fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Quantum Mechanics
JEST-2012
Q1.
The ground state (apart from normalization) of a particle of unit mass moving in a one-

 

dimensional potential V(x) is exp  x 2 / 2 cosh 2 x . The potential V(x), in suitable
units so that h = 1, is (up to an addiative constant.)
 
2 x coth  2 x 
(b)  2 / 2  2 x tanh 2 x
(a) π2/2
 
(c)  2 / 2  2 x tan 2 x
(d)  2 / 2 
Ans. : (b)
Q2.
Consider the Bohr model of the hydrogen atom. If  is the fine-structure constant, the
velocity of the electron in its lowest orbit is
(a)
c
1
(b)
c
or 1    c
1 2
(c)  2 c
(d)  c
Ans. : (d)
Solution: mvr  n
1 ze 2
mv 2
1 ze 2

r
r
4 0 r 2
4 0 mr 2
1 ze 2
mv 
 n
4 0 mv 2
ze 2
e2
and fine structure constant  
v
4 0 n
4 0 c
ze 2
ze 2 c
v
For lowest orbit, v 
4 0 
4 0 c
v  c
Q3.




Define  x  f †  f , and  y  i f †  f , where the   are Pauli spin matrices and


f , f † obey anti-commutation relations  f , f   0, f , f †  1 . Then  z is given by
(a) f † f  1
(b) 2 f † f  1
(c) 2 f † f  1
(d) f † f
Ans. : (c)
Solution:  x y  i z
i z   x  y
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 z   x y 
i †
f f
i


1
i

 f
†

 
 f   f †

2
 f † f  ff †  f 2 

    f † f  1  f † . f    1  2 f † f   2 f † f  1
Q4.
Consider a system of two spin-
1
particles with total spin S  S1  S2 , where S1 and S2
2
are in terms of Pauli matrices  i . The spin triplet projection operator is
(a)
1
 S1  S2
4
(b)
3
 S1  S2
4
(c)
3
 S1  S2
4
(d)
1
 S1  S2
4
Ans. : (c)
Solution:  S  S1  S 2
S 2  S12  S 22  2S1  S 2
3 3

S 2     2.S1  S2   2
4 4

 S  0, 1
3

S 2  2   S1  S2   2 for Triplet projection operator
4

3

s  s  1  2  2   S1  S 2   2
4

3

11  1  2  S1  S 2 
4

Q5.
Consider a spin-
S 1

3
 S1  S 2  I
4
1
particle in the homogeneous magnetic field of magnitude B along z 2
axis which is prepared initially in a state  
1
2


  at time t  0 . At what time
t will the particles be in the state   (  B is Bohr magneton)?
(a) t 

B B
(b) t 
2
B B
(c) t 

2 B B
(d) Never
Ans.: (a)

1 1
 
Solution: E   B  B zˆ  
2 1
  x, t  
1  1  iEtb
   x, t    
 e
2  1
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1 1
 e
2 1
e
 i B Bt

 i B Bt


 1 1
 
2 1
 1
  Bt 
cos B   cos 
  
B B t

Q6.
 t 

B B
The ground state energy of 5 identical spin-
1
particles which are subject to a one2
dimensional simple harmonic oscillator potential of frequency ω is
(a)
15

2
(b)
13

2
(c)
1

2
(d) 5
Ans. : (b)
1
Solution: Degeneracy  2 s  1  2   1  2
2
13
1
3
5
Eground  2    2    1   
2
2
2
2
Q7.
The spatial part of a two-electron state is symmetric under exchange. If  and 
represent the spin-up and spin-down states respectively of each particle, the spin-part of
the two-particle state is
(a)  
(b)  



(c)      / 2

(d)      / 2
Ans. : (c)
Solution: Since, electrons are Fermions and Fermions have anti-symmetric wave function
 spatial part is symmetric then its spin part is antisymmetric to maintain antisymmtric
wave function
 x  
1
2

   

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Q8.
The
wave
function
of
a
free
particle
in
one
dimension
is
given
by
  x   A sin x  B sin 3 x . Then   x  is an eigenstate of
(a) the position operator
(b) the Hamiltonian
(c) the momentum operator
(d) the parity operator
Ans. : (d)   x     x 
   x  {parity (even and odd)
  x   A sin  x   B sin  3 x    A sin x  B sin 3 x 
  x     x   negative parity i.e. parity operator
Q9.
The quantum state sin x   expi  cos x  , where    0 and x,  are, real, is
orthogonal to:
(a) sin x 
(b) cos x   expi sin x 
(c)  cos x   expi sin x 
(d)  exp i  cos x   sin x 
Ans.: (d)
Solution:    0 ,   sin x   expi  cos x 
     exp  i  cos x sin x    exp  i  exp  i  cos x  
 sin 2 x    exp  i  cos x sin x  
  expi  cos x sin x  expi  cos x sin x  0
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JEST-2013
Q10.
A particle of mass m is contained in a one-dimensional infinite well extending from
x
L
L
to x  . The particle is in its ground state given by  0  x   2 / L cosx / L  .
2
2
The walls of the box are moved suddenly to form a box extending from x   L to x  L .
what is the probability that the particle will be in the ground state after this sudden
expansion?
(a) 8 / 3 
2
(c) 16 / 3 
(b) 0
2
(d) 4 / 3 
2
Ans.: (a)
Solution: Probability  0 1
2
, 0 
2
2
x
x
cos , 1
cos
L
L
2L
2L
Since the wall of box are moved suddenly then
2
Probability 

2
1 cos  x cos  x
2 1 L / 2 2 cos  x cos  x


dx 

dx
2L
2L
L
L L
L 2  L / 2 L
L/2
L / 2
2
2
L/2
2 1 L / 2   3 x 
2 1  2L
3 x 2 L
x
  x 

  cos 
 cos 
  sin

dx 
sin




2L
2 L   L / 2
L 2  L / 2   2L 
L 2  3
 2L 
2 1  2 L  3
3

   sin
 sin
4
4
L 2  3 
Q11.
 
