chap 7 prob

# chap 7 prob - Physics 162a Problem Set 6 Miles Ketchum...

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Unformatted text preview: Physics 162a Problem Set 6 Miles Ketchum [email protected] due date: 11/09/07 Sakurai 2.4 For a free particle, V (x) = 0 and the Hamiltonian is simply p2 2m For x and p, the Heisenberg equations of motion are (eq. 2.2.25 and 2.2.26) ˆ ˆ Hfree = 1 dp(t) = [p(t), H (t)] = 0 dt i¯ h → p(t) = p(0) dx(t) 1 1 p(2) p(0) p(t)2 1 = [x(t), H (t)] = (2i¯ p(t)) = h = x(t), = dt i¯ h i¯ h 2m 2mi¯ h m m t p(0) + x(0) m Since we have the necessary parts, we can now calculate the commutator x(t) = [x(t), x(0)] = t t i¯ t h p(0) + x(0), x(0) = [p(0), x(0)] = − m m m Sakurai 2.16 (a) |α = c0 |0 + c1 |1 → |c1 | = 1 − |c0 |2 We know that |c0 |2 + |c1 |2 = 1 but we can also write c0 = |c0 |eiδ0 , c1 = |c1 |eiδ1 = eiδ1 1 − |c0 |2 x = α|x|α = (c∗ 0| + c∗ 1|) x (c0 |0 + c1 |1 ) 1 0 = |c0 |2 0|x|0 + c∗ c1 0|x|1 + c∗ c0 1|x|0 + |c1 |2 1|x|1 0 1 Let me deﬁne β = h ¯ to save some time. Since 2mω x = β (a + a† ) where a and a† are the creation and annihilation operators, respectively, we have x = |c0 |2 0|a + a† |0 + c∗ c1 0|a + a† |1 + c∗ c0 1|a + a† |0 + |c1 |2 1|a + a† |1 0 1 = β (c∗ c1 + c∗ c0 ) = 2β cos(δ1 − δ0 )|c0 | 0 1 1 1 − |c0 |2 Now to maximize x , take the derivative with respect to |c0 | and δ1 , dx = d|c0 | 1 − |c0 |2 + 1 |c0 | = √ 2 1 − |c0 |2 |c0 |2 =0 dx = − sin(δ1 − δ0 ) = 0 dδ1 δ1 = nπ + δ0 But we also need the second derivative to be negative in order for x to be maximized, d2 x 2 = − cos(δ1 − δ0 ) < 0 dδ1 δ1 = 2kπ + δ0 where k is an integer. 1 1 |α = eiδ0 √ |0 + ei(2kπ+δ0 ) √ |1 2 2 But we can set δ0 = 0, so 1 α = √ (|0 + |1 ) 2 (b) iE0 t 1 h |α, t = e−iHt/¯ |α = √ exp − h ¯ 2 1 iωt |α, t = √ exp − 2 2 i) Schrodinger x = α, t|x|α, t = 1 2 exp = iωt 2 0| + exp 3iωt 2 1| x exp − 0|x|1 + exp iωt 2 |0 + exp − 1|x|0 3iωt 2 |1 |0 + exp − 3iωt 2 iE1 t h ¯ |1 |0 + exp − |1 1 exp 2 iωt 3iωt − 2 2 =β 3iωt iωt − 2 2 e−iωt + eiωt = β cos ωt 2 ii) Heisenberg x(t) = x(0) cos ωt + from Problem 2.15 in a previous homework. 1 1 p(0) √ (|0 + |1 ) sin ωt x = √ ( 0| + 1|) x(0) cos ωt + mω 2 2 1 1 = cos ωt [ 0|x|1 + 1|x|0 ] + sin ωt [ 0|p|1 + 1|p|0 ] 2 2mω 1 1 m¯ ω h m¯ ω h 1 β cos ωt − (−i) sin ωt + (i) sin ωt = β cos ωt 2 2mω 2 2mω 2 2 p(0) sin ωt mω =2 (c) I chose the Schroedinger picture. x2 = β 2 a2 + (a† )2 + aa† + a† a x2 = = β2 1 iωt/2 e 0| + e3iωt/2 1| x2 e−iωt/2 |0 + e−3iωt/2 |1 2 1 i(ωt/2−ωt/2) e 0|aa† |0 + ei(3ωt/2−3ωt/2) 1|aa† |1 + 1|a† a|1 2 = β2 using the result from part (b), (∆x)2 = β 2 − β 2 cos2 ωt = β 2 sin2 ωt Sakurai 2.24 (a) The energy spectrum is continuous. For E > V , ψI (x) = A i exp 1/4 h ¯ (E − V ) = 2m(E − V )dx + c1 sin (E − V )1/4 1 h ¯ ′ B i exp − 1/4 h ¯ (E − V ) 2m(E − V )dx 2m(E − V )dx π 1x c1 2m(E − V )dx − sin 1/4 h ¯x 4 (E − V ) √ 1/2 2mλ E /λ E π c1 sin dx − = −x h ¯ λ 4 (E − V )1/4 x 3/2 √ 2E c1 2mλ π sin − −x − = 1/4 3λ h ¯ 4 (E − V ) = For E < V , ψII (x) = 1 c2 exp − 1/4 h ¯ (V − E ) x E/λ x E/λ 2m(V − E )dx = c2 1 exp − 1/4 2m¯ λ h (V − E ) = 2m(λx − E )d(2mλx) 2mλ h ¯2 1/3 2 c2 exp − (V − E )2/3 3 (V − E )1/4 1/3 Let q = α(E − V ), α = 2mλ h ¯2 ψI (x) = ψII (x) = c1 1 π sin q 3/2 − 1/4 3 4 q 2 c2 exp − (−q )2/3 1/4 3 (−q ) 3 (b) If V (x) = λ|x| there are bound states and the energy spectrum is discrete. The energy eigenstates have to satisfy 2m(E − λ|x|)dx = n+ 1 2 π¯ h Sakurai 2.36 (a) The vector potential is A= So we now have Πx = px − eB eAx = px + y c 2c [Πx , Πy ] = px + =− Since we know the commutator [xi , pi ] = i¯ , h = (b) The Hamiltonian is 1 H= 2m H= eA p− c 2 B (−y, x, 0) 2 eAy eB = py − x c 2c Πy = py − eB eB y, py − x 2c 2c eB eB [px , x] + [y, py ] 2c 2c h eB i¯ eB i¯ eB h + = i¯ h 2c 2c c 1 Π2 + Π2 + p2 y z 2m x This can be rewritten using the following substitutions: ρ = Πy H= q= c Πx eB ρ2 e2 B 2 2 p2 z + + q 2m 2m 2mc2 Since [q, ρ] = i¯ , the second two terms are the harmonic oscillator Hamiltonian; h H= p2 ρ2 1 z + + mω 2 q 2 , 2m 2m 2 H = Hpz + Hosc To ﬁnd the energy eigenvalues H |k, n = Ek,n |k, n 4 ω= |eB | mc (Hpz + Hosc )|k, n = (Epz + Eosc )|k, n But we know the energy eigenvalues of the harmonic oscillator as ¯ ω n + h the operator pz is ¯ k, h Ek,n = Epz + Eosc = 1 h ¯ 2 k2 + hω n + ¯ 2m 2 = h h ¯ 2 k2 |eB |¯ + 2m mc n+ 1 2 1 , and since the eigenvalue of 2 Shankar 7.3.1 Equation 7.3.8 reads: ψ ′′ + (2ǫ − y 2 )ψ = 0 We will plug in a power series of the form ∞ ψ (y ) = n=0 ∞ n=0 Cn y n Cn [n(n − 1)y n−2 + 2ǫy n − y n+2 ] I will borrow from Shankar to re-write the ﬁrst term; ∞ n=0 ∞ Cn n(n − 1)y n−2 = Cn+2 (n + 2)(n + 1)y n n=0 For the last term (following Shankar’s example for the ﬁrst term), ∞ ∞ Cn y n=0 n+2 = n=0 Cn−2 y n Now everything is of the power yn, y n [Cn+2 (n + 2)(n + 1) + 2ǫCn − Cn−2 ] = 0 Cn+2 (n + 2)(n + 1) + 2ǫCn − Cn−2 = 0 Cn+2 = Cn−2 − 2ǫCn (n + 2)(n + 1) ∞ n=0 This is a relation between Cn+2 , Cn , and Cn−2 , but I’m not even sure what the question is asking. I think it’s asking to show that plugging in a power series gives the above relation... Shankar 7.3.7 The Hamiltonian for the harmonic oscillator in momentum space is H= using x = i¯ h ∂ . ∂P 1 P2 + kx2 → 2m 2 P2 1 ∂2 − ¯ 2k h 2m 2 ∂P 2 5 1 This is the same as the Hamiltonian in position space with x → p, mω → ; so the momentum space mω eigenfunctions using these substitutions become: ψ0 (x) = mω π¯ h 1/4 exp − mω 2 x 2¯ h → ψ0 (P ) = 1 π ¯ mω h 1/4 exp − P2 2¯ mω h Shankar 7.4.5 Isn’t this the same as what we did in Sakurai 2.16? 