4. Quantum Mechanics NET-JRF June 2011 - June 2014

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QUANTUM MECHANICS SOLUTIONS

NET/JRF (JUNE-2011)

Q1. The wavefunction of a particle is given by

+= 1021 i , where 0 and 1 are the

normalized eigenfunctions with energies E0 and E1 corresponding to the ground state and

first excited state, respectively. The expectation value of the Hamiltonian in the state is (a) 1

0

2EE + (b) 102 E

E (c) 3

2 10 EE (d) 3

2 10 EE + Ans: (d)

0 11 i2

= + 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 l is the orbital angular momentum quantum number and m is 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 each Enlm. (b) There are l(l + 1) different wave functions nlm , for each Enlm. (c) Enlm does not depend on l and m for the Coulomb potential.

(d) There is a unique wave function nlm for each Enlm. Ans: (c)





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

is given by BHGG = .

where is a positive constant and ( )321 ,, =K denotes the Pauli matrices. Let =/B = and I be the 2 2 unit matrix. Then the operator =/tHie simplifies to

(a) 2

sin2

cos tB

BitI KG + (b) t

BBitI sincosKG +

(c) tB

BitI cossinKG + (d) t

BBitI 2cos2sinKG +

Ans: (b)

H .B= GG where ( )321 ,, =K are pauli spin matrices and BG are constant magnetic field. ( )1 2 3i j k = + +K , BG = ( )x y zB i B j B k+ + and Hamiltonion BH GG = in matrices form is given by z x y

x y z

B B iBB iB B

+ . Eigen value of given matrices are given by

B+ and B . H matrices are not diagonals so =/tHie is equivalent

to

i Bt

1i Bt

e 0S S

0 e

=

= where S is unitary matrices and 1

1 12 2S S

1 12 2

= = .

i Bt

1i Bt

e 0S S

0 e

=

==

i Bt

i Bt

1 1 1 1e 02 2 2 2

1 1 1 10 e

2 2 2 2

=

=

where =/B = .

=/tHie =cos t i sin ti sin t cos t

which is equivalent to xI cos t i sin t + can be written

as tB

BitI sincosKG + where x i BB

=KG





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

potential

( ) = otherwise.0for0 xxV The correction to the ground state energy, to first order in a, is

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

4a (d)

2a

Ans: (a)

dxHE = 0

0*0

10 ' dx

xxa = 02sin2

2a=

xsin20 = .

Q5. A particle in one dimension moves under the influence of a potential ( ) 6axxV = , where a is a real constant. For large n the quantized energy level En depends on n as:

(a) En ~ n3 (b) En ~ n4/3 (c) En ~ n6/5 (d) En ~ n3/2

Ans: (d)

( ) 6axxV = , 622

axm

pH x += , 62

2ax

mpE x += and ( )[ ]2162 axEmpx = .

According to W.K.B approximation nhpdx ( )( ) hdxaxEm 2/162

We can find this integration without solving the integration

62

2ax

mpE x += 1

/2

62

=+aE

xmEpx

61

=aEx at 0=xp .

Area of Ellipse = semi major axis semiminor axis

16E2mE n

a =

23

nE .

x( )aE /6/1

mE2Px

6/1

( )aE /

mE2





Q6. (A) In a system consisting of two spin 21 particles labeled 1 and 2, let ( ) ( )11

2G=G =S and

( ) ( )222G=G =S denote the corresponding spin operators. Here ( )zyx ,,G 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)

10 014,1001

4

22 == (b)

iiii 0 04,00

4

22 ==

(c)

000000000

000

4,

000000000

000

4

22

ii

ii

ii

ii == (d)

01001000000000

4,

000000

00010010

4

22 ii

ii

==

Ans: (c)

( ) ( ) 21 2x y

0 1 0 iS S

1 0 i 04 =

= 20 0 0 i0 0 i 00 i 0 04i 0 0 0

=

( ) ( )

=

0110

00

4

221

ii

SS xy= 2

0 0 0 i0 0 i 00 i 0 04i 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)

We have matrix ( ) ( )1 2x yS S and ( ) ( )1 2y 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 ( ) ( )222 421, yxmyxV += is

