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Brijmohan
Banaras Hindu University (BHU), Varanasi
PhD Condensed matter Physics

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fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES QUANTUM MECHANICS SOLUTIONS Q1. Ans: NET/JRF (JUNE-2011) 1 The wavefunction of a particle is given by 0 i 1 , where 0 and 1 are the 2 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 E E E 2 E1 E 2 E1 (a) 0 E1 (b) 0 E1 (c) 0 (d) 0 2 2 3 3 (d) Solution: Q2. 1 0 i 1 and 2 H H E 0 2 E1 3 The energy levels of the non-relativistic electron in a hydrogen atom (i.e. in a Coulomb potential V r 1 / r ) are given by E nlm 1 / n 2 , 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: Q3. (c) The Hamiltonian of an electron in a constant magnetic field B is given by H B . where is a positive constant and 1 , 2 , 3 denotes the Pauli matrices. Let B / and I be the 2 2 unit matrix. Then the operator e i H t / simplifies to t i B t i B sin sin t (a) I cos (b) I cos t 2 B 2 B i B i B cos t cos 2 t (c) I sin t (d) I sin 2 t B B Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 1 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (b) Solution: H B where 1 , 2 , 3 are pauli spin matrices and B are constant magnetic ,B B i , 2 j, 3k field. 1i x B y j Bz k and Hamiltonion H B in matrices Bx iBy . Eigenvalue of given matrices are given by B Bz form is given by Bx iBy B and B . H matrices i Bt e S 1 0 0 S i Bt e i Bt e S 1 0 0 S i Bt e where 1 2 1 2 z are S not diagonals e i H t / is so is unitary matrices and 1 i Bt 2 e 1 0 2 1 0 2 i Bt 1 e 2 equivalent S 1 S to 1 2 . 1 2 1 2 1 2 1 2 where B / . 1 2 cos t i sin t which is equivalent to I cos t i x sin t can be written eiHt / i sin t cos t i B i B sin t where x as I cos t B B Q4. If the perturbation H' = ax, where a is a constant, is added to the infinite square well potential 0 for V x 0 x otherwise. The correction to the ground state energy, to first order in a, is (a) Ans: a 2 (b) a (c) a 4 (d) a 2 (a) 1 * 0 H ' 0 dx Solution: E0 0 a 2 x sin 0 2 x a dx 2 0 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 2 sin x . Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 2 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q5. A particle in one dimension moves under the influence of a potential V x ax 6 , where a is a real constant. For large n the quantized energy level En depends on n as: (a) En ~ n3 Ans: (b) En ~ n4/3 (c) En ~ n6/5 (d) En ~ n3/2 (d) Solution: V x ax 6 , H 2 2 1 px px ax 6 , E ax 6 and p x 2m E ax 6 2 . 2m 2m According to W.K.B approximation pdx nh 2m E ax 6 1/ 2 dx n We can find this integration without solving the integration E 2 x 2 x 6 p x p 1 ax 6 2mE E / a 2m E x a 1 6 Px 1/ 6 at p x 0 . 1/ 6 E / a E / a x Area of Ellipse = semi major axis semiminor axis 1 6 2mE 2mE E 2mE n E n 2 . a Q6. (A) In a system consisting of two spin 3 1 particles labeled 1 and 2, let S 1 1 and 2 2 S 2 2 denote the corresponding spin operators. Here x , y , z and 2 x , y , z are the three Pauli matrices. 