4 Phys46.nb Time-Independent Perturbation Theory.. Overview... General question Assuming that we have a Hamiltonian, H = H + λ H (.) where λ is a very small real number. The eigenstates of the Hamiltonian should not be very different from the eigenstates of H. If we already know all eigenstates of H, can we get eigenstates of H approximately? Bottom line: we are studying an approximate method.... 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, /N expansions, 4-ϵ expansions)...3. Assumptions Assumption #: we know all eigenstates of H, as well as their corresponding eigenenergies H ψ n = E n ψ n (.) Assumption #: we know the perturbation H '. What do we mean by knowing H '? Here, we mean that we can write down H ' using the complete basis of ψ n, i.e., we know the value of ψ n H' ψ m 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 E n = (n + /) ħω)
Phys46.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 (/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 < ). 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 >. For E >, 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 > ). 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... Non-degenerate Perturbation Theory... Assumptions Key assumption: we consider a specific state ψ n. Here, we assume that E n - E m is much larger than λ H for any other eigenstate ψ m... Preparation # wavefunctions Since the eigenstates of H form a compete basis, we can write down any quantum state as a linear superposition of ψ m ψ = m a m ψ m (.3) Now, if we consider an eigenstate of H, ψ n, it can also be written in a similar fashion ψ n = m a m ψ m (.4) As discussion above, if λ is small, an eigenstate of H would be similar to an eigenstate of H. Here, we assume that ψ n is very close to ψ n. This means that a n and for other values of m n, a m ~, To highlight this, we separate the term for ψ n out from the sum ψ n = a n ψ n + m n a m ψ m (.5) It turns out that it is usually more convenient to use unnormalized eigenstates. Now, let us define unnormalized eigenstates of H ψ n = a n ψ n = ψ n + m n a m a n ψ m (.6) For simplicity, we will now call a m /a n = c m ψ n = ψ n + m n c n ψ m (.7) Because a m ~ and a n ~, we know that c m ~ for small λ. Comment #: This state is NOT normalized ψ n ψ n = + m n c m (.8)
6 Phys46.nb But we can easily normalized it, if we want to ψ n = ψ n + m n c m (.9) Comment #: (almost) any quantum states can be written in the form of Eq. (.3). This is because ψ m forms a complete basis. Q: what does the word almost mean here? A: If a state is orthogonal to ψ n, we cannot write the state the form of Eq. (.3). But we don t need to worry about it here, because we are doing perturbation theory and we know that the eigenstates of H is close to eigenstates of H. So it can not be orthogonal to ψ n. 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: c m s are functions of λ, i.e. c m (λ). For small λ, we can use the Taylor series: c m = c () m λ + c () m λ + c (3) m λ 3 + (.) Here, the Taylor series doesn t contain the th order term of λ (i.e. the constant term). This is because when λ =, ψ n = c m (λ) = at λ =. ψ n, and thus As a result, ψ n = ψ n + m n c m (λ) ψ m = ψ n + m n k= cm (k) λ k ψ m = ψ n + k= λ k m n c m (k) ψ m (.) If we define we get ψ n k = m n c m (k) ψ m (.) ψ n = ψ n + λ ψ n + λ ψ n + (.3) This is Eq. [6.5] in the textbook. Important: ψ n, ψ n doesn t contain ψ n. In other words, all corrections are orthogonal to ψ n...3. Preparation # eigenenergies Eigenenergies of H are also functions of λ, and for small λ, we can use the Taylor series: E n (λ) = E n + λ E n + λ E n + (.4) This is Eq. [6.6] in the textbook...4. Schrodinger Equation in the perturbation theory H ψ n = E n ψ n (.5) (H + λ H ') ψ n + λ ψ n + λ ψ n + = E n + λ E n + λ E n + ψ n + λ ψ n + λ ψ n + (.6) H ψ n + λh ψ n + H ' ψ n + λ H ψ n + H ' ψ n + = E n ψ n + λe n ψ n + E n ψ n + λ E n ψ n + E n ψ n + E n ψ n + (.7) In the perturbation theory, we need to compute two sets of quantities () energy corrections at each order E n, E n,... and () wavefunction corrections at each order, ψ n, ψ n, ψ 3 n. 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...5. Zeroth order The leading order terms in the equation is λ = constant H ψ n = E n ψ n (.8)
