4. Quantum Mechanics Quantum_Mechanics_NET-JRF June 2011-Dec 2014

7/23/2019 4. Quantum Mechanics Quantum_Mechanics_NET-JRF June 2011-Dec 2014

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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES

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Head office

fiziks, H.No. 23, G.F, Jia Sarai,

Near IIT, Hauz Khas, New Delhi-16

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Branch office

Anand Institute of Mathematics,

28-B/6, Jia Sarai, Near IIT

Hauz Khas, New Delhi-16

QUANTUM MECHANICS SOLUTIONS

NET/JRF (JUNE-2011)

Q1. The wavefunction of a particle is given by

102

1

i , where 0 and 1 are the

normalized eigenfunctions with energies 0E and 1E corresponding to the ground state

and first excited state, respectively. The expectation value of the Hamiltonian in the state

is

(a) 10

2E

E (b) 1

0

2E

E (c)

3

2 10 EE (d)3

2 10 EE

Ans: (d)

Solution: 0 11

2i and

3

2 10 EEHH

Q2. The energy levels of the non-relativistic electron in a hydrogen atom (i.e. in a Coulomb

potential rrV /1 ) are given by 2/1 nEnlm , where n is the principal quantum

number, and the corresponding wave functions are given by nlm , where lis the orbital

angular momentum quantum number and mis the magnetic quantum number. The spin of

the electron is not considered. Which of the following is a correct statement?

(a) There are exactly (2l+ 1) different wave functions nlm , for eachEnlm.

(b) There are l(l+ 1) different wave functions nlm , for eachEnlm.

(c)Enlmdoes not depend on land mfor the Coulomb potential.

(d) There is a unique wave function nlm for eachEnlm.

Ans: (c)

Q3. The Hamiltonian of an electron in a constant magnetic field B

is given by BH

.

where is a positive constant and 321 ,,

denotes the Pauli matrices. Let

/B andIbe the 2 2 unit matrix. Then the operator /tHie simplifies to

(a)2

sin2

cos t

B

BitI

(b) t

B

BitI

sincos

(c) tB

BitI

cossin

(d) tB

BitI

2cos2sin
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Ans: (b)

Solution: H B

where 1 2 3, ,

are pauli spin matrices and B

are constant magnetic

field. 1 2 3 , ,i j k

, x y zB B i B j B k

and Hamiltonion BH

in matrices

form is given byz x y

x y z

B B iB

B iB B

. Eigenvalue of given matrices are given by

B and B . Hmatrices are not diagonals so /tHie is equivalent to

1 0

0

i Bt

i Bt

eS S

e

where S is unitary matrices and 1

1 1

2 2

1 1

2 2

S S

.

1

1 1 1 1

0 02 2 2 2

1 1 1 10 0

2 2 2 2

i Bt i Bt

i Bt i Bt

e eS S

e e

where /B .

/ cos sin

sin cos

iHt t i t

ei t t

which is equivalent to cos sinxI t i t can be written

as tB

BitI

sincos

where xi B

B

Q4. If the perturbation H' = ax, where a is a constant, is added to the infinite square well

potential

otherwise.0for0 x

xV

The correction to the ground state energy, to first order in a, is

(a)2

a (b) a (c)

4

a (d)

2

a

Ans: (a)

Solution: dxHE

0

0*0

10 ' dx

xx

a

0

2sin2

2

a

xsin

20 .
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Q5. A particle in one dimension moves under the influence of a potential 6axxV , where a

is a real constant. For large nthe quantized energy levelEndepends on nas:

(a)En~ n

3

(b)En~ n

4/3

(c)En~ n

6/5

(d)En~ n

3/2

Ans: (d)

Solution: 6axxV , 62

2ax

m

pH

x , 62

2ax

m

pE

x and 21

62 axEmpx .

According to W.K.B approximation nhpdx

1/ 2

62 m E ax dx n

We can find this integration without solving the integration

62

2ax

m

pE

x 1/2

62

aE

x

mE

px 61

a

Ex at 0xp .

Area of Ellipse = semi major axis semiminor axis

1

6E2mE n

a

2

3

nE .

Q6. (A) In a system consisting of two spin2

1 particles labeled 1 and 2, let 11

2

S and

22

2 S denote the corresponding spin operators. Here zyx ,,

and

zyx ,, are the three Pauli matrices.

In the standard basis the matrices for the operators 21

yx SS and 21

xy SS are respectively,

(a)

1001

4,

1001

4

22

(b)

ii

ii

00

4,

00

4

22

(c)

000000000

000

4,

000000000

000

4

22

ii

i

i

ii

i

i

(d)

0100

1000000

000

4,

000

0000001

0010

4

22

i

i

i

i

Ans: (c)

x aE/

6/1

mE2Px

6/1

aE/

mE2


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Solution: 2

1 2

x y

0 1 0 iS S

1 0 i 04

2

0 0 0 i

0 0 i 0

0 i 0 04

i 0 0 0

2

1 2 0 0 1

0 1 04

y x

iS S

i

2

0 0 0 i

0 0 i 0

0 i 0 04

i 0 0 0

(B)These two operators satisfy the relation

(a) 212121 , zzxyyx SSSSSS (b) 0, 2121 xyyx SSSS

(c) 212121 , zzxyyx SiSSSSS (d) 0, 2121 xyyx SSSS

Ans: (d)

Solution: We have matrix 1 2

x yS S and

1 2

y xS S from question 6(A) so commutation is given by

0, 2121 xyyx SSSS .

