SOLUTION MANUAL COMPILED BY YEMI BUKKY +234(0)8057474928; +234(0)8064974071 (
[email protected])
Department of Physics, Federal University of Technology, Minna, NG Nigeria
PRINCIPLES OF QUANTUM MECHANICS BY R. SHANKAR SECOND EDITION
SOLUTIONS COMPILED BY YEMI BUKKY (
[email protected]) Department of Physics, Federal University of Technology, Minna, NG Nigeria +2348057474928
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Physics 710-712
due October 30, 2009
Problem Set 4 Problem 1: Do exercise 2.5.3 of the text. Solution: The problem asks us to get the equations of motion using the Hamiltonian method for the system shown in Figure 1.5 (p. 46) of the text. Using x1 and x2 shown there as the generalized coordinates, the kinetic energy and potential energy are T =
m 2 m 2 x˙ + x˙ 2 , 2 1 2
V =
k 2 k k x + (x1 − x2 )2 + x22 , 2 1 2 2
so the Lagrangian is L = T − V and thus the generalized momenta are p1
=
p2
=
∂T ∂L = = mx˙ 1 , ∂ x˙ 1 ∂ x˙ 1 ∂L ∂T = = mx˙ 2 , ∂ x˙ 2 ∂ x˙ 2
p1 , m p2 x˙ 2 = , m
⇒
x˙ 1 =
⇒
so the Hamiltonian is H=T +V =
p2 k k k p21 + 2 + x21 + (x1 − x2 )2 + x22 , 2m 2m 2 2 2
and Hamilton’s equations are x˙ 1
=
x˙ 2
=
∂H p1 = , ∂p1 m p2 ∂H = , ∂p2 m
∂H = −2kx1 + kx2 , ∂x1 ∂H = − = −2kx2 + kx1 . ∂x2
p˙1 = − p˙2
Problem 2: Do exercise 2.7.2 of the text. Solution:
(i): X X ∂qi ∂qj ∂qi ∂qj ∂qi ∂qj {qi , qj } := · − · = ·0 − 0· =0 ∂qk ∂pk ∂pk ∂qk ∂qk ∂qk k k X X ∂pi ∂pj ∂pi ∂pj ∂pj ∂pi · − · = 0· − ·0 = 0 {pi , pj } := ∂qk ∂pk ∂pk ∂qk ∂pk ∂pk k k X X ∂qi ∂pj ∂qi ∂pj {qi , pj } := · − · = (δik δjk − 0·0) = δij , ∂qk ∂pk ∂pk ∂qk k
k
and X X ∂qi ∂H ∂qi ∂H ∂H ∂H ∂H {qi , H} := · − · = δik · − 0· = = q˙i , ∂qk ∂pk ∂pk ∂qk ∂pk ∂qk ∂pi k k X X ∂pi ∂H ∂pi ∂H ∂H ∂H ∂H {pi , H} := · − · = 0· − δik · =− = p˙i , ∂qk ∂pk ∂pk ∂qk ∂pk ∂qk ∂qi k
k
where in the last steps I used Hamilton’s equations. (ii): The Hamiltonian given is H = p2x + p2y + ax2 + by 2 .
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If a = b, H has a
symmetry under simultaneous rotations in the x-y and px -py planes, under which `z (the generator) is conserved. Therefore {`z , H} = 0. We check this as follows: X ∂`z ∂H ∂`z ∂H ∂`z ∂H ∂`z ∂H ∂`z ∂H ∂`z ∂H {`z , H} = · − · = · + · − · − · . ∂qk ∂pk ∂pk ∂qk ∂x ∂px ∂y ∂py ∂px ∂x ∂py ∂y k
But ∂H = 2pk , ∂pk ∂`z ∂(xpy − ypx ) ∂`z ∂`z = = , = (−y , x) , ∂pk ∂pk ∂px ∂py
∂H ∂H ∂H = , = (2ax , 2by) , ∂xk ∂x ∂y ∂`z ∂`z ∂`z = , = (py , −px ) , ∂qk ∂x ∂y
so {`z , H} = py ·2px + (−px )·2py − (−y)·2ax − x·2by = 2xy(a − b) which vanishes if a = b.
Problem 3: Do exercise 2.8.1 of the text. Solution:
Since g = p1 + p2 , it generates the infinitesimal transformations ∂g = +, ∂p1 ∂g δx2 = + = +, ∂p2
∂g = 0, ∂x1 ∂g δp2 = − = 0. ∂x2 δp1 = −
δx1 = +
So, to order , these give the canonical transformations xi → x ¯i (xj , pj ) and pi → p¯i (xj , pj ) with x ¯1 = x1 + ,
p¯1 = p1 ,
x ¯2 = x2 + ,
p¯2 = p2 ,
which is precisely a spatial translation of the whole system by an amount .
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Physics 710-711-712
November 16, 2009
Problem Set 5 Problem 1: Do exercise 4.2.1 of the text. (1) The possible outcomes are Lz = {1, 0, −1}, which are the eigenvalues of Lz .
Solution:
(2) Lz |ψi = 1 · |ψi implies
Solution:
1 |ψi = 0 . 0 (Note that I have normalized |ψi!)
Then
0 1 0 1 0 1 1 hLx i = hψ|Lx |ψi = 1 0 0 √ 1 0 1 0 = √ 1 0 0 1 = 0. 2 2 0 1 0 0 0 0 1 0 0 1 0 1 0 1 1 1 2 2 hLx i = hψ|Lx |ψi = 1 0 0 0 1 0 1 = . 1 0 1 1 0 1 0 = 2 2 2 0 1 0 0 1 0 0 0 s q 1 1 2 ∆Lx = hL2x i − hLx i = − 02 = √ . 2 2
(3) Solution:
The characteristic equation for Lx is 0 −λ √12 0 = det(Lx − λ) = det √12 −λ √12 = λ − λ3 , √1 0 −λ 2
⇒
λ ∈ {1, 0, −1}.
The corresponding eigenvectors, |λi, then satisfy −λ √12 −λa + √b2 0 a 0 = (Lx − λ)|λi = √12 −λ √12 b = √a2 − λb + √c2 √1 √b − λa c −λ 0 2 2 where we have parameterized the components of |λi by (a b c). For λ = 1, we can solve √ for b and c in terms of a, giving b = 2a and c = a. We then determine a by normalizing |λ = 1i:
√a |λ = 1i = 2a , a
⇒
1 = hλ = 1|λ = 1i = a∗
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√ ∗ 2a
a √ a∗ 2a = 4|a|2 , a
⇒
a=
1 2
(where I have chosen the arbitrary phase to be 1). λ = 0 and λ = −1, gives 1 1 √ 1 1 |λ = 0i = √ 0 , |λ = 1i = 2 , 2 2 1 −1
Thus, and doing the same thing for
1 √ 1 |λ = −1i = − 2 . 2 1
(4) Solution: The possible outcomes are Lx = {1, 0, −1}, which are the eigenvalues of Lx . |ψi is the normalized eigenstate of Lz with eigenvalue Lz = −1, which is 0 |ψi = 0 . 1 So (here P stands for "probability of"): 2 0 1 P(Lx = 1) 1 0 = , 4 1 2 0 1 1 2 P(Lx = 0) = |hλ = 0|ψi| = 1 0 −1 0 = , 2 2 1 2 0 √ 1 1 2 P(Lx = −1) = |hλ = −1|ψi| = 1 − 2 1 0 = . 2 4 1 1 √ = |hλ = 1|ψi| = 1 2 2 2
(5) Solution: 1 L2z = 0
⇒
,
the possible outcomes are L2z = {0, 1}.
