Quantum Mechanics
Solution Set 6
Due: 7 October 2015
Reading: Shankar, Finish Chapter 2; Lecture notes, Finish Chapter 3, Sections 4.1-4.4
29. Start with the Lagrangian for a relativistic particle moving in an electromagnetic field
B = ∇ × A, E = −∇φ − Ȧ/c,
q
Q
2
(1)
L = −mc 1 − ṙ 2 /c2 + ṙ · A − Qφ
c
a) Obtain the Hamiltonian for this system, making sure to express it as a function of the
canonical variables r, p.
p
Solution: The conjugate momentum is p = mṙ/ 1 − ṙ 2 /c2 + QA/c. Then
q
q
mṙ 2
Q
2 2
2
+ mc 1 − ṙ 2 /c2 + Qφ
H = ṙ · p + mc 1 − ṙ /c − ṙ · A + Qφ = p
c
1 − ṙ 2 /c2
mc2
= p
+ Qφ
(2)
1 − ṙ 2 /c2
2
We still have to eliminate ṙ in favor of p. We have
m2 c2
m2 ṙ 2
2 2
=
(p − QA/c) =
2 2
2 2 −m c
1 − ṙ /c
1 − ṙ /c
p
p
So mc/ 1 − ṙ 2 /c2 = m2 c2 + (p − QA/c)2 , and
p
H = c m2 c2 + (p − QA/c)2 + Qφ
2
b) Show that Hamilton’s equations imply the familiar relativistic Newton equation
!
mṙ
ṙ
d
p
= QE + Q × B
2
dt
c
1 − ṙ /c2
1
(3)
(4)
Solution: We calculate Hamilton’s equations:
ṙ =
ṗ =
=
=
d
(p − QA/c) =
dt
c(p − QA/c)
∂H
=p
,
∂p
m2 c2 + (p − QA/c)2
c(pk − qAk )
−Q∇φ + q(∇Ak ) p
= −Q∇φ + Qṙk (∇Ak )/c
m2 c2 + (p − QA/c)2
Q
Q dA Q ∂A
Q
Q
−
−Q∇φ + ṙ × (∇ × A) + ṙk ∇k A = −Q∇φ + ṙ × B +
c
c
c
c dt
c ∂t
Q dA
qE + q ṙ × B +
c dt
Q
(5)
qE + ṙ × B
c
The first Hamilton equation
just gives the relation between p and ṙ, which of course is
p
just p − qA = mṙ/ 1 − ṙ 2 /c2 ≡ p0 , the relativistic momentum of the particle. Note
that p0 is not the momentum conjugate to r. It is the quantity that appears on the
left of the relativistic Newton equation
Q
dp0
= qE + ṙ × B
dt
c
(6)
30. a) S, Exercise 2.3.1; b) S, Exercise 2.5.2;
Solution:
a) Plug (2.3.3) and (2.3.4) into (2.3.5), calling r CM ≡ R:
2
2
1
1
m2 ṙ
m1 ṙ
L =
m1 Ṙ +
+ m2 Ṙ −
− V (r)
2
m1 + m2
2
m1 + m2
1
1 m1 m22 + m2 m21 2
2
=
ṙ − V (r)
(m1 + m2 )Ṙ +
2
2 (m1 + m2 )2
1 m1 m2 2
1
2
ṙ − V (r)
(m1 + m2 )Ṙ +
=
2
2 m1 + m2
(7)
(The cross terms in the two squared quantities on the first line exactly cancel.)
b) The energy of a harmonic oscillator is H = p2 /(2m) + kx2 /2. Since the energy is
conserved H is a constant for each motion, which implies that p and x are related by
the relation
1=
p2
x2
+
2mE 2E/k
(8)
which describes an ellipse in phase space with axes a, b given by a2 = 2E/k and
b2 = 2mE.
2
31. a) S, Exercise 2.7.1; c) S, Exercise 2.7.2.
Solution:
a) These identities quickly follow from the definition of the P.B.
X ∂ω ∂λ
∂ω ∂λ
−
{ω, λ} ≡
∂q
∂pk ∂qk
k ∂pk
k
(9)
Inspection immediately shows antisymmetry under ω ↔ λ and linearity in each argument of the P.B. The identity involving a product follows from the Leibniz rule for
derivatives of the product of two functioins.
b)(i) When one of the arguments of a P.B. is a qk or pk , only one term in the defining sum
contributes:
X
X ∂ql ∂A
∂A
∂qk ∂A
∂A
{ql , A} =
=
δkl
−
=
∂qk ∂pk ∂pk ∂qk
∂pk
∂pl
k
k
X ∂A
X ∂pl ∂A
∂pk ∂A
∂A
−
=−
(10)
=−
{pl , A} =
δkl
∂qk ∂pk ∂pk ∂qk
∂qk
∂ql
k
k
Plugging in qn or pn for A proves (2.7.4), and putting A = H proves (2.7.5).