1 2 1 2 In the standard basis the matrices for the operators Sx S y and S y Sx are respectively, (a) 2 1 0 2 1 0 , 4 0 1 4 0 1 0 0 2 0 0 (c) 4 0 i i 0 Ans: 0 i 0 0 i 2 0 0 , 0 0 4 0 i 0 (b) 0 0 i 0 0 i i 0 0 0 0 0 2 i 0 2 i 0 , 4 0 i 4 0 i 0 2 1 (d) 4 0 0 1 0 0 0 0 0 2 , 0 0 i 4 0 i 0 0 i 0 0 i 0 0 0 0 0 0 0 1 0 1 0 (c) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 3 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 0 4 1 Solution: S x1 S y2 1 2 Sy Sx 0 i 0 4 0 0 i 1 0 0 i i i 0 0 0 0 i i 0 0 i 0 0 0 0 0 2 i 0 0 0 0 0 0 1 0 0 4 0 i i 0 0 1 0 4 i 2 0 2 0 (B) These two operators satisfy the relation (c) S S , S S iS S (d) S S , S S 0 1 2 1 2 (a) Sx S y , S y Sx Sz 1 Sz 2 1 x 2 y 1 y 2 x 1 z 1 2 1 2 (b) S x S y , S y Sx 0 2 1 x z 2 y 1 y 2 x Ans: (d) Solution: We have matrix S x1 S y2 and S y1 S x2 from question 6(A) so commutation is given by S S , S S 0 . 1 x 2 y 1 y 2 x NET/JRF (DEC-2011) Q7. The energy of the first excited quantum state of a particle in the two-dimensional potential V x, y (a) 2 Ans: 1 m 2 x 2 4 y 2 is 2 (b) 3 (c) 3 2 (d) 5 2 (d) Solution: V x, y 1 1 1 1 1 m 2 x 2 4 y 2 m 2 x 2 m4 2 y 2 , E n x n y 2 2 2 2 2 2 For ground state energy n x 0, n y 0 E First exited state energy n x 1, n y 0 1 3 2 2 2 2 3 5 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 4 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q8. Consider a particle in a one dimensional potential that satisfies V x V x . Let 0 and 1 denote the ground and the first excited states, respectively, and let 0 0 1 1 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 2 (a) 0 0 x 0 12 1 x 1 2 12 (c) 0 Ans: (b) 0 1 0 x 1 1 x 0 (d) 2 0 1 (b) Solution: Since V x V x so potential is symmetric. 0 x 0 0 , 1 x 1 0 x 0 0 1 1 0 0 1 1 0 1 0 x 1 1 x 0 Q9. The perturbation H ' bx 4 , where b is a constant, is added to the one dimensional harmonic oscillator potential V x 1 m 2 x 2 . Which of the following denotes the 2 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 integral x e 2n ax 2 m is 0 dx a n (a) Ans: 3b 2 4m 2 2 (b) 1 2 1/ 4 e m x 2 / 2 . You may use the following (d) 15b 2 4m 2 2 1 n ]. 2 3b 2 2m 2 2 (c) 3b 2 2 m 2 2 (a) Solution: H ' bx 4 , V x 1 m 2 x 2 . 2 Correction in ground state is given by E 0 H ' 0 1 0 m where 0 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com 1/ 4 e m x 2 2 . Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 5 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 m 2 4 * E 0 bx 4 0 dx b x e 1 0 m x 2 1 m 2 2 2 m x dx dx b x e 2 It is given in the equation Thus n 2 and 1 1 2 n n2 x e dx n 1/ 2 n 2 m 1 2 m 2 m m 2 1 2 2 m x E0 dx b b x e 1 m 2 m E b 1 0 Q10. 5 2 2 1 2 1 2 2 5 3 b 2 . 2 4 m 2 2 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 1 2 0 1 , is (a) p m / 2 (c) p m Ans: (b) p m / 2 (d) p 2 m (c) Solution: a p i p2 p m 1 0 1 , p i 2 a a 2 1 2 1 1 2 2 and a 1 2 0 10 m m 2 a a 0 , p2 a a 2 2 N 1 2 2 m 1 m 2 m a a2 2N 1 2N 1 2 1 m 2 2 2 2 p2 p 2 = m . Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 6 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q11. The wave function of a particle at time t = 0 is given by 0 u1 u2 and 1 2 u 1 u 2 , where are the normalized eigenstates with eigenvalues E1 and E2 respectively, E 2 E1 . The shortest time after which t will become orthogonal to 0 is (a) Ans: 2 E 2 E1 (b) E 2 E1 (c) 2 E 2 E1 2 E 2 E1 (b) 1 Solution: 0 2 u1 u2 0 iE1t iE2t 1 u e u e 1 2 2 1 2 t is orthogonal to 0 0 t 0 e e iE1t cos Q12. (d) e iE 2t 0 e iE1t e E2 E1 t cos t iE 2t e i E2 E1 iE1t 1 e 2 iE 2t 0 1 E2 E1 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. V0 V0 2 0 L L/2 the first-order correction to the ground state energy of the particle is (a) Ans: V0 2 (b) 3V0 4 (c) V0 4 (d) 3V0 2 (b) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 7 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: E 1 V p 1 1 1 L 2 V 2 x 2 x dx V0 sin 2 dx = 0 sin 2 2 L L L L L 0 L 2 L V 2 1 2V 1 2 x 2 x 1 E1 0 1 cos dx 0 1 cos dx L 0 2 L L L 2 L L 2 E1 1 V0 L 2V0 L V0 2V0 3V0 L = 2L 2 2L 2 4 4 4 NET/JRF (JUNE-2012) Q13. , with direction cosines n x , n y , n z , of the The component along an arbitrary direction n spin of a spin 1 particle is measured. The result is 2 (b) (a) 0 Ans: nz 2 (c) nx n y nz 2 (d) 2 (d) Solution: S x 0 2 1 i 1 , Sz 0 2 0 1 0 , Sy 0 2 i 0 1 and n 2 n 2 n 2 1 , S S i n nx i n y j nzk x S y j Szk x y z 0 n S nx 2 0 2 n y i 2 i 2 n 2 + z 0 0 0 2 n x in y 2 nz 2 Let is eigen value of n S nz n S 2 n x in y 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 8 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES nz n x in y 2 0 n z 2 2 n x in y 2 2 2 2 nz 2 2 nz nz 2 2 2 2 n x n y 0 nx n y 0. 2 2 4 4 4 2 2 2 nx n y n z2 2 0 . 4 2 Q14. A particle of mass m is in a cubic box of size a. The potential inside the box 0 x a,0 y a,0 z a eigenstate of energy E is zero and infinite outside. If the particle is in an 14 2 , its wavefunction is 2ma 2 2 (b) a 4 x 8 y 2 z sin sin a a a 2 (d) a 3/ 2 sin 3/ 2 2 (c) a 3 x 5 y 6 z sin sin a a a 3/ 2 sin 2 (a) a 3/ 2 sin sin 7 x 4 y 3 z sin sin a a a x a sin 2 y 3 z sin a a Ans: (d) 2 2 ny n z2 Solution: E nx ,n y ,nz n x 2 14 2 2 2ma 2 2ma 2 2 2 2 2 nx ny nz 14 n x 1, n y 2, n z 3 . Q15. Let nlml denote the eigenfunctims of a Hamiltonian for a spherically symmetric potential V r . The wavefunction 1 210 5 21 1 10 211 is an eigenfunction 4 only of (a) H, L2 and Lz Ans: (b) H and Lz (c) H and L2 (d) L2 and Lz (c) Solution: H En L2 l l 1 2 and Lz m . Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 9 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q16. The commentator x 2 , p 2 is (b) 2i ( xp px) (a) 2i xp Ans: (d) 2i ( xp px) (c) 2i px (b) 2 2 2 2 Solution: x , p x x, p x, p x xp x, p x x, p p p x, p x x, p px x Q17. 2 , p 2 xp i x i p p i x i px 2i xp px . A free particle described by a plane wave and moving in the positive z-direction undergoes scattering by a potential V V r 0 0 if r R if r R 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) Solution: V r V0 , 0, r R r R Low energy scattering amplitude f , d given by 1 f d 2 2mV0 R 3 2 3 m 4 V R 3 and differential scattering is 2 0 2 3 2 3 d 2 2m 2V0 R 2mV0 Now V r 2V0 for r R 4 2 d 3 2 3 2mV0 R 3 d 2 2m 2V0 R d 4 1 