1 An eigenbasis of the L2z = 1 eigenspace is {|ai, |bi} with 1 0 |ai = 0 , |bi = 0 . 0 1 Therefore, upon measuring L2z = 1, the state collapses to |ψi −→ |ψ 0 i =
(|aiha| + |bihb|)|ψi . |(|aiha| + |bihb|)|ψi|
But 1 [|aiha| + |bihb|] |ψi = 0 1 0
0
0 0 + 0 0 1
0
1 1 0 1 1 1 1 1 = 0 + 0 √ = 1 1 0 , 2 √ 2 2 √ 2 0 1 2 2
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has norm
v u u u1 t 1 2
√ 1 √ 1 3 = , 0 2 0 √ 2 2 2
so 1 1 1 1 1 0 = √ 0 . |ψ 0 i = √ 3/2 2 √ 3 √ 2 2 The probability of L2z = +1 is P(L2z = 1)
= hψ| (|aiha| + |bihb|) |ψi = |ha|ψi|2 + |hb|ψi|2 2 2 1 1 1 1 1 1 3 = 1 0 0 1 + 0 0 1 √1 = + = . √ 2 2 4 2 4 2 2
If we measured Lz the posible outcomes are the eigenvalues Lz , {0, ±1}, with probabilities 2 0 2 1 1 1 P(Lz = 1) = 1 0 0 |ψ i = √ 1 0 0 √0 = . 3 3 2 P(Lz = 0)
= 0
P(Lz = −1)
= 0
1
0 2 1 0 |ψ i = √ 0 3
1
0
0 2 1 1 |ψ i = √ 0 3
0
2 1 0 √0 = 0. 2 2 1 2 1 √0 = . 3 2
(6) Solution:
In the Lz eigenbasis 1 |Lz = 1i = 0 , 0
0 |Lz = 0i = 1 , 0
0 |Lz = −1i = 0 , 1
write the unknown state as a |ψi = b . c Then 1 P(Lz = 1) = 4
P(Lz = 1) =
1 2
P(Lz = 1) =
1 4
2 a b = |a|2 , 0 0 c 2 a 2 = |hLz = 0|ψi| = 0 1 0 b = |b|2 , c 2 a 2 = |hLz = −1|ψi| = 0 0 1 b = |c|2 . c 2 = |hLz = 1|ψi| = 1
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The most general solution to these three equations is then a=
1 b = √ eiδ2 , 2
1 iδ1 e , 2
c=
1 iδ3 e , 2
for some arbitrary phases δi , which gives the desired answer. The δi phase factors are not irrelevant. For example
P(Lx = 0)
iδ1 2 iδ 2 e 1 1 √e iδ eiδ3 2 = √ √ 0 −1 2e 2 2 − 2 2 2 eiδ3 1 1 − ei(δ3 −δ1 ) − e−i(δ3 −δ1 ) + 1 − e−iδ3 = 8
1 2 = |hλ = 0|ψi| = √ 1 2 = =
1 iδ1 e − eiδ3 e−iδ1 8 1 (1 − cos(δ3 − δ1 )) , 4
so something measurable (a probability) depends on the difference of the phases.
Problem 2: Do exercise 4.2.2 of the text. Solution: Z
∞
hP i = hψ|P |ψi = dxhψ|xihx|P |ψi −∞ Z ∞ Z ∞ dψ(x) d ∗ ψ(x) = −i~ dx ψ(x) = dx ψ (x) −i~ dx dx −∞ −∞ Z i~ ∞ i~ d ∞ = − ψ(x)2 = − ψ 2 −∞ = 0 dx 2 −∞ dx 2 if ψ → 0 as |x| → ∞. Alternatively, use the k-basis: Z ∞ Z hP i = hψ|P |ψi = dkhψ|kihk|P |ψi = −∞
But
∞
Z
−∞
Z
∞
since ψ(x) is real.
1 ψ(−k) = √ 2π
Z
Z
∞
dx eikx ψ(x),
−∞
∞
dx e−ikx ψ(x) = ψ ∗ (k)
−∞
So Z hP i =
∞
dk ~kψ ∗ (k)ψ(k) =
dk ~k ψ ∗ (k)ψ(k).
−∞
1 ψ(k) = hk|ψi = dx hk|xihx|ψi = √ 2π −∞
therefore
∞
dk ~k hψ|kihk|ψi =
Z
−∞
∞
dk ~kψ(−k)ψ(k). −∞
and under the change of variables k → −k, this becomes Z ∞ hP i = dk ~(−k)ψ(k)ψ(−k) = −hP i, −∞
and so hP i = 0.
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Problem 3: Do exercise 2.4.3 of the text. Solution: Z E D = eip0 x/~ ψ P eip0 x/~ ψ
E Z ∞ ∗ d ip0 x/~ dx eip0 x/~ ψ(x) (−i~) dx heip0 x/~ ψ|xihx|P eip0 x/~ ψ = e ψ(x) dx −∞ −∞ Z ∞ ip dψ 0 ip0 x/~ = −i~ dx ψ ∗ (x)e−ip0 x/~ e ψ(x) + eip0 x/~ ~ dx −∞ Z ∞ Z ∞ dψ dx ψ ∗ (x) dx ψ ∗ (x) p0 ψ(x) − i~ = dx −∞ −∞ Z ∞ Z ∞ dx hψ|xihx|P |ψi = p0 hψ|ψi + hψ|P |ψi = p0 + hP i. dx hψ|xihx|ψi + = p0 ∞
−∞
−∞
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Problem 2 In[3]:=
Psi@x_, t_D := HPi H∆ ^ 2 + hbar ^ 2 t ^ 2 ê Hm ^ 2 ∆ ^ 2LLL ^ H−1 ê 2L Exp@H−Hx − Hp0 ê mL tL ^ 2L ê H∆ ^ 2 + hbar ^ 2 t ^ 2 ê Hm ^ 2 ∆ ^ 2LLD; PlotAEvaluate@Table@Psi@x, tD ê. 8p0 → 1, ∆ → 1, hbar → 1, m → 1<, 8t, 0, 14
»yHxL»2
0.5
0.4
0.3
Out[4]=
0.2
0.1
-20
-10
x 10
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Physics 710-711-712
December 4, 2009
Problem Set 7 Problem 1: Exercise 5.3.1 The Hamiltonian is
1 2 P + Vr (X) − iVi 2m where Vr is a real function and Vi a real constant. Therefore H=
H† =
1 2 1 (P † )2 + Vr (X † ) − (i)∗ Vi = P + Vr (X) + iVi 6= H, 2m 2m
so H is not Hermitian. Derivation of the continuity equation. Schrodinger’s equation and its complex conjugate in this case read ∂ψ ~2 2 = − ∇ ψ + Vr ψ − iVi ψ, ∂t 2m ∂ψ ∗ ~2 2 ∗ −i~ = − ∇ ψ + Vr ψ ∗ + iVi ψ ∗ . ∂t 2m i~
Multiplying the first by ψ ∗ and teh second by ψ and taking the difference, then dividing by i~ gives ∂P ~ ~j − 2 Vi P, = −∇· ∂t ~ ~ − ψ ∇)/(2mi) ~ where, as before, P = |ψ|2 and ~j = ~(ψ ∗ ∇ψ are the probability density ~ ~j term vanishes (by and current, respectively. Integrating this over all space, the ∇· the divergence theorem, since we assume ~j → 0 at infinity), giving 2 dP = − Vi P, dt ~ R where P = d3 xP is the total probability. (I can pull Vi out of the integral since it is assumed constant in the problem.) Integrating this differential equation gives P(t) = P(0) e−2Vi t/~ .