Phys46.nb 7 This is identical to the case of λ =, i.e. the unperturbed system...6. First order To the order of λ, we have H ψ n + H ' ψ n = E n ψ n + E n ψ n (.9) Here, we first compute the energy correction E n. This is done by multiplying on both sides ψ n ψ n H ψ n + ψ n H ' ψ n = ψ n E n ψ n + ψ n E n ψ n (.) For the first term on the l.h.s., we use the fact that ψ n H = ψ n E n (.) For the last term on the r.h.s., we use the fact that E n is a number (not a quantum operator), and thus ψ n E n ψ n = E n ψ n ψ n = E n ψ n E n ψ n + ψ n H ' ψ n = ψ n E n ψ n + E n ψ n H ' ψ n = E n (.) (.3) The first order correction in energy is the expectation value of H '. E n = E n + λ ψ n H ' ψ n + Oλ = ψ n H ψ n + ψ n λ H ' ψ n + Oλ = ψ n H + λ H ' ψ n + Oλ = ψ n H ψ n + Oλ (.4) Bottom line: to the first order (or say up to corrections at the order of λ ), we can use the old wavefunction (the zeroth order wavefunction). Then we compute the first order correction for the wavefunction ψ n. To do that, we multiply both sides of the equation with ψ m where m n ψ m H ψ n + ψ m H ' ψ n = ψ m E n ψ n + ψ m E n ψ n (.5) For the first term on the l.h.s., we use the fact that ψ m H = ψ m E m (.6) For the two terms on the r.h.s., we use the fact that E n and E n are both numbers (not quantum operators), so ψ m E n ψ n = E n ψ m ψ n and ψ m E n ψ n = E n ψ m ψ n =. Here, we used the fact that when m n, the two quantum states are orthogonal and thus ψ m ψ n =. E m ψ m ψ n + ψ m H ' ψ n = E n ψ m ψ n (.7) So ψ m ψ n = ψ m H ' ψ n E n - E m (.8) According to the definition of ψ n ψ n = m n c m () ψ m (.9) we have c m () = ψ m ψ n = ψ m H ' ψ n E n - E m (.3) And therefore,
8 Phys46.nb ψ n = m n ψ m ψ m H ' ψ n E n - E m (.3) So ψ n = ψ n + λ ψ m ψ m H ' ψ n + m n (.3) E n - E m..7. Second order H ψ n + H ' ψ n = E n ψ n + E n ψ n + E n ψ n (.33) Here, we first compute the energy correction E n. This is done by multiplying on both sides ψ n ψ n H ψ n + ψ n H ' ψ n = ψ n E n ψ n + ψ n E n ψ n + ψ n E n ψ n (.34) ψ n E n ψ n + ψ n H ' ψ n = ψ n E n ψ n + E n ψ n ψ n + E n (.35) The second term on the r.h.s. is zero, because we required ψ n ψ n = at the beginning. E n = ψ n H ' ψ n (.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. E n = ψ n H ' ψ n = m n ψ n H ' ψ m ψ m H ' ψ n E n - E m = m n ψ n H ' ψ m ψ m H ' ψ n E n - E m ψ n H ' ψ m ψ m H ' ψ n E n = E n + λ ψ n H ' ψ n + λ + Oλ 3 m n (.38) E n - E m (.37) The we compute the second order correction for the wavefunction m n ψ n. To do that, we multiply both sides of the equation with ψ m where ψ m H ψ n + ψ m H ' ψ n = ψ m E n ψ n + ψ m E n ψ n + ψ m E n ψ n (.39) E m ψ m ψ n + ψ m H ' ψ n = E n ψ m ψ n + E n ψ m ψ n + E n ψ m ψ n (.4) E n - E m ψ m ψ n = ψ m H ' ψ n - E n ψ m ψ n (.4) ψ m ψ n = ψ m H ' ψ n E n - E n - E m E n - E ψ m ψ n = m m' n ψ m m' n ψ m H ' ψ m' E n - E m m' n ψ m H ' ψ m' E n - E m ψ m' H ' ψ m' E n - E m H ' ψ n E n - E m' ψ m' H ' ψ n E n - E m' ψ m' - m' n ψ n H ' ψ n E n - E m - m' n ψ n H ' ψ n E n - E m ψ m ψ m' ψ m' H ' ψ n E n - E m' ψ m' H ' ψ n δ m,m' = E n - E m' H ' ψ n ψ n H ' ψ n ψ m H ' ψ n - E n - E m' n m' E n - E m = (.4) According to the definition of ψ n We have ψ n = m n c m () ψ m (.43) c m () = ψ m ψ n (.44)
Phys46.nb 9 and thus ψ n = m n m' n ψ m H ' ψ m' E n - E m ψ m' H ' ψ n - ψ n H ' ψ n ψ m H ' ψ n E n - E m' E n - E m ψ m (.45)..8. Third order Same as the second order, we can use the same method to show that E n 3 = ψ n H ' ψ n (.46) So, ψ n H ' ψ m ψ m H ' ψ m' ψ m' H ' ψ n E 3 n = m n - ψ n H ' ψ n ψ m H ' ψ n ψ n H ' ψ m m' n E n - E m E n - E m' E n - E m (.47) And one can keep doing this for higher and higher order..9. Summary For a Hamiltonian H = H + λ H (.48) assuming that we know all the eigenstates of H ψ n ), and we know the expectation values ψ m H ' ψ m for any two eigenstates of H, ψ m and ψ m ), then we can write down eigenstates of H as a power series expansions of λ ψ n H ' ψ m ψ m H ' ψ n E n = E n + λ E n + λ E n + = E n + λ ψ n H ' ψ n + λ + m n (.49) E n - E m and ψ n = ψ n + λ ψ n + λ ψ n + = ψ n + λ ψ m ψ m H ' ψ n + m n (.5) E n - E m... Second order perturbation always reduces the energy of the ground state One key conclusion from the perturbation theory is that the second order correction always makes the energy of the ground state lower (in comparison to the unperturbed one). This can be seen by looking at E E n = m n ψ n H ' ψ m ψ m H ' ψ n E n - E m = m n ψ n H ' ψ m E n - E m In the numerator, ψ n H ' ψ m is the complex conjugate of ψ m H' ψ n, so it is ψ n H ' ψ m, which is non-negative (.5) ψ n H ' ψ m (.5) The denominator E n - E m <, 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 ψ n H ' ψ m (.53) E n - E m So ψ n H ' ψ m E n = m n E n - E m (.54) The equal sign only arise when ψ n H ' ψ m = for ALL m n. (if this is the case, we don t need to do perturbation theory. The first order perturbation become exact). As long as we ignore this very special case, we find that E n < for the ground state, regardless of details.
