Download Solutions - Stanford University

Physics 143a: Quantum Mechanics I
Spring 2015, Harvard
Section 11:
Review
Solutions
Below are some sample problems to help study for the final. The practice final handed out is a better
estimate for the actual length of a final – 3 of the problems below would be appropriate for a final exam
in this course.
Problem 1 (Metallic Junction): Suppose that we have a junction between two different metals: metal 1
for x < 0, and metal 2 for x > 0. A simple model for the conduction of an electron between these two
metals is as follows. An electron of mass m, propagating perpendicular to the area of contact, sees an
effective 1d Hamiltonian
p2
µ1 x < 0
H=
.
+ αδ (x) −
µ2 x > 0
2m
where α, µ1 and µ2 are real numbers. Crudely speaking, µ1,2 are the Fermi energies of metals 1 and
2. The parameter α represents the fact that the interface might be dirty (e.g., higher concentration of
impurities). Without loss of generality we take µ1 > µ2 .
(a) Under what circumstances will a bound state exist? If so, how many are there? If they exist, find
their energies – up to the solution of a equation for E, expressed only in terms of ~, α, m and µ1,2 .
Solution: If we have a bound state, we need E < −µ1 . So let us define
p
−2m(E + µ1,2 )
κ1,2 =
.
~
The solution to the Schrödinger equation away from x = 0 is
Aeκ1 x
x<0
.
ψ=
Be−κ2 x x > 0
Continuity of ψ demands A = B. The derivative will in general be discontinuous and equal to
0 = αA +
~2
(κ1 + κ2 )A.
2m
√
p
2m p
= −E − µ1 + −E − µ2 .
~
Now both sides must be positive, this gives us that α < 0, and since E < −µ1 :
−α
2mα2
≥ (µ1 − µ2 )2 .
~2
(b) Suppose we have an electron wave packet of energy ≈ E incident on the barrier from x → −∞. What
is the probability P (E) that the electron will, after a long time, be found propagating towards x → ∞
in metal 2? Explain how impurities affect P (E) and justify the effect on physical grounds.
1
Solution: If E < −µ2 then P (E) = 0. Otherwise, we can write the wave function as
ik x
e 1 + Ae−ik1 x x < 0
ψ=
,
Beik2 x
x>0
with
p
2m(E + µ1,2 )
k1,2 =
~
and continuity of the wave function fixing B = 1 + A and the derivative “continuity” requiring:
0 = αB +
~2
(ik1 (1 − A) − ik2 B) .
2m
We re-arrange this to find
A=
i(k1 − k2 ) + b
.
i(k1 + k2 ) − b
where
b≡
2mα
.
~2
The probability of reflection is given by |A|2 (no worries here about relative factors of wave number!
since it is reflected) and so the probability of transmission is
P = 1 − |A|2 = 1 −
b2 + (k1 − k2 )2
4k1 k2
=
.
2
2
b + (k1 + k2 )
(k1 + k2 )2 + b2
Note that P is a strictly decreasing function of b at any E. Thus the impurities always reduce the
probability to transmit through the junction. This makes sense intuitively since even if µ1 = µ2
impurities lead to a finite chance to reflect.
Problem 2 (Particle Mixing): In quantum field theory, it is possible for 2 particles of “species 1” to
“collide” and become a pair of 2 particles of “species 2” – and vice versa. We will explore a toy model of
this effect in this problem. Let a1 and a†1 be the annihilation and creation operators for a 1d harmonic
oscillator, and a2 and a†2 for an independent oscillator:
[a1 , a†1 ] = 1,
[a2 , a†2 ] = 1, [a1 , a2 ] = [a1 , a†2 ] = [a2 , a†1 ] = [a†1 , a†2 ] = 0.
The Hilbert space of this system consists of |n1 n2 i, with n1,2 = 0, 1, 2, . . .. n1 can be interpreted as the
number of type 1 particles, and n2 as the number of type 2 particles, in our universe.