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Problem 2: Exercise 5.3.4 With primes denoting derivative with respect to x, j := =
=
=
=
~ [ψ ∗ ψ 0 − ψ(ψ ∗ )0 ] 2mi ~ ∗ −ixp/~ (A e + B ∗ eixp/~ )(Aeixp/~ + Be−ixp/~ )0 2mi − (Aeixp/~ + Be−ixp/~ )(A∗ e−ixp/~ + B ∗ eixp/~ )0 1 ∗ −ixp/~ (A e + B ∗ eixp/~ )(ipAeixp/~ − ipBe−ixp/~ ) 2mi − (Aeixp/~ + Be−ixp/~ )(−ipA∗ e−ixp/~ + ipB ∗ eixp/~ ) p 2 |A| + AB ∗ e2ixp/~ − A∗ Be−2ixp/~ − |B|2 2m + |A|2 + A∗ Be−2ixp/~ − AB ∗ e2ixp/~ − |B|2 p 2 |A| − |B|2 . m
Problem 3: Exercise 5.4.2 (a) For x < 0, V = 0, so the general is ψ< = √ solution of the energy eigenstate equation ikx −ikx ikx Ae + Be where ~k = 2mE. Similarly, for x > 0, ψ> = Ce + De−ikx . Scattering boundary conditions means we set D = 0 (no incoming particles from x = +∞). Now we need to figure out the boundary conditions at x = 0. Look at the time-independent Schrodinger equation, −
~2 00 ψ + V0 aδ(x)ψ = Eψ. 2m
(1)
Since the potential has an infinite jump in it, ψ will be continuous, but ψ 0 may have a finite jump. To see how big the ψ 0 jump is, integrate (1) from x = − to x = + to get Z ~2 0 0 [ψ (−) − ψ ()] + V0 aψ(0) = E dx ψ. 2m − In the limit as → 0, the right hand side vanishes since ψ is continuous, from which we learn that 0 0 (0) − ψ< (0) = (2maV0 /~2 )ψ(0). ψ>
Applying this boundary condition along with continuity of ψ to ψ< and ψ> gives the two conditions A+B = C ikC − ikA + ikB = (2maV0 /~2 )C.
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Dividing through by A, and solving for B/A and C/A gives B/A = maV0 /(ik~2 − maV0 ) and C/A = ik~2 /(ik~2 − maV0 ). Since R = |B/A|2 and T = |C/A|2 , we get k 2 ~4 m2 a2 V02 , T = . R= 2 4 k ~ + m2 a2 V02 k 2 ~4 + m2 a2 V02 (b) Call x < −a region I, |x| < a region II, and x > a region III. Then solving for the energy eigenstates, ~ψ 00 = 2m(E − V (x))ψ, of energy 0 < E ≤ V0 in each region ikx −κx gives ψI = Ae + Be−ikx , ψII = Ce√ + Deκx , and ψIII = Eeikx + F e−ikx , p with ~κ = 2m(V0 − E) and ~k = 2mE. Scattering boundary conditions means we set F = 0 (no incoming particles from x = +∞). The incoming wave has amplitude A, the reflected has amplitude B, the transmitted amplitude E. Therefore R = |B/A|2 and T = |E/A|2 , so we only need to solve for B/A and E/A. The boundary conditions at x = ±a are that ψ and ψ 0 are continuous, implying Ae−ika + Beika ikAe−ika − ikBeika Eeika ikEeika
= = = =
Ceκa + De−κa , −κCeκa + κDe−κa , Ce−κa + Deκa , −κCe−κa + κDeκa .
Dividing by A and eliminating C/A and D/A gives e−2iak (e4aκ − 1)(k 2 + κ2 ) B = , A (e4aκ − 1)(k 2 − κ2 ) + 2i(e4aκ + 1)kκ 4ie2a(κ−ik) kκ E = , A (e4aκ − 1)(k 2 − κ2 ) + 2i(e4aκ + 1)kκ so R=
(e4aκ − 1)2 (k 2 + κ2 )2 , (e4aκ − 1)2 (k 2 − κ2 )2 + 4(e4aκ + 1)2 k 2 κ2
and T = 1 − R.
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Physics 711
January 15, 2010
Problem Set 8 Problem 1: Exercise 7.3.1 P n 00 2 Plug the power series expansion ψ = ∞ n=0 cn y into the equation ψ +(2ε−y )ψ = 0 to get ∞ X cn n(n − 1)y n−2 + (2ε − y 2 )y n = 0. n=0
Shift n → n + 2 in the first term, and n → n − 2 in the third term to get ∞ X
y n [(n + 2)(n + 1)cn+2 + 2εcn − cn−2 ] = 0
n=0
with the convention that c−2 = c−1 = 0. This implies cn+2 =
cn−2 2εcn − (n + 1)(n + 2) (n + 1)(n + 2)
for all n ≥ 0.
Problem 3: Exercise 7.3.7 In the momentum basis |ψi → ψ(p), P → p, and X → i~(d/dp), so the energy eigenvalue equation 1 2 mω 2 2 P + X |Ei = E|Ei 2m 2 becomes
1 2 mω 2 2 00 p ψ(p) − ~ ψ (p) = Eψ(p). 2m 2 Compare this to the position-basis equation mω 2 2 1 2 00 x ψ(x) − ~ ψ (x) = Eψ(x). 2 2m These are the same equations with the substitutions x ↔ p and m ↔ 1/(mω 2 ).
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Problem 2: Exercise 7.3.5 ∞
Z
dx ψn∗ (x)·x·ψn (x) −∞
hn|X|ni =
Z =
since x is odd and ψn2 (x) is even.
