One of the most important problems in quantum mechanics is the simple harmonic oscillator, in part because its properties are directly applicable to field theory. The treatment in Dirac notation is particularly satisfying.

1 Hamiltonian

1 2 q k Writing the potential 2 kx in terms of the classical frequency, ω = m , puts the classical Hamiltonian in the form p2 mω2x2 H = + 2m 2 Since there are no products of non-commuting operators, there is no ambiguity in the resulting Hamiltonian operator, 1 mω2 Hˆ = Pˆ2 + Xˆ 2 2m 2 We make no choice of basis.

2 Raising and lowering operators

Notice that  ip   ip  p2 x + x − = x2 + mω mω m2ω2 2 1 p2  = mω2x2 + mω2 2 2m so that we may write the classical Hamiltonian as

mω2  ip   ip  H = x + x − 2 mω mω

We can write the quantum Hamiltonian in a similar way. Choosing our normalization with a bit of foresight, we define two conjugate operators,

rmω  i  aˆ = Xˆ + Pˆ 2~ mω rmω  i  aˆ† = Xˆ − Pˆ 2~ mω The operator aˆ† is called the raising operator and aˆ is called the lowering operator. Notice that they are not Hermitian. In taking the product of these, we must be careful with ordering since Xˆ and Pˆ ! ! mω iPˆ iPˆ aˆ†aˆ = Xˆ − Xˆ + 2~ mω mω ! mω i i Pˆ2 = Xˆ 2 + XˆPˆ − PˆXˆ + 2~ mω mω m2ω2 ! mω i h i Pˆ2 = Xˆ 2 + X,ˆ Pˆ + 2~ mω m2ω2

1 h i Using the commutator, X,ˆ Pˆ = i~1ˆ, this becomes !  1  1  Pˆ2 aˆ†aˆ = mω2 Xˆ 2 − ~ + ~ω 2 mω m2ω2 ˆ2 ! 1 1 2 2 1 P = mω Xˆ − ~ω + ~ω 2 2 2m 1  1  = Hˆ − ~ω ~ω 2 and therefore,  1 Hˆ = ω aˆ†aˆ + ~ 2

3 The number operator

This turns out to be a very convenient form for the Hamiltonian because aˆ and a† have very simple properties. First, their commutator is simply " ! !# mω iPˆ iPˆ a,ˆ aˆ† = Xˆ + , Xˆ − 2~ mω mω mω  i   i  = X,ˆ − Pˆ + P,ˆ Xˆ 2~ mω mω 2i mω h i = − X,ˆ Pˆ mω 2~ i = − i~ ~ = 1

Consider one further set of commutation relations. Defining Nˆ ≡ aˆ†aˆ = Nˆ †, called the number operator, we have h i N,ˆ aˆ = aˆ†a,ˆ aˆ =a ˆ† [ˆa, aˆ] + aˆ†, aˆ aˆ = −aˆ

and h i N,ˆ aˆ† = aˆ†a,ˆ aˆ† =a ˆ aˆ†, aˆ† +a ˆ† a,ˆ aˆ† =a ˆ†   ˆ ˆ ˆ 1 Notice that N is Hermitian, hence observable, and that H = ~ω N + 2 .

4 Energy eigenstates 4.1 Positivity of the energy Prove that expectation values of Hamiltonian, hence all energies, are positive.

2 Q1: Consider an arbitrary expectation value of the Hamiltonian,

 1 hψ| Hˆ |ψi = hψ| ω aˆ†aˆ + |ψi ~ 2  1  = ω hψ| aˆ†aˆ |ψi + hψ | ψi ~ 2

Use the fact that the norm of any state is positive, hψ |ψi > 0 to show that the energy of any SHO state is positive:

hψ| Hˆ |ψi = E > 0

Hint: It helps to define |βi ≡ aˆ |ψi.

4.2 The lowest energy state Suppose that |Ei is any energy eigenket with

Hˆ |Ei = E |Ei

Consider the new ket formed by acting on |Ei with the lowering operator aˆ.

Q2: Show that aˆ |Ei is also an energy eigenket with energy E − }ω,

Hˆ (ˆa |Ei) = (E − }ω) (ˆa |Ei) Since aˆ |Ei is an energy eigenket, we may repeat this procedure to show that aˆ2 |Ei is an energy eigenket with energy E − 2~ω. Continuing in this way, we find that aˆk |Ei will have energy E − k~ω. This process cannot continue indefinitely, because the energy must remain positive. Let k be the largest integer for which E − k~ω is positive, k k Hˆ aˆ |Ei = (E − k~ω)a ˆ |Ei with corresponding state aˆk |Ei. Then applying the lowering operator one more time cannot give a new k state. The only other possibility is zero. Rename the lowest energy state |0i = A0aˆ |Ei, where we choose A0 so that |0i is normalized. We then must have

aˆ |0i = 0

This is the lowest energy state of the oscillator, called the ground state.

