here. We don’t need to worry about degenerate perturbation theory,

because ψn l m is already a good set of basis:

〈ψn l m H' ψn l'm' 〉=0 (2.298)

if l≠l' or m≠m'. Phys460.nb 35

2 2  This is because both H0 and H ' commute with L . And we can also show that both H0 and H ' commute with Lz. Here, L and Lz serve as the A 2 operator that we defined in the previous section. For a fixed n, because the degenerate states all have different eigenvalues for L and Lz

(different l and m), 〈ψn l m H' ψn l'm' 〉= 0. So we don’t need to choose any other basis and can start with non-degenerate perturbation. The correction to the energy is (to the first order)

1 En l m = 〈ψn l m H' ψn l m〉= p4 1 1 1 4 2 2 2 2 (2.299) -ψn l m ψn l m=- ψn l m p ψn l m=- ψn l m p p ψn l m=- ψn l m p p ψn l m 8m 3 c2 8m 3 c2 8m 3 c2 8m 3 c2

For ψn l m〉, we know that

0 H0 ψn l m〉=E n ψn l m (2.300)

p2 0 +V ψn l m=E n ψn l m (2.301) 2m p2 0 ψn l m=E n -V ψn l m (2.302) 2m

2 0 p ψn l m=2mE n -V ψn l m (2.303)

2 0 p ψn l m=2mE n -V ψn l m (2.304)

The conjugate of this equation gives

2 0 ψn l m p = ψn l m 2mE n -V (2.305)

So,

2 2 0 0 2 0 2 ψn l m p p ψn l m= ψ n l m 2mE n -V2mE n -V ψn l m=4m ψn l m En -V ψn l m (2.306)

2 Here, V=- e 1 and 4πϵ 0 r

0 2 ψn l m En -V ψn l m=

0 2 0 2 0 2 0 2 ψn l m En  -2E n V+V ψn l m= ψ n l m En  ψn l m-2 ψ n l m En V ψn l m+ ψ n l m V(r) ψn l m= (2.307) e2 1 e2 2 1 0 2 0 En  +2E n ψn l m ψn l m+ ψn l m ψn l m 2 4πϵ 0 r 4πϵ 0 r

Without going into details, we will just show the results here 1 1 ψn l m ψn l m= (2.308) r n2 a 1 1 ψn l m ψn l m= 2 (2.309) r n3l+ 1 a 2 2

Thus, 1 1 e2 1 e2 2 1 1 2 2 2 0 2 0 En l m = - ψn l m p p ψn l m=- 4m En  +E n + = 3 2 3 2 2 1 8m c 8m c 2πϵ 0 n a 4πϵ 0 n3l+ a 2 2 (2.310) 0 2 2 2 1 2E n e 1 e 2 0 2 - 4m En  + +  3 2 2 1 8m c n 4πϵ 0 a n3l+  4πϵ 0 a 2

As we have shown early on 1 1 e2 0 En = - (2.311) 2 n 2 4πϵ 0 a 36 Phys460.nb

e2 2 0 -2n En = (2.312) 4πϵ 0 a

0 2 2 2 1 2E n e 1 e 1 0 2 En l m = - En  + + = 2 2 1 2mc n 4πϵ 0 a n3l+  4πϵ 0 a 2 (2.313) 0 0 2 1 2E n 1 En  4n 0 2 2 0 2 0 2 - En  - 2n En + -2n En  =- -3 2mc 2 n2 n3l+ 1  2mc 2 l+ 1 2 2

So, the eigen-energy in a H atom shall be

0 2 En  4n 0 1 0 En l m =E n +E n l m + … =E n - -3 + … (2.314) 2mc 2 l+ 1 2

The zeroth order term 1 m e2 2 0 En = - (2.315) 2 2 2n ℏ 4πϵ 0

it is proportional to m e2 2 e2 2 0 2 En ∝ = m c (2.316) 2 ℏ 4πϵ 0 4πϵ 0 ℏc

2 The prefactor e is a very important physics constant, known as the fine structure constant. 4πϵ 0 ℏc e2 1 α= ≈ (2.317) 4πϵ 0 ℏc 137.036

So,

0 2 2 En ∝ α m c (2.318)

The first order term

0 2 En  1 4 2 En l m ∝ =α m c (2.319) m c2

If we compare the first order and zeroth order term, 1 4 2 2 En l m α m c 1 1 ∝ =α 2 = ≈ (2.320) 0 2 2 En α m c 137.036 10 000

So indeed, the perturbation theory works, i.e. higher order term is much smaller than the leading order. (remember that Taylor expansions only converge when the small parameter λ is small enough. Here, our small parameter is α, which is smaller than 1%). Relativistic correction is indeed an small correction in a H atom NOTE: the fine structure constant is one of the most important physics constant. It is dimensionless. It involves special relativity (contains the 2 speed of light). It involves quantum mechanics (having ℏ in its definition) and it also involves E&M (having ϵ 0 and e ). In this section, we showed that it is so lucky for us that for a H atom, because α is small, our relativistic correction is indeed small and thus we can do perturbation theory. In QFT (QED), small α means that interactions between particles (quantum electron-dynamics) is a small perturbation. To the leading order, we can treat a particle as a free particle, and then add E&M interactions as a perturbation. Because the small parameter α is so small, in QED, our perturbation theory converge very fast. First order perturbation gives us an accuracy of the order α1 ~10 -2. Second order perturbation increases the accuracy to α2 ~10 -4. By going to 5th order in perturbation, we can get an accuracy of the order α5 ~10 -10. This is the reason why QED is such a successful theory.