Now consider the Hamiltonian
H = a†1 a1 + a†2 a2 − η a†1 a†1 a2 a2 + a†2 a†2 a1 a1 .
with , η > 0 real parameters.
(a) Show that H is Hermitian.
Solution:
† †
†
†
†
H = a1 a1 + a2 a†2
− η ∗ a†2 a†2 a1 a1 + a†1 a†1 a2 a2 = H
†
∗
since (a†1 )† = a1 and , η are real and we just flip the order of all operators when taking Hermitian
conjugates (term by term).
2
(b) Define the total number operator N = a†1 a1 + a†2 a2 . What are the eigenvalues of N ? Show that
[N, H] = 0.
Solution: First, note that [a1 , a†2 a2 ] = [a†1 , a†2 a2 ] = 0 and this implies that N = N1 + N2 where
N1 = a†1 a1 , N2 = a†2 a2 . The eigenvalues of N1 and N2 are all non-negative integers, and this implies
that the eigenvalues of N are the sum of 2 non-negative integers which is a non-negative integer.
The first term in H is just N , and so this clearly commutes with N . So we only have to worry about
the second term:
[a†1 a1 + a†2 a2 , a†1 a†1 a2 a2 + a†2 a†2 a1 a1 ] = a†1 [a1 , a†1 a†1 ]a2 a2 + a†2 a†2 [a†1 , a1 a1 ]a1 + a†1 a†1 [a†2 , a2 a2 ]a2 + a†2 [a2 , a†2 a†2 ]a1 a1 .
We have used liberally in the above the fact that if [A, B] = 0 then [A, BB] = 0 and [A, BC] =
B[A, C] + [A, B]C. Now we see the two pieces arise:
[a1 , a†1 a†1 ] = [a1 , a†1 ]a†1 + a†1 [a1 , a†1 ] = 2a†1
[a†1 , a1 a1 ] = [a†1 , a1 ]a1 + a1 [a†1 , a1 ] = −2a1 .
So
[N, H] = 2a†1 a†1 a2 a2 + a†2 a†2 (−2a1 a1 ) + a†1 a†1 (−2a2 a2 ) + a†2 (2a†2 )a1 a1 = 0.
(c) Prove that if we start in a quantum state |Ψ (0)i with N |Ψ (0)i = n|Ψ (0)i, then at any time t > 0,
N |Ψ (t)i = n|Ψ (t)i.
Solution: Let us consider t = dt, an infinitesimally small time step. Then using the abstract
Schrödinger equation:
H
H
H
|Ψ (0)i = 1 +
N |Ψ (0)i = n 1 +
|Ψ (0)i = n|Ψ (t)i.
N |Ψ (dt)i = N 1 +
i~
i~
i~
Now of course if at time dt |Ψ (t)i has the same eigenvalue, then it will have the same eigenvalue at
time 2dt, etc., and at time t we will still have N |Ψ (t)i = n|Ψ (t)i.
(d) Use the result of part (c) to conclude that if we start in a state with eigenvalue N = 2, while H is
non-diagonal in an infinite-dimensional Hilbert space, we need only worry about the dynamics in a
subspace of dimension 3. Give a simple basis for this subspace.
Solution: The states with N = 2 have N1 + N2 = 2, and there are only 3 possibilities: |20i, |11i, |02i,
using the notation |n1 n2 i. Thus from the theorem in (c) we know that these states can never mix
with any other states if we start in a linear combination of just these 3, since these are the only states
for which N |Ψ i = 2|Ψ i.
(e) Write down H, in Dirac notation, in the basis of part (d), for all states with N = 2. Diagonalize H
and find the eigenvalues and eigenvectors.