1 iE0 t h |ψ (t) = e−iHt/¯ |ψ (0) = √ exp − h ¯ 2 ψ (0)|x|ψ (0) = x 0 |0 + exp − iE1 t h ¯ |1 = 1 2 h ¯ 2mω = 0|a + a† |0 + 1|a + a† |0 + 0|a + a† |1 + 1|a + a† |1 1 2 h ¯ (2) = 2mω h ¯ 2mω x(t) = ψ (t)|x|ψ (t) = β 2 exp = iE0 t h ¯ β exp 2 0| + exp iE1 t h ¯ 1| x exp − 0|x|1 + exp iE0 t h ¯ |0 + exp − 1|x|0 iE1 t h ¯ |1 iE0 t iE1 t − h ¯ h ¯ =β iE1 t iE0 t − 2 h ¯ e−iωt + eiωt = β cos ωt 2 where ω = E1 − E0 . h ¯ Shankar 7.5.4 1) E (i)P (i) = i 1∂ ∂ 1 Ei e−βEi = − Z (β, Ei ) = − ln Z Z Z ∂β ∂β 2) Z= i ∞ ∞ −∞ e−βEi dx dpe−βEi Zcl = −∞ ∞ ∞ = −∞ ∞ −∞ exp exp 0 β p2 β + mω 2 x2 dx dp 2m 2 ∞ =4 Let a = − βmω 2 β ,α=− . 2 2m 1 mωx2 dx 2 exp 0 β p2 2m dp ∞ =4 0 e−ax dx 0 2 ∞ e−αp dp 2 6 =4 =π 1 2 π a 1 2 π α −2 −2m βmω 2 β 2π 2 = βω βω =π The average energy is therefore E =− =− ∂ ∂ ln Z = − ln ∂β ∂β 2π ωβ ∂ ∂ ∂ 1 ln(2π ) + ln ω + ln β = ∂β ∂β ∂β β For the above I just used the properties of logarithms and paid attention to signs, A = ln A − ln B B ln AB = ln A + ln B ln 3) For the harmonic oscillator, En = hω n + ¯ Z= 1 2 ∞ h e−βEn = e−β ¯ ω/2 + n=0 h e−nβ ¯ ω The second term is a series which converges. h eβ ¯ ω h eβ ¯ ω − 1 I’m tired so I’m not going to bother with the algebra, but it reduces to the desired answer h Z = e−β ¯ ω/2 + h h Z = e−β ¯ ω/2 (1 − e−β ¯ ω )−1 E =− =− h ∂ e−β ¯ ω/2 ln h ∂β 1 − e−β ¯ ω ∂ ∂ h h ln e−β ¯ ω/2 + ln(1 − e−β ¯ ω ) ∂β ∂β h 1 h ¯ ωe−β ¯ ω = ¯ω + h h 2 1 − e−β ¯ ω h In the second term, divide the fraction by e−β ¯ ω and mind the sign of the exponent, and you see that E = hω ¯ 4) As T → ∞, Large T does indeed mean 1 1 + β¯ ω h −1 2e E qu → ¯ ω = kT = E h cl 7 T >> 5) The classical average energy is now h ¯ω k E = 3N0 kT this the speciﬁc heat per atom is Ccl (T ) = The quantum average energy is now E = 3N0 ¯ ω h Cqu (T ) = 1 1 + β¯ ω h −1 2e h eβ ¯ ω ∂ 1 (3kN0 ) (T ) = 3k N0 ∂T 1 ∂ (3N0 ) N0 ∂T 1 kT 2 h ¯ω −1 = 3(¯ ω )2 h Using θE = h ¯ω , k h e¯ ω/kT h (e¯ ω/kT − 1)2 Cqu (T ) = 3k In the high temperature limit, Cqu (T ) → 3k Cqu (T ) → 3k θE T 2 eθE /T (eθE /T − 1)2 θE T θE T 2 2 1 (1 + θE /T − 1)2 T θE 2 → 3k In the low temperature limit, the -1 in the denominator can be ignored since it is small compared to the exponential. Cqu (T ) → 3k Cqu (T ) → 3k θE T θE T 2 eθE /T (eθE /T )2 2 e−θE /T Shankar 9.4.3 Following Shankar’s example and treating P and X as one dimensional, H= H= = Since we know that ∆X ∆P ≥ P2 e2 − 2m (X 2 )1/2 1 P2 − e2 2m (X 2 )1/2 (∆P )2 e2 e2 (∆P )2 − − = 2m 2m ∆X [(∆X )2 ]1/2 h ¯ h ¯ , ∆P ≥ , 2 2∆X 8 H= 1 2m h ¯ 2∆X 2 − 1 e2 h ¯2 e2 = − ∆X 4m (∆X )2 ∆X To ﬁnd the minimum, we diﬀerentiate with respect to ∆X , dH h ¯ e2 =0=− + d(∆X ) 2m(∆X )3 (∆X )2 Solving for ∆X . ∆X = Substituting back in, H= h ¯ 2 4m2 e4 2me2 me4 − e2 2 = − 2 4m ¯ 4 h h ¯ h ¯ h ¯2 2me2 9 ...
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