CMI Entrance Exam 2015 Question Paper Solution Ph.D Physics

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CMI 2015
Question Paper
Solution · Ph.D Physics
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CMI 2015 Question Paper Solution Ph.D Physics

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CMI Ph.D. Phy solutions 2015

(1) (a) Show that angular momentum is conserved by calculating Torque.
(b) We have
L2 mṙ2
Veff = + V (r), E = + Veff ,
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2mr2 2
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where L is the conserved angular momentum and f = −dV . Motion can be considered as one
e
dr

m
dimensional (only radial motion).
e l as
as
(c) Circular motion in potential V is equal to particle at rest in the equilibrium point of the
l ag
ag
effective potential. Equilibrium position r0 is defined by the vanishing of the derivative of Veff
2
w.r.t r. ∂V∂reff |r0 = 0 which implies ∂V
∂r
= mLr3 . Together with this, if the initial condition is such
that E = Veff (r0 ), then you can get a circular orbit.
(d) For stable circular orbit, you need r0 to be stable equilibrium point which requires r0 is the
minimum of Veff , hence second derivative of Veff w.r.t r evaluated at r0 should be greater than
3L2
0 which implies m r4
− ∂f
∂r
> 0 at r0 .
(e) Using the form of the force in the above expression, you get n < 3 for the circular orbit to
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be stable.

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s
⃗ ataa distance z along the axis of the ring. From the
l
agthat the field is only along the z-axis. This field is
(2) (a): (i) We want to determine E
symmetry of the problem it is obvious
determined by summing the contributions from all the infinitesimal charge segments along the
ring.
q
2πr
dθ z
dEz (0, 0, z) =
4πϵ0 (z 2 + r2 ) 23
This gives,
⃗ 0, z) = qz
E(0, k̂
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4πϵ0 (r2 + z 2 ) 2
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where k̂ is the unit vector along the z-axis.

m (ii) The force on charge −q along the z-axis is, s e
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a F⃗ = −
q2z
3 k̂
4πϵ0 (r2 + z 2 ) 2
2
which at a distance z ≪ r is F⃗ ≈ − 4πϵ q z
0r
3 k̂. Hence for small displacements from the center

of the ring particle will execute simple harmonic motion.

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(b) Recall : Magnitude of the electric field due to a infinite sheet of uniform charge density is
E = 2ϵσ0 with the direction perpendicular to the sheet and the field lines are directed towards
or away from the sheet depending on the sign of the charge.
Hence for two perpendicular sheets with charge densities ±σ the magnitude of the electric field is
√
⃗ = 2σ with the field lines shown in the figure.
constant everywhere in space and is given by |E| 2ϵ0

(c) (i) Inside the solenoid, B ⃗ is uniform everywhere and is directed along the axis of the
solenoid. It’s magnitude is
B(t) = µ0 N I0 sin(ωt)
⃗ · dl = − ∂ B⃗ · dS where C is a counter-clock wise
∫ ∫
(ii) For this part we just have to use C E A ∂t
circular loop around the axis of the solenoid which encompasses an area A.
Hence inside the solenoid at r < R, (using above equation and axial symmetry)

⃗ · 2πr = −πr2 µ0 N I0 ω cos(ωt)êϕ
E

Similarly, assuming B = 0 outside the solenoid, the electric field is
2
⃗ = − R µ0 N I0 ω cos(ωt)êϕ
E
2r

(3): (a) The Schrodinger equation for the two regions I ≡ (0, a) and II ≡ (a, b) are
√ √
2mE 2m(E − V 0 ) 2mE 2m(E − V0 )
ψI′′ + 2 ψI = 0, ψII ′′
+ 2
ψII = 0; k1 = 2
, k2 = .
ℏ ℏ ℏ ℏ2
The boundary conditions are ψI (0) = 0, ψII (b) = 0, ψI (a) = ψII (a), ψI′ (a) = ψII
′
(a). This
gives the wavefunction

ψI (x) = A1 sin(k1 x), ψII (x) = A2 (sin(k2 x) − tan(k2 b) cos(k2 x)),

subject to boundary conditions at x = a,

A1 sin(k1 a) = A2 (sin(k2 a) − tan(k2 b) cos(k2 a)),
cos(k1 a) cos(k2 a) + tan(k2 b) sin(k2 a)
[quantization condition] k1 = k2 .
sin(k1 a) sin(k2 a) − tan(k2 b) cos(k2 a)
In the limit of E ≫ V0 , we have k2 ∼ k1 . Then the quantization condition after simplifying
becomes (sin2 (k1 a) + cos2 (k1 a)) tan(k1 b) ∼ 0 which is the usual condition sin(k1 b) = 0 for a free
particle in a box with no internal step.
(b) (i) Levels (n1 , n2 ) and (m1 , m2 ) with arbitrary integers ni , mi are degenerate with the same
energy if n1 ω1 + n2 ω2 = m1 ω1 + m2 ω2 , i.e. if ω1 = m 2 −n2
n1 −m1 2
ω ≡ pq ω2 . Thus if ω1 , ω2 are not

2

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rational multiples of each other, there is no degeneracy.
(ii) If ω1 = ω2 , the 4 lowest energy states are (0, 0) [E = ω1 ], {(1, 0), (0, 1)} [E = 2ω1 ],
{(2, 0), (1, 1), (0, 2)} [E = 3ω1 ], {(3, 0), (2, 1), (1, 2), (0, 3)} [E = 4ω1 ].

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(4) (a) S=0,S=1
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(b) S=0 changes sign and S=1 remains the same.
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(c) J=0 the spatial wave function does not change sign. J=1 the wave function changes
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sign.
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(d) For J=0, S=0 is allowed and for J=1, S=1 is allowed
(e) One state
(f) Nine states.
(h) The Boltzmann factor is e−2λ/(k T ) and there are three S=1 states and three J=1 states.
Thus

N (J = 1) ( 2λ )
= 9 e− k T
N (J = 0)
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s em
la are guaranteed to be real. A can be diagonalised
(5)

g
(a) Since A is real symmetric, eigenvalues
a have real eigenvectors that span the vector space.
by an orthogonal transformation, it must
(b) Integration element is area element on sphere dn = sin θdθdϕ. Unit vector is n̂ =
1
r
(x, y, z). Writing it in spherical polar coordinates, n̂ = (sin θ cos ϕ, sin θ sin ϕ, cos θ).
2 2
∫ ∫
(c) ⟨v|A|v⟩ = (v · n) dn. n · ẑ = cos θ and ⟨ẑ|A|ẑ⟩ = cos θ sin θdθdϕ = 4π/3.
(d) Suppose we rotate v → Rv, then we can undo this rotation by making a change of
integration variable n̂ → R−1 n̂. The area element is rotation invariant. So the integral does
not change.
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(e) Since A has the same expectation value in every state it must be a multiple of the
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identity. So 4π/3 is its only eigenvalue with multiplicity three. Since A is a multiple of the

las identity, all non-zero vectors are eigenvectors of A.
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