Solution: The easiest way to find H is just to act H on our 3 states:
√
H|20i = 2|20i − ηa†1 a†1 a2 a2 |20i − ηa†2 a†2 a1 a1 |20i = 2|20i + 0 − ηa†2 a†2 2|00i = 2|20i − 2η|02i
√
H|02i = 2|02i − ηa†1 a†1 a2 a2 |02i − ηa†2 a†2 a1 a1 |02i = 2|20i − ηa†1 a†1 2|00i + 0 = 2|20i − 2η|20i
H|11i = ( + )|11i − ηa†2 a†2 a1 |01i − ηa†1 a†1 a2 |10i = 2|11i.
3
We conclude that
H = 2|11ih11| + (2|20ih20| + 2|02ih02| − 2η|02ih20| − 2η|20ih02|) .
The key thing to note is that H is block-diagonal. We immediately read off that |11i is an eigenvector
of H with eigenvalue 2. In the |02i, |20i block we obtain a Hamiltonian which is exactly analogous
to H = 2 − 2ησx if we label |02i, |20i as “spin states”. We know the eigenvectors of σx (and the
identity) are
|20i + |02i
|20i − |02i
√
√
|+i =
, |−i =
.
2
2
And indeed we can check these are eigenvectors of H:
1
H|+i = √ (2|20i − 2η|02i + 2|02i − 2η|20i) = 2( − η)|+i,
2
1
H|−i = √ (2|20i − 2η|02i − 2|02i + 2η|20i) = 2( + η)|−i.
2
(f) Suppose that at time t = 0, |Ψ (0)i = |20i – namely, we start in a state of 2 type 1 particles. Find
|Ψ (t)i exactly and comment on its physical interpretation.
Solution: Since
|Ψ (0)i =
we conclude that
|Ψ (t)i =
|+i + |−i
√
2
|+ie−2i(−η)t/~ + |−ie−2i(+η)/~
√
.
2
Converting back to our old basis:
"
#
2iηt/~ + e−2iηt/~
2iηt/~ − e−2iηt/~
e
e
2ηt
2ηt
−2it/~
−2it/~
|20i
|Ψ (t)i = e
+ |02i
+ i|02i sin
.
=e
|20i cos
2
2
~
~
Physically, the η term allows us to convert from a pair of type 1 particles to a pair of type 2 particles.
(g) Suppose we measure N2 = a†2 a2 at time t0 , given the initial state of part (f). What values could we
measure and with what probabilities? What is the maximum of this probability, and at what time
does it occur?
Solution: If we measure N2 , we can either measure 0 (from state |20i) or 2 (from state |02i). The
probabilities at time t are:
P(N2 = 0) = |h20|Ψ (t)i|2 = cos2
2ηt
,
~
P(N2 = 2) = |h02|Ψ (t)i|2 = sin2
It is possible at times
t=
π ~ 3π ~ 5π ~
,
,
,...
2η 2η 2η
to be guaranteed to have 2 particles of type 2 if we make a measurement.
4
2ηt
.
~
Problem 3 (Constant Forces): Consider a non-relativistic particle of mass m in the potential
V (x) = −F x.
(a) Does this potential have any bound states?
Solution: No, because there is no global minimum of V (x).
(b) Write down the momentum space time dependent Schrödinger equation for Φ(p, t). Find the exact
solution given initial conditions Φ(p, 0) = Φ0 (p).
Solution: In momentum space the Schrödinger equation reads
p2
∂Φ
p2
∂
∂Φ
Φ=
i~
=
Φ + V i~
Φ − i~F
.
∂t
2m
∂p
2m
∂p
This can be solved exactly by rewriting Φ(p, t) = Ω(p − F t, t) = Ω(r, t) and plugging in:
i~
∂Ω
∂Ω
(r + F t)2
∂Φ
− i~F
=
Ω − i~F
.
∂t
∂r
2m
∂r
This can be solved exactly:
(r + F t)3 − r3
Ω(r, t) = Ω(r, 0) exp
.
6i~mF
At t = 0, r = p, and Φ0 (p) = Ω(r, 0), so we conclude that
3
p − (p − F t)3
Φ(p, t) = Φ0 (p − F t) exp
.