(a) =2 (b) =3 (c) =23 (d) =

25

Ans: (d)

( ) ( )222 421, yxmyxV += 2 2 2 21 1m x m4 y

2 2= + , == 2

21

21

++

+= yx nnE

For ground state energy 0,0 == yx nn == 221

2+= E

23 ==

First exited state energy 0,1 == yx nn == + 23

25 ==

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 operator x in the state is given by (a) 11

2100

20 xx + (b) [ ]011010 xx +

(c) 2120 + (d) 102

Ans: (b)

Since ( ) ( )xVxV = so potential is symmetric. 000 = x , 011 = x

( ) ( )11001100 ++=x [ ]011010 +=





Q9. The perturbation 4' bxH = , where b is a constant, is added to the one dimensional harmonic oscillator potential ( ) 22

21 xmxV = . 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

02xmem

= . You may use the following

integral

+=212

12 2 nadxex

naxn ].

(a) 222

43

mb= (b) 22

2

23

mb= (c) 22

2

23

mb= (d) 22

2

415

mb=

Ans: (a)

4' bxH = , ( ) 2221 xmxV = .

Correction in ground state is given by 0010 ' HE = where ==2

4/1

0

2xm

em

= .

10E

= dxbx 04*0 =

= dxexbm

xm=

=2

421

( ) dxxmexbm ==

2222

1

It is given in the equation

+=

212/12 2 ndxex nnn Thus 2=n and =

m=

( ) dxxmexbmE ==2

2221

10

=

+

=

212

212

21

==

mmb

252

521

10

= ==

mmbE

22

2

43

mb== .





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)

( )1 0 12

= + , ( )mp i a a2 += = ( )2211

21 +=+ a and ( )010

21 +=a

( ) 02

== + aamip = , ( )( )122

222 ++= + Naamp =

[ ]122

222 ++= + Naamp = 122

+= Nm = m 1 3.12 2 2 = + = =m=

22 ppp = = =m .

Q11. The wave function of a particle at time t = 0 is given by ( ) ( )21210 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

2EE = (d)

12

2EE =

Ans: (b)

( ) ( )1 210 u u2 = + ( )1 2iE t iE t

1 21t u e u e2

= + = =





( )t is orthogonal to ( )0 ( ) ( )0 t 0 = 021

21 21 =+

==

tiEtiE

ee

021

=+==

tiEtiE

ee ==tiEtiE

ee21

=( )

112

==

EEie

( )2 1E E tcos cos = = 2 1t E 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 and L.

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)

1111 pVE = = + L

l

L

dxLx

LVdx

Lx

LV

2

20

22

0

0 sin2sin22

+ = dxLx

LVdx

Lx

LVE

L

2cos12122cos1

212

0

0011

1 0 01

V 2VL LE L2L 2 2L 2

= + = 43

42

4000 VVV =+

20V 0V

2/L L0





NET/JRF (JUNE-2012)

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

21 particle is measured. The result is

(a) 0 (b) zn2= (c) ( )zyx nnn ++ 2= (d) 2=

Ans: (d)

=

0110

2=

xS ,

=

00

2 ii

S y= ,

= 1001

2=

zS

x y z n n i n j n k= + +G and 1222 =++ zyx nnn , kSjSiSS zyx ++=

=

2

20

=

=GG

xnSn

+

02

20

=

=

i

i

ny +

2

0

02

=

=zn

( )( )

+

=

22

22==

==GG

zyx

yxz

ninn

innnSn

Let is eigen value of Sn GG

( )

( ) 022

22 =+

==

==

zyx

yxz

ninn

innn

( )2 2 2z z x yn n n n 02 2 4 + + = = = = ( ) 044 222

222

=+

yxz 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





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

undergoes scattering by a potential

( )

>=

RrifRrifV

rV0

0

If V0 is changed to 2V0, keeping R fixed, 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)

( )