Phys46.nb This conclusion is very important in quantum mechanics, because in many systems, the first order perturbation of the ground state happens to be zero. E n =. There, E n = E n + λ E n + (.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 E n < E n (.56) 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. (i.e., if E n > E n, we will violate the special relativity)..3. Brillouin-Wigner Perturbation Theory.3.. 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.. Too complicated to go to higher order (e.g. third order or fourth order correction). 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.3.. 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. It offers a nice and simple physical interpretation (a baby version of Feynman diagrams used in quantum field theory). 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 (H + λ H ') ψ n = E n ψ n (.57) where ψ n represents the same unnormalized eigenstate of H ψ n = ψ n + m n c n ψ m (.58) We can rewrite the equation above as (E - H ) ψ n = λ H ' ψ n (.59) and ψ n = λ (E - H ) - H ' ψ n (.6) Note: here (E - H ) - is the matrix inverse, instead of a number inverse, because H 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 f (x) = a + a x + a x + (.6)
Phys46.nb then, f (Q ) represents the same power series, but with number x substitute by operator Q f (Q ) = a + a Q + a Q + (.6) where Q = Q Q, etc. Here, the inverse function of operator (E - H ) - shall be understood the same way Because E - x = E + x E + x we know that E - H - = E + H E + H E 3 + (.63) Now, back to the derivation above: E 3 + (.64) λ ψ m ψ n = ψ m λ (E - H ) - H ' ψ n = E - E ψ m H ' ψ n (.65) m Remember that from the definition, of ψ n ψ n = ψ n + m n c m ψ m (.66) we have ψ m ψ n = c m (.67) for any m n. And thus, we get ψ n = ψ n + λ ψ m m n E - E ψ m H ' ψ n (.68) m If we define a quantum operator X as R = m n ψ m E - E m ψ m (.69) we get Thus, ψ n = ψ n + λ R H ' ψ n (.7) I - λ R H ' ψ n = ψ n (.7) where I is the identity operator So, ψ n = I - λ R H ' - ψ n (.7) Again, we emphasize that here, I - λ R H ' - represent matrix inverse. For matrix inverse, we can use Taylor expansions to write it out. We know that ( - a) - = + a + a + a 3 + (.73) So similarly, we have I - λ R H ' - = I + λ R H ' + λ R H ' R H ' + λ 3 R H ' R H ' R H ' + (.74) So we have ψ n = I - λ R H ' - ψ n = ψ n + λ R H ' ψ n + λ R H ' R H ' ψ n + λ 3 R H ' R H ' R H ' ψ n + (.75)
Phys46.nb So we find that ψ n k = R H ' k ψ n (.76) Previous, we found that E n = ψ n E n = ψ n E n 3 = ψ n H ' ψ n H ' ψ n H ' ψ n (.77) (.78) (.79) In fact, we can use the same procedure to show that for kth order, E n k = ψ n H ' ψ n k- (.8) Because we have found that ψ n k- = R H ' k- ψ n E n k = ψ n H ' ψ n k- = ψ n H ' R H ' k- ψ n (.8) So, we have E n = ψ n H ' ψ n E n = ψ n H ' R H ' ψ n = ψ n H ' ψ m m n E n - E ψ m H ' ψ n (.83) m E n 3 = ψ n H ' R H ' R H ' ψ n = m n m' n ψ n H ' ψ m E n - E m ψ m (.8) H ' ψ m' E n - E ψ m' H ' ψ n (.84) m'... (.85) From these formula we see a pattern.. For any E nk, if we look at the formula from right to left, one always start from unperturbed state ψ n and eventually goes back to the same state ψ n.. In the path from ψ n to ψ n, we go through several intermediate states ψ m, ψ m'. For kth order perturbation, we have k - intermediate states. 3. To turn from a state to another along the path (e.g. from n to m or from m to m in E n3 ), we use the perturbation H ' 4. For each intermediate state,we have an denominator.3.3. Diagrammatic representation We can represent the E n k using diagrams. En-Em. For each intermediate state, we represent as a solid line with integer m labeling the state. En-Em. For each ψ m H ' ψ m', we represent it as a dot. And we use V m m' to represent ψ m H ' ψ m' 3. Connect everything together in the same order as in E n k 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 ψ n First order:
Phys46.nb 3 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 ω-ϵ where ω is frequency, pretty much the same as energy E n and ϵ 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 photons to get E&M interactions)..3.4. How to compute the energy using Brillouin-Wigner Perturbation Theory? First, let s define some abbreviation to make the formula shorter, V ij = ψ i and thus H ' ψ j (.86) E n = E n + λ V nn + λ V n m V m n E n - E m + λ3 V n m V m m' V m' n E n - E m E n - E m' + λ 4 V n m V m m' V m' m'' V m'' n +... E n - E m E n - E m' E n - E m'' (.87) Here all the ms are summed over but they cannot be the same as n. It may looks like that we can find E n using this formula, but it is not quite the case yet. This is because on the r.h.s., the denominator contains also E n, i.e. IT is a equation for E n and E n 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 E n in the kth order calculation, we just use the (k - )th order E n on the r.h.s.. Here is how it is done First run E n () = E n + λ V nn (.88) Second run V n m V m n E () n = E n + λ V nn + λ E () n - E m (.89) Third run E n (3) = E n + λ V nn + λ V n m V m n E n () - E m + λ3 V n m V m m' V m' n E n () - E m E n () - E m' (.9) Fourth run