 2L  

 sin  sin  
4
4 
  
2
2 2


3 
2
8

3
2
2
A quantum mechanical particle in a harmonic oscillator potential has the initial wave
function  0  x    1  x  , where  0 and  1 are the real wavefunctions in the ground and
first excited state of the harmonic oscillator Hamiltonian. For convenience we take
m      1 for the oscillator. What is the probability density of finding the particle at
x at time t   ?
(a)  1  x   0  x  
2
(b)  1  x     0  x  
2
(c)  1  x    0  x  
2
(d)  1  x     0  x  
2
2
2
Ans.: (a)
Solution:  x    0 x    1  x 
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E0t
Et
 1  x  ei 1


Now probability density at time t
  x, t    0  x  e  i
  x, t    *  x, t   x, t    0  x    1  x   2 Re 0*  x  1  x  cos  E1  E0 
2
2
2
putting t  
 E1  E0    1
  x, t    0  x    1  x   2 Re 0*  x  1  x  cos 
2
2
2
  x, t    0  x    1  x   2 Re 0*  x  1  x    1  x   0  x  
2
Q12.
2
t

2
2
If J x , J y and J z are angular momentum operators, the eigenvalues of the operator
J
x
 Jy 
are:

(a) real and discrete with rational spacing
(b) real and discrete with irrational spacing
(c) real and continuous
(d) not all real
Ans.: (b)
Solution: J x 
Jx 
0 1 
0 0 
1
i
J
,


 J  J  , J y   J  J   J   


1 0 
2
2
0 0 


J  J y 1  0 1 i
 0 1 
i 0 1
, Jy  
 x
 



2 1 0 
2 1 0 
2 1  i 0 

1    1 i
  2  2  0     2

2 1  i   
A simple model of a helium-like atom with electron-electron interaction is replaced by
eigen value
Q13.
Hooke’s law force is described by Hamiltonian
  2
1
 2 2

1   22  m 2 r12  r22 
m 2 r1  r2 .
2m
2
4



What is the exact ground state energy?
3
(a) E   1  1  
2

(c) E 

3
 1  
2



(b) E 
3
 1  
2
(d) E 
3
 1  1  
2


Ans.: (b)
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Q14.
1 / 2 
Consider the state 1 / 2  corresponding to the angular momentum l  1 in the L z basis
1 / 2 
of states with m  1, 0,  1 . If L2z is measured in this state yielding a result 1, what is the
state after the measurement?
 1/ 3 


(b)  0 


 2/3
1
 
(a)  0 
 0
 
1 / 2 


(d)  0 


1 / 2 
 0
 
(c)  0 
1
 
Ans.: (d)
1 0 0 
1 0 0




2
Solution: L z   0 0 0  , L z   0 0 0  , eigenvector
 0 0  1
0 0 1




1
 
 0 ,
 0
 
0
 
1,
0
 
0
 
0
1
 
Corresponding eigenvalue 1, 0, 1
Q15.
1
1
  1  
Now state after measurement yielding 1 1  3   0  
0
2 
1
 
1
 

What are the eigenvalues of the operator H    a , where  are the three Pauli matrices

and a is a vector?

(a) a x  a y and a z
(b) a x  a z  ia y
(c)  a x  a y  a z 
(d)  a
Ans.: (d)
 
Solution: H    a   x .a x   y .a y   z .a z 

az
0 1
 0 i 
1 0 


 ax  
 ay  
 az 

1 0
i 0 
 0 1
  ax  ia y 
a
 ia y  

 az 
x
For eigen value,
  az     ax  ia y  

  0    az    az      ax  ia y  ax  ia y   0
  ax  ia y    az    


  az2   2  ax2  a y2  0   2  ax2  a y2  az2     a
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Q16.

The hermitian conjugate of the operator 
 is
 x 

x
(a)
(b) 

x
(c) i

x
(d)  i

x
Ans.: (a)
†
*



    x 

x 
Solution:   *  x     x    


x
x

 



  x 
 

    x      x   dx   *  x   x    
  x  dx



x
 x

*


 *  x 

x

Q17.
*
  x  dx
If the expectation value of the momentum is p for the wavefunction   x  , then the
expectation value of momentum for the wavefunction eikx /   x  is
(b) p  k
(a) k
(c) p  k
(d) p
Ans.: (c)
Solution:





 *  x   i
 
  x  dx  p
x 
Now



e

 ikx

 ikx 

  ikx
ik ikx

*

  x   i  e   x  dx   e   x  i   e
  x   e   x 

x 

x



ikx


 e

*

ikx


ik  ikx

 
  x   i   x   e   i. e   *  x   x  dx

x



ikx

*





   *  x   i   x    k   *  x   x   p  K


x


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Q18.
Two electrons are confined in a one dimensional box of length L . The one-electron states
are given by  n  x  
2
 n x 
sin 
 . What would be the ground state wave function
L
 L 
  x1 , x2  if both electrons are arranged to have the same spin state?
2
  x1   2 x2  2
 2 x1    x2  
 L sin  L  sin  L   L sin  L  sin  L  

 


 


(a)   x1 , x2  
1
2
(b)   x1 , x2  
1 2
  x1   2 x2  2
 2 x1    x2  
 L sin  L  sin  L   L sin  L  sin  L  
2

 


 

(c)   x1 , x2  
2
  x   2 x2 
sin  1  sin 

L
 L   L 
(d)   x1 , x2  
2
 2 x1    x2 
sin 
 sin 

L
 L   L 
Ans.: (b)
Solution: Electrons are Fermions of spin
1
and its wave functions are anti-symmetric
2
Since, spin part is symmetric, therefore, space part will be anti-symmetric (since as total
wave function is anti-symmetric)
Then,
  x1 , x2  
Q19.
1
2
2
  x1 
 2 x2  2
 2 x1 
  x2  
 L sin  L  .sin  L   L sin  L  .sin  L  









d
 d

The operator   x   x  is equivalent to
 dx
 dx

d2
 x2
(a)
2
dx
d2
(b) 2  x 2  1
dx
d2
d
 x x2 1
(c)
2
dx
dx
d2
d
 2x  x 2
(d)
2
dx
dx
Ans.: (b)
d
 d

