Chapter 6. Q. Suppose we put a delta-function bump in the center of the infinite square well: H = αδ(x a/2) (1)

Transcription

1 Tor Kjellsson Stockholm University Chapter 6 6. Q. Suppose we put a delta-function bump in the center of the infinite square well: where α is a constant. H = αδ(x a/ ( a Find the first-order correction to the allowed energies. Explain why the energies are not perturbed for even n. For the infinite square well we have that the energies are given by: E n = n π ma n =,, 3... ( since each energy level is unique the spectrum is non-degenerate. So when we seek the first order corrections to the energy levels we can use non-degenerate perturbation theory: E n = ψ n H ψ n (3 where ψn are the unperturbed eigenfunctions of Ĥ. For the infinite square well these are: ( nπ ψ n (x = a sin a x. (4 Using eq.(3 we now get: E n = ψ n H ψ n = ψ n αδ(x a/ ψ n = α = a α ( ψ n (x δ(x a/ ψ n (x dx = a α sin ( nπ a ( nπ ( nπ sin a x δ(x a/ sin a x a = { if n = even a α else For even n we see that the correction is. We can understand this, besides trusting the calculations, from the fact that the even wave functions are where the delta-functions is located - they thus never get affected by it. dx

2 b Find the first three nonzero terms in the expansion (eq.(6.3 in the text book of the correction to the ground state ψ. Eq.(6.3 in the book is: n = gives: ψ n = n m ψ = m ψ m(x ψ(x = E m E m ψm H ψn En Em ψm (5 ψm H ψ E ψ E m m (ψ m(x δ(x a/ ψ (x dx ψ (x = ( 3/ α sin ( mπ a x a E m E m ( mπ ( π δ(x a/ sin a x sin a x dx which gives: ψ (x = ψ (x = ( 3/ α sin ( mπ a x a E m sin E m ( 3/ α [ sin ( 3π a x a E E 3 + sin ( 5π a x E E 5 ( mπ ( π sin. }{{ } sin ( 7π a x E E ]. Using eq.( and performing some basic algebra we obtain: ψ (x = a αm π [ ( 3π sin a x 3 ( 5πa sin x + 6 ( 7πa sin x ] +... (6

3 6. Q. For the harmonic oscillator [V (x = (/kx ], the allowed energies are: E n = (n + / ω (n =,,,... (7 where ω = k/m is the classical frequency. Now suppose the spring constant increases slightly: k ( + ɛk. a Q. Find the exact new energies. Expand your formula as a power series in ɛ, up to second order. Obtaining the new energies is trivial: ( + ɛk E n Ẽn = (n + / ω = (n + / m = (n + / ω + ɛ. (8 Expanding this to second order in ɛ: gives: + ɛ + ɛ 8 ɛ +... ( Ẽ n (n + / ω + ɛ 8 ɛ (9 b Q. Now calculate the first-order perturbation in energy using eq.(6.9 in the text book. What is H here? Compare your result with part (a. Hint: do not calculate any integral but use information from p when suitable. First-order energy perturbation is given by: E n = ψ n H ψ n and to calculate it we must of course know the perturbation H. For the Harmonic oscillator (in one dimension we have the Hamiltonian: so with the new spring constant k = ( + ɛk we get: d H = m dx + kx ( d H + H = m dx + kx + ɛkx ( 3

4 and thus: H = ɛkx. ( This gives us the first-order energy perturbation as: En = ψn H ψn = ɛk ψ n x ψn. From pages in the text book we now find x = [ a mω + + a + a + a a + + a ] (3 Using these we obtain: E n = ɛk a + ψ n = n + ψ n+, a ψ n = nψ n. (4 E n = ɛk mω ψ n [ a + + a + a + a a + a ] ψ n. (5 [ ψ mω n a + ψn + ψn a + a ψn + ψn a a + ψn + ψn a ψn ] Focusing on the square brackets we get: [ ψ n a + ψ n + ψ n a + a ψ n + ψ n a a + ψ n + ψ n a ψ n ] = n + ψ n a + ψ n+ + n ψ n a + ψ n + n + ψ n a ψ n+ + n ψ n a ψ n = n + n + ψn ψ n+ + n n ψn ψ n + n + n + ψ }{{}}{{} n ψ n + n n ψ }{{} n ψ n }{{} (6 = + n + (n + + = n + so: E n = ɛk mω (n + = ɛmω ( n + = ɛ ( n + ω (7 mω There is a famous quote that says that physics is just learning to solve the harmonic oscillator problem at ever increasing level. 4