2.5.2. Spin- coupling Phys460.nb 37

In QM I, we treat the spin of an electron as an independent quantity, independent from the orbit agular momentum. For example in a hydrogen atom, for any eigenwavefunction ψn l m(x,y,z), it actually means two degenerate states: (1) one electron with wavefunction ψ n l m(x,y,z) and spin up and (2) one electron with wavefunction ψn l m(x,y,z) and spin down. This conclusion remains the same after we take into accound the relastivistic correction (now the eigenenergy depends on both n and l, but still for every eigenwavefuntion, it means two degenerate state when we taken into account the spins). In this section, we will consider one more effect, which was ignored previously and this effect will tell us the spin of an electron and its orbital motion are coupled togehter. Warning: you may find that the derivtion in this section very disturbing. Because for multiple times, after some derivations, we will say the following without providing much justification, “by the way, this result is in fact not quite right, and we will need to throw in an extra factor of 2 to get the correcti answer.” The reason for these extra factors of 2 is because this section is NOT treating spin-orbit coupling in the rigroius and correct way, which requires Dirac’s equation. Instead, what we are trying to do here is to use various tricks trying to recover Dirac’s finally conclusion without using Dirac’s equation. These tricks (they are not rigrious at all) get some part of the story right, but in many cases, they lead to wrong results. Because we already know the right answer from Dirac’s equation, whenever we find that these tricks fail to get the correct answer, we will correct it by adding some extra factor. Within our deviation, these extra factor looks totally unreasonable and weird, but if one start from Dirac’s equation, the results are all very natural and straghtforward. Bottom line, please don't take these derivations very seriously, because they are not supposed to give (fully) correct descriptoin after all. But the physics, at the end of the day, is correct. Magnetic dipole of an electron If a charge partilce moves in ciricles, it a creates circular , and the circular current will result in an magnetic dipole moment (according to E&M). To see this, we use a simple model to demonstrate this physics. Assuming that we have a ring, and there is a charged particle (with charge q) moving around the ring. The dipole momentu is

→ 1 → → μ = q r ×v (2.321) 2 where q is the charge of the particle. r→ and v→ are the location and velocity of the particle.

→ 1 → → q → → q → μ = q r ×v = m r ×v = L (2.322) 2 2m 2m → where L =mr→ ×v→ is the angular momentum. Simillary, if we have a spinning charged particle, the angular momentum from the spin will also result in a magnetic dipole. The diople moment from spins is also proportional to the angluar momentum of the spin, but with an extra factor known as the g-factor

→ q → μ =g S (2.323) 2m

The charge of an electron is -e (negative charge), so

→ e → μ = -g S (2.324) 2m and g is a number, whose value is really close to 2. In this course, we will say that g= 2 for simplicity, but in reality, g= 2.00231930436182. In Dirac theory, g is exactly 2. The reason the real value of g is a little bit larger than 2 is due to interactions between electrons and photons (light), which wasn’t considered in Dirac’s equation.

Note: in many cases, people absorb the minus sign into the definition of g,

→ e → μ =g S (2.325) 2m where g=-2. But no matter what convetion one adopts,

→ e → μ = - S (2.326) m

We can define the Bohr magneton, which is a fundimental physics constant 38 Phys460.nb

eℏ → -24 μB = = 9.27400968(20)×10 J/T (2.327) 2m

and → → → e → eℏ S S μ =g S =g =gμ B (2.328) 2m 2m ℏ ℏ

For an electron, the magnetic dipole is ±μ B. We demonstrate this by considering the dipole moment along z direction

Sz μz =gμ B (2.329) ℏ

The spin operator Sz has eigenvalues ±ℏ/2, and g = -2. For an eigenstate of Sz,

-μB ifS z eigenvalue is+ℏ/2,i.e. spin up μz = (2.330) +μB ifS z eigenvalue is-ℏ/2,i.e. spin down

Effective B field from the nucleon If we stand on an electron (using the electron as our reference frame), we will find that the nucleon is moving around us in a circle. Because the nucleon has postive charge +e, when it moves around us, it generate a circular current and thus leads to a magnetic field. According to the “Biot–Savart law” in E&M, the B field generated by a wire with current I is → → → μ0 I ⅆl ×r B =  (2.331) 4π r3 → where ⅆl is a small section of the wire and the direction is parrella to the wire. r→ is the distance between the wire and the place at which we want to measure the B field. For a circular motion of the nucleon, the wire here is a circle and we want to know the B field at the center of the circle

μ0 I 2π rⅆθr μ0 I 2π μ0 I μ0 I B=  =  ⅆθ= 2π= (2.332) 4π 0 r3 4π 0 4πr 2r

The current I here is e I= (2.333) T

where e is (the absolute value of) the charge of an electron (remember that the nucleon in a hydrogen atome is +e). T is the how long it takes for the nucleon to go around a ciricle. e e e ω I= = = T 2π/ω 2π (2.334)

Here, ω is the angular velocity. Notice that the angular velocity here is the same as the angular velocity of the electron ω (in the rest frame). eω e L I= = (2.335) 2π 2πmr 2

So

μ0 I μ0 e L e L B= = =ϵ 0 μ0 3 3 (2.336) 2r 2 2πmr 4πϵ 0 m r → → 2 Notice that ϵ0 μ0 =1c and in addition, it is easy to realize that B //L , so we get → → → e L e L B =ϵ 0 μ0 = (2.337) 3 2 3 4πϵ 0 m r 4πϵ 0 m c r

Magnetic dipole in an B field

We proved above that using the frame of the electron, the electron feels a B field, which is generated by the nucleon Phys460.nb 39

→ → e L B = (2.338) 2 3 4πϵ 0 m c r and the electron has a magnetic diople

→ e → μ = - S (2.339) m

We have a dipole in a B field, we shall have energy 2 → → e 1 → → H'=-μ ·B = S ·L (2.340) 2 2 3 4πϵ 0 m c r

Here, our naive tricks miss a factor of 1/2 in comparison to the correct result (from Dirac’s equation). The right result should be 2 e 1 → → HSO = S ·L (2.341) 2 2 3 8πϵ 0 m c r

Therefore, we shall add one extra term to the Hamiltonian

H=H 0 +H SO (2.342)

Here, H0 is what we learned in QM1. And HSO is this new term. Here, we treat H0 as unperturbed Hamiltonian, and treat HSO as a small perturbation.