d
 d

Solution:    x   x  f  x     x   f  x   xf  x 
 dx
 dx

 dx
  dx

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
d d
d

f x   xf x   x
f x   x 2 f x 

dx  dx
dx


df  x 
d2
d
f  x  f  x  x
x
f  x   x2 f  x 
2
dx
dx
dx


 d2
d2
2
2
 f x 







f
x
x
f
x
f
x




x

1
 dx 2
dx 2


JEST-2014
Q20.
Suppose a spin 1 / 2 particle is in the state
1 1  i 


6 2 
 
If S x ( x component of the spin angular momentum operator) is measured what is the
probability of getting   / 2 ?
(b) 2 / 3
(a) 1 / 3
(c) 5 / 6
(d) 1 / 6
Ans.: (c)
Solution: S x 
1 1
 0 1 


with eigenvalues  and eigenvector corresponding to is
 


2 1 0 
2
2
2 1
Now probability getting 

2


p  

2  
Q21.
2
1  i 
1
2

1 1  2 
1 i  2
2 6
5


 12

1
6
1  i 
1
6
1  i 2  2 
6
6


1
1
The Hamiltonian operator for a two-state system is given by
H    1 1  2 2  1 2  2 1 ,
where  is a positive number with the dimension of energy. The energy eigenstates
corresponding to the larger and smaller eigenvalues respectively are:

(c) 1  
(a) 1 

 2  1 2
2  1 2 ,  2  1 1  2
2 1 2 , 1 

1 
(b) 1 
(d)

 2  1 2
2  1 2 ,  2  1 1  2
2 1 2 , 1 
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Ans.: (b)
Solution: H    1 1  2 2  1 2  2 1   H 1    1  2  , H 2    1  2

 H 1 

Lets check for option (b): 1 
Now H    
H 1 



2 1 2   H  1  

  1  2  1 1   1 

 2  1 




 2  1 2
2  1 2   H 1  H  2  1 2

2 1 2 , 1 


2 1 H 2    1  2  

2 1   1  2



2 1 2 



   1  2   

Q22.



2 1  2   2 1   2  2 2

Now H 1  2  1 2  H  1 






2 1 2   H 1  H





2 1 2



2  1  1  2    1  2  1 1   1  2  1 2




  2 1  2  2  2   2  1  1  2 2 


2

Consider an eigenstate of L and Lz operator denoted by l, m . Let A  nˆ  L denote an
operator, where n̂ is a unit vector parametrized in terms of two angles
as  nx , n y , nz    sin  cos  ,sin  sin  , cos   . The width  A in l, m state is:
(a)
l l  1  m 2
 cos 
2
(b)
l l  1  m 2
 sin 
2
(c)
l l  1  m 2  sin 
(d)
l l  1  m 2  cos 
Ans.: (c)

x
y
z
Solution: A  nˆ  L  A  Lx   Ly   Lz 
r
r
r
 A  Lx 
r sin  sin 
r sin  cos 
r cos 
 Ly 
 Lz 
r
r
r
 A  Lx sin  cos   Ly sin   sin   Lz cos 
Now A 
A2  A
2
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A  Lx sin  cos   Ly sin  sin   Lz cos 
A   m  cos 
 Lx  0, Ly  0
A 2  L2x sin 2  cos 2   L2y sin 2  sin 2   L2z cos 2 
L
 L


 sin  
2
x
 L2y sin 2   L2z cos 2 
2
 L2z
2
L2z cos 2 
 A2  l  l  1  m 2   2 sin 2   m 2  2 cos 2 
 A2  l  l  1  m 2   2 sin 2   m 2  2 cos 2 
A 
A2  A
2

 l  l  1  m  
2
2
sin 2   m 2  2 cos 2   m 2  2 cos 2 
A  l  l  1  m 2   sin 
Q23.
Consider a three-state system with energies E , E and E  3g (where g is a constant) and
1
1
1
1  
1  
1  
respective eigenstates  1 
1 ,  2 
1 and  3 
1
2  
6  
3  
 2 
0
1
1
 
If the system is initially (at t  0 ), in state  i   0 
0
 
what is the probability that at a later time t system will be in state  f
(a) 0
(c)
4
 3 gt 
cos 2 

9
 2 
(b)
4 2  3 gt 
sin 

9
 2 
(d)
4 2  E  3gt 
sin 

9
 2 
0
 
 0
1
 
Ans.: (b)
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Q24.
The lowest quantum mechanical energy of a particle confined in a one-dimensional box
of size L is 2 eV . The energy of the quantum mechanical ground state for a system of
three non-interacting spin
1
particles is
2
(b) 10 eV
(a) 6 eV
(c) 12 eV
(d) 16 eV
Ans.: (c)
Solution: E1 
 22
2ml 2
Spin, spin is
 2eV , E2  4 E1  8 eV
1
1
, therefore, degeneracy gi  2S  1  2   1  2
2
2
 ground state energy = 2  2 eV  1  8 eV  12 eV
Q25.
A ball bounces off earth. You are asked to solve this quantum mechanically assuming the
earth is an infinitely hard sphere. Consider surface of earth as the origin implying
V 0   and a linear potential elsewhere (i.e. V  x    mgx for x  0 ). Which of the
following wave functions is physically admissible for this problem (with k  0 ):
(a)   e  kx / x
(b)   xe  kx
2
(c)    Axekx
(d)   Ae  kx
2
Ans.: (b)
Solution:   xe  kx
2
For given potential, at x  0 and x   wave function must vanish.
Q26.
The operator A and B share all the eigenstates. Then the least possible value of the
product of uncertainties AB is
(a) 
(b) 0
(c)  / 2
(d) Determinant (AB)
Ans.: (b)
Solution: A  B 
A  B  0
AB
2
[ A and B have share their eigen values, so AB   0 ]
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Q27.
Consider a square well of depth  V0 and width a with V0 as fixed. Let V0   and
a  0 . This potential well has
(a) No bound states
(b) 1 bound state
(c) 2 bound states
(d) Infinitely many bound states
Ans.: (b)
Solution: It forms delta potential, so it has only one bound state.
JEST-2015
Q28.
Consider a harmonic oscillator in the state   e

2


e a 0 , where 0 is the ground
2
state, a  is the raising operator and  is a complex number. What is the probability that
the harmonic oscillator is in the n th eigenstate n ?
(a) e
(c) e
2n

2
(b) e
n!