NET/JRF (DEC-2011)

Q7. The energy of the first excited quantum state of a particle in the two-dimensional

potential 222 42

1, yxmyxV is

(a) 2 (b) 3 (c) 2

3 (d)

2

5

Ans: (d)

Solution: 222 42

1, yxmyxV 2 2 2 2

1 1m x m4 y

2 2 , 2

2

1

2

1

yx nnE

For ground state energy 0,0 yx nn

2

2

1

2 E

2

3

First exited state energy 0,1 yx nn

2

3

2

5


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Q8. Consider a particle in a one dimensional potential that satisfies xVxV . Let 0

and 1 denote the ground and the first excited states, respectively, and let

1100 be a normalized state with 0 and 1 being real constants. The

expectation value x of the position operatorxin the state is given by

(a) 112

100

2

0 xx (b) 011010 xx

(c) 212

0 (d) 102

Ans: (b)

Solution: Since xVxV so potential is symmetric.

000 x , 011 x

11001100 x 0 1 0 1 1 0 x x

Q9. The perturbation 4' bxH , where b is a constant, is added to the one dimensional

harmonic oscillator potential 222

1xmxV . Which of the following denotes the

correction to the ground state energy to first order in b?

[Hint: The normalized ground state wave function of the one dimensional harmonic

oscillator potential is

2/

4/1

0

2xmem

. You may use the following

integral

2

12

1

2 2

nadxexn

axn ].

(a)22

2

4

3

m

b (b)

22

2

2

3

m

b (c)

22

2

2

3

m

b (d)

22

2

4

15

m

b

Ans: (a)

Solution: 4' bxH , 222

1xmxV .

Correction in ground state is given by 0010 ' HE where

2

4/1

0

2xm

em

.


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1

0E

dxbx 04*

0

dxexbm

xm

2

42

1

dxxmexbm

222

2

1

It is given in the equation22 1/ 2 1

2

n n nx e dx n

Thus 2n and

m

dxxmexbmE

222

2

1

1

0

1 12

2 2 12

2

m mb

1 5

2 21

0

5

2

m m

E b 22

2

4

3

m

b

.

Q10. Let 0 and 1 denote the normalized eigenstates corresponding to the ground and first

excited states of a one dimensional harmonic oscillator. The uncertainty p in the

state 102

1 , is

(a) 2/mp (b) 2/mp

(c) mp (d) mp 2

Ans: (c)

Solution: 1

0 12

, 2

mp i a a

1 1 1 2 22

a and 0102

1a

02

mp i a a ,

22 22 1

2

mp a a N

22 22 1

2

mp a a N 12

2 N

m m 12 1

2 2

m

22 ppp = m .


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Q11. The wave function of a particle at time t= 0 is given by 212

10 uu , where

1u and 2u are the normalized eigenstates with eigenvalues E1 and E2

respectively, 12 EE . The shortest time after which t will become orthogonal to

0 is

(a) 122 EE

(b)

12 EE

(c)12

2

EE

(d)12

2

EE

Ans: (b)

Solution:

1 2

1 2 1 2

1 10 0

2 2

iE t iE t

u u u e u e

t is orthogonal to 0 0 0t 02

1

2

1 21

tiEtiE

ee

021

tiEtiE

ee tiEtiE

ee21

112

EEi

e

2 12 1

cos cosE E t

tE E

Q12. A constant perturbation as shown in the figure below acts on a particle of mass m

confined in an infinite potential well between 0 andL.

the first-order correction to the ground state energy of the particle is

(a)2

0V (b)4

3 0V (c)4

0V (d)2

3 0V

Ans: (b)

20

V 0V

2/L L0
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Solution: 111

1 pVE =2

2 200

0

2

2 2sin sin

2

L

L

L

V x xdx V dx

L L L L

21 0 01

0

2

21 2 1 21 cos 1 cos

2 2

L

L

L

V Vx xE dx dx

L L L L

1 0 01

V 2VL LE L

2L 2 2L 2

=4

3

4

2

4

000 VVV

NET/JRF (JUNE-2012)

Q13. The component along an arbitrary direction n , with direction cosines zyx nnn ,, , of the

spin of a spin2

1 particle is measured. The result is

(a) 0 (b)zn

2

(c) zyx nnn

2

(d)

2

Ans: (d)

Solution:

01

10

2

xS ,

0

0

2 i

iSy

,

10

01

2

zS

x y z n n i n j n k

and 1222 zyx nnn , kSjSiSS zyx

2

20

xnSn

02

20

i

i

ny +

2

0

02

zn

22

22

zyx

yxz

ninn

innn

Sn

Let is eigen value of Sn
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0

22

22

zyx

yxz

ninn

innn

2

2 2z zx y

n nn n 0

2 2 4

044

222

222

yx

z nnn

.