Z
∞
d dx ψn∗ (x)(−i~) ψn (x) dx −∞
i~ ∞ = − ψn2 −∞ = 0 2 Z
2
h1|X |1i =
dx x·ψn2 (x) since ψn (x) is real
−∞
= 0 hn|P |ni =
∞
∞
dx ψ1∗ x2 ψ1 = −∞ Z 2 mω 3/2 ∞
Z = (−i~)
∞
dx ψn ψn0 −∞
i~ = − 2
Z
∞
dx (ψn2 )0
−∞
since ψn → 0 as |x| → ∞. mω 1/2 Z 4π~
∞
dx x
−∞ 2
2
2x
mω 1/2 2
2 /~
e−mωx
~ √ 2 mω 3/2 3 π mω −5/2 = √ 4 ~ π ~
= √ dx x4 e−mωx /~ π ~ −∞ 3~ = . 2mω Z ∞ mω 3/2 Z ∞ 00 2 2 2 ∗ 2 00 2 2 h1|P |1i = dx ψ1 (−i~) ψ1 = −~ √ dx xe−mωx /2~ xe−mωx /2~ π ~ −∞ −∞ Z ∞ h mω i 3/2 mω mω 2 2 dx x2 x2 − 3 e−mωx /~ = −~2 √ ~ ~ π ~ −∞ √ 2 mω 5/2 3 π mω −3/2 3mω~ . = −~2 √ ·(−1) = 4 ~ 2 π ~
∆X
∆P
2
2
r
2
= h0|X |0i =
2
= h0|P |0i = −~
2
∴ ∆X ∆P =
∞
2 −mωx2 /~
dx x e −∞
mω 1/2 √π mω −3/2 ~ = = . ~ 2 ~ 2mω
Z 00 mω ∞ 2 2 dx e−mωx /2~ e−mωx /2~ π~ −∞ mω 3/2 Z ∞ mω 2 2 dx x − 1 e−mωx /~ ~ ~ −∞ √ mω 3/2 mω~ π mω −1/2 ·(−1) = . ~ 2 ~ 2
r
~2 √ = − π 2 ~ = −√ π r
Z
mω π~
r ~ mω~ ~ · = . 2mω 2 2
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Physics 828: Homework Set No. 3 Due date: Friday, January 27, 2011, 1:00pm in PRB M2043 (Biao Huang’s office) Total point value of set: 60 points + 10 bonus points Problem 1 (20 pts.): Exercise 10.3.5 (Shankar, p. 278) Problem 2 (10 pts.): Exercise 10.3.6 (Shankar, p. 278) Problem 3 (5 pts.): Exercise 11.2.2 (Shankar, p. 283) Problem 4 (5 pts.): Exercise 11.4.1 (Shankar, p. 300) Problem 5 (5 pts.): Exercise 11.4.2 (Shankar, p. 300). If you correctly derive in closed ˆ you receive 10 bonus points. form the explicit expression for [Pˆ , H] Problem 6 (10 pts.): Exercise 11.4.3 (Shankar, p. 300) Problem 7 (5 pts.): Exercise 11.4.4 (Shankar, p. 300)
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Physics 828
Sketch of Solution to Set 2
Shankar 12.3.7 (1) The two-dimensional harmonic oscillator is obviously invariant under rotations about the z-axis: the magnitude of the position and momenta are unaltered by rotations around the z-axis. Therefore, the Hamiltonian commutes with the generator of rotations about the z-axis, Lz . (2) So we write ψ(ρ, φ) = eimφ REm (ρ) where m is an integer, positive or negative. In two dimensions since we know the Laplacian we have Ã
h ¯2 1 0 00 REm − + REm 2µ ρ
!
à 2
+
1 h ¯ m2 + µω 2 ρ2 2 2µρ 2
!
REm (ρ) = EREm .
For small ρ assuming that REm (ρ) ∝ ρk (derivatives decrease the power by unity increasing its importance for small ρ etc.) we can neglect the potential energy term and the constant term on the right-hand side. Thus we have 00 REm +
0 m2 REm ∝ 2 REm ⇒ [k(k − 1) + k] = m2 ⇒ k 2 = m2 . ρ ρ
For the wave function to be normalizable, for ρ→0 limit k ≥ 0. Thus we have REm (ρ) −→ ρ|m| .
R∞ 0
2 dρ ρ REm (ρ) to be finite at the lower
(3) For large ρ, terms with inverse powers of ρ including the centrifugal term and also the constant term on the right-hand side can be neglected. So we have µ2 ω 2 ρ2 REm ; h ¯2 this is identical to the one-dimensional oscillator equation. See the careful analysis on page 191. Up to powers of ρ the solution is1 00 REm =
ρ→∞
µω 2
REm (ρ) −→ e− 2¯h ρ . So we write as instructed 1
µω − µω ρ2 µω ³ µω 2 ´ − µω ρ2 00 ρ e 2¯h and REm −1 + ρ e 2¯h . = ¯h ¯h ¯h 00 Thus we can neglect the constant term in R and recover the solution. One can use this to check that terms we have neglected are indeed small compared to the terms we have retained. When you neglect terms it is a good idea to substitute the solution you have obtained and check that the terms you have neglected are indeed smaller. 0 REm = −
1
185
µω 2
REm (ρ) = Um (ρ) ρ|m| e− 2¯h ρ . (4) Use the dimensionless variable ² = ¯hEω and y 2 = equation by h ¯ ω we have dropping the annoying subscripts "
1 − 2
Ã
d2 1 d m2 + − dy 2 y dy y2
!
µω 2 ρ. h ¯
Dividing the radial
#
1 + y 2 R(y) = ² R(y) . 2
(5) We do the substitution and do elementary calculus and obtain the result given: "Ã 00
U +
2|m| + 1 y
!
#
− 2y U 0 + (2² − 2|m| − 2) U = 0 .
P
r r (6) We substitute U (y) = ∞ r=0 Cr y and collect the coefficient of y . The second derivative reduces the power by two and so we use the Cr+2 term etc.
(r + 2)(r + 1)Cr+2 + (2|m| + 1)(r + 2)Cr+2 − 2rCr + (2² − 2|m| − 2)Cr = 0 yielding a two-term recursion relation. (7) We write this as (2(² − |m| − r − 1) Cr+2 = − . Cr (r + 2)(2|m| + r + 2) First if C0 is given C2 and the other even terms can be computed. As r → ∞ we have 2 Cr+2 → . Cr r 2
2
This implies that U (y) grows as ey which overwhelms the e−y /2 in R pushing it out of the Hilbert space. So the series must terminate. Thus the boundary condition at infinity leads as usual to energy quantization. What about the odd terms? A series only with odd terms (set C0 = 0 so that all even terms vanish) is inconsistent since then U (y) ∼ y for small y and thus R(ρ) ∼ ρ|m|+1 inconsistent with our earlier result in (2). This appears to be suggested as an argument. What if one starts with C0 and C1 non-zero? Substituting into the equation for U we find that the (1/y)(dU/dy) leads to the term C1 /y and there is no other source of y −1 terms. Thus C1 = 0 and therefore, all odd terms vanish.
2
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Therefore, we have r = 2k and the termination of the series condition yields ² = 1 + r + |m| = |m| + 2k + 1 ≡ (n + 1) with k = 0, 1, , 2, · · ·. (8) Since |m| = n − 2k for a given n (i.e., for a given energy) the maximum value of m is n which occurs for k = 0, The azimuthal quantum number m decreases by steps of 2 until m reaches the value −n. It is easy to se that there are n + 1 allowed values of m yielding a degeneracy of n + 1. In Cartesian coordinates the energy is nx + ny + 1 in units of h ¯ ω. So the degeneracy corresponds to the number of ways in which we can choose two non-negative integers to add up to n. We can choose nx to be any integer from 0 to n and ny = n − nx . This yields the same degeneracy. Shankar 12.6.1 (1) Since there is no angular dependence ` = 0. (b) We have R(r) ∝ e−r/a0 and for large r (retaining only the dominant terms) −
h ¯ 2 00 R = ER(r) . 2m 2
h ¯ Substituting the given form we obtain E = − 2ma 2 . 0
(c) Clearly the R00 term and the energy term cancel for all r. If the equation is valid for all r we must have h ¯2 2 0 R + V (r) R(r) = 0 . − 2m r Substituting R(r) = e−r/a0 we obtain h ¯2 h ¯2 + V (r) = 0 ⇒ V (r) = − . ma0 r ma0 r Shankar 13.1.1 and 13.1.3 You should be able to fill in the steps. Here are some steps dropping some subscripts for notational simplicty. v =
∞ X
Ck ρk+`+1 .
k=0
3
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Substituting into
à 00
0
v − 2v +
`(` + 1) e2 λ − ρ ρ2
!
