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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 1
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 2
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 3
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 4
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 5
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 6
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 7
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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Website:www.physicsbyfiziks.com
Email: [email protected] 8
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 9
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 10
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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Email: [email protected] 11
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziksInstitute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES
Website:www.physicsbyfiziks.com
Email: [email protected] 12
Head office
fiziks, H.No. 23, G.F, Jia Sarai,
Near IIT, Hauz Khas, New Delhi-16
Phone: 011-26865455/+91-9871145498
Branch office
Anand Institute of Mathematics,
28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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Head office
fiziks, H.No. 23, G.F, Jia Sarai,
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Phone: 011-26865455/+91-9871145498
Branch office
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28-B/6, Jia Sarai, Near IIT
Hauz Khas, New Delhi-16
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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fiziks, H.No. 23, G.F, Jia Sarai,
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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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fiziks, H.No. 23, G.F, Jia Sarai,
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Phone: 011-26865455/+91-9871145498
Branch office
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28-B/6, Jia Sarai, Near IIT
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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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Branch office
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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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Branch office
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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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fiziks, H.No. 23, G.F, Jia Sarai,
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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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Branch office
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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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Branch office
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28-B/6, Jia Sarai, Near IIT
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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.