6i~mF
(c) Evaluate hp(t)i and comment on your result. From this result, compute hx(t)i.
Solution: In momentum space we have
Z
Z
Z
hp(t)i = dp p|Φ(p, t)|2 = dp p|Φ0 (p − F t)|2 = dr (r + F t)|Φ0 (r)|2 = hpit=0 + F t.
This is not surprising, from Ehrenfest’s Theorem. We also know that
hp(t)i
dhxi
=
dt
m
and so – using that
Z
hx(0)i =
dp Φ0 (p)∗ i~
we obtain
hx(t)i = hx(0)i +
∂Φ0 (p)
,
∂p
hp(0)i
F t2
t+
.
m
2m
(d) Without performing an explicit computation, do you think that σx σp will increase or decrease in
time? Explain your answer qualitatively.
Solution: Up to a phase, the wave function just shifts in momentum space with time, so we know
that σp is time independent. However, we expect that (like with the free particle), the initial wave
packet spreads out because the “fast modes” move faster than the “slow modes”. So we expect σx is
increasing with time, and thus so is σx σp .
5
Problem 4: Consider a non-relativistic particle of mass m in an eigenstate of an unknown Hamiltonian
in three spatial dimensions. The state is
ψ = N (x + y + 2z)rs−1 exp[−r/ξ],
where N is a normalization factor, and s and ξ are positive constants. You may find helpful:
r
r
3
x ± iy
3
z
±1
0
p
p
Y1 = ∓
,
Y1 = ∓
.
8π x2 + y 2 + z 2
4π x2 + y 2 + z 2
(a) If we make a measurement of L2 , what could we measure, and with what probability?
Solution: We can write the wave function as
ψ = N 0 2Y10 + aY11 + bY1−1 rs e−r/ξ
and we need that
1=
or
−a + b
−ia − ib
√ , 1= √
,
2
2
1+i
b= √ ,
2
i−1
a= √ .
2
This implies that all states are l = 1 and so we are guaranteed to get 2~2 if we measure L2 .
(b) If we make a measurement of Lz , what could we measure, and with what probability?
Solution: The relative weights of the angular pieces of the wave function (modulus squared) are 4 to
1 to 1 for the m = 0, +1, −1 states. So we conclude
1
4
P(Lz = ~) = P(Lz = −~) = , P(Lz = 0) = .
6
6
(c) Suppose that we now learn that the potential is spherically symmetric. What is the potential V (r)
associated with this state, and what is its energy? Assume that V (r → ∞) = 0.
Solution: We write out the wave function in the form
ψ = N 0 2Y10 + aY11 + bY1−1
u(r)
,
r
u(r) = rs+1 e−r/ξ .
We know that u(r) obeys the radial Schrödinger equation with l = 1 centrifugal potential:
−
~2 d2 u
~2 × 2
+
V
(r)u
+
u = Eu.
2m dr2
2mr2
So now let’s just plug in u (we don’t need to worry about normalization since the differential equation
is linear):
~2 s(s + 1) s−1 −r/ξ
~2 s+1 −r/ξ ~2 (s + 1) s −r/ξ
~2
−
r e
−
r e
+
r e
+ V (r) +
rs+1 e−r/ξ = Ers+1 e−r/ξ .
2m
2mξ 2
mξ
mr2
−
~2 (s(s + 1) − 2) ~2 (s + 1)
~2
+
−
+ V (r) = E
2mr2
mξr
2mξ 2
6
and we conclude that
E=−
V (r) = −
~2
2mξ 2
~2 (s + 1) ~2 (s(s + 1) − 2)
.
+
mξr
2mr2
(d) If we put 3 bosons in a state with this wave function, what are the possible values of L2tot and Ltot,z
that we could measure? You do not need to determine the probabilities.