4 2 2 d 3 d 3 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 10 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q18. A variational calculation x is done with the normalized trial wavefunction 15 2 a x 2 for the one-dimensional potential well 5/ 2 4a 0 V x if x a if x a The ground state energy is estimated to be (a) Ans: 5 2 3ma 2 (b) 3 2 2ma 2 (c) 3 2 5ma 2 (d) 5 2 4ma 2 (d) Solution: x 15 4a 5 2 a 2 x2 , V x 0 , x a and V x , x a a E H dx where H a 2 2 2m x 2 a 15 2 2 d 2 15 2 15 2 2 2 a x a x dx a 2 x2 2 dx 5/ 2 2 5/ 2 5 4 a 2 m dx 4 a 16a 2m a a a 15 2 2 15 2 4a 3 5 2 2 2 = E a x dx a 16a 5 m 3 4ma2 16a 5 2m a E Q19. A particle in one-dimension is in the potential V x V0 0 if x 0 if 0 x l if x l If there is at least one bound state, the minimum depth of potential is (a) Ans: 2 2 8ml 2 (b) 2 2 2ml 2 (c) 2 2 2 ml 2 (d) 2 2 ml 2 (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 11 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: For bound state V0 E 0 Wave function in region I, I 0 , II A sin kx Bcos kx , III ce x Where k 2m E 2m V0 E , 2 2 V0 o . l Use Boundary condition at x 0 and x l (wave function is continuous and differential at x 0 and x l ) one will get k cot kl kl cot kl l cot where l , kl . 2mV0l 2 2 2 2 1/ 2 2mV0l 2 For one bound state 2 Q20. 2 V0 2 2 . 8ml 2 o 2 3 2 Which of the following is a self-adjoint operator in the spherical polar coordinate system r , , ? (a) Ans: i sin 2 (b) i (c) i sin (d) i sin (c) Solution: i is Hermitian. sin 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) dE dp constant (c) v cp Ans: p m c 2 2 (b) p = mv (d) E = mc2 2 (c) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 12 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: p mv m0 v 1 v2 c2 2 2 m0 v p 2v 2 2 2 2 m v p 0 c2 v2 1 2 c p2 2 p2 p2 2 2 v m0 c 2 p v m2c 2 p 2 v 0 c2 pc 2 Q22. 2 2 p m0 c 2 The wave function of a state of the Hydrogen atom is given by, 200 2 211 3 210 2 21 1 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) Ans: 15 6 (b) 11 6 3 8 (d) 8 (d) Solution: Firstly normalize , P 0 1 2 3 2 200 211 210 21 1 16 16 16 16 1 9 10 . 16 16 16 Probability of getting (i ) i.e. P i Now, Q23. (c) Lz 4 2 and P i . 16 16 Lz 10 4 2 4 2 2 0 1 1 16 16 16 16 16 16 8 The energy eigenvalues of a particle in the potential V x 1 m 2 x 2 ax are 2 1 a2 (a) En n 2 2m 2 1 a2 (c) En n 2 m 2 Ans: 1 a2 (b) En n 2 2 m 2 1 (d) En n 2 (a) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 13 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: Hamiltonian H of Harmonic oscillator, H 2 px 1 m 2 x 2 2m 2 1 Eigenvalue of this, E n n 2 2 2 px px 1 1 2ax a2 a2 2 2 2 2 m x ax H But here, H m x 2m 2 2m 2 m 2 m2 4 2m 2 2 p2 1 a a2 H x m 2 x 2m 2 m 2 2m 2 1 a2 Energy eigenvalue, E n n 2 2 m 2 Q24. If a particle is represented by the normalized wave function 15 a 2 x 2 x 4a 5 / 2 0 for a x a otherwise the uncertainty p in its momentum is (a) 2 / 5a Ans: (b) 5 / 2a (c) 10 / a (d) 5 / 2a (d) Solution: p p p2 p i x 2 and a a 15 a 2 x 2 2 i 15 a x 2 dx 5/ 2 5/ 2 4a 4a x 15 i a 2 x 2 2 x dx ih 2 155 5 16 a a16a a