Basis without HSO For the commutation relations for the orbital angular momentum, we know that

[Lx,L y] =ⅈℏL z (2.343)

[Ly,L z] =ⅈℏL x (2.344)

[Lz,L x] =ⅈℏL y (2.345) or we can write the same formular as

[Li,L j] =ⅈℏϵ i,j,k Lk (2.346) where ϵi,j,k is the Levi-Civita symbol.

For L, we know that we can define the opertor L2

2 2 2 2 L =L x +L y +L z (2.347)

2 And we know that it commute with Lz, L ,L z= 0. As a result, we cannot measure all the three components of the angular momentum due to 2 the uncertainly principle. But, we can measure L and Lz at the same time, by defining common eigenstates for these two opertoators

2 2 L ψl m(x,y,z) =l(l+1)ℏ ψl m(x,y,z) (2.348)

Lz ψl m(x,y,z) =mℏψ l m(x,y,z) (2.349)

Here, l is a non-negative integer, l= 0, 1, 2,…, and m is an integer between + l and -l. The eigenwavefunctions is very easy to write down in spherical corrediate

ψl m(r,θ,ϕ) =R(r)Y lm(θ,ϕ) (2.350) where R(r) is an arbitary function of r (the function doesn’t depends on θ or ϕ), and Y lm(θ,ϕ) are a set of special functions known as the spherical harmonics. For spins, we have the same communtation relation,

[Sx,S y] =ⅈℏS z (2.351)

[Sy,S z] =ⅈℏS x (2.352)

[Sz,S x] =ⅈℏS y (2.353) 40 Phys460.nb

And again, we can define

2 2 2 2 S =S x +S y +S z (2.354)

2 2 And same as above, S ,S z= 0, so we can measure S and Sz at the same time.

S2 s,m=s(s+1)ℏ 2 s,m (2.355)

Sz s,m〉=mℏ s,m〉 (2.356)

where s is a non-negative integer or half-integer, s= 0, 1/2, 1, 3/2,… Once s is determined, m=-s,-s+ 1,…,s- 1,s. For electrons, s=1/2, and thus m=-1/2 or +1/2 → → In addition, we know that S and L commute with each other, → → S,L =0 (2.357)

→ → Without spin-orbit coupling (i.e. the unperturbed Hamiltonian H0), we can easily prove that H0,S =H 0,L = 0, as a result, we find that the

2 2 following operators commute with one another, H0, L , Lz, S , Sz. So we can request our quantum states to be common eigenstates of all these

operators: n,l,m,s,s z〉 13.6 eV H0 n,l,m,s,s z〉=- n,l,m,s,s z (2.358) n2

2 2 L En,l,m,s,s z=l(l+1)ℏ n,l,m,s,s z (2.359)

Lz En,l,m,s,s z〉=mℏ n,l,m,s,s z〉 (2.360) 3 2 2 S En,l,m,s,s z= ℏ n,l,m,s,s z (2.361) 4

Sz En,l,m,s,s z〉=s z ℏ n,l,m,s,s z〉 (2.362)

where sz = +1/2 or -1/2. For an electron, we know that s is always 1/2, so we don’t really need to writeit out: n,l,m l,m s〉. Compare to the

results without spins (ψlmn), the only thing we get here is an extra index sz = ±1/2. This quantum number tells me whether my spin is pointing up or down. At the end of the day, we didn’t get anything beyond what we have already known, except that we now need to specify whether the spin of the electron is up or down.

With SO coupling, the basis desribed above is NOT a good option, because 〈n,l,m,s z HSO n',l',m',s z '〉 ≠ 0, i.e. to do degenerate perturbation theory, we will need a new basis.

Basis with HSO To get the proper basis, we can go throw the derivation that we demonstrated for degenerate perturbation theory. Here, instead, we will use a  trick to get the correct basis direction. The trick is what we proved early on. We know that if we can find a quantum operator, A, which  commutes with both H0 and the perturbation HSO, we can use common eigenstates of A and H0 as a set of basis. If in this set, every state has a  different eigenvalue for A, then it is a good state for degenerate pertubation theory.

2 In the previous section (relastivisctic correction), we used L and Lz to serve as the A operator. Here, after taking into account spins and for the 2 2 2  perturbatoin HSO, we will need to use L , S , J and Jz as A. If an electron have both orbit and spin angular momenta, we can add them up to get the total angular momentum → → → J =L +S (2.363)

or equivallently,

Jx =L x +S x (2.364)

Jy =L y +S y (2.365)

Jz =L z +S z (2.366) Phys460.nb 41

For Js, we have the same commutation relation

[Jx,J y] =ⅈℏJ z (2.367)

[Jy,J z] =ⅈℏJ x (2.368)

[Jz,J x] =ⅈℏJ y (2.369)

And we can also define J2 as

2 2 2 2 J =J x +J y +J z (2.370)

2 2 Same as L and S, we know that J ,J z= 0. So we can measure J and Jz at the same time

J2 j,m=j(j+1)ℏ 2 s,m (2.371)

Jz j,m〉=mℏ s,m〉 (2.372) where j is an integer or half-integer, j = 0, 1/2, 1, 3/2,… Once j is determined, m=-s,-s+ 1,…,s- 1,s. For electrons, s=1/2, and thus m=-1/2 or +1/2.