5 6.4 a Q. Find the second-order correction to the energies (E n for the potential in problem 6.. Comment: You can sum the series explicitly, obtaining m(α/π n for odd n. The second-order energy correction (eq.(6.5 in the text book is: E n = n m ψ m H ψ n E n E m (8 On p. in this chapter of solutions we found so: ψm H ψn = ψm αδ(x a/ ψn = ( mπ ( nπ a α sin sin ψ m H ψ n = 4α a ( mπ sin ( { sin nπ = for even n or m else. 4α a Since E n E m = π ma (n m we obtain: E n = 8mα π n m n=odd n m = 8mα π ( α 4n = m (9 π n note also that the m outside the sum is the mass, not the summation index in the sum. The explicit expression for the sum can be looked up in a table or derived if you think that is fun - it is not important for the understanding of quantum mechanics (in my humble opinion. b Q. Calculate the second-order correction to the ground state energy (E for the potential in problem 6.. Check that your result is consistent with the exact result obtained in problem 6. a (eq.(9 in these solutions!. If you have a hard time seeing this I suggest that you redo the steps from problem where we computed ψ n H ψ n. 5

6 Once again we need to use eq.(8 and we note that we already computed a lot of it in eq.(6 from the previous problem: [ ψm H ψn = ɛk n + n + ψ mω m ψn+ + }{{} δ m,n n n ψ m ψ n + n + n + ψm ψ n + n n ψ }{{} m ψn }{{} δ m,n δ m,n+ + }{{} δ m,n Now, before we start computing this using the brute force method to obtain 6 terms we pause a moment and think. The 6 products will all consist of two delta functions and if you look at the product of two such that are not equal, say δ m,n δ m,n+ this has to be. For fun, try to prove this - it is a one line argument that I provide in the following (don t peek! footnote 3. So: ψm H ψn = ] = ɛ k [(n+(n+ 6 m ω (δ m,n+ +n (δ m,n +(n+ (δ m,n +n(n (δ m,n. Inserting this into eq.(6, noting that we are interested in n = and that the sum does not allow m = n we obtain: E = ɛ k 6 m ω m E E m E = ɛ k 6 m ω [( + ( + (δ m,+ + ( (δ m,+ ] m one of the delta functions collapses the sum: E = ɛ k 6 m ω δ m, E E E (δ m, ( E m = ɛ k 8 m ω E E while the other one can be thrown away since it does nothing - there are no more m to force to the value. Now we use: ]. to obtain: E E = ω ( and k = m ω 4 ( E = ɛ m ω 4 8 m ω ( ω = ɛ ω ( 6 3 This product is non-zero if and only if n = n + which can not happen. 6

7 note that both this result and eq.(7 are in agreement with the exact result obtained in eq.(9! 6.7 Q. Consider a particle of mass m that is free to move in a one-deminsional region of length that closes on itself. An example of this would be problem.46 in the text book, where a bead that slides frictionlessly on a circular wire. a Q. Show that the stationary states can be written in the form: ψ n (x = e πin/, ( /, < x < / (3 where n =, ±, ±,... and the energies are given by: E n = m ( nπ. (4 Notice that the spectrum is doubly degenerate for all n but n=. To find the eigenstates and eigenvalues we solve the eigenvalue equation: Ĥ ψ = E ψ (5 Notice now that our system is a particle that moves freely. The Hamiltonian must thus only contain a kinetic energy term: Ĥ = m dx (6 and since our operator explicitly depends on the x-basis we naturally work in this representation. The eigenvalue equation to solve is thus: d Ĥψ(x = Eψ(x ψ(x = Eψ(x m dx which has the most general solution: d ψ(x = A e ikx + B e ikx (7 with k = me/. As always when we are working with functions we find the eigenvalues by applying the boundary conditions - which in this case must be that the functions are periodic 4 : 4 The problem statement says that the region closes on itself. 7