v = 0
(1)
we extract the coefficient of ρk+l carefully. Since two derivatives reduce the power of ρ by two we should start from the term with ρk+`+2 with coefficient Ck+1 for the first term and similarly for the last term. For the second and third terms which reduce the power of ρ by unity we can start with the ρk+`+1 term with coefficient Ck . Thus we have (k + ` + 2)(k + ` + 1)Ck+1 − 2(k + ` + 1)Ck + e2 λCk − `(` + 1)Ck+1 = 0 which yields
Since λ2 =
−q 2 λ + 2(k + ` + 1) Ck+1 = Ck (k + ` + 2)(k + ` + 1) − `(` + 1) 2m ¯ 2W h
(2)
(3)
from 13.1.9. For the numerator to vanish we have e4 λ2 = −
2me2 = 4(k + ` + 1)2 . 2 h ¯ E
This yields 13.1.14. Mathematical aside: Here is a different representation using classical mathematP k ical physics. Note that the function L(ρ) ≡ ∞ k=0 Ck ρ obeys the equation ρL00 (ρ) + [ 2(` + 1) − 2ρ ]L0 (ρ) + (q 2 λ − 2(` + 1))L(ρ) = 0 . Substituting q 2 λ = 2n (where n is an integer eventually) we have 1 00 ρL + [ (` + 1) − ρ ]L0 + (n − (` + 1))L = 0 . 2 Let z = 2ρ we have z
d2 L dL + [2(` + 1) − z] − (` + 1 − n)L = 0 . 2 dz dz
We know (with a good mathematical methods course) that the general solution to zw00 + (c − z) w0 − aw = 0 is given by the confluent hypergeometric function w = 1 F1 (a; c; z). Thus we find that the solution L(ρ) is 1 F1 (`+1−n; 2`+2; 2ρ) . The particular terminating (for integer n) 4
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confluent hypergeometric function can be related to the associated Laguerre polynomial. Shankar 13.3 We are considering the case n = 2 and ` = 1 and thus k = 0 in the notation of the 4 from 13.1.4 and form 13.1.6 text. So we have W = me 8¯ h2 s
ρ =
me2 r 2mW r = 2 2 r = 2a0 h h ¯ 2¯
using 13.1.24. From 13.1.10 since k = 0 and ` = 1 we have v = C0 ρ2 and thus R(r) =
U (r) − r = C e 2a0 r r q
3 cos θ from 12.5.39 which is where C is an overall constant. We know that Y10 = 4π −r/(2a0 ) normalized. So all we need is that Ce r is notmalized when integrated over the radial coordinate. We have
C
2
Z ∞ 0
− ar
dr r2 r2 e
0
= C 2 24 a50 .
q
1 Thus C = 24a 3 × 1/a0 and including the normalization from the spherical harmonic 0 yields the quoted answer.
Shankar 13.5 Since we are asked to compute hΩi for stationary states |n`mi its time derivative vanishes. Thus we have h[Ω, H]i = 0 by Ehrenfests’ theorem. So we compute the ~ · P~ as ordered. We calculate (using the summation convention) commutator for Ω = R ·
Pi Pi Rj P j , 2m
¸
·
¸ Pi Pi P~ · P~ = Rj , Pj = 2i¯h = 2i¯h T . 2m 2m
We have used [Rj , Pi Pi ] = Pi [Rj , Pi ] + [Rj , Pi ]Pi = 2i¯h Pj . We consider the potential energy term next: Ã
[Rj Pj , V (R)] = Rj [Pj , V (R)] = Rj 5
189
!
d V (R) . −i¯ h dRj
Note that [Pi , V (R)] can be evaluated in the coordinate representation (in Cartesian coordinates) by acting on a function f (r): Ã
#
∂ ∂ ∂ ~ ∇V. ~ [Pi , V (r)]f (r) = −i¯ h V (r)f (r) − V (r) V (r)f (r) = −i¯h V (r) = −i¯ hR· ∂ri ∂ri ∂ri Thus we can write formally [Pi , V (R)] = −i¯h∂V (R)/∂Ri . So we need to evaluate ~ · ∇V ~ (R). In the coordinate representation using spherical coordinates for central R potentials this is just rV 0 (r). For the Coulomb potential we obtain −V (r) and including the factor of −i¯h we obtain i¯h V (r). Substituting into the basic relation we have h2T + V i = 0 as asserted. If V (R) ∝ Rn , rV 0 (r) = nV and thus we obtain hT i =
6
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n 2
hV i.
Physics 710
March 12, 2010
Problem Set 15 Problem 12.6.1: ψE = Ae−r/a0 . (1) No (θ, φ)-dependence implies ψE ∝ Y00 , so we must have ` = 0 and m = 0. (2) Therefore ψE = RE,`=0 = 1r UE,0 which satisfies eqn. (12.6.5) with ` = 0: (rψE )00 +
2µ [E − V (r)](rψE ) = 0. ~2
(1)
As r → ∞, V (r) → 0, which implies in this limit (rψE )0 = A(1 − ar0 )e−r/a0 ≈ −(A/a0 )re−r/a0 , so (rψE )00 ≈ (A/a20 )re−r/a0 . Therefore, eqn. (1) reads in this limit A −r/a0 2µE = − 2 Are−r/a0 , re 2 a0 ~ from which it follows that ~2 E=− . 2µa20 (3) Now plug E into (1) and use (rψE )00 = A(− a20 + ar2 )e−r/a0 to get 0 2 r ~2 2µ −r/a0 A − + 2 e − V (r) Are−r/a0 = 0, + 2 − a0 a0 ~ 2µa20 which gives V (r) = −
~2 . µa0 r
Problem 12.6.4: R (1) δ 3 (r − r0 ) is defined by the property that d3 rδ 3 (r − r0 )f (r) = f (r0 ). So simply check: Z 1 r2 dr sin θdθdφ 2 δ(r − r0 )δ(θ − θ0 )δ(φ − φ0 )f (r, θ, φ) r sin θ Z = drdθdφδ(r − r0 )δ(θ − θ0 )δ(φ − φ0 )f (r, θ, φ) = f (r0 , θ0 , φ0 ). ∂ ∂ ∂ ∂ 1 (2) If r 6= 0 then ∇2 ( 1r ) = r12 ∂r (r2 ∂r ( r ))+(angular parts) = r12 ∂r (r2 ( −1 )) = r12 ∂r (−1) = r2 0. When r = 0 the above calculation breaks down since terms are singular So consider an Rarbitrary R 3there. R continuous function f (r) and the integral d x∇2 ( 1r )f (r) = lim→0 0 r2 dr dΩ∇2 ( 1r )f (r), since ∇2 ( 1r ) = 0 for r > 0. Then, integrating by parts we get Z Z Z 1 1 1 3 2 2 ~ ~ ~ ~ d x∇ f (r) = lim r dr dΩ ∇· f (r)∇ − ∇f (r) ·∇ . →0 0 r r r
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~ = rb ∂ + θb1 ∂ + φb 1 ∂ , implying Now recall that in spherical coordinates ∇ ∂r r ∂θ r sin θ ∂φ ~ r/r2 , so that ∇(1/r) = rb ∂ (1/r) = −b ∂r
Z Z Z 1 1 f 2 3 2 ~ −b ~ r dr dΩ ∇· f = lim r 2 + rb·∇f . d x∇ →0 0 r r r2 ~ = (∂f /∂r)|r=0 = The second term on the right side vanishes, because as → 0, rb·∇f const., so: Z Z Z Z 1 1 2 2 ~ lim r dr dΩ rb·∇f 2 = const.· lim r dr dΩ 2 = const.· lim 4π = 0. →0 0 →0 0 →0 r r Therefore, Z Z Z Z f f 1 2 2 3 2 ~ −b r 2 r dr dΩ∇· r 2 = lim dΩr rb· −b f = lim d x∇ →0 0 →0 r r r r= Z Z = − lim dΩ f |r= = −f (0) lim dΩ = −f (0) lim 4π →0
→0
→0
= −4πf (0). R 3 ~ g = In the second step I used the divergence theorem which states d x∇·~ R R 2 d an b·~g where R is any region, ∂R is its boundary, n b is the normal unit vector ∂R 2 to ∂R (pointing out of R) and d a is the surface area element. In our case R = {r < }, d2 a = dΩ, n b = rb, and ~g = −b rR(f /r2 ). Thus we have shown that ∇2 (1/r) = 0 for r 6= 0 and for any f (~r) that d3 x∇2 (1/r)f = −4πf (0). This is the definition of the delta function, so 1 2 ∇ = −4πδ 3 (r). r Problem 12.6.9: Since ` = 0, ψ = R(r)Y00 (θ, φ) = R(r). So the radial equation becomes, with ψ(r) = (1/r)U (r), r 2 d 2µ(E + V0 ) 2 + k U = 0 r ≤ r , k ≡ , in 0 dr2 ~2 r 2 d −2µE − κ2 Uout = 0 r ≥ r0 , κ≡ , 2 dr ~2 where k and κ are defined to be the positive root. The solutions of these equations are e cos kr, Uin = A sin kr + A e +κr . Uout = Be−κr + Be
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e = 0, and Uout → 0 The boundary conditions are that Uin → 0 at r = 0, implying A e at r = ∞, implying that B = 0. The matching conditions at r = r0 are the continuity of ψ and its first derivative: ψin = ψout 0 0 ψin = ψout
⇒ ⇒
Uin = Uout |r=r0 , (Uin /r)0 = (Uout /r)0 |r=r0 .