Solution: The total l in this state is 1 and we have all possible Lz accounted for. So after adding 2
bosons together we can measure l = 0, 1, 2 and Lz = 0, ±~, ±2~. After adding a third boson, we can
get l = 0, 1, 2, 3 and Lz = 0, ±~, ±2~, ±3~.
Problem 5 (Measuring Quantum Light): One of the ways to detect the quantum nature of light in an
experiment is to rely on a trick called photon antibunching. Experimentally, we consider the following
set-up:
E2
E3
E1
E4
In classical electrodynamics, the first beam of light has electric field E1 = E1 (t)x̂, and the second
beam has electric field E2 = E2 (t)ŷ.1 We pass these beams through a 50:50 beam splitter, which converts
these beams into beams 3 and 4 with electric fields
E3 =
E1 + E2
√
,
2
E4 =
E1 − E2
√
.
2
We then measure the intensity of light I3,4 (t) associated with each one of these beams at time t at a
photodetector (the orange boxes in the figure). The object we will focus on is the correlation function
g≡
hI3 (t)I4 (t)i
,
hI3 (t)ihI4 (t)i
with h· · · i denoting the appropriate averages (either classical or quantum).
(a) In classical electrodynamics the intensity of a beam of light is given by (up to an unimportant
dimensional factor Icl ):
Icl = Icl E2 .
1
Note for simplicity we have dropped the spatial variations of E1,2 . These can rigorously be included but play no role in
the physics we’re interested in.
7
Show that for any arbitrary (real-valued) E1,2 (t), I3 = I4 . Conclude that g ≥ 1, classically.
Solution: The intensity of beams 3 and 4 are given by
E1 x̂ + E2 ŷ 2 Icl
E1 x̂ − E2 ŷ 2 Icl
2
2
√
√
I3 = Icl
=
E1 + E2 , I4 = Icl
=
E12 + E22 = I3 .
2
2
2
2
For an arbitrary random variable, hX 2 i ≥ hXi2 (since σX ≥ 0). We conclude g ≥ 1.
Let us now revisit this problem from a quantum perspective. In quantum electrodynamics, we simply
associate a (1d) quantum harmonic oscillator to each mode of the electromagnetic field; thus if E1,2
correspond to simple propagating waves, we simply have a pair of oscillators with creation operators a†1,2
and annihilation operators a1,2 , obeying
[a1 , a†1 ] = [a2 , a†2 ] = 1, [a1 , a†2 ] = [a2 , a†1 ] = 0.
The creation operators create single photons in that mode; the Hilbert space is spanned by |n1 , n2 i with
n1,2 = 0, 1, 2, . . .. After passing through the beam splitter, we have a new pair of oscillators with creation
operators a†3,4 with
a1 + a2
a1 − a2
a3 = √
, a4 = √
.
2
2
It turns out that the proper quantum mechanical definition of g is
g≡
ha†3 a†4 a4 a3 i
ha†3 a3 iha†4 a4 i
.
(b) Find the commutation relations between a3,4 and a†3,4 . Is the definition of g sensitive to the labeling
of beams 3 and 4?
Solution: Since [a1 , a2 ] = 0 and [a†1 , a†2 ] = 0, it’s easy to see that [a3 , a4 ] = [a†3 , a†4 ] = 0.
1
[a† , a1 ] − [a†2 , a2 ]
[a†3 , a4 ] = [a†1 + a†2 , a1 − a2 ] = 1
=0
2
2
[a†1 , a1 ] + [a†2 , a2 ]
1 †
†
†
[a3 , a3 ] = [a1 + a2 , a1 + a2 ] =
= −1
2
2
1
[a† , a1 ] + [a†2 , a2 ]
[a†4 , a4 ] = [a†1 − a†2 , a1 − a2 ] = 1
= −1.