4.3 The complete spectrum Q3: Now that we have the ground state, we reverse the process, acting instead with the raising operator. Show that, acting any energy eigenket, |Ei, the raising operator gives another energy eigenket, aˆ† |Ei, with energy E + ~ω,

†
†
Hˆ aˆ |Ei = (E + ~ω) aˆ |Ei

3 Q4: Find the energy of the state aˆ† |0i.

† Q5: Define the normalized state to be |1i ≡ A1aˆ |0i. Find the normalization constant, A1. There is nothing to prevent us continuing this procedure indefintely. Continuing n times, we have states †
n |ni = An aˆ |0i satisfying  1 Hˆ |ni = n + ω |ni 2 ~ This gives the complete set of energy eigenkets.

4.4 Normalization th †
n Now, consider the expectation of Nˆ in the n state, |ni = An aˆ |0i.

Q6: Develop a recursion relation for the normalization constants, An, starting from 1 = hn | ni. 2 2 Specifically, show that |An−1| = n |An| . Iterating the recursion relation gives

1 |A |2 = |A |2 n n n−1 1 = |A |2 n (n − 1) n−2 . . 1 = |A |2 n! 1 so that, choosing all of the coefficients real, we have the complete set of normalized states

1 n |ni = √ aˆ†
|0i n! √ Q7: Show that aˆ† |ni = n + 1 |n + 1i and find the corresponding relation for aˆ |ni.

5 Wave function

Choosing a coordinate basis to express the condition

aˆ |0i = 0 gives 0 = hx| aˆ |0i rmω  i  = hx| Xˆ + Pˆ |0i 2~ mω rmω  i  = hx| Xˆ |0i + Pˆ |0i 2~ mω rmω  i  = x hx | 0i + hx| Pˆ |0i 2~ mω

4 ˆ d and since we have shown that hx| P |0i = −i~ dx hx | 0i this becomes a differential equation, d mωx hx | 0i + hx | 0i = 0 dx ~

Setting ψ0 (x) = hx | 0i and integrating, dψ mωx 0 = − dx ψ0 ~ ψ (x) mωx2 ln 0 = − ψ0 (0) 2~ 2 − mωx ψ0 (x) = ψ0 (0) e 2~ we find that the wave function of the ground state is a Gaussian, with the magnitude of the remaining coefficient ψ0 (0) determined by normalizing the Gaussian,

1/4 2 mω  − mωx ψ0 (x) = e 2~ π~

Now consider the wave function, ψn (x), for the eigenstates. We have already found that the ground state is given by a normalized Gaussian,To find the wave functions of the higher energy states, consider

ψn (x) = hx | ni 1 n = hx| √ aˆ†
|0i n! 1 1 n−1 = √ hx| aˆ† aˆ†
|0i n p(n − 1)! 1 rmω  i  = √ hx| Xˆ − Pˆ |n − 1i n 2~ mω 1 rmω  i  d   = √ x hx | n − 1i − −i~ hx | n − 1i n 2~ mω dx r   mω ~ d = x − ψn−1 (x) 2n~ mω dx Therefore, we can find all states by iterating this operator,

n  n 0  mω  2 ~ ∂ ψn (x ) = x − ψ0 (x) 2n~ mω ∂x The result is a series of polynomials, the Hermite polynomials, times the Gaussian factor.

Q8: Find ψ1 (x) and ψ2 (x).

6 Time evolution of a mixed state of the oscillator

Consider the time evolution of the most general superposition of the lowest two eigenstates

|ψi = cos θ |0i + eiϕ sin θ |1i

5 Applying the time translation operator with t0 = 0, |ψ, ti = U (t) |ψi − i Htˆ = e ~ |ψi − i Htˆ iϕ − i Htˆ = cos θe ~ |0i + e sin θe ~ |1i − i ωt iϕ − 3 iωt = cos θe 2 |0i + e sin θe 2 |1i − i ωt iϕ −iωt
= e 2 cos θ |0i + e sin θe |1i

Now look at the time dependence of the expectation value of the position operator, which we write in terms q ˆ ~ †
of raising and lowering operators as X = 2mω aˆ +a ˆ . Remembering that the states are orthonormal,