a a a 2 x x 3 dx odd funcn =0 p 2 15 16a 5 2 2 2 a x 2 a x 2 dx x a a 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 14 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES a a 2 15 x3 15 2 2 2 2 a x 2 a x dx 3 a 16a 5 16a 5 a 2 2 p2 3 2a 3 15 1 15 2 2 15 2 3 2 2 a 1 2 2 a 2 16 3 16a 5 3 3 4a 5 2 2a 2 Now, p p 2 p 2 5 2 0 2a 2 5 2a Q25. Given the usual canonical commutation relations, the commutator A, B of A i xp y yp x and B yp z zp y is (a) xp z p x z (c) xp z p x z Ans: (b) xp z p x z (d) xp z p x z (c) Solution: A, B ixp y iyp x , yp z zp y A, B i xp y , yp z i yp x , yp z i xp y , zp y i yp x , zp y A, B i xp y , yp z 0 0 i yp x , zp y i xp y , yp z i yp x , zp y A, B ix p y , yp z i x, yp z p y iy p x , zp y i y, zp y p x A, B ix p y , yp z 0 0 i y, zp y p x ix p y , yp z i y, zp y p x A, B ix i p z izi p x A, B xp z p x z 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 E J S1 S 2 S 2 S 3 S 3 S1 where J is a positive constant. The minimum energy and the corresponding number of spin configuration are, respectively, (a) J and 1 Ans: (b) 3J and 1 (c) 3J and 2 (d) 6J and 2 (c) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 15 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Solution: If we take S1 S2 S3 1 i.e. S1 S2 S3 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 S1 S2 S3 Then we get Maximum energy E J . Q27. The energies in the ground state and first excited state of a particle of mass m 1 in a 2 potential V x are 4 and 1 , respectively, (in units in which 1). If the corresponding wavefunctions are related by 1 x 0 x sinh x, then the ground state eigenfunction is (a) 0 x sec hx (c) 0 x sec h 2 x Ans: (b) 0 x sec hx (d) 0 x sec h 3 x (c) Solution: Given that ground state energy E0 4 , first excited state energy E1 1 and 0 , 1 are corresponding wave functions. Solving Schr dinger equation (use m 1 and 1) 2 2 2 0 V 0 E0 0 2m x 2 2 0 V 0 4 0 ..(1) x 2 2 2 1 V 1 E1 1 2m x 2 2 1 V 1 1 1 ..(2) x 2 Put 1 0 sinh x in equation (2) one will get 2 20 .sinh x 2 0 cosh x 0 sinh x V 0 sinh x 0 sinh x x x Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 16 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 0 2 2 0 coth x 0 V 0 0 x x 2 0 0 2 0 V 0 4 0 using relation V 2 coth x 0 0 0 2 x 2 x x 4 0 2 0 coth x 0 0 x d 0 0 2 tanh xdx 0 sec h 2 x . NET/JRF (JUNE-2013) Q28. In a basis in which the z - component S z of the spin is diagonal, an electron is in a spin 1 i / 6 state . The probabilities that a measurement of S z will yield the values 2/3 / 2 and / 2 are, respectively, (a) 1/ 2 and 1/ 2 Ans: (b) 2 / 3 and 1/ 3 (c) 1/ 4 and 3/ 4 (d) 1/ 3 and 2 / 3 (d) 1 0 Solution: Eigen state of S z is 1 and 2 corresponds to Eigen value and 2 2 0 1 respectively. 1 P 2 Q29. 2 2 2 1 i 2 1 , P 2 6 3 6 2 2 3 Consider the normalized state of a particle in a one-dimensional harmonic oscillator: b1 0 b2 1 where 0 and 1 denote the ground and first excited states respectively, and b1 and b2 are real constants. The expectation value of the displacement x in the state will be a minimum when (a) b2 0, b1 1 Ans: (b) b2 1 2 b1 (c) b2 1 b1 2 (d) b2 b1 (d) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 17 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 Solution: x b12 0 x 0 b2 1 x 1 2b1b2 0 x 1 Since 0 x 0 0 and 1 x 1 0 x 2b1b2 0 x 1 . 