RrRrVrV

>==

,0,0

Low energy scattering amplitude ( ) 302 34

2, RVmf == and differential scattering is

given by23

2 012

2mV Rd fd 3

= = =

Now ( ) 02VrV = for Rr < ( )23

022

2m 2V Rdd 3

= =

= 2

30

32

4 =RmV 1d4

d=

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

( ) ( )222/54 15 xaax = for the one-dimensional potential well ( )

>=

axif

axifxV

0

The ground state energy is estimated to be

(a) 22

35ma= (b) 2

2

23ma= (c) 2

2

53ma= (d) 2

2

45ma=





Ans: (d)

( ) ( )2225

4

15 xaa

x = , ( ) 0=xV , ax and ( )V x = , ax >

=a

a

dxHE where 222

2 xmH

= =

( ) ( )

=a

a

dxxaadx

dm

xaa

E 222/5222

222/5 4

1524

15 = ( )( )

=a

a

dxxama

2216

15 2225

=

( )a2 2 25a

15 2E a x dx16a 2m

= = = 34161532

5

ama=

2

2

45ma==

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

( ) llxifxifV

xifxV

>





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)

sin=i

Ans: (c)

isin = is Hermitian.

NET/JRF (DEC-2012)

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

free particle of rest mass m. Then

(a) =dpdE constant (b) p = mv

(c) 222 cmp

cpv += (d) E = mc2

Ans :( c)

2

2

0

1cv

vmmvp

==

2

2

2202

1cvvmp

= 2

22222

0 cvppvm = 22

220

2 pcpmv =

+

2

2220

22

cpcm

pv += 220

2 cmppcv+

=

Q22. The wave function of a state of the Hydrogen atom is given by,

121210121200 232 +++= where nlm is the normalized eigen function of the state with quantum numbers n, l, m in the usual notation. The expectation value of Lz in the state is (a)

615= (b)

611= (c)

83= (d)

8=





Ans: (d)

Firstly normalize , 121210121200 162

163

161

161

+++=

( )169

1610 +==P

1610= .

Probability of getting (i) i.e. ( )164==iP and ( )

162= =iP .

Now, z

z

LL = ( )

1621

1641

16100 ++= === ==

162

164 = =

162=

8==

Q23. The energy eigenvalues of a particle in the potential ( ) axxmxV = 2221 are

(a) 22

221

manEn

+= = (b) 2

2

221

manEn +

+= =

(c) 22

21

manEn

+= = (d) =

+=

21nEn

Ans: (a)

Hamiltonian (H) of Harmonic oscillator, 222

21

2xm

mpH x +=

Eigenvalue of this, =

+=21nEn

But here, axxmm

pH x += 222

21

2

2 2 22 2

2 2 4 2

1 22 2 2

xp ax a aH m xm m m m

= + +

2

22

22

2

221

2 ma

maxm

mpH x

+=

Energy eigenvalue, 22

221

manEn

+= =





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

( ) ( )





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)

[ ] ( ) ( )[ ]yzxy zpypiypixpBA += ,,[ ] [ ] [ ] [ ] [ ]yxyyzxzy zpypizpxpiypypiypxpiBA ,,,,, +=[ ] [ ] [ ] [ ] [ ]yxzyyxzy zpypiypxpizpypiypxpiBA ,,,00,, =+=[ ] [ ] [ ] [ ] [ ] xyyxyzzy pzpyizppiypypxiyppixBA ,,,,, += [ ] [ ] [ ] [ ] [ ] xyzyxyzy pzpyiyppixpzpyiyppixBA ,,,00,, =+= [ ] ( ) xz pizipiixBA = ==, [ ] ( )zpxpBA xz += =,

Q26. Consider a system of three spins S1, S2 and S3 each 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) J and 1 (b) -3J and 1 (c)-3J and 2 (d) -6J and 2

Ans :(b)

If we take S1= S2 = S3 +1 i.e. 321 SSS

Then energy, E = - J [1 1 + 1 1 + 1 1]= - 3 J

Again S1= S2 = S3 -1, then Energy (E) = - 3 J

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

If we take 321 SSS

Then we get Maximum energy E = J.





Q27. The energies in the ground state and first excited state of a particle of mass 21=m 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 30 sec=

Ans : (c)

Given that ground state energy 0 4E = , first excited state energy 1 1E = and 0 1, are corresponding wave functions.