(2) (3)
Eqn. (2) implies A sin kr0 = Be−κr0 , or, B = A sin kr0 eκr0 . Eqn. (3) implies
d [(A/r) sin kr dr
−
(4)
− (B/r)e−κr ]r=r0 = 0, which gives
A Bκ −κr0 Ak B = 0. sin kr0 + cos kr0 + 2 e−κr0 + e 2 r0 r0 r0 r0
Plugging (4) into this gives (Ak/r0 ) cos kr0 + (Aκ/r0 ) sin kr0 = 0, or − tan kr0 =
k , κ
(5)
which is what we wanted to show. If V0 < π 2 ~2 /(8µr02 ) and −V0 < E < 0 (for a bound state), then k 2 = 2µ (E + V0 ) < ~2 2µ 2 2 2 2 2 π ~ /(8µr0 ) = π /(4r0 ). Thus 0 < kr0 < (π/2), which implies that − tan kr0 < 0. ~2 On the other hand, by definition, k/κ > 0. So there is no solution to eqn. (5).
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Physics 710-712
April 2, 2010
Problem Set 16 Problem 13.3.1: To say the pion has a range of λ ' 10−5 angstroms is to say that a single pion can be localized on this scale: ∆X ∼ λ. From the uncertainty principle ∆X ∆P & ~, we then deduce ∆P & ~/λ. Since we assume λ is the smallest scale on which we can localize the pion, it is plausible that the inequality is saturated, so P ∼ ∆P ∼ ~/λ. The relation between the energy and momentum is (from special relativity) E 2 = m2 c4 + c2 P 2 ∼ m2 c4 + c2 ~2 /λ2 . Now, for the notion of a “single 2 pion” to exist, we must have E . 2mc . (See discussion in text on p. 363). So √ 4m2 c4 & m2 c4 + c2 ~2 /λ2 , or mc2 & c~/( 3λ). Again, since λ is the smallest scale, it is plausible that the inequality is saturated, giving ◦
2000 eV A c~ mc ∼ √ ∼ ◦ ∼ 100 MeV. 3λ 1.7 × 10−5 A 2
(In reality mπ c2 = 140 MeV.) Problem 13.3.2: Since the kinetic energy is T = 200 eV 0.5 MeV ' mc2 , the 2 2 2 2 electron √ is non-relativistic, so we can use T = p /(2m) = (p c )/(2mc ) which implies pc = 2mc2 T . Then √ √ ◦ ◦ 2π~c 2π~c 2π~ 2π(2000 eV A) 2π ◦ = =√ '1A. λ= 'p = A p pc 10 (0.5 × 106 eV)(200 eV) 2mc2 T Problem 13.3.3: Recalling En = −Ry/n2 ' −13eV/n2 , we have P (n = 2) 5 = 4e−(E2 −E1 )/kT = 4e−[(−1/4)−(−1)]13 eV/(kB T ) ' 4e−10 /(T /K) P (n = 1) where I used (kB /eV) ∼ 9 × 10−5 K−1 . So it is clear that we need T & 105 K so that the exponent is not very small. For example, if T = 6000 K, then P (n = 2) 5 ' 4e−10 /(6000) ' 4e−16 ∼ 2 × 10−7 1. P (n = 1)
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Physics 710-712
May 14, 2010
Problem Set 21 Problem 17.2.1: With H 0 = p X = ~/(2mω)(a + a† ), we have: (1): En1 = hn|H 1 |ni =
1 P2 2m
+ 12 mω 2 X 2 and H 1 = λX 4 , and using that
λ~2 hn|(a + a† )4 |ni 4m2 ω 2
λ~2 hn|(a2 a†2 + aa† aa† + a† a2 a† + aa†2 a + a† aa† a + a†2 a2 )|ni 4m2 ω 2 λ~2 = hn|(6a2 a†2 − 12aa† + 3)|ni 4m2 ω 2 λ~2 {6hn + 2|(n + 1)(n + 2)|n + 2i − 12hn + 1|(n + 1)|n + 1i + 3hn|ni} = 4m2 ω 2 λ~2 3~2 λ = {6(n + 1)(n + 2) − 12(n + 1) + 3} = (2n2 + 2n + 1), 4m2 ω 2 4m2 ω 2 =
where in the first line we dropped the zero superscripts from the unperturbed eigenstates; in the second line we kept only terms with equal numbers of a’s and a† ’s; in the √ third line we used [a, a† ] = 1; and in the fourth line we used a† |ni = n + 1|n + 1i. (2): For any finite value of λ, as n gets large ~2 λn2 /(m2 ω 2 ) En1 ~λ ∼ ∼ n2 2 3 1. ∆E ~ω mω Physically, at large x, no matter how small λ is, V (x) =
mω 2 2 x + λx4 ∼ λx4 2
for λ 6= 0. ~ = −γ S· ~B ~ = H 0 + H 1 with H 0 = −γB0 Sz and Problem 17.2.2: H = −~µ·B 1 0 0 H = −γBSx . The H eigenvalues are E± = ∓γB0 ~/2 with eigenstates |±i0 , the Sz eigenstates. Then E±1 = h±|0 H 1 |±i0 = −γBh±|0 Sx |±i0 = −
γB h±|0 (S+ + S− )|±i0 = 0, 2
and |h±|0 H 1 |∓i0 |2 γ 2 B 2 |h±|0 (S+ + S− )|∓i0 |2 = 0 E±0 − Em E±0 − E∓0 4 ∓γB0 ~ m q γB 2 γB 2 γ~B 2 = ∓ |h±|0 S± |∓i0 |2 = ∓ |h±|0 ~ ( 12 ∓ (∓ 12 ))( 32 ± (∓ 12 ))|∓i0 |2 = ∓ , 4B0 ~ 4B0 ~ 4B0
E±2 =
X0 |h±|0 H 1 |mi0 |2
=
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and |±i0 =
X0 |mi0 hm|0 H 1 |±i0 0 E±0 − Em
m
=
γB |∓i0 h∓|0 S∓ |±i0 |∓i0 h∓|0 H 1 |±i0 B = − = ± |∓i0 , 0 0 E± − E∓ 2 ∓γB0 ~ 2B0
where we used that h∓|0 S∓ |±i0 = ~, as computed in the previous equation. Therefore γB0 ~ B2 γB0 ~ γB 2 ~ 0 1 2 3 , ∓ =∓ 1+ E± = E± + E± + E± + O(B ) = ∓ 2 4B0 2 2B02 and
B |∓i0 . 2B0 p ~ where b·S, Now compare to the exact answer. H = −γ(B B02 + B 2 n 0 Sz + BSx ) = −γ p 2 2 ~ are ±~/2 with eigenvectors |±i n b = (B0 b k + Bbı)/ B0 + B . The eigenvalues of n b ·S with |±i = |±i0 + |±i1 + O(B 2 ) = |±i0 ±
|+i = cos 2θ e−iφ/2 |+i0 + sin 2θ eiφ/2 |−i0 , θ 2
|−i = − sin e
−iφ/2
0
θ 2
|+i + cos e
iφ/2
(1) 0
|−i ,
where |±i0 are the Sz eigenvectors. Eqn. (1) can be rewritten as |±i = cos 2θ e∓iφ/2 |±i0 ± sin 2θ e±iφ/2 |∓i0 , and r
1 + cos θ = 2
r
1 + nz = 2
r
1 − cos θ = 2
r
1 − nz 2
cos 2θ = sin
θ 2
=
sp
B02 + B 2 + B0 p = 1 + O(B 2 ), 2 2 2 B0 + B sp B02 + B 2 − B0 B p = + O(B 3 ), = 2 2 2B0 2 B0 + B
and φ = 0, since n b is in the x-z plane. This implies |±i = |±i0 ±
B |∓i0 + O(B 2 ), 2B0
in agreement with the perturbation theory result. Similarly, the exact eigenvalues are γ~ E± = ∓ 2
1/2 q γ~B0 B2 γ~B0 B2 4 2 2 B0 + B = ∓ 1+ 2 =∓ 1+ + O(B ) , 2 B0 2 2B02
again in agreement with the perturbation theory result.