2
n
(d) e
n!


2


2

n
n!
2

2
2n
n!
Ans.: (a)
Solution:   e

2

2
 a
0 e
e


2
2

 a 

 e


2

2
 
n
n
n
n

0 and n
n
n
n    e
 e


2

 
n
n
n
 n e

n
2

n

2


n

n

 
0  a
n
  
*
 n
n
n
n n e
2

2
n
0 

n
 n

n
n
n n
2
2
 e  e   1
2
1
n
n
2
1
e
n  n 
nn 
n
n
2
n
2
 
2
n
2

n e


2
n
Probability that  is in n state is,
2
a 

n
2
e

2

2n
n
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Q29.
A particle of mass m moves in 1  dimensional potential V  x  , which vanishes at infinity.
The exact ground state eigenfunction is   x   A sec h   x  , where A and  are
constants. The ground state energy eigenvalue of this system is,
(a) E 
 2 2
m
(b) E  
Ans.: (d)
Solution:   x   A sec h   x  

 2 2
m
(c) E  
 2 2
2m
(d) E 
 2 2
2m
d
  A sec h   x  tanh   x 
dx
d 2
  A   sec h   x  tan 2 h   x     sec h   x  sec 2 h   x  
dx 2
  A 2 sec h   x    tan 2 h   x   sec 2 h   x   
  A 2 sec h   x  sec 2 h   x   tan 2 h   x   
  A 2 sec h   x  sec 2 h   x   1  sec 2 h   x    


 tan 2 h   x   1  sec 2 h   x 
  A 2 sec h   x  sec 2 h   x   1  sec 2 h   x   

d 2
  A 2  2sec3 h   x   sec h   x  
dx 2
Now put the value

d 2
 2 d 2
in
equation

 V  x   x   E  x 
dx 2
2m dx 2
2 2
 A  2sec3 h   x   sec h   x    V  x  A sec h   x   EA sec h   x 
2m
V  x   0

as x  
2 2
2 2
 A sec h   x  
2 A sec3 h   x   EA sec h   x 
2m
2m
Now we have to do approximation i.e. sec3 h   x  dacays very fastly as x   so second
term
2 2
2 2
2 2
3
2 A sec h   x   0 . Thus
A sec h   x   EA sec h   x   E 
2m
2m
2m
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Q30.
Consider a spin 
1
particle characterized by the Hamiltonian H  S z . Under a
2
perturbation H   gS x , the second order correction to the ground state energy is given by,
g2
(a) 
4
g2
(b)
4
g2
(c) 
2
g2
(d)
2
Ans.: (a)
 1 0 
2 0 1
  1 0 
g  1 0 
H 

 and H   gsx 


2  0 1
2  0 1 
0

Ground state energy is 
with eigenvector 1   
2
1
1

and first excited state energy is
with eigenvector 2   
2
0
Solution:  H   sz
and
sz 
m H  1
Second order correction in ground state E12  
2
E10  Em0
m 1
m H  1

 

2

2
2
2
0 10
1
0



 
g 22
g2
g 22
1 01
2



 E1 
2 
4

4


4

2
Q31.
Given that  1 and  2 are eigenstates of a Hamiltonian with eigenvalues E1 and E2
respectively, what is the energy uncertainty in the state  1   2  ?
(a)  E1 E 2
(c)
(b)
1
E1  E 2 
2
1
E1  E 2
2
(d)
1
2
E 2  E1
Ans.: (b)
Solution:
E
2

E12  E22
1 2 1 2
 E1  E2 
2
2
2

and
E 
1
1
E1  E2
2
2
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Q32.
2

E
2
1
 E22
  1 E  E 
 E 
E2  E
 E 
E12  E22  2 E1 E2 1
 E1  E2
4
2
2
4
1
2
2

2 E12  2 E22  E12  E22  2 E1 E2
4
A particle moving under the influence of a potential V r  
kr 2
has a wavefunction
2
  r , t  . If the wavefunction changes to   r , t  , the ratio of the average final kinetic
energy to the initial kinetic energy will be,
(a)
1

(b) 
2
(c)
1
(d)  2

Ans.: (c)

 2  2
2
2
Solution: For   r , t  the average kinetic energy T    *  r , t   
     r dr ,  is
0
 2m 
written in spherical polar coordinate, which is dimension of  length 
2
For wave function   r , t 

 2  2
2
T    *  r , t   
     r , t   r dr
0
 2m 
r
dr 
Put  r  r  or r   dr 
and  2r   2  r2


T  3

 T
Q33.

   2
1
2
0   r , t    2m     r , t  r  dr   
T
T
1

  


T
2

*
2
 2  2
2

0   r , t    2m     r , t  r  dr 

*
If a Hamiltonian H is given as H  0 0  1 1  i  0 1  1 0  , where 0 and 1 are
orthonormal states, the eigenvalues of H are
(a)  1
Ans:
(b)  i
(c)  2
(d)  i 2
(c)
Solution: H  0 0  1 1  i  0 1  1 0 
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H 0  0 i 1
and
H 1   1 i 0
The matrix representation of H is
1  

 i
Eigenvalue of H
Q34.
0 H 0
0 H 1
1H 0
1H 1
1 i 


 i 1
i 
2
  0   1     1  0     2
1   
A particle of mass m is confined in a potential well given by V  x   0 for
L
L
x
2
2
L
and V  x    elsewhere. A perturbing potential H  x   ax has been applied to the
2
system. Let the first and second order corrections to the ground state be E 01 and E02  ,
respectively. Which one of the following statements is correct?
(a) E 01  0 and E 02   0
(b) E 01  0 and E 02   0
(c) E 01  0 and E 02   0
Ans.: (d)
 L / 2  x  L / 2
and H   x    x
elsewhere
0
Solution: V  x   

For ground state 0 
E01 
(d) E01  0 and E 02  0
2
x
cos
L
L
0 H  0
2 L/2
x
   x cos 2
0
L

