Chapter 2: Time-Independent Perturbation Theory
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4 Phys460.nb 2 Time-Independent Perturbation Theory 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 physics. Perturbation theories is in many cases the only theoretical technique that we have to handle various complex systems (quantum and classical). Examples: in quantum field 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 energy 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 hydrogen atom. But if we look at states with positive energies, 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 quantum field theory, 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 + λH 0 ψn +H' ψn +λ H0 ψn +H' ψn +…=E n ψn + λE 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 '.