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Problem 17.3.2: H = ASz2 + B(Sx2 − Sy2 ) on the spin-1 Hilbert space. In the Sz basis, |mi (m = ±1, 0), we have 0 1 0 0 −1 0 1 0 0 i~ ~ Sy = √ 1 0 −1 S z = ~ 0 0 0 , Sx = √ 1 0 1 , 2 0 1 0 2 0 1 0 0 0 −1 (see p. 328 of the text). This implies
A 0 B H = ~2 0 0 0 . B 0 A
(2)
Clearly 0 |mi = |0i = 1 0 is the eigenvector with E = 0. So we only need to look at the |mi = |±1i subspace where H = H 0 + H 1 with 1 0 0 1 0 2 1 2 H =~ A , H =~ B on the |mi = {|±1i} subspace. 0 1 1 0 H 0 is degenerate: H 0 |mi = ~2 A|mi for m = ±1. The basis stable under H 1 is the one which diagonalizes H 1 . Since 0 1 1 1 0 1 1 1 = and =− , 1 0 1 1 1 0 −1 −1 the eigenvectors of H 1 are 1 ≡ √1 (|1i + |−1i) 2
−1 ≡ √1 (|1i − |−1i) . 2 To order O(B), the energy shifts of 1 and −1 are
1 0 1 1 1 1 1 2 √ = ~2 B, E1 = 1 H 1 = ~ B √ 1 1 1 0 1 2 2
0 1 1 1 1 1 √ = −~2 B. E−1 = −1 H 1 −1 = ~2 B √ 1 −1 1 0 −1 2 2 and
So, the eigenvalues, to O(B), are E1 = ~2 (A + B),
E−1 = ~2 (A − B),
E0 = 0.
Compare this to the exact eigenvalues of (2) given by 0 = det(H − λ) = λ(~2 (B − A) + λ)(~2 (B + A) − λ), which implies λ = {0, ~2 (A − B), ~2 (A + B)}. So, the O(B) results are exact.
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Physics 710-712
May 26, 2010
Problem Set 22 Problem 18.2.2: To first order, the amplitude d2`m (∞) for the atom to be in the |n = 2, `, mi state is Z i ∞ 2 2 h2`m|(eEZ)|100ie−t /τ eiω21 t dt d2`m (∞) = − ~ −∞ ~ = eZEe−t2 /τ 2 . Here ω21 = (E2 − E1 )/~, which from since the potential H 1 (t) = −~µ·E now on we’ll just call ω. So we need to evaluate h2`m|Z|100i. Z is a component of a vector irreducible tensor operator, Z = T10 , so by angular momentum selection rules, only the h210|Z|100i matrix element is non-vanishing. This matrix element is easily evaluated using the wave functions for the the |210i and |100i states, and that Z = r cos θ: 1/2 1/2 Z Z ∞ 1 r −r/2a0 1 2 e−r/a0 r dr dΩ 5 3 e cos θ (r cos θ) h210|Z|100i = 3 2 πa a πa 0 0 0 0 Z ∞ Z 1 1 dr r4 e−3r/2a0 · 5/2 4 d(cos θ) cos2 θ · = 2π 2 πa0 0 −1 5 Z ∞ 2 2a0 1 = 2π· · dρ ρ4 e−ρ · 5/2 4 = 215/2 ·3−5 ·a0 . 3 3 2 πa0 0 Therefore Z ∞ −i 2 2 15/2 −5 d2`m (∞) = δ`1 δm0 eE·2 ·3 ·a0 e−t /τ eiωt dt ~ −∞ √ −ieE 2 2 = δ`1 δm0 215/2 ·3−5 ·a0 πτ 2 e−ω τ /2 , ~
and so the probability for the transition is 2 15 2 X eE 2 a0 2 2 2 P (n=2) = |d2`m | = πτ 2 e−ω τ /2 . 10 ~ 3 `,m This answer does not depend on the electron spin: since Sz is conserved by H 0 and H 1 , there is still only a single final state it can go into. Problem 18.2.4: The kinetic energy of the emitted electron is 16 keV = 12 mve2 = 1 mc2 (ve /c)2 = 21 (511keV)(ve /c)2 , which implies that ve /c ∼ 1/4. Therefore the time, 2 τ , for emission is τ ∼ a0 /ve = (a0 /c)/(ve /c) ∼ 4a0 /c, since the typical size of the 1s electron orbit is r ∼ a0 , the Bohr radius. In comparison, the characteristic time scale of the 1s electron, T , is T ∼ a0 /vs = (a0 /c)/(vs /c) = (a0 /c)/α ∼ 140a0 /c, since
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the typical velocity of an electron bound in the hydrogen atom is vs /c = α, the fine structure constant. Therefore, T τ , and the sudden approximation is appropriate. In the sudden approximation, right after emission the 1s electron will be in the same state, the |100i(Z=1) state of hydrogen. Therefore, the amplitude for the electron to be in the |100i(Z=2) state of (He3 )+ is given by the overlap Z (Z=2) h100|100i(Z=1)
∞ 2
Z
Z3 πa30
1/2
−rZ/a0
r dr dΩ e Z 23/2 ∞ 2 −3r/a0 27/2 = 4π· 3 r dre = πa0 0 a30
=
0
1/2 1 · e−r/a0 πa30 a 3 Z ∞ 0 ρ2 dρe−ρ = 29/2 ·3−3 , 3 0
where in the first line Z = 2 and I used the fact that under changing the hydrogen nucleus charge from 1 to Z, all that changes is a0 → a0 /Z. Finally, (Z=2) h16, 3, 0|100i(Z=1) = 0 since ` = 3 states are orthogonal to ` = 0 ones (and the radial part, and therefore Z, does not affect this).