If we have a particle with spin quantum number s and orbit angular momentum quantum number l, then j=l+s,l+s- 1,… l-s . NOTE: j cannot be negative. For spin s=1/2, this means that j =l-1/2 or j =l+1/2 for l≥ 1. And j=1/2 if l= 0. ◼ If we put an electron on an s-wave state (l=0), the total angular momentum j=1/2 ◼ If we put an electron on an p-wave state (l=1), the total angular momentum j=1/2 or 3/2 ◼ If we put an electron on an d-wave state (l=2), the total angular momentum j=3/2 or 5/2 ◼ If we put an electron on an f-wave state (l=3), the total angular momentum j =5/2 or 7/2 ◼ ...

2 2 2 2 2 2 In our homework, we will show that J , Jz, L , S compute with HSO. It is also straightforward to see that J , Jz, L , S all commute with H0, so  we can use them as our A operator. In addition, it is also easy to verify that these four operators commute with each other, so we can define 2 2 2 common eigenstates for H0, J , Jz, L , S and using this common eigenstates as our basis for perturbation theory -13.6 eV H0 n,l,s,j,j z〉= n,l,s,j,j z (2.373) n2

2 2 L n,l,s,j,j z=l(l+1)ℏ n,l,s,j,j z (2.374) 3 2 2 2 S n,l,s,j,j z=s(s+1)ℏ n,l,s,j,j z= ℏ n,l,s,j,j z (2.375) 4

2 2 J n,l,s,j,j z=j(j+1)ℏ n,l,s,j,j z (2.376)

Jz n,l,s,j,j z〉=j z ℏ n,l,s,j,j z〉 (2.377)

Notice that → → → → → → → → → → → → → → J2 =J ·J =L +S ·L +S =L ·L +S ·S +2S ·L =L 2 +S 2 +2S ·L (2.378)

As a result, 2 2 2 → → J -L -S S ·L = (2.379) 2

So we can write our pertubation as 2 2 2 2 2 e 1 → → e 1 J -L -S HSO = S ·L = (2.380) 2 2 3 2 2 3 8πϵ 0 m c r 8πϵ 0 m c r 2

The first order perturbation theory e2 1 J2 -L 2 -S 2 1 En,l,s,j,j z =〈n,l,s,j,j z HSO n,l,s,j,j z〉= n,l,s,j,j z n,l,s,j,j z= 2 2 3 8πϵ 0 2m c r 42 Phys460.nb

e2 1 j(j+1)-l(l+1)-s(s+1) 2 n,l,s,j,j z ℏ n,l,s,j,j z= 2 2 3 8πϵ 0 2m c r e2 ℏ2 j(j+1)-l(l+1)-s(s+1) 1 n,l,s,j,j z n,l,s,j,j z= 2 2 3 16πϵ 0 m c r

The average value for 1r 3 is known for the unperturbed Hamiltonian

1 → 1 1 * n,l,s,j,j z n,l,s,j,j z=  ⅆr ψn,l,m (r,θ,ϕ) ψn,l,m(r,θ,ϕ) = (2.382) r3 r3 l(l+1/2)(l+1)n 3 a3

So e2 ℏ2 j(j+1)-l(l+1)-s(s+1) 1 1 En,l,s,j,j z = n,l,s,j,j z n,l,s,j,j z= 2 2 3 16πϵ 0 m c r 3 3 (2.383) e2 ℏ2 j(j+1)-l(l+1)- 1 e2 1 ℏ2 1 j(j+1)-l(l+1)- 4 = 4 n 2 2 3 3 4 2 2 2 16πϵ 0 m c l(l+1/2)(l+1)n a 16πϵ 0 n a a m c l(l+1/2)(l+1)

Remember, the Bohr radius is 2 ℏ 4πϵ 0 a= (2.384) m e2

and the eigenenergies for the unperturbed Hamiltonian is 1 e2 En = - (2.385) 2 n 8πϵ 0 a e2 ℏ2 j(j+1)-l(l+1)-s(s+1) 1 1 En,l,s,j,j z = n,l,s,j,j z n,l,s,j,j z= 2 2 3 16πϵ 0 m c r 3 3 e2 ℏ2 j(j+1)-l(l+1)- 1 e2 1 m e2 1 j(j+1)-l(l+1)- 4 = 4 n= 2 2 3 3 4 2 2 2 (2.386) 16πϵ 0 m c l(l+1/2)(l+1)n a 16πϵ 0 n a 4πϵ 0 m c l(l+1/2)(l+1) 3 3 2 2 2 e 1 1 j(j+1)-l(l+1)- En j(j+1)-l(l+1)- 4 n= 4 n 2 2 2 8πϵ 0 n a m c l(l+1/2)(l+1) m c l(l+1/2)(l+1)

Recall that in the previous section, we find that the relastivistic correction is (at the first order)

0 2 En  4n - -3 (2.387) 2mc 2 l+ 1 2

If we combine both effects together, to the first order, the energy is

0 2 3 En  (j+1)-l(l+1)- 4n 0 4 En,j,l,s,j z =E n + 2 n- +3 + … (2.388) 2mc 2 l(l+1/2)(l+1) l+ 1 2

Notice that j=l+1/2 or l-1/2, so we have l=j-1/2 or j+1/2. For l=j-1/2, we find that

0 2 En  4n 0 En,j,l,s,j z =E n - -3 + … (2.389) 2mc 2 j+1/2

for l=j+1/2, we find exactly the same result

0 2 En  4n 0 En,j,l,s,j z =E n - -3 + … (2.390) 2mc 2 j+1/2

So we conclude, no matter what, we have Phys460.nb 43

0 2 En  4n 0 En,j,l,s,j z =E n - -3 + … (2.391) 2mc 2 j+1/2

After taken into account both SO coupling and relastivisitic correction, we find that the energy of a quantum state only depends on n and j,

