CMI Entrance Exam 2016 Question Paper Solution Ph.D Physics

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CMI 2016
Question Paper
Solution · Ph.D Physics
EXAM YEAR TYPE SUBJECT

CMI 2016 Question Paper Solution Ph.D Physics

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.
.co s e m

s em l a
a ag

m
omPhd. Entrance 2016 : Solutions
Physics . co
. c e m
e m l as
l as ag
ag April 19, 2017

1 Classical Mechanics
Problem 1

m
Errata : Constant gravtitational force is in the negative y direction. There

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is no z-direction in the problem.

e m
l as
Problem 1(a) : Generalized co-ordinates, x, ẋ.

Problem 1(b) : L = m
2
(ẋag
2
+ 2α xẋ) + mgαx . 2

Problem 1(c) : Equillibrium position is : 2mpgαx = 0 =⇒ x = 0, y =
0.

Problem 1(d) : For small oscillations, one can neglect x2 ẋ2 term and
whence eq. of motion in this case is

d2 x
o m(1)
c
= −2 g α x
m dt2
√.
m .co s
m
e 2gα.
Whence x executes simple Harmonic oscillation with frequency,

s e g la
g la a
a

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c. o s e m
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Page 1 of 10

Problem 2
P Gmj mi
Problem 2(a) : mi~ai = j6=i r̂
|rij |2 ij

1 1
P P
Problem 2(b) : As ai = d2 mi j6=i Gmj mi r̂ij =⇒ i~
ai × ~ri = 0,
Hence the proof.

Problem 3

For small oscillation around the axis passing through the
qHinge, we have
2 2 2
following equation for τ = I ddt2θ . M2R ddt2θ = M GRθ =⇒ 2G
R
is frequency
of small oscillation.

2

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2 Quantum Mechanics
a)
m
o m . co
. c e m
s
∞ ∞ ∞
√
m
X √
e a
X X
â | α >=
a s
c ân| n >= c n | n − 1 >=
n c n+1|n> n+1

g l
l a
n=0 n=0 n=0

ag
Also
√ ∞
√ 2 αn+1 n + 1 X
cn+1 n + 1 = e−|α| /2 √ = αcn ⇒ â | α >= α cn | n >= α | α >
n + 1! n=0

b)

2 √ 2 √
Pn =|< n | α >|2 =| cn |2 =| e−|α| /2 αn / n! |2 = e−|α| (| α |2 )n / n!
m
c)
m .co
â s
2~ â + a
e
l √
r
† †
â − â
mω g2
x̂ = , p̂ = 2m~ω

Also
a 2i

â | α >= α | α >⇒< α | a | α >= α, < α | a† | α >= α∗
It follows that

√
r
2~
< x̂ >= Re(α), < p̂ >= 2m~ωIm(α)
m
mω

m d)
.co
m .co √ p
s em
lato get
∆x = < x2 > − < x > 2 ∆p = < p 2 > − < p >2
s e g
la a
†
Substitute for x̂ and p̂ in terms of â and â and calculate

ag r
~
r
m~ω ~
∆x = , ∆p = ⇒ ∆x∆p =
2mω 2 2
e)

3

m . c
c. o s e m
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∞
X
| ψ(t) >= U (t) | ψ(0) >= ∞∞
n=0 cn U (t) | n >= cn e−iEn t/~ | n >
n=0

2 √
Also En = ~ω(n + 1/2) and cn = e−|α| /2 αn / n!.
Substituting in the previous equation, we get

| ψ(t) >= e−iωt/2 | α(t) >
where α(t) = αe−iωt .
The phase e−iωt/2 comes from the zero-point energy.
f)
r
2~
<| ψ(t) | x̂ | ψ(t) >=<| α(t) | x̂ | α(t) >= Re(αe−iωt )
mω
Similarly

√
<| ψ(t) | p̂ | ψ(t) >=<| α(t) | p̂ | α(t) >= 2m~ωIm(αe−iωt )

g)