4 Phys46.nb V n m V m n E (4) n = E n + λ V nn + λ E (3) n - E + V n λ3 m V m m' V m' n m E (3) n - E m E (3) n - E m' + λ 4 V n m V m m' V m' m'' V m'' n E n (3) - E m E n (3) - E m' E n (3) - E m'' (.9)... Another option is using analytic methods, as will be discussed below..3.5. Preparation Consider the following function f (x) = x g(x) = x + a x + b x + c x 3 +... (.9) If we want to keep f (x) to O(x n ), we only need to keep g(x) to Ox n-. Similarly, for the following function f (x) = x g(x) = x + a x + b x + c x 3 +... (.93) If we want to keep f (x) to O(x n ), we only need to keep g(x) to Ox n-. This will be something that useful for us latter.3.6. iterative method E n = E n + λ V nn + λ V n m V m n E n - E m + λ3 V n m V m m' V m' n E n - E m E n - E m' + λ 4 V n m V m m' V m' m'' V m'' n +... E n - E m E n - E m' E n - E m'' (.94) Zeroth order (no E n on the l.h.s., so job done) E n = E n + O(λ) (.95) First order (no E n on the l.h.s., so job done) E n = E n + λ V nn + Oλ (.96) Second order, we use E n obtained at zeroth order for the λ term E n = E n + λ V nn + λ V n m V m n E n - E + Oλ3 = E n + λ V nn + λ V n m V m n m E n - E + Oλ3 (.97) m This is because the third term on the r.h.s. already has a λ prefactor. Thus to keep to Oλ, we only need to keep the denominator to Oλ. Third order, E n = E n + λ V nn + λ V n m V m n E n - E m + λ3 V n m V m n E n + λ V nn + λ E n + λ V nn - E m V n m V m m' V m' n + Oλ 4 = E n - E m E n - E m' + λ 3 V n m V m m' V m' n + Oλ 4 E n - E m E n - E m' (.98) In the λ term, we now need to keep to O(λ). In the λ 3 term, we just need to keep E n to the zeroth order. For the λ term, we can expand it for small λ λ V n m V m n E n + λ V nn - E m = λ V n m V m n E n - E m - λ3 V n m V m n E n - E m V nn E n - E m + (.99) So E n = E n + λ V nn + λ V n m V m n E n - E + V n m V m m' V m' n λ3 m E n - E m E n - E m' - V n m V m n V n n E n - E m + Oλ4 (.) Fourth order,
Phys46.nb 5 V n m V m n E n = E n + λ V nn + λ + E n + λ V nn + λ Vn m' Vm' n - E En -Em' m λ 3 V n m V m m' V m' n E n + λ V nn - E m E n + λ V nn - E m' + λ 4 V n m V m m' V m' m'' V m'' n + Oλ 5 E n - E m E n - E m' E n - E m'' (.) E n = E n + λ V nn + λ V n m V m n E n - E + V n m V m m' V m' n λ3 m E n - E m E n - E m' - V n m V m n V n n E n - E m + V n m V λ 4 m m' V m' m'' V m'' n E n - E m E n - E m' E n - E m'' - V n m V m n E n - E m V n m' V m' n E n - E m' + V n m V m n V V n m V E n - E m 3 nn m m' V m' n V n n - + E n - E m E n - E m' (.) Oλ 5.4. Degenerate Perturbation Theory In the previous section, we studied the effect of a small perturbation λ H ' on an eigenstate of H, ψ n. The key assumption there is that before we turn on the perturbation (i.e. at λ = ), the eigenenergies of all other eigenstates of H are very far away from E n E n - E m > > ψ i λ H ' ψ j (.3) This section, we will consider the opposite situation, where there is at least one other eigenstate of H which has the same eigenenergy as ψ n. Two states having the same eigenenergy is known as degeneracy. So this perturbation theory is known as the degenerate perturbation theory..4.. 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 ψ n H ' ψ m ψ m H ' ψ n E n = E n + λ E n + λ E n + = E n + λ ψ n H ' ψ n + λ + m n (.4) E n - E m Here, we focus on the second order correction: λ m n ψ n H ' ψ m ψ m H ' ψ n E n - E m (.5) If H has another eigenstate ψ n' with the same eigenenergy, at least one term in this sum will have zero in the denominator and thus will diverge, i.e., when E n = E n', En -En', and thus the theory becomes ill-defined. NOTE: the same divergence will arise also in higher order corrections. But there is no divergence in the first order correction E n. 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 f (x) = a + a x + a x + (.6) Now, assume that I made a mistake in the Taylor expansion for the coefficient a. Instead of the correction value, a, I used a wrong coefficient for the linear term, say b. f (x) = a + b x + (a - b ) x + a x + (.7) In other words, here I missed part of the linear term, (a - b ) 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 x term to correct this error, i.e. f (x) = a + b x + a - b + a x + (.8) x
6 Phys46.nb Let me define b = a + a-b x f (x) = a + b x + b x + (.9) Now, once again, I wrote my function as a power-law expansion. Because I used a wrong coefficient for the linear term, b, my second order term needs to use this new coefficient. This new coefficient b is infinite at small x. This is transparent if we notice that when x b = a + a - b 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..4.. What to do? Here, let s first take another look at the second order correction E n = m n ψ n H ' ψ m ψ m H ' ψ n E n - E m (.) As we know, the problem arises because E n = E m for certain m, and thus we get =. To avoid this singularity, the only thing that we need to do is to request that the numerator also vanish whenever the denominator is zero. i.e., if E n = E m, we must make sure that ψ m H ' ψ n =. NOTE: the two factors in the numerator are complex conjugate to each other: ψ n H ' ψ m = ψ m H ' ψ n *, 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 make sure that for any degenerate states, ψ m H ' ψ n =.