2
2
So the 3,4 creation annihilation operators are like independent harmonic oscillators! Furthermore,
using the commutation relations: it’s easy to see that
a†3 a†4 a4 a3 = a†4 a†3 a4 a3 = a†4 a†3 a3 a4 ,
and we conclude that since the numbers in the denominator can be multiplied in either order, that g
is insensitive to the labeling of beams 3 and 4.
(c) In a seminal experiment [1], only beam 1 was turned on. What is the value of g – express it in terms
of the quantum state |ψi and the operator N1 ≡ a†1 a1 ? Show explicitly that it is possible to obtain
g < 1; this was observed quite conclusively in [1].
Solution: If only beam 1 is turned on, then the eigenstates must be of the form
X
|ψi =
cn |n1 = n, n2 = 0i.
n
8
We need to evalute expectation values in g with respect to this state:
1
1
1X
hψ|a†3 a3 |ψi = hψ|(a†1 + a†2 )(a1 + a2 )|ψi = hψ|a†1 a1 |ψi =
n|cn |2 .
2
2
2 n
We’ve used in the above equation that a2 |ψi = 0 since all states have n2 = 0; similarly, hψ|a†2 = 0.
Similarly:
1
1
1X
hψ|a†4 a4 |ψi = hψ|(a†1 − a†2 )(a1 − a2 )|ψi = hψ|a†1 a1 |ψi =
n|cn |2 .
2
2
2 n
1
1X
n(n − 1)|cn |2 .
hψ|a†3 a†4 a4 a3 |ψi = hψ|a†1 a†1 a1 a1 |ψi =
4
4 n
Consider, e.g., the state with cn = δ nm , with m > 0. Then we obtain that
g=
1
(m2 − m)/4
=1−
< 1.
2
(m/2)
m
Problem 6: A pair of interacting spin-1/2 fermions of mass m feels an effective Hamiltonian of the form
p21
p22
B
H=
+
+ A + 2 S1 · S2 (x1 − x2 )2
2m 2m
~
For simplicity, we only consider the (position) dynamics in one spatial dimension. These sorts of Hamiltonians arise when considering nucleon-nucleon interactions inside of the nucleus.
(a) Show that S1 + S2 ≡ S commutes with H. What does this imply about the eigenvectors of H?
Solution: We write
S1 · S2 =
(S1 + S2 )2 3 2
1
(S1 + S2 )2 − S21 − S22 =
− ~ .
2
2
4
And since the only place that spin shows up in H is in the form (S1 + S2 )2 , which commutes with
S1 + S2 , we conclude that [H, S1 + S2 ] = 0. This implies that the eigenstates of H are states that are
eigenvalues of (S1 + S2 )2 , which will have (for the addition of two spin-1/2 states) either eigenvalue
0 or 2~2 .
(b) Under what conditions will there be a well defined ground state?2
Solution: We know that
S1 · S2
1
= , (s = 1),
~2
4
S1 · S2
3
= − , (s = 0).
~2
4
Let us denote henceforth α ≡ S1 · S2 /~2 which can be either 1/4 or −3/4. If we further write out H
in center of mass (P, X) and relative (p, x) coordinates:
H=
P2
p2
+
+ (A + αB) x2 .
4m m
We need that the potential is bounded from below, which means that A + αB > 0 for the ground
state to be well defined.
2
We need that V (x) is bounded from below!
9
(c) Give an example of an A and B for which the particles bind together in a spin-1 state, but unbind in
a spin-0 state.
Solution: When the coefficient of x2 is positive, we have that the particles will be bound together,
and when the coefficient is negative, the particles will be unbound. So take A = 0 and B > 0. In this
case A + αB is positive when α > 0 and negative when α < 0.
(d) Is it possible to pick an A and B such that the spin-0 state is bound but the spin-1 state is unbound?
Solution: Take A > 0, and B = −8A < 0. In this case A + αB = −A when s = 1 (unbound), and
A + αB = 7A when s = 0 (bound).
For the remainder of the problem, assume that the condition of part (b) holds, but do not make any
further assumptions on the values of A or B.