8 ψ( / = ψ(/ = A e ik + B e ik = A e ik + B e ik e This means that: = e ik = e = k = nπ nz. k = nπ = me = nπ = E n = m ( nπ n Z. (8 Recall now that the we are asked to give the stationary states, that is, we are asked to give the eigenfunctions of the operator. The solution found in eq.(7 is actually a linear combination of the functions: ψ n (x = C e i nπx n Z. (9 which in themselves are eigenfunctions of Ĥ. Note that eq.(7 of course is an eigenfunction as well but it is not on the simplest form and we normally want to express the general formula using the smallest constituent part. Thus, eq.(9 is on the form we normally want (and also note that we can obtain eq.(7 from eq.(9. Finally, we just need to normalize this: / = ψ n ψ n = C e / nπx i / e i nπx dx = C / dx = C = (3 and as always, note that even though mathematically the solution is C = e iφ (where the last exponential is called a phase we are free to set this phase to. Thus: with: ψ n (x = e i nπx n Z (3 E n = m ( nπ n Z. (3 b Q. Now suppose we introduce the perturbation: H = V e x /a (33 8

9 where a <<. Find the first-order correction to E n using eq.(6.7 in the text book. Hint: to evaluate the integral, exploit the fact that a << to extend the limits from ±/ to ±. The first-order correction for degenerate perturbation theory is given by Eq.(6.7 in the text book: E± = [ ] W a,a + W b,b ± (W a,a W b,b + 4 W a,b (34 where: W a,b ψ a H ψ b (35 and the indices a, b denote that indices for the degenerate energy levels. In this example the energy levels are degenerate for ±n, so a = n, b = n. Now we calculate W n,n : W n,n = ψ n H ψ n = V = V / / / / e nπx i e x /a dx V e x /a e i nπx dx e x /a dx = V a π (36 Now we proceed by calculating W n,-n : ψ n H ψ n = V / / Completing the square 5 gives: e nπx i ψ n H ψ n = V = V / nπa e ( e ( x nπa a +i / e x /a nπx i V / e dx = / / / e ( x a dx V +i nπa ( nπa dx nπa e ( Computing this integral is not easy but manageable. substitution: x a + inπa which gives: = y, dx = ady, + inπa e ( x a e x a 4nπx i nπa +i dx dx Start with a variable < y < + inπa W n,-n V nπa +i nπa e ( a +i nπ a e y dy. (37 5 In Swedish: kvadratkomplettering. 9

10 This is a very famous integral which is tabulated in most places (Physics Handbook for instance.. However, the limits are now complex which might be a first sight for most of you following this solution. On this level it suffices to say that you can prove, using something called the Cauchy residue theorem 6, that this integral is the same as if you had computed it on the real line. Thus, you can just ignore the imaginary parts in the integration limits: W n,-n V and we finally arrive at: nπa ae ( e y dy = V nπa ae ( π (38 E± = V ( a π e (πna/ (39 c Q. What are the good linear combinations of ψ n and ψ n for this problem? Show that with these states you get the first-order correction using eq.(6.9 in the text book: What commonly happens during a perturbation of the Hamiltonian is that the energies that are degenerate split in as many components (see figure 6.4 in the text book as the degeneracy. So during the perturbation we get a nondegenerate spectrum for the new eigenstates 7. The problem now is that we don t know what these eigenstates look like more than that they must be a linear combination of the unperturbed degenerate eigenstates. We call such a combination a good linear combination. In the case of a two-fold degeneracy, as this problem, the good states must be of the form: ψ = αψ a + βψ b (4 and we seek the coefficients α, β. Eq.(6. in the text book gives: αw aa + βw ab = αe = β = α E W aa W ab (4 where W aa, W ab, E were calculated in the previous problem (a=n, b=-n. Note that we get one combination for E + and one for E which is precisely what we expect - one for each degeneracy-lift (compare with fig. 6.4 in the book. First note from the previous problem that since W n,n = W -n,-n we have: E ± = W nn W n,-n (4 6 In the beautiful theory of complex analysis you learn how to overcome the many limits in real-valued mathematical analysis and you also explain a lot of the weird trigonometric substitutions people do when solving indefinite integrals computed on the real line. 7 Remember, we have a new Hamiltonian now.