04

22222

zyx nnn

2

.

Q14. A particle of mass m is in a cubic box of size a. The potential inside the box

azayax 0,0,0 is zero and infinite outside. If the particle is in an

eigenstate of energy2

2

214

maE , its wavefunction is

(a)a

z

a

y

a

x

a

6sin

5sin

3sin

2 2/3

(b)a

z

a

y

a

x

a

3sin

4sin

7sin

2 2/3

(c)a

z

a

y

a

x

a

2sin

8sin

4sin

2 2/3

(d)

a

z

a

y

a

x

a

3sin

2sinsin

2 2/3

Ans: (d)

Solution: 222

222,,

2mannnE zyxnnn zyx 2

22

214

ma

14222 zyx nnn 3,2,1 zyx nnn .

Q15. Letlnlm

denote the eigenfunctims of a Hamiltonian for a spherically symmetric

potential rV . The wavefunction 211121210 1054

1 is an eigenfunction

only of

(a)H,L2andLz (b)HandLz (c)HandL

2 (d)L

2andLz

Ans: (c)

Solution: nEH

22 1 llL and mLz .


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Q16. The commentator 22 ,px is

(a) xpi2 (b) )(2 pxxpi (c) pxi2 (d) )(2 pxxpi

Ans: (b)

Solution: 2 2 2 2, , , , , , ,x p x x p x p x xp x p x x p p p x p x x p px

pxixippixixppx 22 , pxxpi 2 .

Q17. A free particle described by a plane wave and moving in the positive z-direction

undergoes scattering by a potential

Rrif

RrifVrV

0

0

If V0is changed to 2V0, keepingRfixed, then the differential scattering cross-section, in

the Born approximation.

(a) increases to four times the original value

(b) increases to twice the original value

(c) decreases to half the original value

(d) decreases to one fourth the original value

Ans: (a)

Solution: Rr

RrVrV

,0,0

Low energy scattering amplitude 302 3

4

2, RV

mf

and differential scattering is

given by

23

2 01

2

2

3

mV Rdf

d

Now 02VrV for Rr

23

0 02

2

2 2 24

3

m V R mVd

d

23

02

2

2 2

3

m V Rd

d

2

3

0

3

24

RmV14

d

d


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Q18. A variational calculation is done with the normalized trial wavefunction

222/54

15xa

ax for the one-dimensional potential well

axif

axifxV

0

The ground state energy is estimated to be

(a)2

2

3

5

ma

(b)

2

2

2

3

ma

(c)

2

2

5

3

ma

(d)

2

2

4

5

ma

Ans: (d)

Solution: 22

2

5

4

15

xaa

x , 0xV , ax and V x , ax

a

a

dxHE where2

22

2 xmH

2 2

2 2 2 2

5 / 2 2 5 / 2

15 15

4 2 4

a

a

dE a x a x dx

a m dx a

a

a

dxxama

2216

15 222

5

a2

2 2

5

a

15 2E a x dx

16a 2m

=

3

4

16

15 32

5

a

ma

2

2

4

5

ma

Q19. A particle in one-dimension is in the potential

l

lxif

xifV

xif

xV

0

0

0

0

If there is at least one bound state, the minimum depth of potential is

(a)2

22

8ml

(b)

2

22

2ml

(c)

2

222

ml

(d)

2

22

ml

Ans: (a)
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Solution: For bound state 00 EV

Wave function in region I, I 0 , II A sin kx Bcos kx , x

III ce

Where 2

02

EVmk ,

2

2m E

.

Use Boundary condition at 0x and x l (wave function is continuous and differential

at 0x andx l ) one will get

cotk kl cotkl kl l cot where l , kl .

22 2 0

2

2mV l

For one bound state2

2

2/1

2

20

lmV

2 2

0 28V

ml .

Q20. Which of the following is a self-adjoint operator in the spherical polar coordinate

system ,,r ?

(a)

2sin

i (b)

i (c)

sin

i (d)

sini

Ans: (c)

Solution: sin

i

is Hermitian.

NET/JRF (DEC-2012)

Q21. Let v, p and Edenote the speed, the magnitude of the momentum, and the energy of a

free particle of rest mass m. Then

(a) dp

dE constant (b)p= mv

(c)222

cmp

cpv

(d)E= mc2

Ans: (c)

lo0V

o2

2

3


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Solution: Hamiltonian H of Harmonic oscillator, 222

2

1

2xm

m

pH x

Eigenvalue of this,

2

1

nEn

But here, axxmm

pH x 22

2

2

1

2

2 2 22 2

2 2 4 2

1 2

2 2 2

xp ax a aH m xm m m m

2

22

2

22

22

1

2

m

a

m

axm

m

pH x

Energy eigenvalue,2

2

22

1

m

anEn

Q24. If a particle is represented by the normalized wave function

otherwise0

for4

152/5

22

axaa

xa

x

the uncertainty p in its momentum is

(a) a5/2 (b) a2/5 (c) a/10 (d) a2/5

Ans: (d)

