Assignment 1 with Solutions - Introduction to Quantum Mechanics 1 | PHYS 5260, Assignments of Quantum Mechanics

Download Assignment 1 with Solutions - Introduction to Quantum Mechanics 1 | PHYS 5260 and more Assignments Quantum Mechanics in PDF only on Docsity! Quantum Mechanics II: HW 1 Solutions Leo Radzihovsky February 11, 2008 1 Harmonic Oscillator Let us examine the harmonic oscillator with the variational theory. We use a gaussian for the variational state: Ψα (x) = Ne −αx2 , (1) with α our variational parameter and normalization N = (2α) 1/4 π1/4 . Our Hamiltonian is: H = p2 2m + 1 2 mω2x2, (2) so our expectation value is: 〈Ψα|H |Ψα〉 = E(α) = ∫ ∞ −∞ dx [ ~ 2 2m |∂xΨ|2 + 1 2 mω2x2 |Ψ|2 ] , (3) where we have used integration by parts for the p2 term, noting that Ψ must vanish at infinity. So: E(α) = ∫ ∞ −∞ dx [ ~ 2 2m 4α2 + 1 2 mω2 ] N2x2e−2αx 2 . (4) But 〈Ψα|x2 |Ψα〉 = N2 ∫ ∞ −∞ dx x2e−2αx 2 = 1 4α , (5) as can be easily verified. This leads to our final expression for E(α): E(α) = ~ 2 2m α + 1 8α mω2. (6) Minimizing, we have: ∂E ∂α = ~ 2 2m − mω 2 8α20 = 0 (7) ⇒ α0 = mω 2~ = 1 2x20 , (8) with x0 = √ ~ mω . Plugging α0 back into E(α), we have our bound state energy: E(α0) = ~ 2 2m mω 2~ + 1 8 mω2 2~ mω = ~ω 2 = Egs. (9) 1 E(α) 21.51α00 7 6 5 4 3 2 1 0 Figure 1: A plot of E(α), with ~ = m = ω = 1. The minimum is 0.5, in agreement with our results. 2 δ-function potential Now, we study a delta function potential with the variational theory. Let us use our same gaussian for the variational state: Ψα (x) = Ne −αx2 , (10) with α our variational parameter and normalization N = (2α) 1/4 π1/4 . Our Hamiltonian is: H = p2 2m − V0aδ(x), (11) so our expectation value is: 〈Ψα|H |Ψα〉 = E(α) = ∫ ∞ −∞ dx [ ~ 2 2m |∂xΨ|2 − V0aδ(x) |Ψ|2 ] , (12) So: E(α) = ~ 2 2m α − V0 a 2 ( 2α π )1/2 . (13) Minimizing, we have: ∂E ∂α = ~ 2 2m − V0 a 2 ( 2 πα0 )1/2 = 0 (14) → α0 = 2m2a2V 20 π~4 , (15) 2 where H1 and H2 are the electron-nucleus energies, and Hee describes the electron-electron repulsion. In the absence of Hee, we have for the ground state: Ψσ1σ2(~r1, ~r2) = Ψ100(~r1)Ψ100(~r2)χsinglet, (33) With χsinglet = 1√ 2 ( χ (1) ↑ χ (2) ↓ − χ (1) ↓ χ (2) ↑ ) , and Ψ100(~r) the Hydrogen ground state: Ψ100(~r) = ( Z3 πa30 )1/2 e−Zr/a0 , (34) where normally Z = 2. Thus, dropping the unimportant spinor (the Hamiltonian is spin-independent), our ground state is: Ψσ1σ2(~r1, ~r2) = Ψ(r1, r2) = Z3 πa30 e−Z/a0(r1+r2). (35) Now, we treat Z not as a constant, but as our variational parameter. The physical justification for this is the partial screening of the nucleus by each electron. 