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Problem S18.4.3 (Shankar, page 494): (1)
Show that a gauge transformation on potentials φ 0 and A 0 using t
Λ 0( r , t ) = −c ∫ φ (r , t ′)dt ′ gives φ 1( r , t ) = 0 and A1(r , t ) = A 0 − ∇Λ 0 Solution:
(2)
(1.1)
−∞
This is problem 1 of HW8.
Show that if we transform once more to φ 2 and A 2 using Λ 1= −
1 ∇′i A1(r ′, t ) 3 d r′ 4π ∫ r − r ′
(1.2)
then ∇i A 2 (r , t ) = 0 . Solution:
Using equation (18.4.12) from Shankar ( A′ = A − ∇Λ ),we have ⎡ ∇′i A1 (r ′, t ) ⎤ 3 1 ∇⎢ A2 (r , t ) = A1 (r , t ) + ⎥d r ′ ∫ 4π ⎣ r − r′ ⎦
(1.3)
Note that the integrand is the gradient of a scalar function, which produces a vector. If we now take the divergence of this, we get ⎡ ∇′i A1 (r ′, t ) ⎤ 3 1 ∇i A2 (r , t ) = ∇i A1 (r , t ) + ∇2 ⎢ (1.4) ⎥d r ′ ∫ 4π ⎣ r − r′ ⎦ Since the only r-dependence in the integrand comes from the denominator, we can use the identity ∇ 2 (1/ r − r ′ ) = −4πδ 3 ( r − r ′ ) (1.5) Substituting this into (1.4) and doing the integrals over d 3r gives 1 ∇i A2 (r , t ) = ∇i A1 (r , t ) + [ −4π∇′i A1 (r ′, t )]r′=r = ∇i A1 (r , t ) − ∇i A1 (r , t ) = 0 4π
(3)
(1.6)
Verify that φ2 is also zero using ∇i E0 = 0 .
Solution:
φ2 (r , t ) = φ1 +
1 ∂Λ 1 ∂ ⎪⎧ ∇′i A1 (r ′, t ) 3 ⎪⎫ = φ1 − d r ′⎬ ⎨ c ∂t 4π c ∂t ⎪⎩ ∫ r − r ′ ⎪⎭
Incorporating the results of part (1), this becomes
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(1.7)
⎧ ⎫ ⎡ ∂A (r ′, t ) ⎤ ∇′i ⎢ 0 + c∇′φ0 ( r ′, t ) ⎥ ⎪ ⎪ 1 ∂Λ 1 ∂⎪ ⎣ ∂t ⎦ d 3 r ′⎪ = 0− φ2 (r , t ) = φ1 + ⎨∫ ⎬ 4π c ∂t ⎪ c ∂t r − r′ ⎪ (1.8) ⎪⎩ ⎭⎪ 1 ∂ ⎧⎪ ∇′i ⎡⎣ −cE0 ( r ′, t ) ⎤⎦ 3 ⎫⎪ = d r ′⎬ ⎨ 4π c ∂t ⎪⎩ ∫ r − r′ ⎪⎭ If we are in free space, so ρ = 0 (a condition that was implied but not explicitly stated), then Maxwell’s first equation tells us that the numerator of the integrand is zero; so φ2 (r , t ) = 0 .
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Physics 215C Homework #1 Solutions Richard Eager Department of Physics University of California; Santa Barbara, CA 93106
Shankar 20.1.1 Derive the continuity equation ∂P +∇·j =0 ∂t where P = ψ † ψ and j = cψ † αψ The Dirac equation takes the equivalent forms ∂|ψi = cα · P + βmc2 |ψi ∂t i ∂ψ = −cα · ∇ − βmc2 ψ ∂t ~
i~
The conjugate equation is ∂ψ † = ∂t
−cα · ∇ +
i βmc2 ψ † ~
∂ψ ∂ψ † ∂P = ψ† + ψ ∂t ∂t ∂t = −c ψ † α · ∇ψ + (α · ∇ψ † )ψ ∇ · j = c ∇ψ † αψ + ψ † α · ∇ψ
Adding both equations together yields the desired result ∂P +∇·j =0 ∂t Both terms in the Hamiltonian, cα · P and βmc2 are Hermitian, so where did the relative minus sign come from? To show that P is a Hermitian operator you need to integrate by parts. If you are working with the L2 inner product you can drop boundary terms, but when working locally to derive the continuity equation we don’t integrate by parts and get a relative minus sign.
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2 Show that the probability current j of the previous exercise reduces in the nonrelativistic limit to Eq.(5.3.8) [which is the same as Sakurai Eq.(2.4.16)]. 0 σ † The probability current j = cψ αψ and α = We can write the wave σ 0 χ function ψ = in terms of its relativistic and non-relativistic components, Φ Φ and χ respectively. 0 σ χ j = c χ† Φ† (0.1) σ 0 Φ = c χ† σΦ + Φ† σχ (0.2) In the non-relativistic limit (20.2.13) Φ≈
σ·π 2mc
pˆ † pˆ χ+( χ )χ 2m 2m which is the non-relativistic current (5.3.8) j = χ†
j=
~ (ψ ∗ ∇ψ − ψ∇ψ ∗ ) 2mi
from the identification pˆ = −i~∇.
Shankar 20.2.1 Show that π×π = where π = P −
iq~ B c
qA c .
π×π =P ×P −
qA qA qA qA ×P −P × + × c c c c
q = − (A × P + P × A) c iq~ = (A × ∇ + ∇ × A) c
The simplest way to manipulate operators is to act on a test function ψ iq~ iq~ (A × ∇ + ∇ × A) ψ = (A × ∇ψ + ∇ × (ψA)) c c iq~ (A × ∇ψ + (∇ψ)A + (∇ × A)ψ) = c iq~ = (∇ × A) ψ c iq~ = Bψ c Therefore π×π =
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iq~ B c
Shankar 20.1.1 Solve for the 4 spinors w that satisfy Shankar Eq. 20.3.3. You may assume that the 3- momentum p~ is along the z-axis. Normalize them to unity, and show that they are mutually orthogonal. Equation (20.3.3) is Ew = (α · p + βm)w In terms of the relativistic and non-relativistic components, E − m −σ · p χ 0 = −σ · p E + m Φ 0 The solutions are given in equations (20.3.7) and (20.3.8). Choosing the basis 1 0 and for Φ and letting p be in the z direction, 0 1 0 p/(±E − m) p/(±E − m) 0 w = w1,3 = 2,4 1 0 0 1 Orthogonality of the spinors is easy to see using E 2 = p2 + m2 .
5 The five terms in 20.2.28 are
P2 2m V
Hermitian Hermitian
4
P 8m3 c2 iσ · P × [P, V ] 4m2 c2 P · [P, V ] 4m2 c2 −
Hermitian Hermitian anti-Hermitian
Recall that Pˆ is a Hermitian operator, the potential V is assumed to be Hermitian (for conservation of probability) and the Pauli matrices σ are Hermitian. The commutator [X, Y ] of two Hermitian operators is anti-Hermitian since [X, Y ]† = (XY )† − (Y X)† = Y † X † − X † Y † = Y X − XY = −[X, Y ]. The cross product of two Hermitian (vector) operators is again Hermitian since (X × Y )† = X † × Y † = X × Y.
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