(e) Write down all ground state wave functions ψ0 (x1 , s1 , x2 , s2 ). Be careful to consider all possibilities
in A and B.
Solution: Let’s suppose that B > 0 first. In this case the s = 0 state has lowest energy and this
state is antisymmetric in spin, so we need a coordinate space function that is symmetric. We take
the n = 0 state of the resulting harmonic oscillator, which is associated with frequency:
s
r
2(A + αB)
A + αB
ω=
=2
.
m/2
m
Also define
r
ξ≡
~
.
mω
We then find that the ground state is
ψ = φ0
x1 − x2
ξ
| ↑↓i − | ↓↑i
√
2
where φ0 is the ground state of the harmonic oscillator in natural ~ = m = ω = 1 units – it’s just a
Gaussian but this isn’t too concerning to us now. This state has
r
A − 3B/4
E0,s = ~
.
m
If B < 0, then the s = 1 state may have lower energy. In this case, the ground state is 3-fold
degenerate and since the spin wave function is symmetric we take an antisymmetric coordinate space
wave function, which would be the n = 1 harmonic oscillator state:
x1 − x2
x1 − x2
x1 − x2 | ↑↓i + | ↓↑i
√
ψ = φ1
| ↑↑i or φ1
| ↓↓i or φ1
ξ
ξ
ξ
2
with energy
r
A + B/4
.
m
< E0,s , then we instead take the triplet wave function, and this occurs when
B
3B
9 A+
<A−
4
4
E0,a = 3~
If E0,a
12B
= 8A + 3B < 0.
4
This can happen, for example if B = −3A (note that A + αB > 0 in both cases still).
8A +
10
Problem 7 (Mass of a Meson): Consider two quarks of mass m, which interact via the potential
V (x1 , x2 ) = T |x1 − x2 |.
For simplicity, we do not consider the dynamics in three dimensions, but only in one. T is equal to the
tension of the “flux tube” that must stretch from one quark to the other. If these two quarks (really,
one of them is an antiquark) make up a meson, then a crude model for the mass M of the meson is that
M c2 is the ground state energy of the quantum mechanical problem. Do not worry about the spin of the
quarks in this problem.
(a) Suppose that the quarks can be treated non-relativistically. Argue that the ground state energy is
determined entirely by the one particle Schrödinger equation in the potential V = T |x|. What is the
mass of this particle?
Solution: The total Hamiltonian is
H=
p21
p2
+ 2 + T |x1 − x2 |
2m 2m
and so in reduced coordinates we have
H=
p2rel
P2
+
+ T |xrel |.
2(2m) 2(m/2)
So we have a particle of mass m/2 in the potential V = T |x| – the center of mass dynamics, as usual,
decouples.
(b) Explain why hxi = 0 for a stationary state. What does this imply about hpi?
Solution: We must have hxi = 0 for a stationary state because the potential is even. Therefore the
eigenstates are going to be either even functions or odd functions (there areR also no degeneracies
in 1d...), and in either case |ψ|2 is an even function, meaning that hxi = dx x|ψ|2 = 0. And
dhxi/dt = hpi/m = 0.
(c) Estimate M by using Heisenberg’s uncertainty principle to estimate the ground state energy.
Solution: Since hpi = 0 and hxi = 0, we estimate the typical scale of x and p in the ground state by
√
√
σx and σp respectively. Using Heisenberg’s uncertainty principle we estimate that
σx σp ∼ ~.
The energy is therefore
E = M c2 =
σp2
σp2
~
+ T σx =
+T .
m
m
σp
We now find the minimum of E (remember σp must be positive!)
0=
2σp T ~
dE
=
− 2
dσp
m
σp
which gives
σp ∼ (mT ~)1/3 ,
and
1 σp2
1
M∼ 2
∼ 2
c m
c
11
~2 T 2
m
1/3
.
