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fiziks InstituteforNET/JRF,GATE,IITJAM,JEST,TIFRandGREinPHYSICALSCIENCES
Website:www.physicsbyfiziks.comEmail:fiziks.physics@gmail.com1
Headofficefiziks,H.No.23,G.F,JiaSarai,NearIIT,HauzKhas,NewDelhi16Phone:01126865455/+919871145498
BranchofficeAnandInstituteofMathematics,28B/6,JiaSarai,NearIITHauzKhas,NewDelhi16
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)
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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
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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
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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 .
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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 +=
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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== .
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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
= + = =
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( )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
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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== .
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Q14. A particle of mass m is in a cubic box of size a. The potential inside the box
( )azayax
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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=
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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
>
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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=
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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
+= =
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Q24. If a particle is represented by the normalized wave function
( ) ( )
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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.
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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 = .
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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 =
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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 == ==
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Q32. Consider a two-dimensional infinite square well
( )
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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 + + +
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[ ]=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.
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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
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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
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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=
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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
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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)
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 = =
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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
( )
>
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( ) ( ) ( )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
= + =
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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 =
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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
( )
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( )
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