/2
0 0
L
L
E0  
 2
m0
m H  0
E E
0
0
0
m
2
 E0   0
2
 E00  Em0
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Q35.
The wavefunction of a hydrogen atom is given by the following superposition of energy

eigen functions  nlm  r  ( n, l , m are the usual quantum numbers):
2
3
1



 100  r  
 210  r  
 322  r 
7
14
14
The ratio of expectation value of the energy to the ground state energy and the

 r  
expectation value of L2 are, respectively:
229
12 2
and
7
504
101
(c)
and  2
504
101
12 2
and
7
504
229
(d)
and  2
504
(a)
(b)
Ans.: (a)
2 E
9 E
1 E
229
Solution: E   0   0   0 
E0
7 1 14 4 14 9 504
2
9
1
24
12
L2   0 2   2 2   6 2   2   2
7
14
14
14
7
1
particle in a uniform external magnetic field has energy eigenstates 1 and 2 .
2
 1  2  at time t  0 . It evolves to the state
The system is prepared in ket-state
2
 1  2  in time T . The minimum energy difference between two
described by the ket
2
levels is:
h
h
h
h
(a)
(b)
(c)
(d)
6T
4T
2T
T
Q36. A spin-
Ans.: (c)

1

 E1t 
 E t 
i
 2 exp  i 2  


1  2    t t
 
 


Solution:   t  0  
 
2
2

  E2  E1  t  
 1  2 exp  i
 

 E1t  


  t  t    i

 
2

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  E  E1  t 
exp  i 2
  1



 E2  E1  T    E  E    
 2 1

Q37.
T
h
2T
The energy of a particle is given by E  p  q where p and q are the generalized
momentum and coordinate, respectively. All the states with E  E0 are equally probable
and states with E  E0 are inaccessible. The probability density of finding the particle at
coordinate q , with q  0 is:
(a)
 E0  q 
(b)
2
0
E
q
E02
(c)
 E0  q 
(d)
2
0
E
1
E0
Ans.: (c)
Solution: For condition, E  p  q total number of accessible state upto energy E0 for q  0
1
is area under the curve   2  E02  E02
2
The probability density of finding the particle at coordinate q , with q  0
 E  q  dq
dpdq pdq
 2  0 2
2
E0
E0
E0
For probability at point q , dq is insignificant so p  q  
 E0  q 
E02
Q38. Consider a quantum particle of mass m in one dimension in an infinite potential well, i.e.,
a
a
x
2
2
V  x   0 for
V  x 
2 x
and V  x    for
x
a
. A small perturbation,
2
is added. The change in the ground state energy to O  is:
a
(a)

2 4
2
2

(b)

2 4
2
2
(c)
 2 2
 4
2
(d)
 2 2
 4
2







Ans.: (a)
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a
2
a
2 2 2
2 x
Solution: E11   1*V '  x 1dx 
dx
 x cos
a
a
a
a
a



2
2
a
2
a
4 2
a
4 2
2
2
x
2
2 x 
2 x 

 1dx  2  x  cos
 1dx
.2.  x cos 2
dx  2  x  cos
a
a
a
a
a
a 0 2
a 0 


0
a
4 2 
2 x 

 2  x  cos
 1dx 
2 4
2
a
a 0 
2


Q39.
If Yxy 

1
Y2,2  Y2,2  where Yl ,m are spherical harmonics then which of the following
2
is true?
(a) Yxy is an eigenfunction of both L2 and Lz
(b) Yxy is an eigenfunction of L2 but not Lz
(c) Yxy is an eigenfunction both of Lz but not L2
(d) Yxy is not an eigenfunction of either L2 and Lz
Ans.: (b)
Solution: The L2Yxy  l  l  1  2Yxy , where l  2 and LzYxy  mYxy
So, Yxy is an eigenfunction of L2 but not Lz
Q40.
A spin-1 particle is in a state  described by the column matrix
2 
1  
 2  in the S z
10  
 2i 
basis. What is the probability that a measurement of operator S z will yield the result 
for the state S x  ?
(a)
1
2
(b)
1
3
(c)
1
4
(d)
1
6
Ans.: (c)
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 2 
0 1 0
1  
 

Solution: S x 
1 0 1 ,  
 2
2 
10  

0 1 0
 2i 
 1 
2 

Sx  
 1  i 
10 

 1 
1 0 0 


Sz    0 0 0 
 0 0 1 


1
The eigen state corresponding to eigen value  of S z is  0 
0
 
2
 1 
2 

 1  i 
1 0 0 
10 

1
 1 
 P  

 1  4
2 2


 1 1  i 1 1  i 
10
 1 


Q41.
The Hamiltonian of a quantum particle of mass m confined to a ring of unit radius is:
2  

H
 
 i
2m  

2
where  is the angular coordinate,  is a constant. The energy eigenvalues and
eigenfunctions of the particle are ( n is an integer):
ein
2
2
(a)  n   
and En 
n  
2m
2
(c)  n   
cos  n 
2
and En 
(b)  n   
sin  n 
2
2
2
and En 
n  
2m
2
ein
2
2
2
and En 
 n    (d)  n   
n  
2m
2m
2
Ans.: (a)
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2
Solution: H 

2  
 2   2


i




  2   E


  2  2i
2m  
2m  



 n   
By inspection,
 n   2    n  
2  n   
E
2m
Q42.
ein
, which will also satisfy boundary condition
2
and satisfies the eigen value equation with eigen value
2
The adjoint of a differential operator
d
acting on a wavefunction   x  for a quantum
dx
mechanical system is:
(a)
d
dx
(b) i
d
dx
(c) 
d
dx
(d) i
d
dx
Ans.: (c)
Q43.
In the ground state of hydrogen atom, the most probable distance of the electron from the
nucleus, in units of Bohr radius a0 is:
(a)
1
2
(b) 1
(c) 2
(d)
3
2
Ans.: (d)
Solution:  100 
1
 a03
e
r
a0
r
dP
1
 0  rp  a0
P     3 e a0  rp 
 a0
dr
*
Q44.
For operators P and Q , the commutator  P, Q 1  is
(a) Q 1  P, Q  Q 1
(b) Q 1  P, Q  Q 1
(c) Q 1  P, Q  Q
(d) Q  P, Q  Q 1
Ans.: (b)
Solution:  P, Q 1   PQ 1  Q 1 P
Q 1  P, Q  Q 1  Q 1  PQ  QP  Q 1  Q 1  PQQ 1  QPQ 1   Q 1 P  PQ 1   P, Q 1 
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Q45.
A spin
1
particle is in a state
2
     where
2
 and  are the eigenstates of S z
operator. The expectation value of the spin angular momentum measured along x
direction is:
(b) 
(a) 
(c) 0
(d)