Solving Schrdinger equation (use 21=m and 1== )

220

0 0 022V E

m x + =

= 2

00 02 4Vx

+ = ..(1) 22

11 1 122

V Em x

+ ==

21

1 12 1Vx + = ..(2)

Put 1 0 sinh x = in equation (2) one will get 2

0 00 0 02 .sinh 2 cosh sinh sinh sinhx x x V x xx x

+ + + = 2

0 00 0 02 2 coth x Vx x

+ + + =

2

0 00 0 02 2 cothV xx x

+ = using relation 2

00 02 4Vx

+ = 0

0 0 04 2 coth xx =

0

0

2 tanhd xdx = 2

0 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 ( )

+=/

3/26/1 i . The probabilities that a measurement of zS will yield the values

2/= and 2/= are, respectively, (a) 1/2 and 1/2 (b) 2/3 and 1/3 (c) 1/4 and 3/4 (d) 1/3 and 2/3

Ans: (d)

Eigen state of zS is 110

= and 201

= corresponds to Eigen value 2= and

2=

respectively.

21

2Q

P =

=31

62

61

61

61 2 ==+=+= i ,

32

61

22 ==

QP

Q29. Consider the normalized state / of a particle in a one-dimensional harmonic oscillator: 10 21 bb +=/ 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) 12 21 bb = (c) 12 2

1 bb = (d) 12 bb =

Ans: (d)

1021100 2122

21 xbbxbxbx ++=

Since 000 =x and 011 =x 102 21 xbbx = . Min of x means min 212 bb . We know that 1

22

21 =+ bb .

( )[ ] 102221221min xbbbbx ++= ( )[ ] 101221 xbb += ( )[ ] 101 221 xbb 1 2b b =





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

given by

( ) ( )rzfr =/ G where ( )rf is a function of the radial variable r . The eigenvalue of the operator

2LG

(namely the square of the orbital angular momentum) is

(a) 4/2= (b) 2/2= (c) 2= (d) 22= Ans: (d)

( ) ( )rzfr = ( )rfr cos= ( )( )01 ,r Y = , ( ) ( )2 2 01 ,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 15351 //+/=/ is

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

2222zyx LLLL =+ 2222 zyx LLLL =+ 22 zLL =

22 zLL =

++ 2222 125150

2511

2592 ====

22zLL 22 25

242 == = 222526

252450 == ==





Q32. Consider a two-dimensional infinite square well

( )





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

function potential, namely

( )xadxd

mH = 2

22

2=

using the variational principle with the trial wavefunction ( ) ( )2exp bxAx = is ( )

+= 1:Note

0

adtte at

(a) 22 4/ =ma (b) 22 2/ =ma (c) 22 / =ma (d) 22 5/ =ma

Ans: (c)

For given wavefunction mbT

2

2== and baV 2=

bambE 2

2

2

= =

For variation of parameter 0=db

Ed0

212

22

12 == bamdt

Ed

=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) [ ] IBe A =,

Ans: (a)

[ ] IBA =, and 21 .......1 2

A A Ae = + + +

[ ]=Be A , 21 .......,1 2A A B

+ + + =[ ] [ ]

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

2 3A B A B

B A B + + +





[ ]=Be A , [ ] [ ] [ ] [ ] ...........!3

,,!2

,,022

++++++ ABABAAABABAAI

[ ] AA eAABe =+++= ........!2

1,2

where [ ] IBA =, , [ ] ABA 2,2 = and [ ] 23 3, ABA = .

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

( ) .21 22 xmxV = 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 ( ) == 241

0

2xm

emx

=/ . The ground state

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

(a) 02V+= (b) + 0V=

(c) 12

0 1 2V

+ + = (d)

++ 10V=

Ans: (c)

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 21 2

12

2 20

m x m xmx 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 ( )[ ]=4/exp 221012 xxmVV = = ( )21 20 (1 ...)4x xV +=

( )2 21 2 1 212 0

2 .(1 ...)