11 so we now get: β = α E ± W n,n W n,-n which gives the good linear combinations: ψ + = α + ψ n α + ψ -n = α ψ = α ψ n + α ψ -n = α = α ± W n,-n W n,-n = α (43 ( e πin/ e πin/ = α + sin (πnx/ ( e πin/ + e πin/ = α cos (πnx/. From the fact that we have added two orthonormal functions to create ψ ± we immediately see that the normalisation constant is α + = α =. Since E ± are non-degenerate and we have found the corresponding eigenstates we should be able to get the values E ± using non-degenerate perturbation theory with ψ ± and indeed this is the case: E+ = ψ + H ψ + = V ψ + e x /a ψ + = = V / / e x /a sin (πnx/ dx V Note now that using the Euler formula we obtain: e x /a sin (πnx/ dx = e x /a sin (πnx/ dx. (44 e x /a ( 4i e i4πnx/ + e i4πnx/ dx = 4 e x /a e i4πnx/ dx 4 e x /a e i4πnx/ dx + e x /a dx. }{{} a π/ Note that the first two terms are very reminiscent of W n,-n, which we computed between eq.(37 and eq.(39. As an exercise you can prove the following: int e x /a e i4πnx/ dx = int e x /a e i4πnx/ dx by making the variable substitution x = x. Thus we now have: E+ = ψ + H ψ + = V [ e x /a e i4πnx/ dx + a ] π where the first term now is exactly W n,-n. Referring back to eq.(37 through eq.(39 where we computed this we obtain: E + = V [ ae nπa ( π + a ] π

12 E+ = V [ ] a nπa π e ( (45 which is in agreement with eq.(4, the result obtained using degenerate perturbation theory. Now, you can produce the steps for ψ H ψ yourself, either by expressing cos (πnx/ as sin (πnx/ or by using the Euler formula for cosine. You should then also verify that you actually get E as stated in eq.(4.

13 6.9 Q. Consider a quantum system with just three linearly independent states. Suppose the Hamiltonian in matrix form is: ɛ H = V ɛ (46 ɛ where V is a constant and ɛ <<. a Q. Write down the eigenvectors and eigenvalues of the unperturbed Hamiltonian (ɛ =. This is very straightforward. If ɛis then: H = V (47 and we immediately see that the eigenvectors are: α = α = α 3 = (48 with the three eigenvalues: λ = V λ = V λ 3 = V (49 b Q. Solve for the exact eigenvalues of H. Expand each of them as a power series in ɛ up to second order. This is something that we have done a lot. Start by setting up the eigenvalue matrix equation: H α = λ α = ɛ a a V ɛ b = λ b ɛ c c V ( ɛ λ V λ V ɛ V ɛ V λ = 3

14 = (V ( ɛ λ(v λ(v λ V ɛ V ɛ (V ( ɛ λ = (V ( ɛ λ [(V λ(v λ V ɛ V ɛ] = which has the solutions: (V ( ɛ λ [ λ 3V λ + V ( ɛ ] = (V ( ɛ λ = λ = V ( ɛ λ 3V λ+v ( ɛ λ = V (3 + 4ɛ λ 3 = V ( ɛ expanding + 4ɛ + ɛ +... gives: λ V ɛv (5 λ V ɛ V (5 λ 3 V + ɛ V (5 ( N.B: we chose λ = V 3 + 4ɛ out of convenience. We could have chosen the other square root instead if we wanted to. c Q. Use first- and second-order nondegenerate perturbation theory to find the approximate eigenvalue for the state that grows out of the nondegenerate eigenvector of H. Compare this to the exact value obatined in (a. For nondegenerate perturbation theory we get the first order correction from: E n = α n H α n (53 and since our perturbation (on matrix form is: H = ɛv (54 and the state that is nondegenerate is the one with the unperturbed eigenvalue V we find: E3 = α3 H α3 = ( ɛv = (55 4

15 (which means that there are no first-order energy corrections due to the perturbation. The second-order correction is given by: so: E 3 = m α m H α 3 E 3 E m = α H α3 E3 E + α H α3 E3 E (56 α H α3 = ( ɛv = (57 α H α3 = ( ɛv = ɛv (58 E 3 = α H α 3 E 3 E which is the same as the result you obtained in (a. d = ɛ V = ɛ V (59 V V Q. Use degenerate perturbation theory to find the first-order correction to the two initially degenerate eigenvalues. Compare to the exact results. First order degenerate perturbation theory gives the energy correction: E ± = [W aa + W bb ± (W aa W bb + 4 W ab ] (6 with W ab = αa H αb. Our degenerate states are a= and b= so we obtain: W = ( ɛv = ɛv W = ( ɛv = W = ( ɛv = assuming that ɛv > we obtain: E ± = [ ɛv ± ] (ɛv + = { for E+ ɛv for E (6 5