0 2 En  4n 0 En,j =E n - -3 + … (2.392) 2mc 2 j+1/2

This is the fine structure correction in a hydrogen atom. According to this formular, the fine structure correctio always reduces the energy of a state by a very smalll fraction (the correction is α2 ~10 -4, which is a 0.01% change). The smaller the j is the bigger this correction is. So s-wave states (with l= 0 get the largest modification). For states with l> 0, e.g. p-wave, d-wave, etc., they splitts into two different energy levels (with j=l-1/2 and j =l+1/2, and the former has lower energy than the latter). NOTE: the fine structure correction can also be written as 0 En 4n 0 En,j =E n 1- -3 + … (2.393) 2mc 2 j+1/2 because 1 e2 2 α2 0 2 2 En = - m c = - m c (2.394) 2 2 2n 4πϵ 0 ℏc 2n wehre α is the fine structure constant e2 1 α= ≈ (2.395) 4πϵ 0 ℏc 137.036 we can rewrite the formular as α2 4n α2 n 3 13.6 eV α2 n 3 0 0 En,j =E n 1+ -3 + …=E n 1+ - + …=- 1+ - + … (2.396) 4n 2 j+1/2 n2 j+1/2 4 n2 n2 j+1/2 4

2.6. The

In the previous section, we found that after considering relativistic effects (i.e., fine structure), the eigenenergies in a hydrogen atom only depends on the quantum numbers n and j. In particular, the energy is independent of jz.

For a fixed j, j z = -j,-j+ 1,…,j- 1,j, all have the same energy, i.e. 2j+ 1-fold degeneracy. In this section, we will show that in the presence of an external B field, these 2j+ 1-fold degeneracy will be lifted. → In a magnetic field, the energy of a magnetic dipole is E = -μ→·B. So for an atom

→ → → HZ '=-μ L +μ S·B (2.397)

→ where μL is the magnetic dipole moment from orbit motion → → e → L μL = - L = -μB (2.398) 2m ℏ

→ and μS is the magnetic dipole moment from electron spin → → e → e → S μS = -2× S = - S = -2μ B (2.399) 2m m ℏ → → where μ = eℏ = 5.788×10 -5 eV/ T is Bohr magneton. Here, L and S are angular momenta from orbit motion and electron spin respectively B 2m 44 Phys460.nb

→ → L +2S → HZ '=μ B ·B (2.400) ℏ

Without loss of genericity, we will set B to be along the z direction, so the total energy is → B =Bz (2.401)

As a result,

Lz +2S z HZ '=μ B B (2.402) ℏ

Consider

H=H 0 +H r '+H SO '+H z ' (2.403)

where H0 is the Hamiltonian that we studied in QM I (kinetic energy+1/r attraction), and Hr ' is the relativistic correction. HSO ' is the SO coupling effect, and H '=μ B Lz+2S z . Z B ℏ

2.6.1. Difficulty

For the Hamiltonian above H, the key difficulty lies in the fact that the last two terms, H SO ' and Hz ', do not commute with each other. For Hz ',

we must know Lz and Sz. However, we have learned early on HSO ' doesn’t commute with Lz and Sz (HSO ' commutes with jz, but not with Sz or

Lz, as we showed in our homework). So, we cannot measure HSO ' with Lz and Sz at the same time, but HZ ' needs information about Lz and Sz. This is the confliction

→ → NOTE: this problem comes from the g factor for electrons. μ→ = -μ L and μ→ = -2μ S , the prefactor for them are DIFFERENT! (differ L B ℏ S B ℏ by a factor of 2, i.e., the g-factor). If there is no this extra factor g = 2, things would be very easy. There, H '=μ B Lz+Sz =μ B Jz , so we only Z B ℏ B ℏ

need Jz. But unfortunately, Hz ' is not proportional to Jz.

2.6.2. Strong field

When Hz >>H SO ', we can treat HSO ' and Hr ' (they two are comparable as we learned in the previous section) as perturbation, and thus our unperturbed Hamiltonian is

H0 +H z ' (2.404)

The eigenstates of this Hamiltonian is the same as the eigenstates of H0: n,l,m l,m s〉. Here, n,l,m l〉 are the eigenwavefunctions that we

learned in QM I. Here, we add back the spin Sz quantum state ms

2 L n,l,m l,m s=ℏl(l+1) n,l,m l,m s (2.405)

Sz n,l,m l,m s〉=m s ℏ n,l,m l,m s〉 (2.406)

Lz n,l,m l,m s〉=m l ℏ n,l,m l,m s〉 (2.407) 13.6 eV H0 n,l,m l,m s〉=- n,l,m l,m s (2.408) n2

μB B μB B μB B μB B Hz ' n,l,m l,m s〉= (Lz +2S z) n,l,m l,m s= Lz n,l,m l,m s+2 Sz n,l,m l,m s= ℏm l n, ℏ ℏ ℏ ℏ (2.409) μB B l,m l,m s+2 ℏm s n,l,m l,m s=μ B B(ml +2m s) n,l,m l,m s ℏ

So our zeroth order eigenenergy is 13.6 eV 0 En,l,ml,ms = - +μ B B(ml +2m s) (2.410) n2

At B= 0, we know that energy is independent of m l and ms, i.e., all quantum states are degenerate with (at least 2×(2l+1)-fold degenerate). For finite B however, these states splits. Phys460.nb 45

Example: if we consider states n= 2 and l= 1 (first excited states with orbit angular moment quantum number l= 1). There, m l = -1, 0,+1 and m = - 1 or + 1 . At B= 0, all these six states are degenerate (E=-13.6/4=-3.4 eV). In the presence of strong B field, s 2 2

-3.4 eV+2μ B B ml = +1 andm s = +1/2 -3.4 eV+μ B B ml = 0 andm s = +1/2 E 0 = -3.4 eVm = -1 andm = +1/2, or,m = +1 andm = -1/2 n,l,ml,ms l s l s (2.411) -3.4 eV-μ B B ml = 0 andm s = -1/2 -3.4 eV-2μ B B ml = -1 andm s = -1/2