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. H = p m = - ħ m d d x (.) The eigenstates of H arises in pairs (i.e. there is a degeneracy for any excited states). The static Schrodinger equation here is - ħ m d dx ψ(x) = E ψ(x) (.3) It is a second order differential equation and we know the solution are just plane waves ψ = A e i k x + B e -i k x (.4) The eigenenergy for this state is E = p m = (ħ k) m, i.e. the kinetic energy. Here, A and B are two arbitrary coefficients. For each fixed k, we have one eigenenergy E = (ħ k) m, but infinite number of eigenstates ψ = A e i k x + B e -i k x, 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, e i k x and e -i k x, all other eigenstates can be written as linear superposition of these two. Bottom line: two-fold degeneracy means that any linear combination of these two states is an eigenstate of H with the same eigenenergy. Now, let s look at the same second order differential equation again. - ħ m d dx ψ(x) = E ψ(x) (.5) we know that we can also write the solution for this equation as ψ = C cos k x + D sin k y (.6)
Phys46.nb 7 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 cos k x or sin k x. 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 e i k x and e -i k x, but now for the two degenerate states, we use cos k x or sin k x. These different choices are just different basis to represent all the eigenstates. We can choose to use e i k x and e -i k x or cos k x or sin k x. There is no difference between them. In fact, we can choose any two linear independent states ψ = A e i k x + B e -i k x ψ = A e i k x + B e -i k x (.7) (.8) And then, we can say that we have two degenerate states ψ (x) and ψ (x). and then, we can represent any other eigenstates (with the same eigenenergy) as ψ(x) = X ψ (x) + Y ψ (x) (.9) Q: Why do we usually use e i k x and e -i k x or cos k x or sin k x? Why not use e i k x and cos k x. A: There is no problem (mathematically) if we choose to use e i k x and cos k x. However, for convenience, it is usually better using orthonormal bases dx ei k x * e i k x = π δ(k - k ) (.) dx (cos k x)* sin k x = (.) Q: Why do we use e i k x and e -i k x more often than cos k x or sin k x in quantum mechanics? A: For H, there is little difference between the two choices. However, if we consider other quantum operators, like momentum, e i k x and e -i k x is a better choice. This is because e i k x and e -i k x 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). cos k x or sin k x don t have well defined momentum. They have 5% chance having momentum ħ k and another 5% chance having momentum -ħ k. 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 H, there are two degenerate eigenstates: H ψ a = E ψ a (.) and H ψ b = E ψ b (.3) We assume that these two states are orthogonal to each other (otherwise, we make them orthogonal, using Gram-Schmidt procedure). We assume that ψ a H ' ψ b. Here we first prove a fact: if ψ a and ψ b are both eigenstates of H and they have the same eigenvalue E, then any linear superposition of ψ a and ψ b is also an eigenstate of H with the same eigenenergy. Let's define ψ = α ψ a + β ψ b H ψ = H α ψ a + β ψ b = α H ψ a + β H ψ b = α E ψ a + β E ψ b = E α H ψ a + β H ψ b = E ψ (.4) Bottom line: if H has two degenerate eigenstates ψ a and ψ b, we have infinite eigenstates with the same eigenvalue ψ = α ψ a + β ψ b To represent these infinite eigenstates, we need to choose two states as basis, e.g. ψ a and ψ b. Then, any eigenstates with eigenenergy E 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 change to a different set of basis): for example, we can define ψ and ψ, where ψ = α ψ a + β ψ b (.5)
8 Phys46.nb ψ = α ψ a + β ψ b (.6) Here, we request ψ and ψ to be orthogonal to each other (otherwise, we make them orthogonal, using Gram-Schmidt procedure) ψ ψ = ψ a α * + ψ b β * α ψ a + β ψ b = α * α + β * β = (.7) and we assume that ψ and ψ are normalized ψ ψ = ψ a α * + ψ b β * α ψ a + β ψ b = α * α + β * β = α + β = (.8) ψ ψ = ψ a α * + ψ b β * α ψ a + β ψ b = α * α + β * β = α + β = (.9) Bottom line, instead of our old states ψ a and ψ b, we can use ψ and ψ instead as our basis..4.4. Which basis shall we use? As mentioned above, in general, ψ a H ' ψ b. If so, the native perturbation theory will have singularity at the second order. Now, we learned that we can choose to use a different set of unperturbed eigenstates ψ and ψ, so can we make ψ H ' ψ =? 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 matrix W = ψ a H ' ψ a ψ a H ' ψ b ψ b H ' ψ a ψ b H ' ψ b To make the formula shorter, we define (.3) W ij = ψ i H ' ψ j (.3) So W = W aa W ba W bb (.3). W is a Hermitian matrix (W = W), so its eigenvalues are real This is pretty straightforward to prove, because ψ X ψ * = ψ X ψ. In particular, if X is an Hermitian operator (the quantum operator of any physics observable is Hermitian), we have X = X and thus ψ X ψ * = ψ X ψ. Thus it is easy to notice that W aa = W aa * and W bb = W bb * and W ba = * (.33) W = W * = Waa* W ba * * W bb * = Waa W bb = W (.34). W has two eigenvalues E + and E -, and each of them has a vector, α β for E + and α β for E - Waa α = E W ba W bb β + α β (.35) and Waa α = E W ba W bb β - α β (.36) where E + and E - are the two eigenvalues. 3. We can use these two eigenvectors to define our ψ and ψ as ψ = α ψ a + β ψ b ψ = α ψ a + β ψ b (.37) (.38) As will be shown below, these two states are precisely what we should use for the perturbation theory.