Solution:22 ppp and

x

i

p

dxxaxa

ia

xaa

a

22

2/52/5

22

4

15

4

15

a

a

dxxxaia

216

15 225

a

a

dxxxaa

ih

nfuncodd

32

516

152

= 0

a

a

dxxax

xaa

p 222

222

5

22

16

15


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a

a

dxxaa

22

5

2 216

15

a

a

xxa

a

3

216

15 325

2

3222

1615

33

5

2 aaa

31122

1615 32 a

32

415

2

2

a

2

22

2

5

ap

Now,22 ppp 0

2

52

2

a

a2

5

Q25. Given the usual canonical commutation relations, the commutator BA, of

xy ypxpiA and yz zpypB is(a) zpxp xz (b) zpxp xz

(c) zpxp xz (d) zpxp xz

Ans: (c)

Solution: yzxy zpypiypixpBA ,,

yxyyzxzy zpypizpxpiypypiypxpiBA ,,,,,

yxzyyxzy zpypiypxpizpypiypxpiBA ,,,00,,

xyyxyzzy pzpyizppiypypxiyppixBA ,,,,,

xyzyxyzy pzpyiyppixpzpyiyppixBA ,,,00,,

xz pizipiixBA ,

zpxpBA xz ,

Q26. Consider a system of three spins S1, S2and S3each of which can take values +1 and -1.

The energy of the system is given by 133221 SSSSSSJE where J is a positive

constant. The minimum energy and the corresponding number of spin configuration are,

respectively,

(a) and 1J (b) 3 and 1J (c) 3 and 2J (d) 6 and 2J

Ans: (c)


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Solution: If we take 1 2 3 1S S S i.e.321 SSS

Then energy, 1 1 1 1 1 1 3E J J

Again 1 2 3 1S S S , then

Energy 3E J

So, minimum energy is 3J and there are two spin configuration.

If we take321 SSS

Then we get Maximum energy E J .

Q27. The energies in the ground state and first excited state of a particle of mass

2

1m in a

potential xV are 4 and 1 , respectively, (in units in which 1 ). If the

corresponding wavefunctions are related by ,sinh01 xxx then the ground state

eigenfunction is

(a) hxx sec0 (b) hxx sec0

(c) xhx 20 sec (d) xhx 3

0 sec

Ans: (c)

Solution: Given that ground state energy 0 4E , first excited state energy 1 1E and 0 1,

are corresponding wave functions.

Solving Schrdinger equation (use2

1m and 1 )

22

00 0 022

V Em x

2

00 02

4Vx

..(1)

22

11 1 12

2

V E

m x

2

11 12

1V

x

..(2)

Put 1 0sinhx in equation (2) one will get

2

0 00 0 02

.sinh 2 cosh sinh sinh sinhx x x V x xx x
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2

0 00 0 02

2 cothx Vx x

2

0 00 0 02

2 cothV xx x

using relation

2

00 02

4Vx

00 0 04 2 cothx

x

0

0

2tanhd

xdx

20 sec h x .

NET/JRF (JUNE-2013)

Q28. In a basis in which the z- component zS of the spin is diagonal, an electron is in a spin

state 1 / 62 / 3i

. The probabilities that a measurement of zS will yield the values

2/ and 2/ are, respectively,

(a) 1/ 2 and 1/ 2 (b) 2 / 3and 1/ 3 (c) 1/ 4 and 3/ 4 (d) 1/ 3 and 2 / 3

Ans: (d)

Solution: Eigen state of zS is 11

0

and 2

0

1

corresponds to Eigen value

2

and

2

respectively.

2

1

2P

21 2 1

6 36

i ,

2

2 2

2 3P

Q29. Consider the normalized state of a particle in a one-dimensional harmonic oscillator:

1 20 1b b

where 0 and 1 denote the ground and first excited states respectively, and 1b and 2b

are real constants. The expectation value of the displacement x in the state will be a

minimum when

(a) 1,0 12 bb (b) 122

1bb (c) 12

2

1bb (d) 12 bb

Ans: (d)


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Solution: 1021100 212

2

2

1 xbbxbxbx

Since 000 x and 011 x 102 21 xbbx .

Min of x means min 212 bb . We know that 122

21 bb .

2 2 2

1 2 1 2min 0 1x b b b b x 101

2

21 xbb 101 2

21 xbb

for min value of 2

1 21 b b there must be maximum of

2

1 2b b so 1 2b b

none of answer is matched but if we consider about magnitude then option (d) is correct .

Q30. The un-normalized wavefunction of a particle in a spherically symmetric potential is

given by

r zf r

where rf is a function of the radial variable r. The eigenvalue of the operator

2L

(namely the square of the orbital angular momentum) is

(a) 4/2 (b) 2/2 (c)2

(d) 22

Ans: (d)

Solution: rzfr rfr cos

0

1 ,r Y , 2 2 0

1 ,L r L Y where 1l

2222 21111 llL

Q31. If nlm denotes the eigenfunciton of the Hamiltonian with a potential rVV then the

expectation value of the operator 22 yx LL in the state

121210211 1535

1

is

(a) 25/39 2 (b) 25/13 2 (c) 22 (d) 25/26 2

Ans: (d)