2
Ans.: (d)
1 
  
2 
Solution:  
,
1 
2

2

1   0 1
 1
Sx  

 
2  2 1 0
 2










 0 1
Sx  

2 1 0
1 
2  

1  2

2
JEST 2017
Q46.
What is the dimension of
(a) kg m 1s 2

, where  is a wavefunction in two dimensions?
ix
(b) kg s 2
(c) kg m 2 s 2
(d) kg s 1
Ans. : (d)
 dim of  kg  m  sec 2  sec
Solution: Dimension of


 kg sec 1
dim of x
ix
m
Q47.
Suppose the spin degrees of freedom of a 2 - particle system can be described by a 21 dimensional Hilbert subspace. Which among the following could be the spin of one of the
particles?
(a)
1
2
(b) 3
(c)
3
2
(d) 2
Ans. : (b)
Solution: Dimension of Hilbert space   2 s1  1   2 s2  1  7  3  21
So, s1  3, s2  1
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Q48.
If the ground state wavefunction of a particle moving in a one dimensional potential is
proportional to exp   x 2 / 2  cosh


2 x , then the potential in suitable units such that
  1 , is proportional to
 2x
2 x coth  2 x 
(b) x 2  2 2 x tanh
(a) x 2
(c) x 2  2 2 x tan

2x

(d) x 2  2
Ans. : (b)
Solution: From figure, we can conclude that option (b) is the correct answer.
V

x
Q49.
x
A particle is described by the following Hamiltonian
pˆ 2 1
 m 2 xˆ 2   xˆ 4
Hˆ 
2m 2
where the quartic term can be treated perturbatively. If E0 and E1 denote the energy
correction of O    to the ground state and the first excited state respectively, what is the
fraction E1 / E0 ?
Ans. : 5
Pˆ 2 1
ˆ
Solution: H 
 m 2 xˆ 2   xˆ 4
2m 2
Now, energy correction of O    to ground state is
2
2
  
  
2
E0  0 xˆ 0  
 3
 0 6n  6n  3 0  
 2m 
 2m 
4
And energy correction of O    to first excited state is
2
  
2
E1  1 xˆ 4 1  
 1 6n  6 n  3 1
 2m 
2
2
E1 15
  
  
6
6
3
15
 5





 
 
 . Hence,
E0 3
 2m 
 2m 
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Q50.
If x̂  t  be the position operator at a time t in the Heisenberg picture for a particle
pˆ 2 1
ˆ
described by the Hamiltonian, H 
 m 2 xˆ 2 what is ei  t 0 xˆ  t  xˆ  0  0 in units of
2m 2

where 0 is the ground state?
2m
Solution: Operator X̂  t  in Hisenburg picture is written as
Xˆ  t   eiHt /  Xˆ  0  eiHt / 
Thus, 0 Xˆ  t  Xˆ  0  0  0 eiHt /  X  0  e  iHt /  X  0  0
Here, Xˆ  0  0 

1
2m
So, above equation reduces as,
0 Xˆ  t  Xˆ  0  0 

0 eiHt /  Xˆ  0  e  iHt /  1
2m
In integral form,
0 Xˆ  t  Xˆ  0  0 
i  t


0*  t  Xˆ  0  1  t  dx
2m 

0* e 2  Xˆ  0  1e

2m
 i 3  t
2
  it *
e  0 x1 dx
2m
dx 
2

 
†
Therefore, e 0 Xˆ  t  Xˆ  0  0  
 0 a  a 1
2
m




eit 0 Xˆ  t  Xˆ  0  0 
2m
Consider a particle confined by a potential V  x   k x , where k is a positive constant.
i t
Q51.
The spectrum En of the system, within the WKB approximation is proportional to
3/ 2
1

(a)  n  
2

Ans. : (b)
kx
Solution: V  x   
 kx
1

(b)  n  
2

2/3
1/ 2
1

(c)  n  
2

1

(d)  n  
2

4/3
x0
x0
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1

 2m  E  V  x dx   n     2 2m
2

0
b
1
 2 2mE  1  t
0
E/k

E/k
E  kxdx  2 2m
0
1

E  1
0
E
2E
dt 
2mE  1  t dt  2 E 3 / 2
k
k
0
k
x dx
E
2m 2

k
3
1
3 k 
1

  n     En3 / 2 
n  
2
2
4 2m 

 3 k 
1 
En  
 n  
2 
 4 2m 
Q52.
2/3
Consider the Hamiltonian
1 0 0
0 0 1 




H t     0 2 0    t  0 0 0 
 0 0 3
 1 0 2 




The time dependent function   t    for t  0 and zero for t  0 . Find
 t  0  t  0
2
, where   t  0  is the normalised ground state of the system at
a time t  0 and   t  0  is the state of the system at t  0 .
(a)
1
1  cos  2 t  
2
(b)
1
1  cos  t  
2
(c)
1
1  sin  2 t  
2
(d)
1
1  sin  t  
2
Ans. : (a)
1 0 0
0 0 1 


Solution: H  t     0 2 0     t   0 0 0 
 0 0 3
 1 0 2 




 , t  0
Time dependent function   t   
0 , t  0
When t  0
1 0 1


H t     0 2 0 
1 0 1


Eigen value are 0 , 2 , 2 .
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1
1  
For Eigen value zero, the ground state wave function is   t  0  
0 .
2  
 1
1
0
1   i t
1    i 3 t
And   t  0  
0 e 
0 e
2  
2  
0
1
Now,   t  0    t  0 
2
 i 3 t
1 i t
 e e 
4
2

1 
t
3 t  
t
3 t 
 cos  cos
    sin  sin

4 

  

 