Phys46.nb 9 4. The two eigenvalues are the first order corrections to the eigenenergy E = E + λ E + + Oλ E = E + λ E - + Oλ (.39) (.4) Q: Why we have two eigenenergies here? A: Because we started from two degenerate eigenstates. At λ =, the two states ψ and ψ have the same energy. Now, if we turn on a small perturbation λ H ', we find that these two states (in general) have different eigenenergies. One of them is E = E + λ E + + Oλ and the other E = E + λ E - + Oλ. NOTE #: We say that the perturbation lifted the degeneracy. NOTE #: After we lift the degeneracy, ψ and ψ no longer have the same energy. If the perturbation is small enough, we can now do non-degenerate perturbation theory, i.e. problem solved..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 H real to represent the full Hamiltonian of a real system and H model 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. H real = H model + δh (.4) 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 H model, 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 W aa is exactly the same as W bb and = W ba = 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 W aa and W bb is unmeasurably small and and W ba is unmeasurably small)..4.6. Prove: ψ H ' ψ = Here the proof contains two steps: first ψ H ' ψ = ( α * β * ) Waa W ba W bb α β (.4) and then we will show ( α * β * ) Waa W ba W bb α β = (.43) The first step is very straightforward (it is from the definition of ψ and ψ ) ψ = α ψ a + β ψ b ψ = α ψ a + β ψ b (.44) (.45) so
Phys46.nb ψ H ' ψ = ψ a α * + ψ b β * H ' α ψ a + β ψ b = ψ a α * + ψ b β * α H ' ψ a + β H ' ψ b = α * α ψ a H ' ψ a + α * β ψ a H ' ψ b + β * α ψ b H ' ψ a + β * β ψ b H ' ψ b (.46) The r.h.s. of the equation is in fact exactly the same, if we remember the definition of the W matrix W ij = ψ i H ' ψ j ( α * β * ) Waa W ba W bb α β = α * α W aa + α * β + β * α W ba + β * β W bb (.47) So, we proved that ψ H ' ψ = ( α * β * ) Waa W ba W bb α β For the second step, we first consider the situation that E + E -. According to the eigenequations, we have W aa α = E W ba W bb β - α β (.48) If we multiply on both sides (α *, β * ), we get (α *, β * ) Waa W ba W bb α β = (α *, β * ) E - α β = E - (α *, β * ) α β (.49) Similarly, if we start from the other eigenequation Waa α = E W ba W bb β + α β (.5) we get (α *, β * ) Waa W ba W bb α β = (α *, β * ) E + α β = E + (α *, β * ) α β (.5) If we take a conjugate on both sides (α *, β * ) Waa W ba W bb α β = E + (α *, β * ) α β (.5) here we used the fact that W is Hermitian and thus W = W. Notice that we have shown and (α *, β * ) Waa W ba W bb α β = E + (α *, β * ) α β (.53) (α *, β * ) Waa W ba W bb α β = E - (α *, β * ) α β (.54) If E + E - the only way that these two equations can both be valid is that (α *, β * ) Waa W ba W bb α β = E + (α *, β * ) α β = E - (α *, β * ) α β = (.55) So we proved that ψ H ' ψ =. Q: what will happen is E + = E -? A: Turns out that this is the simple case. If E + = E -, as will be shown below, = ψ a H ' ψ b =. So, there is no divergence from the beginning. We can start the perturbation theory without worrying about these divergence..4.7. First order perturbation The calculation described above provides to us the zeroth order wavefunctions (i.e., we should use ψ or ψ ), instead of ψ a or ψ b as our unperturbed wavefunction). As we learned early on (in non-degenerate perturbation theory), the first order correction of energy is just E n = ψ n H ' ψ n (.56) 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 them, ψ and ψ, so we need to compute the first order energy correction for each of them. And we will prove in this section
Phys46.nb E = ψ H ' ψ = E + (.57) and E = ψ H ' ψ = E - (.58) i.e., the first order energy corrections for ψ and ψ are precisely the two eigenvalues of the W matrix. ψ H ' ψ = ψ a α * + ψ b β * H ' α ψ a + β ψ b = ψ a α * + ψ b β * α H ' ψ a + β H ' ψ b = α * α ψ a H ' ψ a + α * β ψ a H ' ψ b + β * α ψ b H ' ψ a + β * β ψ b H ' ψ b (.59) If we remember the definition of the W matrix W ij = ψ i H ' ψ j, we realized immediately that this formula is exactly the same as ( α * β * ) Waa W ba W bb α β = α * α W aa + α * β + β * α W ba + β * β W bb (.6) So we found E = ψ H ' ψ = ( α * β * ) Waa W ba W bb α β (.6) Because α β is an eigenvector of B Waa α = E W ba W bb β + α β (.6) E = ψ H ' ψ = ( α * β * ) Waa W ba W bb α β = E + ( α * β * ) α β (.63) Because we have required α β to be normalized, ( α * β * ) α β = α * α + β * β = E = ψ H ' ψ = ( α * β * ) Waa W ba W bb α β = E + ( α * β * ) α β = E + (.64) Similarly, we can show that E = ψ H ' ψ = ( α * β * ) Waa W ba W bb α β = E - ( α * β * ) α β = E - (.65).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 W aa W ba W bb α β = E α β (.66) This means that W aa α + β = E α W ba α + W bb β = E β (.67) (.68) These two equations have an obvious and trivial solution α = β =. 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 α = E - W aa β (.69) Using the second equation, we get α = E - W bb W ba β (.7) The first relation means