Solution: 2222 zyx LLLL 2222

zyx LLLL 22

zLL


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22

zLL =

2222 125

150

25

11

25

92

22

zLL 22

25

24

2 22

25

26

25

2450

Q32. Consider a two-dimensional infinite square well

otherwise

0,00,

ayaxyxV

Its normalized Eigenfunctions are

a

yn

a

xn

ayx

yxnn yx

sinsin

2,,

where ,3,2,1, yx nn ..If a perturbation

otherwise0

20,

20

' 0

ay

axV

H

is applied, then the correction to the energy of the first excited state to order 0V is

(a)4

0V (b)

2

0

9

641

4

V

(c)

2

0

9

161

4

V (d)

2

0

9

321

4

V

Ans: (b)

Solution: For first excited state, which is doubly degree

1

2 2sin sin

x y

a a a

, 2

2 2sin sin

x y

a a a

11 1 1H H 2/

0

2/

0

22

0

2sin

2sin

2 a ady

a

y

adx

a

x

aV

42

1

2

1 00

VV

12 1 2H H 2/

0

2/

00 sin

2sin

22sinsin

2 a ady

a

y

a

y

adx

a

x

a

x

aV

20012 9

16

3

4

3

4

VVH

, 21 2 1 0 216

9H H V

and0

22 2 2 4

V

H H

.

Thus 0

49

169

16

4

0

2

0

2

00

VV

VV

09

16

4

2

2

0

2

0

VV
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2

0

2

00

9

641

49

16

4

VVV

Q33. The bound on the ground state energy of the Hamiltonian with an attractive delta-

function potential, namely

xadx

d

mH

2

22

2

using the variational principle with the trial wavefunction 2exp bxAx is

1:Note

0

adtte at

(a)

224/

ma (b)22

2/ ma

(c)22

/ ma (d) 22 5/ ma

Ans: (c)

Solution: For given wavefunctionm

bT

2

2

and

baV

2

ba

m

bE

2

2

2

For variation of parameter 0db

Ed 2 12

2 10

2 2

d Ea b

db m

2

4

8mb

.

2

2

min

maE .

Q34. If the operators A and B satisfy the commutation relation IBA , , where I is the

identity operator, then

(a)AA eBe , (b) AeBe BA ,,

(c) AeBe BA ,, (d) IBeA ,

Ans: (a)

Solution: IBA , and2

1 .......1 2

A A Ae

BeA ,2

1 .......,1 2

A AB

=

2 3, ,1, , .........

2 3

A B A BB A B


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BeA ,

...........!3

,,

!2

,,0

22

ABABAAABABAA

I

AA

e

A

ABe ........!21,

2

where IBA , , ABA 2,2

and23

3, ABA .

Q35. Two identical bosons of mass m are placed in a one-dimensional potential

.2

1 22xmxV The bosons interact via a weak potential,

4/exp 221012 xxmVV

where 1x and 2x denote coordinates of the particles. Given that the ground state

wavefunction of the harmonic oscillator is 2

4

1

0

2xm

em

x

. The ground state

energy of the two-boson system, to the first order in 0V , is

(a) 02V (b)

0V

(c)

1

2

0 12

V

(d)

10V

Ans: (c)

Solution: There is two bosons trapped in harmonic oscillator so energy for ground state without

perturbation is 2.2

.

If perturbation is introduced we have to calculate 1,2V where

4/exp 221012 xxmVV .

But calculating 1,2V on state 2 2

1 2

1

22 2

0

m x m xm

x e e

is very tedious task.

So lets use a trick i.e perturbation is nothing but approximation used in Taylor series. So

just expand 4/exp 221012 xxmVV and take average value of first term
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4/exp 221012 xxmVV =

2

1 2

0 (1 ...)4

x xV

2 2

1 2 1 212 0 2 .(1 ...)

4m x x x xV V

1,2V = 2 21 2 1 2

0

2 .(1 ...)

4

m x x x xV

=

( 0)2 2(1 )...

4o

mm mV

12V (1 )4

oV

1

2

0 12

V

.

So

1

2

0 1 2E V

.

NET/JRF (DEC-2013)

Q36. A spin -2

1particle is in the state

3

1

11

1 i in the eigenbasis of 2S and zS . If we

measure zS , the probabilities of getting2

h and

2

h , respectively are

(a)21 and

21 (b)

112 and

119 (c) 0 and 1 (d)

111 and

113

Ans: (b)

Solution: 2

1110

32 11

iP

11

22

2

1 1

2

11 901

32 1111

iP

i.e. probability of zS getting2

and

2


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Q37. The motion of a particle of mass m in one dimension is described by the

Hamiltonian xxmm

pH 22

2

2

1

2. What is the difference between the (quantized)

energies of the first two levels? (In the following, x is the expectation value of x in the

ground state.)