16 Is this consistent with the results in (b? Yes, because there we found: E = V λ ɛv (6 E = V λ ɛ V (63 and in this problem we found that one state gets forced down by ɛv while the other state is unaffected in first order of ɛ. 6

17 6.3 Q. Suppose the Hamiltonian Ĥ for a particular quantum system is a function of some parameter λ; let E n (λ and ψ n (λ be the eigenvalues and eigenfunctions of H(λ. The Feynman-Hellmann theorem states that E n = H ψ n ψ n (64 assuming either that E n is nondegenerate or if degenerate that the ψ n s are the good linear combinations of the degenerate eigenfunctions. a Q. Prove the Feynman-Hellmann theorem. Hint: use eq.(6.9 in the text book for this. First of all, what are the consequences of H being a function of λ? Well, since H is a function it means that we can see it as H(λ = H(λ + V (λ = H + H (λ (65 and the new eigenfunctions are ψ n (λ. Then we note that: E n (λ = ψ n H(λ ψ n = E n = [ ] ψ n H(λ ψ n. (66 There are at least two ways to proceed from here. I will start by solving it following the hint and then solve it in the way that Feynman did it. In the first way to solve this we note that E E n (λ + dλ E n (λ E n(λ = lim = lim n(λ + dλ dλ dλ dλ dλ and from eq.(6.9 in the text book we find: (67 E n(λ + dλ = ψ n H (λ + dλ ψ n. (68 Note that ψ n = ψ n (λ are the unperturbed eigenfunctions while the ones used in eq.(64 and eq.(66 are the perturbed eigenfunctions ψ n (λ. Now, for small changes in λ we can write H as: so eq.(68 becomes: En(λ + dλ = H(λ + dλ H + H(λ dλ = H + H (λ + dλ (69 ψ n H(λ dλ ψ n = ψ n H(λ ψ n dλ. (7 7

18 and combining eq.(67 and eq.(7 we obtain: E n (λ E = lim n(λ + dλ = dλ dλ ψ n H(λ ψ n (7 So we have now shown the relation for small values close to λ. To generalize this we first make a small recap. We started at λ = λ and increased the value with dλ to obtain λ = λ + dλ. To calculate E(λ we used first order nondegenerate perturbation theory using the eigenfunctions of H(λ in eq.(68 and arrived at eq.(7. Now consider a further increase in λ to obtain λ = λ +dλ. To calculate E(λ we just repeat the same steps as before but now the unperturbed eigenfunctions are ψ n (λ. Thus: E n (λ E = lim n(λ + dλ = ψ n (λ H(λ dλ dλ ψ n (λ (7 and the same pattern would continue with further increase of λ. By this we have shown that for any λ we have: E n (λ = ψ n (λ H(λ ψ n (λ (73 In the second way we do not use perturbation theory but rather a direct method. From eq.(66 we obtain: E n = [ ] ψ n H ψ n ψn H H ψ n = Hψ n + ψ n ψ n + ψ n ψn ψn H ψn = E n + ψ n ψ n + Hψ n ψn H ψn ψn = E n ψ n + ψ n ψ n + E n ψn H = E n ψ n + ψ n ψ n + E ψn n ψ n ψn H = E n ψ n + ψ n ψ n + En ψ n ψ n but remember now that the energies are real, so En = E n. Thus: E n = H ψ n ψ n + E n [ ψ n ψ n ] }{{} where we have used that ψ n ψ n =, so the last term is. (74 8

19 b Q. Apply the theorem to the one-dimensional harmonic oscillator, (i using λ = ω (which yields a formula for V, (ii using λ = (which yields a formula for T and (iii using λ = m (which yields a relation between V and T. Compare your answers to problem. and the Virial theorem predictions. Here we are going to use the formula we proved in the previous problem: E n = H ψ n ψ n for the harmonic oscillator. The Hamiltonian and the energies are given by: d (i Using this for λ = ω we get: H = m dx + mω x E n = ω(n + (75 From the energies we then see: and thus: E n ω = ψ n H ω E n ω = ψ n mωx ψ n = ψ n ψ n V ω E n ω = (n + = E n ω which is familiar from problem. and 3.3. (ii Using λ = instead we now get: E n = E n = d ψn ψ n m dx ψ n E n = V (76 H ψ n ψ n = ψ n T ψn From the expression for the energies you get (check this: = T. (77 so eq.(77 gives: E n = E n E n = T. (78 Now you can do the third exercise (iii to verify what you can get from (i and (ii: V = T. 9