α < x | α >=< x | â | α >=< x | Ax̂ + iB p̂ | α >= (Ax + Bd/dx) < x | α >

0
⇒ ()ψ + ()ψ = αψ
√ 2mω √
(αx−x2 mω )
⇒ ψ = ψ0 e ~ 8~

4

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Electro-Magnetism

m
om
1. h20i Recall that under a gauge transformation in magnetostatics, the
. co
. c
vector potential is transformed to A0 (r) = A(r) − ∇χ(r) where χ(r) is
e m
em
a scalar function.
l as
a s
l form for both A and A . ag
ag (a) h2i Show that Ampere’s law (in differential
0
form) takes the same

(b) h5i Give the differential equation that χ must satisfy so that A0
is in Coulomb gauge. Write an integral expression for its solu-
tion, assuming A → 0 sufficiently fast as |r| → ∞. Proceed by
first writing Poisson’s equation for the electrostatic potential φ(r)
due to a localized charge distribution ρ(r) and its solution via an
integral. Give an appropriate analogy between the two problems.

m
(c) h5i Find the magnetic field B(r) represented by the vector poten-
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tial A = − 21 (r × b) where b is a constant vector. Hint: For two
m
e
vector fields we have ∇ × (A × B) = A(∇ · B) − B(∇ · A) + (B ·
s
∇)A − (A · ∇)B.
g l a
a
1
(d) h4i Find whether the above vector potential A = − (r × b) is in 2
Coulomb gauge or not. Hint: ∇·(A×B) = B·(∇×A)−A·(∇×B).
(e) h4i Suppose the constant vector b = b0 ẑ points along the z axis.
Find the vector potential A. Roughly plot the resulting vector
field A in the x − y plane.

Solutions

1. ANS: with B = ∇ × A, Ampere’s law ∇ × B = µ0 j becomes ∇ × (∇ × A) = µ0 j.
m
m Now the curl of a gradient vanishes ∇ × (∇χ) = 0, so Ampere’s law for A0 is also
.co
.co ∇ × (∇ × A0 ) = µ0 j.
s e m
s em l a
ag equation
2. ANS: For A to be in Coulomb gauge ∇ · A = 0 we must have
0 0
∇ χ = ∇ · A, 2

g la which is Poisson’s equation for χ. We compare it with Poisson’s for the

a 2
R 1 ρ(r0 ) 3 0
electrostatic potential ∇ φ = −ρ/ whose solution is φ(r) =
0 d r . So
4π0 |r−r0 |
χ is like the electrostatic potential while the source ∇ · A plays the role of −ρ/0 .
1
R (∇·A)(r0 ) 3 0
Thus we must have χ(r) = − 4π |r−r0 | d r . 1 mark for each part.

5

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3. ANS: B = b is a uniform magnetic field. We evaluate B = ∇ × A = − 12 ∇ × (r × b).
Using the fact that b is a constant vector, we get ∇ × A = − 21 [−b(∇ · r) + (b · ∇)r].
Now r = (x, y, z) so ∇ · r = ∂i xi = 3 and (b · ∇)r = bx x̂ + by ŷ + bz ẑ = b.
Thus B = − 21 (−3b + b) = b. Alternate approach: take b = (b1 , b2 , b3 ) with b1
constants. Then −2A = r × b = (yb3 − zb2 , zb1 − xb3 , xb2 − yb1 ). Then calculate
∇ × (r × b) = (−2b1 , −2b2 , −2b3 ). Thus we see that B = − 12 ∇ × (r × b) = b.

4. ANS: Yes, it is in Coulomb gauge. We calculate ∇ · A = − 12 ∇ × (r × b) =
− 12 (b · ∇ × r − r · ∇ × b) = − 21 b · ∇ × ( 12 ∇r2 ) = 0. This is because b is a constant
vector and the curl of the gradient r = 12 ∇r2 is zero. Thus we have a Coulomb
gauge vector potential for a constant/uniform magnetic field.