2 t  1 
2 t 
1
t
3 t
t
3   1 

1  1  2  cos .cos
 sin cos
 1  cos
    2  2.cos





  4 
  2
 
4

2
2



JEST-2018
Q53.
If   x  is an infinitely differentiable function, then D̂  x  , where the operator
 d 
Dˆ  exp  ax  , is
 dx 
(b)   ae a  x 
(a)   x  a 
(c)   e a x 
(d) e a  x 
Ans. : (c)
Q54.
A one dimensional harmonic oscillator (mass m and frequency  ) is in a state  such
that the only possible outcomes of an energy measurement are E0 , E1 or E2 , where En is
the energy of the n -th excited state. If H is the Hamiltonian of the oscillator,
3
 H 
and
2
11 2 2
 H  
, then the probability that the energy
4
2
measurement yields E0 is
(a)
1
2
(b)
1
4
(c)
1
8
(d) 0
Ans. : (b)
Solution:   a 0  b 1  c 2 let us assume a, b, c is real
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H 
a2 

3
5
 b2 
 c2 
2
2
2
2
2
2  3   a  3b  5c  3 
4
2
2
2
4
a 2  b2  c 2
2
2
.…(i)
2
  
2  3 
2  5  
a 
  b 
  c 

2 2
 2 
 2 
 2   11 
H 
4
a 2  b2  c2
2

a 2 9b 2 25c 2 11 2 2



4
4
4
4
.....(ii)
a 2  b2  c 2  1
.…(iii)
1
1
1
Solving a 2  , b 2  , c 2 
4
2
4
a2
1
  
P

 a2 

2
2
2
4
 2  a b c
Q55.
A quantum particle of mass m is moving on a horizontal circular path of radius a . The
particle is prepared in a quantum state described by the wavefunction

4
cos 2  ,
3
 being the azimuthal angle. If a measurement of the z -component of orbital angular
momentum of die particle is carried out, the possible outcomes and the corresponding
probabilities are
1
1
1
(a) Lz  0,  , 2 with 0 P  0   , P      and P  2  
5
5
5
(b) Lz  0 with P  0   1
1
1
and P     
3
3
2
1
(d) Lz  0, 2 with P  0   and P  2  
3
6
(c) Lz  0,  with P  0  
Ans. : (d)
Solution:  
4
4
cos 2  
3
3
4 1  2
2  exp 2i  exp 2i  
 1  cos 2 
. 



  
2
3 2  2
2
2



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
2
1
1
0 
2 
2
3
6
6
2
1
and P  2  
3
6
Consider two canonically conjugate operators X̂ and Ŷ such that  Xˆ , Yˆ   iI , where I
is identity operator. If Xˆ  11Qˆ1  12Qˆ 2 , Yˆ   21Qˆ1   22Qˆ 2 , where  ij are complex
Lz  0, 2 with P  0  
Q56.
numbers and Qˆ1 , Qˆ 2   zI , the value of 11 22  12 21 is
i
(c) i
(b)
(a) iz
z
Ans. : (b)
(d) z
Solution:  Xˆ , Yˆ   iI , 11Qˆ1  12 Qˆ 2 ,  21Qˆ1   22 Qˆ 2   i I
 11Qˆ1 ,  22Qˆ 2   12Qˆ 2 ,  21Qˆ1   11 22 Qˆ1 , Qˆ 2   12 21 Qˆ 2 , Qˆ1 
11 22  12 21  zI  iI  11 22  12 21  
Q57.
i
z
Suppose the spin degree of freedom of two particles (nonzero rest mass and nonzero spin)
is described completely by a Hilbert space of dimension twenty one. Which of the
following could be the spin of one of the particles?
(a) 2
(b)
3
2
(c) 1
(d)
1
2
Ans. : (c)
Solution:  2 s1  1   2 s2  1  21  7  3  s1  3, s2  1
Q58.
The normalized eigenfunctions and eigenvalues of the Hamiltonian of a Particle confined
to move between 0  x  a in one dimension are
n 2 2  2
2
n x
 n  x 
and En 
sin
2ma 2
a
a
respectively. Here 1, 2,3... . Suppose the state of the particle is
x 
  x 
 1  cos 

 a 
 a 
  x   A sin 
where A is the normalization constant. If the energy of the particle is measured, the
x
 2 2
probability to get the result as
is
. What is the value of x ?
2
2ma
100
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Ans. : 80
a  2
x 
  x 
x  2 2
x 
  x 
1
cos
A
sin
sin
cos
Solution:   x   A sin 













2  a
 a 
 a 
 a  2 a
 a 
 a 
  x 
 
a  2
x  2 2
x 
  x 
A
sin 
sin 

 cos 

2  a
 a  2 a
 a 
 a 
a 
1

A  1  2 
2 
2

  1
 
a 2  1
a 5
8
A 1    A2   1  A 
2  4
2 4
5a
a 8 
1
4
1

.
1  2  
1 
2

2 5a 
2
5
5

  22  4
x
4
P
 
 x   100  80
2 
5
 2ma  5 100
Q59.
A harmonic oscillator has the following Hamiltonian
pˆ 2 1
 m 2 xˆ 2
2m 2
It is perturbed with a potential V   xˆ 4 . Some of the matrix elements of x̂ 2 in terms of
H0 
its expectation value in the ground state are given as follows:
0 xˆ 2 0  C
0 xˆ 2 2  2C
1 xˆ 2 1  3C
1 xˆ 2 3  6C
where
n
is
the
normalized
eigenstate
of
H0
corresponding
to
the
1

eigenvalue En    n   . Suppose E0 and E1 denote the energy correction of
2

O    to thee ground state and the first excited state, respectively. What is the fraction
E1
?
E0
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Ans. : 5
Solution: For nth state En  n X 4 n 
E0  0 X 4 0 
2
6n 2  6n  3
2 2
4m 


2
3 2
15 2
4
2




1
X
1
6.1
6.1
3
4m 2 2
4m 2 2
4m 2 2


E1
5
E0
Q60.
Consider a wavepacket defined by

  x    dkf  k  exp i  kx  

Further, f  k   0 for k 
K
K
and f  k   a for k  . Then, the form of normalized
2
2
  x  is
8 K
Kx
sin
x
2
(a)
(b)
8 K
Kx
cos
x
2
(c)
(d)
2
K
2
K
Kx
2
x
sin
Kx
2
x
sin
Ans. : (b)