4

x x x xV V

+ = +=

1,2V = ( )2 21 2 1 20 2 .(1 ...)4x x x x

V + += =

( 0)2 2(1 )...

4om mV

+

= =

=

12V (1 )4oV =

12

0 1 2V

+ .

So 1

2

0 1 2E V

= + + = .

NET/JRF (DEC-2013)

Q36. A spin -21 particle is in the state

+=

31

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

measure zS , the probabilities of getting 2h+ and

2h , respectively are

(1) 21 and

21 (2)

112 and

119 (3) 0 and 1 (4)

111 and

113

Ans: (2) ( )2

31

10111

2

+=

ibP

1122

21 == 1=

( )119

31

01111

2

2

=

+=

ibP

i.e. probability of zS getting

2b and

2b





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

Hamiltonian xxmm

pH ++= 222

21

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.)

(1) x = (2) x += (3) 22

2 m

+= (4) =

Ans: (4) 2

2 212 2pH m x xm

= + + ( ) 2 212

V x m x x = +

( ) 2 2 21 22V x m x xm = +

++= 42

2

42

2

222 2

21

mmmxxm

( ) 222

22

221

mmxmxV

+=

2

2

12 2n

E nm

= + = 1 03 12 2

E 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 2010561 +++= is (1) =

185 (2) =

65 (3) = (4) =

185

Ans: (4)

zz LL = = 1 5 10 200 0 ( 1 ) (1 )36 36 36 36 + + += = = = =10 536 18

== = 1=

Q39. If ( ) ( )4exp xAx = is the eigenfunction of a one dimensional Hamiltonian with eiggenvalue 0=E , the potential ( )xV (in units where 12 == m= ) is (1) 212x (2) 616x (3) 26 1216 xx + (4) 26 1216 xx





Ans: (4)

Schrodinger equation

02 =+ V (where 2 1m= == and 0=E )

( ) 04422 =+ xx VAeAex 4 434 0x xe x Vex + = ( ){ }[ ] 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,

( )

=otherwisefor 0 axa

xV

The probability to find the particle in the interval between 2a and

2a is

(1) 21 (2)

121 + (3)

121 (4)

1

Ans: (2)

The probability to find the particle in the interval between 2a and

2a is

dxax

ax

aa

a

a 2cos

2cos

22

222/

2/

=

+==

2/

2/

22/

2/ 22cos1

211

2cos1

a

a

a

a

dxax

adx

ax

a

2/

2/

sin21 a

aaxax

a

+= ( )

+++= 11222

1aaa

a

+=

+= 1

212

21 aaa

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

state [ ]1,1,20,1,21,1,2 115351 + += (where nlm are the eigenfunctions in usual notation), is

(1) ( )3112510 = (2) 0 (3) ( )311

2510 += (4) 2=





Ans: (1)

( ) ( ), 1 1 , 1L l m l l m m l m = + = and ( ) ( ), 1 1 , 1L l m l l m m l m+ = + + +=

2xL LL + +=

2xL L

L + + =

210 21 11 2 2 5 25

L + = + = = 1 1 1.2 10 2 10 4 1025 25 25

L + = + == = =

21 1 2101 2 5 2 115

L = = = 1 1.2 10 10 1125 25

L = = =

2xL L

L + += = 1 1 14 10 .2 10 10 11

50 50 50L + = + = = =

1 1.3 10 10 1125 25

L + = = == ( )3112510 = Q42. A particle is prepared in a simultaneous eigenstate of 2L and zL . If ( ) 21 =A +l and =m are

respectively the eigenvalues of 2L and zL , then the expectation value 2xL of the particle

in this state satisfies

(1) 02 =xL (2) 2220 =A xL

(3) ( ) 22 10

2xL

+ A A = (4) ( ) 22 2 12 2x

L+ A A =A=

Ans: (4)

( )( )2 2 2 21 12xL l l m= + = = For max value 0m = and for min m l=

( )21

2

22

2 == + llLl x





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)

ax

ax

ax

ax

a2121 sin2sin2sinsin2

(b)