(a) x (b) x (c)2

2

2

m (d)

Ans: (d)

Solution:2

2 21

2 2

pH m x x

m 2 2

1

2V x m x x

2 2 21 2

2V x m x x

m

42

2

42

2

2

22 22

1

mmmxxm

2

22

2

2

22

1

mmxmxV

2

2

1

2 2nE n

m

1 0

3 1

2 2E E

Q38. Let nlm denote the eigenfunctions of a Hamiltonian for a spherically symmetric

potential rV . The expectation value of zL in the state

211121210200 201056

1 is

(a) 18

5 (b)

6

5 (c) (d)

18

5

Ans: (d)

Solution: zz LL =1 5 10 20

0 0 ( 1 ) (1 )36 36 36 36

=10 5

36 18 1

Q39. If 4exp xAx is the eigenfunction of a one dimensional Hamiltonian with

eiggenvalue 0E , the potential xV (in units where 12 m ) is

(a) 212x (b) 616x (c) 26 1216 xx (d) 26 1216 xx


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Ans: (d)

Solution: Schrodinger equation

02 V (where 2 1m and 0E )

0442

2

xx VAeAex

4 434 0x xe x Ve

x

0434 444 332 xxx Veexxex 01612 444 62 xxx Veexex 26 1216 xxV

Q40. A particle is in the ground state of an infinite square well potential given by,

otherwise

for0 axaxV

The probability to find the particle in the interval between2

a and

2

ais

(a)2

1 (b)

1

2

1 (c)

1

2

1 (d)

1

Ans: (b)

Solution: The probability to find the particle in the interval between2

a and

2

ais

dxax

ax

aa

a

a 2

cos2

cos22

222/

2/

2/

2/

22/

2/ 2

2cos1211

2cos1

a

a

a

a

dxax

adx

ax

a

2/

2/

sin2

1 a

aa

xax

a

11222

1

aaa

a

1

2

12

2

1 aa

a

Q41. The expectation value of the x - component of the orbital angular momentum xL in the

state 1,1,20,1,21,1,2 11535

1

(where nlm are the eigenfunctions in usual notation), is

(a) 31125

10

(b) 0 (c) 311

25

10

(d) 2


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Ans: (a)

Solution: , 1 1 , 1L l m l l m m l m and , 1 1 , 1L l m l l m m l m

2x

L LL 2

xL LL

210 211

13 2 5 2

5L

1 1 1.3 10 110 10(3 11)

25 25 25L

21 1 210

12 5 2 11

5L

1 1.3 10 10 1125 25

L

2x

L LL

=1

10(3 11)25

1 1.3 10 10 11

25 25xL = 311

25

10

Q42. A particle is prepared in a simultaneous eigenstate of 2L and zL . If 21 l and m are

respectively the eigenvalues of 2L andz

L , then the expectation value 2x

L of the particle

in this state satisfies

(a) 02 xL (b)2220 xL

(c) 22 10

2xL

(d)

22 2 1

2 2xL

Ans: (d)

Solution: 2 2 2 21

1

2

xL l l m

For max value 0m and for min m l

2

1

2

22

2

ll

Ll

x


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CBA ,, are Non zero Hermitian operator.

CAbABBAABCBA 0,

but 0C if BAAB ie CBA , false (2)

NET/JRF (JUNE-2014)

Q43. Consider a system of two non-interacting identical fermions, each of mass m in an

infinite square well potential of width a . (Take the potential inside the well to be zero

and ignore spin). The composite wavefunction for the system with total energy

2

22

25

maE is

(a)

a

x

a

x

a

x

a

x

a

2121 sin2

sin2

sinsin2

(b)

a

x

a

x

a

x

a

x

a

2121 sin2

sin2

sinsin2

(c)

a

x

a

x

a

x

a

x

a

2121 sin

2

3sin

2

3sinsin

2

(d)

a

x

a

x

a

x

a

x

a

2221 cossincossin2

Ans: (a)

Solution: Fermions have antisymmetric wave function

1 2 1 21 22 22

sin sin sin sinx x x x

x xa a a a a

2

22

25

maEn

1 2

1, 2x xn n


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Q44. A particle of mass m in the potential 222 42

1, yxmyxV , is in an eigenstate of

energy 2

5

E . The corresponding un-normalized eigen function is

(a)

2222

exp yxm

y

(b)

2222

exp yxm

x

(c)

222

exp yxm

y

(d)

222

exp yxm

xy

Ans: (a)

Solution: 222 42

1, yxmyxV ,

2

5E

2 2 2 21 1, 2

2 2V x y m x m y

Now, yyxxn nnE

2

1

2

1

2

12

2

1yx nn

32

2n x yE n n

5

2

nE when 0xn and 1yn .

Q45. A particle of mass m in three dimensions is in the potential

ar

arrV

0

Its ground state energy is

(a)2

22

2ma

(b)

2

22

ma

(c)

2

22

2

3

ma

(d)

2

22

2

9

ma

Ans: (a)

Solution:

22

2 2

1

2 2

d u r l l V r u r Eu r

m dr mr

2

2 2

2, 0, 0

d u r mEKu r K l V r

dr


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sin cosu r A Kr B Kr

Using boundary condition, 0,B

2 2

2sin , , 0 sin 0 1

2u r A Kr r a u r Ka Ka n E n

ma

Q46. Given that

rr

ipr1

, the uncertainty rp in the ground state.