〈Z|H |Z〉 = Egs(Z) = 〈Z|H1 |Z〉 + 〈Z|H2 |Z〉 + 〈Z|Hee |Z〉 . (36) Rewriting H1, we have: H1 = p21 2m − 2e 2 r1 = ( p21 2m − Ze 2 r1 ) + (2 − Z)e2 r1 . (37) Thus, the first term in Egs(Z) becomes: 〈Z|H1 |Z〉 = −ERyZ2 ︸ ︷︷ ︸ recall ERy=− e 4m 2~2 +(Z − 2)e2 〈Z| 1 r1 |Z〉 . (38) The last term in equation 38 can be readily calculated: 〈Z| 1 r1 |Z〉 = ( Z3 πa30 ) 4π ∫ ∞ 0 dr1r 2 1 1 r1 e−2Zr/a0 ︸ ︷︷ ︸ ( a02Z ) 2 (39) = Z a0 (40) ⇒ 〈Z|H1 |Z〉 = −Z2ERy + (Z − 2) e2Z a0 ︸︷︷︸ 2ERyZ = (Z2 − 4Z)ERy. (41) So the ground state energy can be expressed as: 〈Z|H |Z〉 = 2ERy(Z2 − 4Z) + 〈Z|Hee |Z〉 (42) 5 We must now calculate 〈Z|Hee |Z〉: 〈Z|Hee |Z〉 = ( Z3 πa30 )2 e2 ∫ dφ1 ︸︷︷︸ 2π dθ1 sin θ1dr1r 2 1 ∫ dφ2 ︸︷︷︸ 2π dθ2 sin θ2 ︸ ︷︷ ︸ 2 dr2r 2 2 e−2Z/a0(r1+r2) (r21 + r 2 2 − 2r1r2 cos θ12)1/2 (43) = ( Z3 πa30 )2 e28π2 ∫ ∞ 0 dr1r 2 1 ∫ ∞ 0 dr2r 2 2e −2Z/a0(r1+r2) ∫ 1 −1 dµ 1 (r21 + r 2 2 − 2r1r2µ)1/2 . (44) The integral over µ is readily found: ∫ 1 −1 dµ 1 (r21 + r 2 2 − 2r1r2µ)1/2 = − (r 2 1 + r 2 2 − 2r1r2µ)1/2 r1r2 ∣ ∣ ∣ ∣ 1 −1 = − 1 r1r2 [|r1−r2|−(r1+r2] = 2 min [r1, r2] r1r2 . (45) So our ground state energy becomes: Egs(Z) = 2ERy(Z 2 − 4Z) + ( Z3 a30 )2 8e2 × ∫ ∞ 0 dr1r 2 1 ∫ ∞ 0 dr2r 2 2e −2Z/a0(r1+r2) 2 min [r1, r2] r1r2 ︸ ︷︷ ︸ I . (46) To evaluate I , we first notice that it is symmetric in r1 and r2, I is twice the integral just over the region r2 < r1. So, I = 4 ∫ ∞ 0 dr1r1e −2Z/a0r1 ∫ r1 0 dr2r 2 2e −2Z/a0r2 (47) = 4 ( a0 2Z )5 ∫ ∞ 0 dx1x1e −x1 ∫ x1 0 dx2x 2 2e −x2 ︸ ︷︷ ︸ −x21e−x1−2x1e−x1+2−2e−x1 (48) = 1 8 (a0 Z )5 ∫ ∞ 0 dx1 [ (−x31 − 2x21 − 2x1)e−2x1 + 2x1e−x1 ] (49) = 1 8 (a0 Z )5 [ − 3! 16 − 4 8 − 2 4 + 2 ] (50) = 1 8 (a0 Z )5 5 8 . (51) Plugging this back into our ground state energy, we find: Egs(Z) = 2ERy(Z 2 − 4Z) + ERy 5 4 Z = ERy[2Z 2 − 8Z + 5 4 Z], (52) which matches up with Shankar’s result. Minimizing, we have: ∂Egs ∂Z = 0 = 4Z0 − 8 + 5 4 ⇒ Z0 = 2 − 5 16 . (53) As noted earlier, Z < 2 implies screening of the nucleus. The numerical result for the ground state energy is thus: Egs(Z0) ≡ E0 = −2Z20ERy = −2 ( 2 − 5 16 )2 ERy ' −77.5 eV. (54) 6 5 Finite Triangular Well (WKB) We want to compute the tunnelling probability through V (r) =    ∞ r < 0 x 0 ≤ r < d 0 d < x . (55) In the WKB approximation, we have the following relation: Ψ(x = d) = Ψ(x0)e −γ/2, (56) with x0 the classical turning point of the particle, and γ 2 = 1 ~ ∫ d x0 [2m (V (x) − E)]1/2 dx. (57) To find x0, we require that the total energy is equal to the potential energy at x0: E = V (x0) = x0 (58) ⇒ x0 = E  . (59) So, our expression for γ/2 becomes: γ 2 = (2m)1/2 ~ ∫ d x0 [x − E]1/2 dx (60) = (2m)1/2 ~ 2 3 (z − E)3/2 ∣ ∣ ∣ ∣ d x0 (61) ⇒ γ(E) = 4(2m) 1/2 3~ (d − E)3/2 . (62) Therefore, the probability for tunneling is given by: √ 2m(E − d2 ) 2md exp [−γ(E)], (63) with γ given in equation 62. 6 3D Short-Ranged Potential (WKB) With l 6= 0, our effective potential becomes: Veff = V (r) + ~ 2l(l + 1) 2mr2 . (64) 7