5. ANS: A = − 12 b0 (r×ẑ) = − 12 b0 (xx̂+y ŷ+z ẑ)×ẑ = − 12 b0 (−xŷ+yx̂) = 12 b0 (xŷ−yx̂) =
1 1
2 rb0 (cos θŷ − sin θx̂) = 2 rb0 θ̂. So A circulates around the z axis or the origin of the
x − y plane. It points in the tangential/azimuthal direction counter-clockwise. The
field lines (integral curves) of A are counter-clockwise directed concentric circles.
The magnitude of A grows linearly with radial distance. Here we used x/r = cos θ
and y/r = sin θ in plane polar coordinates.

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3 Mathematical Physics
m
om . co
2
1. Let z = eiθ (unit circle) (and hence dθ = −i dz ) so that cos θ = 1+z .
c m
z 2z

.
Hence the integral to be evaluated now reads
m s e
s e I
dz
I
dz
g l a
g l a I = −i z 2 +1
z 5 + 4 2z )


= −i
(2z + 1)(z + 2)
,
a
a where the integration is now over the unit circle. As is evident, there
are two singularities at z = −1/2 and z = 2. The second singularity
is outside the unit circle and the first one (z = −1/2) is a simple pole
inside the unit circle. Defining
1 1
φ(z) = (z + 1/2) = ,
(2z + 1)(z + 2) 2(z + 2)

m
.co
the residue corresponding to this pole is given by

e m lim φ(z) = 1/3

l as z→−1/2

ag one finds
Hence using residue theorem
2π
I = (−i) (2πi)(1/3) =
3

m
m .co
m .co s e m
s e g la
g la a
a

7

m . c
c. o s e m
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2. (a) The Fourier transform of λ(t) is defined as
Z ∞
λ̃(ω) = λ(t)e−iωt dt
−∞
Z 1
= λ(t)e−iωt dt
−1
Z 0 Z 1
−iωt
= (1 + t)e dt + (1 − t)e−iωt dt
−1 0
sin ω2 2
 
= ω
2

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4 Statistical Mechanics
1. Enaergy levels of a single particle in SHO is
m
om . co
. c E = ~ ω(n + 1/2) n = 0, 1, 2, ...
e m
s
n

m
equantum numbers for the two particles. Further en = E /(~ω) gl
For our case n = 0 , 1 , 2 , 3. Thus the possible arrangements are with n and
a
s
1

a
lsl.n n n en
n as the
2

g a n

a 1. 0 0 1 1 2

2. 0 1 2
3. 0 2 3
4. 0 3 4
5. 1 0 2
6. 1 1 3
7. 1 2 4

m
8. 1 3 5

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9. 2 0 3
10. 2 1 4
e m
11. 2 2 5
l as
ag
12. 2 3 6
13. 3 0 4
14. 3 1 5
15. 3 2 6
16. 3 3 7
Out of these
Sl.nos. 2 and 5 represent the same state
Sl.nos 3 and 9 represent the same state
Sl.nos.4 and 13 represent the same state
Sl.nos. 7 and 10 represent the same state
m
m Sl.nos. 8 and 14 represent the same state
.co
m .co Sl.nos. 12 and 15 represent the same state
s e m
e la
Thus the total number of states are 16-6=10.

las (b) Energy = 3~ ω occurs twice and the state is doubly degenerate.
ag
ag (c) Wave functions are
u0 (x1 )u2 (x2 ) + u2 (x1 )u0 (x2 )
and
u1 (x1 )u1 (x2 )

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2. The energy of a magnetic moment in a magnetic field is −~µ.B ~ =
−gSz B0 if the magnetic field is along z- axis with strength B0 .
Thus the energies of the three states are ( for values of Sz = ~ , 0 , −~)
are g~ , 0 and −g~.
(a)The average value of magnetic moment is
gB0 ~ −gB0 ~
!
(e kT + 0 − e kT )
< µz > = g~ gB0 ~ −gB0 ~
(e kT + 1 + e kT )
For small fields this becomes

2g 2 B0
< µz > =
3
the susceptitibilty is
2 2
g χ =
3
(b) The probability in the state with Sz =1 is
g~B0
e kI
P = gB0 ~ −gB0 ~
(e kT + 1 + e kT )

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