0/3

0

0

1 are

ar

of the hydrogen atom is

(a)0a

(b)

0

2a

(c)

02a

(d)

0

2a

Ans: (a)

Solution:1

,rp ir r

0/3

0

0

1 are

ar

22

r r rP P P

Now

0

2

3

0

//

3

0

411 0

0 drra

e

rrieaP

arar

r

0

2/

0

//

3

0

000114

drrera

eea

i ararar

0 0

/22/2

0

3

0

0014

drredrreaa

i arar

3 230 0 0 0

4 1 2 ! 1 !

2 / 2 /

i

a a a a

2 2

0 0

3

0

40 0

4 4 r

a aiP

a
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drrerrr

ea

P ararr2

0

/

2

22/

3

0

2 421

00

0 0 02 / / / 2

3 2

0 0 00

4 1 2 1r a r a r ae e e r dra a r a

0 0

22 / 2 /2

3 2

0 0 00 0

4 1 2r a r ar e dr re dr a a a

2

3 23 2

0 0 00 0

4 1 2 ! 2 1 !

2 / 2 /a a aa a

3 22 20 0 0 0

3 2 3

00 0 0

2 !4 2 4

8 4 4 2

a a a a

aa a a

2

0

2

0

3

0

2

4

4

a

a

a

22

2 2

0 00r rP P P a a

Q47. The ground state eigenfunction for the potential xxV where x is the delta

function, is given by xAex , where A and 0 are constants. If a perturbation

2bxH is applied, the first order correction to the energy of the ground state will be

(a)22

b (b)

2

b (c)

2

2

b (d)

22

b

Ans: (d)

Solution: ,V x x xAex

1 xx e

dxebxeHE xx

21111

dxbxe x 22

2 2xb e x dx

0

2 2 2 2

0

x xb x e dx x e dx

0

222 dxexb x

dxbxe x 22

233 28!2

22

!22

bbb
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Q48. An electron is in the ground state of a hydrogen atom. The probability that it is within the

Bohr radius is approximately equal to

(a) 0.60 (b) 0.90 (c) 0.16 (d) 0.32

Ans: (d)

Solution: Probability:0

0

2

/ 2

30 0

14

a

r ae r dr

a

0

02 /2

2

0

4 a

r a

o

r e dr a

0

0

2

/ 2

30 0

14

a

r ae r dr

a

0 0

0 02 / 2 /2 0 0 0

3

0 0 0

42

2 2 2

a a

r a r aa a ar e r e

a

0

02 / 0 0 0

0

22 2 2

a

r a a a ae

0

0 0 0 0 0

2 2 3 32 / 2 /2 00 0 0 0

0 03

0

42 2

2 4 4 8

a

a a a a aa a a aa e a e e e

a

3 3 3 3

0 0 0 0

3 2 2 2

0

4 1 1

2 2 44

a a a a

a e e e

2

5 14

44e

21

5 1e

5 0.137 1 0.685 1 0.32

Q49. A particle in the infinite square well

otherwise

00 axxV

is prepared in a state with the wavefunction

otherwise0

0sin 3 axa

xA

x

The expectation value of the energy of the particle is

(a)2

22

2

5

ma

(b)

2

22

2

9

ma

(c)

2

22

10

9

ma

(d)

2

22

2ma

Ans: (c)
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Solution:

otherwise

00 axxV

,

otherwise0

0sin3 axa

xA

x

3 33 1 3

sin sin sin sin 3 3sin 4sin4 4

x x xx A A A A A A

a a a

3 1 3sin sin

4 4

x xA A

a a

2 2 33sin sin

4 2 2

A a x a x

a a a a

1 334 2 2

A a ax x x

2 21 9 132 32a aA A 210 32132 10a A A a

1 31 32 32

3.4 2 10 2 10

a ax x x

a a

xxx 3110

1

10

3

Now,2

22

22

22

12

9,

2 maE

maE

n nE a P a

Probably

2

1

1

9,

10P E

2

2

2

1

10P E

2 2 2 2

2 2

9 1 9

10 2 10 2E

ma ma

2 2

2

9

10E

ma


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* 0 0T dx 1 21 1

02 2

iE iE T T

e e

1 2 2 1

1iE iE T

T T i E E

e e e

Equate real part 2 1cos 1T

E E

1

2 1 2 1cos 1T E E E E

Q52. Consider the normalized wavefunction

113102111 aaa

where lm is a simultaneous normalized eigenfunction of the angular momentum

operators 2L and zL , with eigenvalues 21ll and m respectively. If is an

eigenfunction of the operator xL with eigenvalue , then

(a)2

1,21

231 aaa (b)2

1,21

231 aaa

(c)2

1,

2

1231 aaa (d)

3

1321 aaa

Ans: (b)

Solution:xL

2

L L

For L , 1 11 2 10 3 1 1L a a a 1 12 2 11 3 100 2 2a a a

2 11 3 102 2a a

For L , 1 11 2 10 3 1 1L a a a 1 10 2 1 12 2a a

Given2

L L

2 11 1 3 10 2 1 11

2 2 22 2

L La a a a

2

L L

1 11 2 10 3 1 1a a a (Given)

Thus 2 1 2 122

aa a a


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1 3

22

a aa

1 3 21 32

2 2

a a aa a

22 2 21 2 12

aa a

1 3

1

2a a , 2

1

2a

Q53. Let x and p denote, respectively, the coordinate and momentum operators satisfying the

canonical commutation relation ipx , in natural units 1 . Then the commutator

ppex , is

(a) pepi 1 (b) pepi 21 (c) pei 1 (d) pipe

Ans: (a)

Solution: ,x p i

, px pe 2 3

, , ,1 ....2 3

p p p p px p e p x e ie p x p

2

,1 , , ....2

p pie p x x p x

22 30 ......