Phys46.nb α β = E - W aa but the second relation requires α β = E - W bb W ba So we have α β = = E - W bb E - W aa W ba (.7) (.7) (.73) In general, E-Waa E-W bb, so we find an contradiction. This contradiction means that for a general value of E, we will only have the trivial W ba solution α = β =. To get a nontrivial solution, we have to request E-Waa = E-W bb. This equation is often written in a different form W ba E - W aa = E - W bb W ba (E - W aa ) (E - W bb ) = W ba (E - W aa ) (E - W bb ) - W ba = - det E - Waa = -W ba E - W bb or equivalently det E E - Waa W ba W bb = det (E - W) = (.74) (.75) (.76) (.77) (.78) (.79) Here, W is the matrix that we define above W = W aa W ba W bb (.8) and number E here means E times the identity matrix E* = E E (.8) det (E - W) = means (W aa - E) (W bb - E) - W ba = (.8) And thus E - (W aa + W bb ) E + (W aa W bb - W ba ) = (.83) By definition, tr W = W aa + W bb and det W = W aa W bb - W ba. Therefore, we can write the same equation as E - tr WE + det W = (.84) This equation has two solutions E ± = tr W ± (tr W) - 4 det W (.85) 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 E = E + E + λ + Oλ (.86) and
Phys46.nb 3 E = E + E - λ + Oλ (.87) at small λ. Comment #. tr W and det W are both real. This is straightforward to prove, if we notice that W is Hermitian. Because W aa = W aa* and W bb = W bb* and W ba = *, W aa and W bb are real. And W ba = is also real, so tr W = W aa + W bb and det W = W aa W bb - W ba are both real. Comment #. (tr W) - 4 det W. Therefore, E ± are both real. (tr W) - 4 det W = (W aa + W bb ) - 4 W aa W bb + 4 W ba = (W aa - W bb ) + 4 (.88) here we used the fact that * = W ba. Comment #3. There are in general two possible situations (a) If (tr W) - 4 det W >, E + > E -. i.e. the two eigenvalues are NOT the same. (b) If W aa = W bb and = W ba =, (tr W) - 4 det W =, and thus E + = E - = tr W. The situation (b) is the easy case, because = W ba = means ψ a H ' ψ b =. Remember that the problem we had from the beginning is that the second order correction will diverge, ψa H' ψ b ψ b H' ψa, because E Ea -E a = E b. For situation (b), the numerator is zero, so there is no b 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..4.9. Eigenvectors of the matrix W In this section, we will assume that (tr W) - 4 det W >, i.e. situation (a) discussed in the previous section. We have two eigenvalues. For each eigenvalue, we can solve for the corresponding eigenvector. For the eigenvalue E + = tr W+ (tr W) -4 det W, we have Waa α = E W ba W bb β + α β (.89) and for the other eigenvalue E - = tr W- (tr W) -4 det W, we have Waa α = E W ba W bb β - α β (.9) We will use the first one as example (equation for E + ). There, the matrix equation can be written as two separate equations W aa α + β = E + α W ba α + W bb β = E + β (.9) (.9) Using the first equation, we get α = E + - W aa β (.93) Using the second equation, we get α = E + - W bb W ba β (.94) These two relations are actually identical, because for any eigenvalue E, we have E-Waa In addition, we know that α * α + β * β =, i.e. the normalization condition. So we have = E-W bb W ba as we proved early on. α = β = W ab +(E + - W aa ) E + - W aa W ab +(E + - W aa ) (.95) (.96) Similarly, we have