Now, we consider HSO ' and Hr ' . Because we assumed that they are much smaller than H0 and HZ ', we treat them as perturbation and compute the first order correction to the eigenenergy

1 En,l,ml,ms =〈n,l,m l,m s Hr '+H SO ' n,l,m l,m s〉 (2.412)

The realistic correction is same as what we learned before

0 2 En  4n 〈n,l,m l,m s Hr ' n,l,m l,m s〉=- -3 (2.413) 2mc 2 l+ 1 2

Because 1 m e2 2 13.6 0 En = - = - eV (2.414) 2 2 2 2n ℏ 4πϵ 0 n e2 1 α= ≈ (2.415) 4πϵ 0 ℏc 137.036 we know that

0 2 2 En 1 1 e 1 = - = - α2 (2.416) 2 2 2 2 2 m c 2n ℏ c 4πϵ 0 2n so

0 0 0 2 2 En En 4n En α 4n α n 3 〈n,l,m l,m s Hr ' n,l,m l,m s〉=- -3 = - -3 = -13.6 eV - (2.417) 2 m c2 l+ 1 2 2n 2 l+ 1 n4 l+ 1 4 2 2 2

For spin-orbit coupling,

〈n,l,m l,m s HSO ' n,l,m l,m s〉= 2 2 e 1 → → e 1 → → (2.418) n,l,m l,m s S ·L n,l,m l,m s= n,l,m l,m s S ·L n,l,m l,m s 2 2 3 2 2 3 8πϵ 0 m c r 8πϵ 0 m c r

Notice that in the zeroth order wavefunctions, n, l, ml, ms〉, spin and orbit angular momenta are independent of each other, so → → → → n,l,m l,m s S ·L n,l,m l,m s=S ·L =〈S x〉〈Lx〉+〈S y〉〈Ly〉+〈S z〉〈Lz〉 (2.419)

2 In QM I, we learned that for eigenstates of L and Lz, 〈Lx〉=〈L y〉= 0. And similarly, 〈S x〉=〈S y〉= 0. And thus → → → → 2 n,l,m l,m s S ·L n,l,m l,m s=S ·L =〈S z〉〈Lz〉=m s ℏm l ℏ=m s ml ℏ (2.420)

As a result,

〈n,l,m l,m s HSO ' n,l,m l,m s〉= → → 2 2 2 e 1 S ·L e ms ml ℏ 1 (2.421) n,l,m l,m s n,l,m l,m s= n,l,m l,m s n,l,m l,m s 2 2 3 2 2 3 8πϵ 0 m c r 8πϵ 0 m c r and

1 → 1 1 * n,l,s,j,j z n,l,s,j,j z=  ⅆr ψn,l,m (r,θ,ϕ) ψn,l,m(r,θ,ϕ) = (2.422) r3 r3 l(l+1/2)(l+1)n 3 a3 46 Phys460.nb

2 where a is the Bohr radius a= ℏ 4πϵ 0 m e2 So

〈n,l,m l,m s HSO ' n,l,m l,m s〉= 2 2 2 2 2 2 e ms ml ℏ 1 1 1 e e m ms ml = (2.423) 2 2 2 3 3 2 3 8πϵ 0 m c  ℏ 4πϵ 0  l(l+1/2)(l+1)n 2 4πϵ 0 cℏ 4πϵ 0 ℏ l(l+1/2)(l+1)n m e2

Because 1 m e2 2 13.6 0 En = - = - eV (2.424) 2 2 2 2n ℏ 4πϵ 0 n e2 1 α= ≈ (2.425) 4πϵ 0 ℏc 137.036

we can rewrite the formula as ms m 2 l 〈n,l,m l,m s HSO ' n,l,m l,m s〉= 13.6 eVα l(l+1/2)(l+1)n 3 (2.426)

So our first order correction is

2 α n 3 ms m 1 2 l En,l,ml,ms =〈n,l,m l,m s Hr '+H SO ' n,l,m l,m s〉=-13.6 eV - + 13.6 eVα = n4 l+ 1 4 l(l+1/2)(l+1)n 3 2 (2.427) 13.6 eV 3 1 ms ml 13.6 eV 3 l(l+1)-m s ml α2 - + = α2 -  n3 4n l+ 1  l(l+1/2)(l+1) n3 4n ll+ 1 (l+1) 2 2

So

13.6 eV 13.6 eV 3 l(l+1)-m s m 0 1 2 l En,l,m ,ms =E n,l,m ,ms +E n,l,m ,ms = - +μ B B(ml +2m s) + α  -  l l l 2 3 (2.428) n n 4n ll+ 1 (l+1) 2

Bottom line: at very strong field (second term much larger than the last one), the energy splittings between the levels are proportional

to B and the slop is proportional to μB(ml +2m S). The eigenstates are (almost) n, l, ml, ms〉, where ms and ml are good quantum numbers. (we should arrange the states according to the orbit and spin angular moment, NOT the total angular momentum j).