2 3

p ip ipie p i

, px pe 3

2 ..... 12

p p p ppie i p p ie ipe i p e

Q54. Let 321 ,,

, where 321 ,, are the Pauli matrices. If a

and b

are two

arbitrary constant vectors in three dimensions, the commutator

ba , is equal to (in

the following I is the identity matrix)

(a) 321 ba

(b)

bai2

(c) Iba

(d) Iba

Ans: (b)

Solution:1 2 3

a a i a j a k

,1 2 3

b b i b j b k

, x y zi j k
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,a b

1 2 3 1 2 3,x y z x y za a a b b b

,a b

1 1 1 2 1 3, , ,x x x y x za b a b a b 2 1 2 2, ,y x y ya b a b

2 3 ,y za b 3 1 3 2 3 3, , ,z x z y z za b a b a b

1 1 1 2 1 3 2 1 2 3 3 1 3 20 2 2 2 0 2 2 2 0z y z x y xa b a b i ia b a b i a b i a b i a b i

,a b 2i a b

Q55. The ground state energy of the attractive delta function potential

xbxV ,

where 0b , is calculated with the variational trial function

otherwise,,0

,for,2

cos axaa

xA

x

is

(a)22

2

mb (b)

22

22

mb (c)

22

2

2

mb (d)

22

2

4

mb

Ans: (b)

Solution: ; 0V x b x b and cos ;2

xx A a x aa

Normalized2

cos2 2

x

a a

2 2 2 2*

2 22 8

a

aT dx

m x ma

* 2

2

a

a

bV b x dx b

a a

2 2

28

bE

ama

2 2

3 2

20

8

E b

a ma a

2 2 2 2

04 4

b ama ma


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Put the value of a in equation:2 2

28

bE

ma a

22 2 2

2 2 22 22 2

4 4 2

8

mb b mb mb

m

Q56. Let 10 10 cc (where 0c and 1c are constants with 12120 cc ) be a linear

combination of the wavefunctions of the ground and first excited states of the one-

dimensional harmonic oscillator. For what value of 0c is the expectation value x a

maximum?

(a)2

1, 0 c

mx

(b)

2

1,

2 0 c

mx

(c) 2

1

,2 0 cmx

(d) 2

1

, 0 cmx

Ans: (c)

Solution:0 10 1c c

X X

2 22 2

0 1 0 1 0 1 0 12 0 1 0 1 1 0 1X c c X c c c c X c c X

For max 2 20 1 0 1 1X c c c c 01

2c

1 12 0 1

2 2X X 0 1X

0 12

a am

2X

m

Q57. Consider a particle of mass m in the potential 0, axaxV . The energy eigen-

values ....,2,1,0nEn , in the WKB approximation, are

(a)

3/1

2

1

24

3

nm

a (b)

3/2

2

1

24

3

nm

a

(c)

2

1

24

3n

m

a (d)

3/4

2

1

24

3

nm

a
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Ans: (b)

Solution: , 0V x a x x

According to W.K.B., 21

12

pdq n

where 1 and 2 are positive mid point

2

2

PE a x

m 2P m E a x

/

/

12

2

E

Em E a x dx n

0 /

/ 0

12 2

2

E

Em E ax dx m E ax dx n

/

0

12 2

2

E

m E ax dx n

2m E ax t At 0, 2 ; / , 0x t mE x E a t

2madx dt

21/ 2

0

12

2

mE

ma t dt n

2

0

2 12

3 2

mE

ma t n

23/2

0

4 1

3 2

mE

ma t n

3/ 24 1

23 2ma mE n

3/ 2 5/ 2 3/ 24 123 2

am E n

2/ 3

3 1

24 2

aE n

m

Q58. The Hamiltonian 0H for a three-state quantum system is given by the matrix

200

020

001

0H . When perturbed by

010

101

010

H where 1 , the resulting shift

in the energy eigenvalue 20 E is

(a) 2, (b) 2, (c) (d) 2

Ans: None of the answer is correct.

/E /E


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Solution: 0 0

1 0 0 0 1 0

0 2 0 , 1 0 1

0 0 2 0 1 0

H H

2 0

0 2

in 0H is not 00 1

1 0

in H because H is not in block diagonal form. So we

must diagonalised whole H . The Eigen value at 0 00, 2 , 2H .

After diagonalisation0

0 0 0

0 2 0 , 0

0 0 2

H

is correction for Eigenvalue at 0H .

So 02 is the correction for eigenvalue of 0 2H None of the answer is correct.

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4. Quantum Mechanics Quantum_Mechanics_NET-JRF June 2011-Dec 2014