4 Phys46.nb α = β = W ab +(E - - W aa ) E - - W aa W ab +(E - - W aa ) (.97) (.98) In conclusion, we found that ψ = α ψ a + β ψ b = +(E + - W aa ) ψ a + E + - W aa W ab +(E + - W aa ) ψ b (.99) ψ = α ψ a + β ψ b = +(E - - W aa ) ψ a + E - - W aa W ab +(E - - W aa ) ψ b (.).4.. 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 E ± = tr W ± (tr W) - 4 det W (.) we know that E + = E - can only arise when (tr W) - 4 det W =, i.e. (tr W) - 4 det W =. As shown above (tr W) - 4 det W = (W aa - W bb ) + 4 (.) 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 (W aa - W bb ) = (.3) and 4 = (.4) i.e., W aa = W bb and =. With W aa = W bb and =, W is actually proptional to an identity matrix. W = W aa W aa = W aa (.5) 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, W bb, the real part of and the imaginary part of (note : W aa and W bb are real, so they don t have imaginary part. note : W ba is the complex conjugate of, 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 W aa matches exactly W bb without any error bar, and both the real and imarginaly parts of 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 H 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..4.. 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- matrix). For a complete set of basis, { ψ i }, we can write any quantum states as ψ = i c i ψ i (.6)
Phys46.nb 5 where c i are complex numbers. Here, we find that if we want to describe a state, we just need to know all the coefficient c i. We can write these c i as a vector c c c 3 (.7) These coefficients are c i = ψ i ψ (.8) To see this, we multiply ψ j for both sides of ψ = i c i ψ i ψ j ψ = i c i ψ j ψ i = i c i δ ij = c j Bottom line: a quantum state is a column vector ψ c c c 3 = ψ ψ ψ ψ ψ 3 ψ (.9) (.) Conjugate states is the represented by the conjugate vector. By definition, we know that ψ = i ψ i c i * (.) so, we can write all these c * i as a row vector ( c * c * c * 3 ) = ( ψ ψ ψ ψ ψ ψ 3 ) (.) Here, we used the fact that ψ ψ i is the complex conjugate of ψ i ψ Inner produce of two states are product of a row vector and a column vector If we know two quantum states ψ c c c 3 (.3) and ϕ d d d 3 (.4) then, we know so ϕ ( d * d * d 3 * ) (.5) ϕ ψ ( d * d * d 3 * ) Q: How about a quantum operator? c c c 3 = d * c + d * c + d 3 * c 3 + (.6) 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 (.7) 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
6 Phys46.nb vector? We know that a matrix can do such a job x x x 3 x x x 3... x 3 x 3 x 33 c c c 3 = x c + x c + x 3 c 3 + x c + x c + x 3 c 3 + x 3 c + x 3 c + x 33 c 3 + = d d d 3 (.8) So a quantum operator is really similar to a matrix. In fact, the matrix elements x ij s are very easy to compute x ij = ψ i X ψ j (.9) Q: How about eigenvalues and eigenstates? A: Matrices also have eigenvalues and eigenstates x x x 3 x x x 3... x 3 x 3 x 33 c c c 3 = W c c c 3 (.). The matrix of a Hermitian operator is a Hermitian matrix. 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 4. Each eigenvector corresponds to a eigenstate, i.e. If c c c 3 is an eigenvector with eigenvalue W, then ψ = i c i ψ i is an eigenstate of X 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 million by million). So this approach doesn t make our life easier..4.. Degenerate perturbation theory H = H + λ H ' (.) Using eigenstates of H as basis, then H corresponds to a diagonal matrix H ψ i = E i ψ i (.) where i =,, 3, and we request this is an orthonormal basis ψ i ψ j = δ ij (.3) a matrix element of the matrix is So, ψ i H ψ j = ψ i E j ψ j = E j ψ i ψ j = E j δ i, j (.4) H E E... E 3 (.5) 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. λ H ' λ ψ H ' ψ ψ H ' ψ ψ H ' ψ 3 ψ H ' ψ ψ H ' ψ ψ H ' ψ 3... ψ 3 H ' ψ ψ 3 H ' ψ ψ 3 H ' ψ 3 (.6)
Phys46.nb 7 In general, H ' is NOT an diagonal matrix H = H + λ H ' E E... E 3 + λ ψ H ' ψ ψ H ' ψ ψ H ' ψ 3 ψ H ' ψ ψ H ' ψ ψ H ' ψ 3... ψ 3 H ' ψ ψ 3 H ' ψ ψ 3 H ' ψ 3 E + λ ψ H ' ψ λ ψ H ' ψ λ ψ H ' ψ 3 λ ψ H ' ψ E + λ ψ H ' ψ λ ψ H ' ψ 3... ψ 3 H ' ψ λ ψ 3 H ' ψ E 3 + λ ψ 3 H ' ψ 3 This is a very large matrix and is very hard to handle in general. However, if there are degenerate states, H =... E...... E = (.7) (.8) i.e., two of the eigenvalues of H coincides, or say two of two numbers along the diagonal line of the matrix of H 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 ψ i with i =,,. As a result, the matrix of a quantum operator has dimension, ψ i X ψ j with i =,, and j =,,. Now, if we only limit ourselves to the two degenerate states ψ a and ψ b, then the matrix of our quantum operator only has dimensions, because my i and j here can only be a or b X ψ a X ψ a ψ a X ψ b ψ b X ψ a ψ b X ψ b For H, its matrix is H E E (.9) (.3) and for H ', the matrix is H ' ψ a H ' ψ a ψ a H ' ψ b ψ b H ' ψ a ψ b H ' ψ b (.3) So our H is H = H + λ H ' E + λ ψ a H ' ψ a λ ψ a H ' ψ b λ ψ b H ' ψ a E + λ ψ b H ' ψ b (.3) The eigenvalues of this matrix are E + λ E + and E + λ E -. 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..4.3. n-fold degeneracy If H 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. H E E... E n n (.33)