Solution: Given   x    dkf  k  eikx

  x  
K /2
K / 2
q
 eikx
ix
dK a eiKx
K /2
K / 2
2
x
  x   sin
K
K
2
K
K
2
K
i x
q i x
 e 2 e 2
ix

f K   0
a
kx
2
22 Kx
dx  1
 x 2 2
A2 
K 

K
2
K
2
h 2 Kx / 2
1

x2
4 A2 

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4 A2
x
1
2
A2 
1
 A
2 K
1
2 K
2
1
Kx
 sin
x 2 K
2
  x 
   x 
2
K
Kx
2
x
sin
JEST-2019
Q61.
What is the binding energy of an electron in the ground state of a He  ion?
(a) 6.8eV
(b) 13.6 eV
(c) 27.2 eV
(d) 54.4 eV
Ans. : (d)
Solution: E  
13.6 2
z  eV 
n2
He  : z  2
E 
13.6  4
 eV 
n2
The binding energy of an electron in ground state is
E
Q62.
13.6  4
1
2
 eV   54.4 eV
 b2 x 2 
The wave function   x   A exp  
 (for real constants A and b ) is a normalized
2 

eigen-function of the Schrodinger equation for a particle of mass m and energy E in a
one dimensional potential V  x  such that V  x   0 at x  0 . Which of the following is
correct?
(a) V 
 2b 4 x 2
m
(b) V 
 2b 4 x 2
2m
(c) E 
 2b 2
4m
(d) E 
 2b 2
m
Ans. : (b)
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1/ 4
 m x 2 
 m 
Solution: Comparing with harmonic oscillator   x   
 the potential is
 exp  
2 
  

1

V  x   m 2 x 2 and energy is E 
2
2
2
2 2
 b x 
b4 2 x 2
b2 2
b 


so V  x  
and energy E 

  x   A exp  

2m
2
2m
m
2 

Q63.
A quantum particle of mass m is in a one dimensional potential of the form
1
2 2
 m x , if x  0
V  x   2
if x  0


where  is a constant. Which one of the following represents the possible ground state
wave function of the particle?

1
(a)
0

0
x

(b)
0
1

(c)
0
x

0
x

1

1
0
1
0
x

(d)
1
0
1
1
Ans. : (b)
Q64.
1
particle placed in a magnetic field B , the Hamiltonian is
2
H   BS y   S y , where S y is the y -component of the spin operator. The state of the
For a spin
system at time t  0 is   t  0    , where S z   

measured then what is the probability to get a value  ?
2
(a) cos 2 t 
(b) sin 2 t 
(c) 0

 . At a later time t , if S z
2
 t 
(d) sin 2  
 2 
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Ans. : (d)
Solution: H   BS y   S y Eigen value is
and 2 
   
1
,
with eigen vector 1 
    
2
2
2
1
     respectively.
2
  t  0     I   1 1   2 2  
 t  t  
1
1
1 
2
2
2
1
 it  1
 it 
1 exp 
2 exp  


2
2
 2 
 2 
If S z is measured on   t  then probability to find 

is
2
2
2
  t 
1
 
 it 
 it  
2 t
P  
  exp 
  exp  
   sin
4
2
 2   t   t 
 2 
 2 
Q65.
Consider a quantum particle in a one-dimensional box of length L . The coordinates of
the leftmost wall of the box is at x  0 and that of the rightmost wall is at x  L . The
particle is in the ground state at t  0 . At t  0 , we suddenly change the length of the box
to 3L by moving the right wall. What is the probability that the particle is in the ground
state of the new system immediately after the change?
(a) 0.36
(b)
9
8
(c)
81
64 2
(d)
0.5

L
Ans. : (c)
 2
x
sin
. 0  x  3a

Solution: 1   3a
3a
 0,
otherwise

  22 
 
 1
P
2
 2m  3a  



2
a

0
x
2
sin
3a
3a
 2
x
sin
. 0 xa

  a
a
 0,
otherwise

x
2
81
sin
dx 
a
a
64 2
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Q66.
Consider a quantum particle of mass m and a charge e moving in a two dimensional
potential given as:
V  x, y  
k
2
2
 x  y  k  x  y
2

The particle is also subject to an external electric field E   iˆ  ˆj , where  is a


constant iˆ and ĵ corresponds to unit vectors along x and y directions, respectively. Let
E1 and E0 be the energies of the first excited state and ground state, respectively. What
is the value of E1  E0 ?
(a) 
2k
m
2k
 e 2
m
(b) 
(c) 3
2k
m
(d) 3
2k
 e 2
m
Ans. : (a)
Solution: For constant electric field we know there is not any change in frequency and energy of
each level is changed by constant value.
The total potential is
V  x, y  
k
3
3
2
2
 x  y   k  x  y    x   y  V  x, y   kx 2  ky 2  kxy   x   y
2
2
2
m 0 
T 

 0 m
 3k k 
and V  

 k 3k 
Secular equation is given by

V   2 m  0  3k   2 m

2
 k 2  0  x 
4k
2k
,y 
m
m
1
1


The equivalent quantum mechanical energy is Enx ,n y   nx    x   n y    y  V0
2
2


Where nx  0,1, 2,3... and n y  0,1, 2,3...
The ground state energy E0  E0.0 
 4k  2k

2 m 2 m
The first excited state energy E1  E0.1 
E1  E0  
 4k 3 2k

2 m
2 m
2k
m
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Q67.
A one-dimensional harmonic oscillator is in the state

 
n 0
1
n
n!
1

where n is the normalized energy eigenstate with eigenvalue  n    . Let the
2

expectation value of the Hamiltonian in the state  be expressed as
1
  . What is
2
the value of  ?
Ans. : 3
1

 n    1
 n
2
1 
Solution: H   
        e    3.2
2
n
n 0
n 1 n
2 

Q68.
Consider a system of 15 non-interacting spin-polarized electrons. They are trapped in a
two dimensional isotropic harmonic oscillator potential V  x, y  
1
m 2  x 2  y 2  . The
2
angular frequency  is such that   1 in some chosen unit. What is the ground state
energy of the system in the same units?
Ans. : 55
Solution: Non-interacting spin-polarized electrons means direction of spin is fixed
1   2  2  3  3  4  4  5  5  55
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