+

ax

ax

ax

ax

a2121 sin2sin2sinsin2

(c)

ax

ax

ax

ax

a2121 sin

23sin

23sinsin2

(d)

ax

ax

ax

ax

a2221 cossincossin2

Ans: (a)

Fermions have antisymmetric wave function

( ) 1 2 1 21 2 2 22 sin sin sin sinx x x xx x a a a a a =

2

22

25

maEn

= =1 2

1, 2x xn n = =





Q44. A particle of mass m in the potential ( ) ( )222 421, yxmyxV += , is in an eigenstate of

energy =25=E . The corresponding un-normalized eigen function is

(a) ( ) + 2222exp yxmy = (b) ( ) + 2222exp yxmx = (c) ( ) + 222exp yxmy = (d) ( ) + 222exp yxmxy =

Ans: (a)

( ) ( )222 421, yxmyxV += , =

25=E

( ) ( )2 2 2 21 1, 22 2

V x y m x m y = +

Now, yyxxn nnE ==

++

+=21

21 ==

++

+=

212

21

yx nn

322n x y

E n n = + + =

52n

E = = when 0=xn and 1=yn . Q45. A particle of mass in three dimensions is in the potential

( )

>





( ) ( ) ( )2 2 22 , 0, 0d u r mEKu r K l V rdr = = = = = ( ) sin cosu r A Kr B Kr= +

Using boundary condition, 0,B =

( ) ( ) 2 22sin , , 0 sin 0 12u r A Kr r a u r Ka Ka n E nma= = = = = = ==

Q46. Given that

+=

rripr

1 = , the uncertainty rp in the ground state.

( ) 0/30

01 area

r =/

of the hydrogen atom is

(a) 0a= (b)

0

2a= (c)

02a= (d)

0

2a=

Ans: (a)

1 ,rp i r r = + =

( ) 0/30

01 area

r =

22r r rP P P =

Now

+=

0

2

30

//

30

4110

0 drra

err

iea

Par

arr =

+

=

0

2/

0

//30

000114 drrera

eeai ararar

=

+= 0 0

/22/2

030

0014 drredrreaa

i arar =

( ) ( )3 230 0 0 0

4 1 2 ! 1 !2 / 2 /

ia a a a

= + =





2 20 0

30

4 0 04 4 ra ai P

a

= + = = =

drrerrr

ea

P ararr2

0

/2

22/

30

2 421 00

+

= =

0 0 0

2/ / / 2

3 20 0 00

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

= +

=

=

0 0

/2

0

/2220

30

200

214 drrea

dreraa

arar

=

( ) ( )

= 2

003

020

30

2

/2!12

/2!214

aaaaa=

3 22 20 0 0 0

3 2 30 0 0 0

4 2 ! 2 48 4 4 2a a a a

a a a a

= = = =

20

20

30

2

44

aa

a== =

=

222

20 0

0r 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)

22b (b) 2

b (c) 22

b (d) 22b

Ans: (d)

( ) ( ) ,V x x= ( ) xAex = 1= ( ) xx e =





dxebxeHE xx

== 21111 dxbxe x 22

2 2xb e x dx

=

02 2 2 2

0

x xb x e dx x e dx

= +

= 0

222 dxexb x

dxbxe x 22

( ) 233 28!22

2!22

bbb ==

=

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)

Probability: 0

0

2

/ 2

30 0

1 4a

r ae r dra

0 02 /22

0

4 a r ao

r e dra

=

0

0

2

/ 2

30 0

1 4a

r ae r dra

( )0 00 02 / 2 /2 0 0 03

0 0 0

4 22 2 2

a ar a r aa a ar e r e

a =

0

02 / 0 0 0

0

22 2 2

ar a a a ae

+

0

0 0 0 0 0

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

0 030

4 2 22 4 4 8

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

a

= +

3 3 3 30 0 0 0

3 2 2 20

4 1 12 2 44a a a a

a e e e = + 2

5 1444e

= + 215 1e

= + [ ]5 0.137 1= + [ ]0.685 1 0.32= + =

Q49. A particle in the infinite square well

( )





( )

Documents

4. Quantum Mechanics NET-JRF June 2011 - June 2014