2.6.3. Weak field

When Hz <

section. There, we know that eigenstates are n,j,l,s,m j〉 and the eigenenergy is α2 4n α2 n 3 13.6 eV α2 n 3 0 0 En,j =E n 1+ -3 + …=E n 1+ - + …=- 1+ - + … (2.429) 4n 2 j+1/2 n2 j+1/2 4 n2 n2 j+1/2 4

In first order perturbation theory, the energy correction is

1 En,j,l,s,m j =〈n,j,l,s,m j Hz ' n,j,l,s,m j〉= → → L +2S → μB → → → → (2.430) n,j,l,s,m j μB ·B n,j,l,s,m j= n,j,l,s,m j B·L +2B ·S n,j,l,s,m j ℏ ℏ → → 2 The key is to compute expectation values of L and S for eigenstates of J and Jz. Here, we use the fact that

→ → → L//S //J  (2.431)

So Phys460.nb 47

→ → → → J L=L ·J  (2.432) J2

Here, → → → → → → → → B·J → → B·J B·L =L ·J =n,j,l,s,m j L·J n,j,l,s,m j (2.433) J2 J2 → → For B//z, we have B·J =BJ z

→ → → → BJz B·L =n,j,l,s,m j L·J n,j,l,s,m j (2.434) J2

Here, we use the fact that

Jz n,j,l,s,m j〉=m j ℏ n,j,l,s,m j〉 (2.435)

2 2 J n,j,l,s,m j=j(j+1)ℏ n,j,l,s,m j (2.436) so

→ → → → BJz B·L =n,j,l,s,m j L·J n,j,l,s,m j= J2 (2.437) → → B m j ℏ B m j → → n,j,l,s,m j L·J n,j,l,s,m j= n,j,l,s,m j L·J n,j,l,s,m j j(j+1)ℏ 2 j(j+1)ℏ → → For L ·J , we use the fact that → → → S =J -L (2.438)

→ → → → → → → → → → → → S ·S =J -L ·J -L =J .·J +L ·L -2L ·J (2.439)

Thus, → → → → → → → → J.·J +L ·L -S ·S L·J = (2.440) 2

So, → → n,j,l,s,m j L·J n,j,l,s,m j= → → → → → → J.·J +L ·L -S ·S ℏ2 j(j+1) +ℏ 2 l(l+1)-ℏ 2 s(s+1) n,j,l,s,m j n,j,l,s,m j=n,j,l,s,m j n,j,l,s,m j= 2 2 (2.441) ℏ2 j(j+1) +ℏ 2 l(l+1)-ℏ 2 s(s+1) ℏ2 = [j(j+1) +l(l+1)-s(s+1)] 2 2 we know that s=1/2 for electrons, 2 → → ℏ n,j,l,s,m j L·J n,j,l,s,m j= [j(j+1) +l(l+1)-3/4] (2.442) 2

Therefore, → → B·L =

2 B m j → → B m j ℏ B m j ℏ (2.443) n,j,l,s,m j L·J n,j,l,s,m j= [j(j+1) +l(l+1)-3/4] = [j(j+1) +l(l+1)-3/4] j(j+1)ℏ j(j+1)ℏ 2 2j(j+1)

Similarly, we can prove that → → B·S =

2 B m j → → B m j ℏ B m j ℏ (2.444) n,j,l,s,m j S ·J n,j,l,s,m j= [j(j+1)-l(l+1) +3/4] = [j(j+1)-l(l+1) +3/4] j(j+1)ℏ j(j+1)ℏ 2 2j(j+1) 48 Phys460.nb

And thus

1 En,j,l,s,m j =〈n,j,l,s,m j Hz ' n,j,l,s,m j〉= → → L +2S → μB → → → → n,j,l,s,m j μB ·B n,j,l,s,m j= n,j,l,s,m j B·L +2B ·S n,j,l,s,m j= ℏ ℏ μB B m j ℏ B m j ℏ (2.445)  [j(j+1) +l(l+1)-3/4] +2 [j(j+1)-l(l+1) +3/4]= ℏ 2j(j+1) 2j(j+1)

μB B m j ℏ 1 [3j(j+1)-l(l+1) +3/4] =μ B B m j [3j(j+1)-l(l+1) +3/4] ℏ 2j(j+1) 2j(j+1)

We can define an atomic g-fact, l+ 3  l- 1  1 3 l(l+1)-3/4 3 2 2 g j = [3j(j+1)-l(l+1) +3/4] = - = - (2.446) 2j(j+1) 2 2j(j+1) 2 2j(j+1)

we know that j =l+1/2 or j=l-1/2. If j=l+1/2, we find that

3 1 3 1 l+  l-  l+  l-  l- 1 l+ 1 -1 3 2 2 3 2 2 3 2 3 2 3 1 1 1 g j = - = - = - = - = - + =1+ (2.447) 2 2j(j+1) 2 2l+ 1  l+ 3  2 2l+ 1  2 2l+ 1  2 2 2l+ 1  2l+1 2 2 2 2 2

If j =l-1/2

3 1 3 1 l+  l-  l+  l-  l+ 3 l+ 1 +1 3 2 2 3 2 2 3 2 3 2 3 1 1 1 g j = - = - = - = - = - - =1- (2.448) 2 2j(j+1) 2 2l- 1  l+ 1  2 2l+ 1  2 2l+ 1  2 2 2l+ 1  2l+1 2 2 2 2 2

With this g j - factor,

1 En,j,l,s,m j =gμ B B m j (2.449)

The energy correction (first order), is proportional to m j (total angular momentum along the field). Total energy: 13.6 eV α2 n 3 En,j = - 1+ - +gμ B B m j (2.450) n2 n2 j+1/2 4

2.6.4. Intermediate-field

H=H 0 +H r '+H SO '+H z ' (2.451)

When HSO '~H z ', we should treat the last three terms as perturbation. Here, we can treat the problem using degenerate perturbation theory.

For H0, we consider n= 2 states (first excited states). In QM I, we learned that there are 4 degenerate states: one s-wave state with l= 0 and

three p-wave states (l= 1, and m l = -1, 0, 1). If we consider spins, there are 4×2= 8 states. In first order degenerate perturbation theory, we can ignore all other states except n= 2 states, and only focus on these 8 states. So the perturbation Hamiltonian is now a 8×8 matrix, which was shown in textbook (page 248). The eigenvalues of this matrix give us the first order corrections in energy.