−iϕ iωt
i ωt − i ωt iϕ −iωt
hψ, t| Xˆ |ψ, ti = cos θ h0| + e sin θe h1| e 2 Xeˆ 2 cos θ |0i + e sin θe |1i r = ~ cos θ h0| + e−iϕ sin θeiωt h1|
aˆ +a ˆ†
cos θ |0i + eiϕ sin θe−iωt |1i
2mω r   √  = ~ cos θ h0| + e−iϕ sin θeiωt h1|
cos θ |1i + eiϕ sin θe−iωt |0i + 2 |2i 2mω r   = ~ cos θ sin θe−i(ωt−ϕ) + sin θ cos θei(ωt−ϕ) 2mω r = ~ sin 2θ cos (ωt − ϕ) 2mω √ where we have used aˆ |0i = 0, aˆ |1i = |0i , aˆ† |0i = |1i and aˆ† |1i = 2 |2i. We see that the expected position q ~ oscillates back and forth between ± 2mω sin 2θ with frequency ω. Superpositions involving higher excited states will bring in harmonics, nω, and will then allow for varied traveling waveforms.

√ ˆ 1 †
Q9: Find the expectation value of the momentum, P = 2i 2~mω aˆ − aˆ

 D E2 D E D E2 Q10: Find the uncertainty in position, (∆X)2 = Xˆ − Xˆ = Xˆ 2 − Xˆ

7 Coherent states and the correspondence principle

Any alternative physical theory must approach, in some suitable limit, the previously tested and accepted standard model. Such a limit was stated concretely for quantum mechanics by Niels Bohr in 1920. The correspondence principle states that in the limit of large numbers of quanta, quantum systems should replicate classical behavior. We can see this for the simple harmonic oscillator by finding states that oscillate with the natural frequency ω and its overtones. Schrödinger found that such states are given by minimum uncertainty wave packets called coherent states.

7.1 The uncertainty in position and momentum Schrödinger asked for simple harmonic oscillator states, |λi, for with the dimensionless forms of the position and momentum operators, r mω 1 †
Xˆ0 ≡ Xˆ = aˆ +a ˆ 2~ 2

6 r 1 1 †
Pˆ0 ≡ Pˆ = aˆ − aˆ 2~mω 2i have equal uncertainty, ∆Xˆ0 = ∆Pˆ0 and which achieve the minimum value in the uncertainty relation,

∆Xˆ ∆Pˆ ≥ ~ 0 0 2 The uncertainty in any operator in a state |λi is defined to be ∆Aˆ where  2  D E2 ∆Aˆ ≡ hλ| Aˆ − Aˆ |λi  D E D E2 = hλ| Aˆ2 − 2Aˆ Aˆ + Aˆ |λi

D E D E2 = hλ| Aˆ2 |λi − 2 hλ| Aˆ |λi Aˆ + Aˆ D E D E2 = Aˆ2 − Aˆ D E where we define Aˆ ≡ hλ| Aˆ |λi. The uncertainty for position is then found from the expection value of Xˆ0

D E 1 Xˆ = hλ| aˆ +a ˆ†
|λi 0 2 1 = haˆi + aˆ†
2 ˆ 2 and the expectation value of X0 ,

D E 1 2 Xˆ 2 = aˆ +a ˆ†
0 2 1 = aˆ2 +a ˆaˆ† +a ˆ†aˆ +a ˆ†2
4 1 D E = aˆ2 + a,ˆ aˆ† + 2Nˆ +a ˆ†2 4 1   1  = aˆ2 + 2 Nˆ + + aˆ†2 4 2 1  2 D E  = aˆ2 + Hˆ + aˆ†2 4 ~ω to be given by

2     1 2 2 D E †2 1  2 † † 2 ∆Xˆ0 = aˆ + Hˆ + aˆ − haˆi + 2 haˆi aˆ + aˆ 4 ~ω 4 1  2 D E 2 = aˆ2 + Hˆ + aˆ†2 − haˆi2 − 2 haˆi aˆ† − aˆ† 4 ~ω 1  2 D E 2  = Hˆ + (∆ˆa)2 + ∆ˆa†
− 2 haˆi aˆ† 4 ~ω D E D E  2 ˆ ˆ2 ˆ Similarly, we find P0 , P0 and ∆P0 , D E 1 Pˆ = hλ| aˆ − aˆ†
|λi 0 2i