(ii) When a = b the potential is independent of angles corresponding to a central force,
which is known to conserve angular momentum l˙z = 0. But since lz has no explicit
time dependence, l˙z = {lz , H}. So the P.B. should vanish. To verify this explicitly we
use lz = xpy − ypx and calculate
{xpy − ypx , x2 + y 2 } = −2xy{px , x} + 2xy{py , y} = 2xy − 2xy = 0
{xpy − ypx , p2x + p2y } = 2px py {x, px } − 2px py {y, py } = 2px py − 2px py = 0
(11)
the desired result follows.
P
P
32. As we know the total momentum k pk and total angular momentum L = k r k × pk
are important physical properties of any closed system. For the case of a single particle
evaluate all of the Poisson brackets of px , py , pz , Lx , Ly , Lz with each other. Start with the
Poisson brackets for the canonical variables r, p and exploit the algebraic identities satisfied
by the P.B.’s. Use the antisymmetric symbol ǫklm to present your final answers in compact
form.
Solution:The fundamental Poisson brackets are {rk , rl } = {pk , pl } = 0 and {rk , pl } = δkl .
We can also write Lk = ǫklm rl pm . Then
{pj , Lk } = ǫklm {pj , rl pm } = ǫklm ({pj , rl }pm + rl {pj , pm }) = −ǫkjm pm = ǫjkm pm
{rj , Lk } = ǫklm {rj , rl pm } = ǫklm ({rj , rl }pm + rl {rj , pm }) = ǫklj rl = ǫjkl rl
{Lj , Lk } = ǫklm {Lj , rl pm } = ǫklm ({Lj , rl }pm + rl {Lj , pm }) = −ǫklm ǫljn rn pm − ǫklm ǫmjn pn rl
= (δkj δmn − δkn δjm )rn pm − (δkj δln − δkn δlj )pn rl
= rj pk − rk pj = ǫjkm ǫmns rn ps = ǫjkm Lm
(12)
3
The desired Poisson Brackets are then
{pk , pl } = 0,
{pk , Ll } = ǫklm pm ,
{Lk , Ll } = ǫklm Lm
(13)
33. Show that each of the following transformations is canonical:
a) Translations
Pi = pi ;
Q i = q i + ai
(14)
Solution: It is immediate that {Pk , Pl } = 0 and that {Qk , Ql } = {qk + ak , ql + al } =
{qk , ql } = 0. Also
{Qk , Pl } = {qk + ak , pl } = δkl
(15)
since the P.B. of a constant with any function is immediately 0. Thus the transform
is canonical.
b) Rotations
Pi =
X
Rij pj ;
Qi =
j
X
Rij qj
(16)
j
where RRT = RT R = I, i.e. R is an orthogonal (real unitary) matrix.
Solution: Again it is immediate that the P.B.’s of Qk among themselves and Pk among
themselves is zero. So we compute
X
X
X
{Qk , Pl } = {
Rkj qj ,
Rli pi } =
Rkj Rli {qj , pi }
j
=
X
i
Rkj Rli δij =
i,j
i,j
X
Rki Rli = (RRT )kl = δkl
(17)
i
which completes the proof that rotations are a canonical transformation.
34. An important time dependent canonical transformation in nonrelativistic mechanics is a
Galilei boost: For a single free particle in three dimensions, it is R = r + V t, P = p + mV .
a) Find the generating function for this canonical transformation for finite V .
Solution: We guess an F2 = (r + V t) · P + aV · r + bt. Then it would follow that
R =
∂F2
= r + V t,
∂P
4
p=
∂F2
= P + aV
∂r
(18)
Comparing to the desired transformation we see that choosing a = −m does the trick.
We can fix b by requiring H̄ = P 2 /(2m) where
p2
(P − mV )2
P2
∂F2
=
+b+V ·P =
+b+V ·P =
+ mV 2 /2 + b(19)
H̄ = H +
∂t
2m
2m
2m
So b = −mV 2 /2, and
1
F2 = (r + V t) · P − mV · r − mV 2 t
2
(20)
b) By taking V small determine the infinitesimal generator K of Galilei transformations.
Solution: The infinitesimal generator of the transformation is the term in F2 linear
in V , with P replaced by p:
V · K = V · (pt − mr)
(21)
So we can identify three independent components K = pt − mr.
c) Show that K is a constant of the motion, even though its Poisson bracket with the
Hamiltonian H = p2 /(2m), which you should determine, is not zero.
Solution: We have the general formula
K̇ = {K, H} +
∂K
1
1
=
{K, p2 } + p = pk {K, pk } + p = −p + p = 0 (22)
∂t
2m
m
In the process we have shown that {K, H} = −p 6= 0. K is a constant because of its
explicit time dependence.
d) Find the Poisson brackets of the three components Kk with each other.
Solution: We simply calculate
{Kk , Kl } = {pk t, −mrl } + {−mrk , pl t} = +mtδkl − mtδkl = 0
(23)
Although this wasn’t asked, we can display the full Galilei Poisson bracket algebra
including rotations Jk , boosts Kk , translations in space pk and translations in time
H = p2 /2m:
{pk , pl } = 0,
{pk , H} = 0,
{H, Jk } = 0,
{H, Kk } = pk ,
{Jk , Jl } = ǫklm Jm
{pk , Jl } = ǫklm pm ,
{Kk , Kl } = 0,
{pk , Kl } = mδkl
{Kk , Jl } = ǫklm Km
(24)
5