(d) In reality, the quarks may be ultra-relativistic, so that E ≈ c|p|. (Continue to approximate, however,
that the one particle description is valid.) Again use Heisenberg’s uncertainty principle to estimate
M.
Solution: Now the energy is
E = M c2 = σp c + T σx = σp c + T
~
.
σp
We now find the minimum of E (remember σp must be positive!)
0=
T~
dE
=c− 2
dσp
σp
which gives
r
T~
,
c
r
T~
.
c3
σp ∼
and
M∼
(e) A typical meson has M ∼ 10−27 kg. Using c = 3 × 108 m/s and ~ ≈ 10−34 J/s, estimate the tension
T of the flux tubes binding quarks, assuming they behave ultrarelativistically. Compare to best
estimates of T ∼ 105 N [2].
Solution: A simple calculation gives us T ∼ 3 × 105 N. This is quite close to the best estimate of the
tension in a flux tube.
Problem 8 (Nuclear Magic Numbers): It is an empirical fact that, at least for light nuclei consisting
of A nucleons (either protons or neutrons), nuclei are particularly stable if A = 2, 8, 20, 28, 50, etc. [3]
For simplicity, suppose that all the nucleons were neutrons, which are spin-1/2 fermions. A simple model
to try and understand this effect would be to approximate the nucleus as a quantum mechanical system
with Hamiltonian
A 2
X
pi
1
2 2
H=
+ mω ri .
2m 2
i=1
(a) Describe the eigenvectors and eigenvalues of H in the case A = 1. What is the degeneracy of each
energy level?
Solution: Because
H=
p2y
mω 2 x2
mω 2 y 2
p2
mω 2 z 2
p2x
+
+
+
+ z +
= Hx + Hy + Hz
2m
2
2m
2
2m
2
we can find separable eigenstates of the form ψx ψy ψz . Each of the ψi s will be eigenstates of the 1d
harmonic oscillator, which are denoted with an (independent) integer ni = 0, 1, 2, . . .:
ψ = ψnx (x)ψny (y)ψnz (z).
The energy levels are
3
E = nx + ny + nz +
~ω.
2
12
The degeneracy of level n is the number of ways to combine together nx + ny + nz = n. Call it dn .
We can compute that
dn =
n n−n
X
Xx
nx =0 ny =0
1=
n
X
(n − nx + 1) = (n + 1)2 −
nx =0
n(n + 1)
(n + 1)(n + 2)
=
.
2
2
(b) Describe how to write down the ground state of H for any A. (You do not need to carry out the
procedure explicitly.)
Solution: We need to write down a totally antisymmetric wave function. We find the A lowest energy
states φ1 , . . . , φA , and then write the total wave function as a totally antisymmetric combination of
the form
X
1
ψ(r1 , s1 , . . . , rA , sA ) = √
(−1)# ψ1 (rσ(1) , sσ(1) , . . . , rσ(A) , sσ(A) )
A! permutations
(c) A heuristic argument – similar to that used for atoms – is that we might expect the nucleus to be
particularly stable when the ground state is unique. What are the nuclear magic numbers that we
predict? Is this a good model?
Solution: The nuclear magic numbers should correspond to 2d0 , 2d0 + 2d1 , etc. So we predict:
2d0 = 2
2d0 + 2d1 = 8
2d0 + 2d1 + 2d2 = 20
2d0 + 2d1 + 2d2 + 2d3 = 40.
The first 3 magic numbers work, but the larger magic numbers cannot be described by this simple
model.
[1] R. Hanbury-Brown and R. Q. Twiss. “Correlation between photons in two coherent beams of light”,
Nature 177 27 (1956).
[2] G. S. Bali, K. Schilling and C. Schlichter. “Observing long colour flux tubes in SU(2) lattice gauge
theory”, Physical Review D51 5165 (1995), arXiv:hep-lat/9409005.
[3] J-L. Basdevant, J. Rich and M. Spiro. Fundamentals in Nuclear Physics (Springer, 2005).
13