7 1 1 = haˆi − aˆ† 2i 2i D E 1 2 Pˆ2 = − hλ| aˆ − aˆ†
|λi 0 4 1   = − hλ| aˆaˆ − 2Nˆ − 1 +a ˆ†aˆ† |λi 4 1  2 D E  = Hˆ − aˆ2 − aˆ†2 4 ~ω 2     1 2 D E 2 †2 1  2 † † 2 ∆Pˆ0 = Hˆ − aˆ − aˆ + haˆi − 2 haˆi aˆ + aˆ 4 ~ω 4 1  2 D E 2 = Hˆ − aˆ2 − aˆ†2 + haˆi2 − 2 haˆi aˆ† + aˆ† 4 ~ω 1  2 D E 2  = Hˆ − aˆ2 + haˆi2 − aˆ†2 + aˆ† − 2 haˆi aˆ† 4 ~ω 1  2 D E 2  = Hˆ − (∆ˆa)2 − ∆ˆa†
− 2 haˆi aˆ† 4 ~ω 7.2 Equality of uncertainties and the minimum uncertainty state  2  2 We demand equality of the uncertainties, ∆Xˆ0 = ∆Pˆ0 . Imposing this,

2 D E 2 2 D E 2 Hˆ + (∆ˆa)2 + ∆ˆa†
− 2 haˆi aˆ† = Hˆ − (∆ˆa)2 − ∆ˆa†
− 2 haˆi aˆ† ~ω ~ω 2 2 (∆ˆa)2 + ∆ˆa†
= − (∆ˆa)2 − ∆ˆa†

2 and since both (∆ˆa)2 and ∆ˆa†
are nonnegative, both must vanish. Setting

(∆ˆa)2 = 0 aˆ2 = haˆi2

There are many ways to accomplish this equality, but the simplest choice is to let |λi be an eigenstate of aˆ,

aˆ |λi = λ |λi since this always makes the dispersion vanish. It is easy to show that this choice also gives miniumum uncertainty. 2 We also need ∆ˆa†
= 0, but this follows automatically,

2 2 aˆ†2 − aˆ† = hλ| aˆ† − aˆ†
|λi 2 = hλ| aˆ†aˆ† |λi − aˆ† 2 = hλ| aˆaˆ |λi∗ − aˆ† = λ∗2 − haˆi∗2 = λ∗2 − λ∗2 = 0

The condition aˆ |λi = λ |λi is therefore sufficient to guarantee that the uncertainties in position and momentum will be equal.

8 When this condition holds, the uncertainties are

2     1 1 D E † ∆Xˆ0 = Hˆ − haˆi aˆ 2 ~ω 2     1 1 D E † ∆Pˆ0 = Eˆ − haˆi aˆ 2 ~ω with product

 2  2 1  1 2 ∆Xˆ ∆Pˆ = λ∗λ + − λ∗λ 0 0 4 2 1 ∆Xˆ ∆Pˆ = 0 0 4 and putting the units back in terms of position and momentum,

∆Xˆ∆Pˆ = ~ 2 so this condition also gives a minimum uncertainty wave packet.

7.3 Coherent states We define a coherent state of the harmonic oscillator to be an eigenstate of the lowering operator,

aˆ |λi = λ |λi

To find such states explicitly, expand in the complete number basis,

∞ X |λi = cn |ni n=0

Q11: Find the form of coherent states. Writing X X aˆ cn |ni = λ cn |ni n n work out the effect of aˆ on the left to show that ∞ ∞ X √ X √ cn n |n − 1i = cm+1 m + 1 |mi n=1 m=0

This gives a recursion relation for the coefficients cn. Shifting the summation index to m = n − 1 on the left, ∞ ∞ X √ X √ cn n |n − 1i = cm+1 m + 1 |mi n=1 m=0 and renaming m = n on the right,

∞ ∞ X √ X cm+1 m + 1 |mi = λ cm |mi m=0 m=0 so that term by term equality gives λ cm+1 = √ cm m + 1

9 Iterating this recursion relationship, we find λn cn = √ n! for all n. The coherent state is therefore given by

X λn |λi = √ |ni (1) n n!

n Finally, substituting the expression for the nth number state, |ni = √1 aˆ†
|0i, we express the eigenstates n! of the lowering operator as

n X λ n |λi = aˆ†
|0i n! n † = eλaˆ |0i (2)

These are the coherent states of the simple harmonic oscillator. They correspond to a Poisson distribution of number states.

Q12: Normalize the states. Show using eq.(1) that normalized coherent states are given by

− 1 |λ|2 λaˆ† e 2 e |0i

7.4 Time dependence of coherent states

Q13: Show that the time dependence of a coherenet state, given by acting with Uˆ (t, t0).

Setting t0 = 0 and |λ, t0i = |λi show that

− iωt −iωt |λ, ti = e 2 λe

that is, up to an overall phase the complex parameter λ is just replaced by λe−iωt in the original state.

Q14: Show that the expectation values of Xˆ and Pˆ oscillate harmonically just as their classical D E counterparts. (If you like, show that the wave function is a Gaussian centered on Xˆ !)

10