2.7. Summary

Objective: compute eigenvalues for the Hamiltonian  H ψn=E n ψn (2.452)

Key assumption:    H =H 0 +λH ' (2.453)  where the second term λH ' is dramatically smaller than the first part. Phys460.nb 49

2.7.1. nondegenerate perturbation theory  Step 1: solve for eigenstates for H0

 0 0 0 H0 ψ n=E n ψ n (2.454)

If the state that we consider has no degeneracy, we use nondegenerate perturbation theory Step 2: first order correction

1 0  0 En = ψ n H ' ψ n (2.455)

Step 3: second order correction 0 0 2 1 ψm H' ψn  2 0 0 0 0 En =  ψn H' ψm  ψm H' ψn =  (2.456) m≠n 0 0 m≠n 0 0 En -E m En -E m

Step 4: eigenenergy

0 1 2 2 En =E n +λE n +λ En + … (2.457)

Wave functions: 1 0 0 0 ψn〉= ψn +λ  ψm  ψm H' ψn+… m≠n 0 0 (2.458) En -E m

2.7.2. degenerate perturbation theory  Step 1: solve for eigenstates for H0

 0 0 0 H0 ψ n=E n ψ n (2.459)

If the state that we consider has degeneracy (i.e. there is at least one other state has the same eigenenergy), we use degenerate perturbation theory:

0 0 0 H0 ψa =E ψa  (2.460) and

0 0 0 H0 ψb =E ψb  (2.461)

Step 1: Create a n×n matrix (if there is an n-fold degeneracy), 0 0 0 0 ψa H' ψa  ψa H' ψ  W= b 0 0 0 0 (2.462) ψb H' ψa  ψb H' ψb 

Step 2: The eigenvalues of the matrix is the first order correction

0 2 E1 =E +λE + +Oλ  (2.463)

0 2 E2 =E +λE - +Oλ  (2.464)

Wavefunctions: eigenvectors

Waa Wab α1 α1    =E +  (2.465) Wba Wbb β1 β1 and

Waa Wab α2 α2    =E -  (2.466) Wba Wbb β2 β2 where E+ and E- are the two eigenvalues.

0 0 0 ψ1 =α 1 ψa +β 1 ψb  (2.467) 50 Phys460.nb

0 0 0 ψ2 =α 2 ψa +β 2 ψb  (2.468)

2.7.3. Perturbation theory in matrix formula (example: homework 2.3)

This case usually uses eigenstates of H0 as basis,

 0 0 0 H0 ψ n=E n ψ n (2.469)

With one complete set of basis, we can write an operator as a matrix

0  0 (H0)mn = ψ m H0 ψ n (2.470)

If we use eigenstates of H0 as basis, the matrix is diagonal and the diagonal components are eigenvalues of the H0 0 E1 0 0… 0 0E 2 0 ... H0 → 0 (2.471) 0 0E 3 … ⋮ ⋮ ⋮ ⋱

Using the same basis, we can write H ' as a matrix

〈ψ1 H' ψ1〉 〈ψ1 H' ψ2〉 〈ψ1 H' ψ3〉… 〈ψ2 H' ψ1〉 〈ψ2 H' ψ2〉 〈ψ2 H' ψ3〉 ... λH'→λ (2.472) 〈ψ3 H' ψ1〉 〈ψ3 H' ψ2〉 〈ψ3 H' ψ3〉… ⋮ ⋮ ⋮ ⋱

In many cases, we only need keep a small number of states (e.g. only the tree states with lowest energy) 0 E1 0 0 0 H0 = 0E 2 0 (2.473) 0 0 0E 3

and

〈ψ1 H' ψ1〉 〈ψ1 H' ψ2〉 〈ψ1 H' ψ3〉 λH'=λ 〈ψ2 H' ψ1〉 〈ψ2 H' ψ2〉 〈ψ2 H' ψ3〉 (2.474) 〈ψ3 H' ψ1〉 〈ψ3 H' ψ2〉 〈ψ3 H' ψ3〉

Objective: compute eigenstates for the H matrix 0 E1 0 0 〈ψ1 H' ψ1〉 〈ψ1 H' ψ2〉 〈ψ1 H' ψ3〉 0 H= 0E 2 0 +λ 〈ψ2 H' ψ1〉 〈ψ2 H' ψ2〉 〈ψ2 H' ψ3〉 (2.475) 0 0 0E 3 〈ψ3 H' ψ1〉 〈ψ3 H' ψ2〉 〈ψ3 H' ψ3〉

Assumption: the second matrix is much smaller than the first Nondegenerate perturbation:

0 0 0 0 Among the three unperturbed eigenvalues, E1 , E2 and E3 , if one of them is different from the other two (e.g. E1 is different), then we can use nondegenerate perturbation theory. First order perturbation:

1 E1 = 〈ψ1 H' ψ1〉 (2.476)

Notice that it is just one element in the second matrix in H. Second order perturbation: 0 0 0 0 0 0 0 0 1 ψ1 H' ψ2  ψ2 H' ψ1  ψ1 H' ψ3  ψ3 H' ψ1  2 0 0 0 0 E1 =  ψ1 H' ψm  ψm H' ψ1 = + (2.477) m≠n 0 0 0 0 0 0 E1 -E m E1 -E 2 E1 -E 3

Notice that the denominator are just elements from the first matrix in H and the numerators are from the second matrix. Degenerate perturbation:

0 0 0 0 0 Among the three unperturbed eigenvalues, E1 , E2 and E3 , if two (or more) of them are identical (e.g. E2 and E3 have the same value, then we can use degenerate perturbation theory. Phys460.nb 51

Step one: create the W matrix ψ 0 H' ψ 0 ψ 0 H' ψ 0 W= 2 2 2 3 0 0 0 0 (2.478) ψ3 H' ψ2  ψ3 H' ψ3 

Step two: compute eigenvalues of W, which are the first order correction