2 1. Min of x means min 2b1b2 . We know that b12 b2 x min 2 2 2 2 b1 b2 b12 b2 0 x 1 b1 b2 1 0 x 1 1 b1 b2 0 x 1 2 2 for min value of 1 b1 b2 there must be maximum of b1 b2 so b1 b2 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 f r is a function of the radial variable r . The eigenvalue of the operator L2 (namely the square of the orbital angular momentum) is (a) 2 / 4 (b) 2 / 2 (c) 2 (d) 2 2 Ans: (d) Solution: r zf r r cos f r r Y10 , , L2 r L2Y10 , where l 1 L2 l l 1 2 1 1 1 2 2 2 Q31. If nlm denotes the eigenfunciton of the Hamiltonian with a potential V V r then the 2 expectation value of the operator L2 x L y in the state 1 3 211 210 15 21 1 5 is (a) 39 2 / 25 Ans: (b) 13 2 / 25 (c) 2 2 (d) 26 2 / 25 (d) 2 2 2 Solution: L2 x Ly L Lz 2 2 2 2 L2 L2 x L y L Lz L z Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 18 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 15 9 2 1 2 0 2 1 2 L2 L2 z = 2 25 25 25 2 L2 L2 z 2 Q32. 24 2 25 50 24 2 26 2 25 25 Consider a two-dimensional infinite square well 0 V x, y 0 x a, 0 y a otherwise Its normalized Eigenfunctions are nx ,n y x, y 2 n x x n y y sin sin a a a V where n x , n y 1,2,3, ..If a perturbation H ' 0 0 a a , 0 y 2 2 otherwise 0 x is applied, then the correction to the energy of the first excited state to order V 0 is (a) (c) Ans: V0 4 (b) 16 1 2 9 V0 4 64 1 2 9 (d) V0 4 V0 4 32 1 2 9 (b) Solution: For first excited state, which is doubly degree 2 a 1 sin x a sin 2 y 2 2 x y , 2 sin sin a a a a H11 1 H 1 V0 1 1 V0 2 a / 2 2 x 2 a / 2 2 2 y V sin dx sin dy 0 2 2 4 a 0 a a 0 a 2 x 2 a / 2 2 y y 2 a/2 x sin sin dx sin sin dy 0 0 a a a a a a 16 V 16 4 4 V0 V0 , H 21 2 H 1 V0 2 and H 22 2 H 2 0 . 2 9 9 4 3 3 H12 1 H 2 V0 H 12 16V0 V0 2 2 V0 16V0 9 2 0 Thus 4 2 0 4 16V0 V0 9 2 4 9 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 19 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 16V0 V V 0 0 2 4 9 4 Q33. 64 1 2 9 The bound on the ground state energy of the Hamiltonian with an attractive deltafunction potential, namely H 2 d 2 a x 2m dx 2 using the variational principle with the trial wavefunction x A exp bx 2 is Note : e t t a dt a 1 0 (a) ma 2 / 4 2 (b) ma 2 / 2 2 (c) ma 2 / 2 (d) ma 2 / 5 2 Ans: (c) Solution: For given wavefunction T For variation of parameter E Q34. m in d E db 2b 2b 2b 2b E a and V a 2m 2m 0 d E 2 2 1 12 4 a b 0 b . db 2m 2 8m 2 ma 2 . 2 If the operators A and B satisfy the commutation relation A, B I , where I is the identity operator, then (c) e , B e (d) e , B I A Ans: B (b) e A , B e B , A (a) e A , B e A A ,A (a) A A2 ....... Solution: A, B I and e A 1 2 1 A2 , B A3 , B A A2 ......... e , B 1 ......., B = 1, B A, B 2 3 2 1 A Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 20 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES e A ,B 0 I e , B 1 A A ........ e 2! 2 A Q35. A A, B A, B A A A 2 , B A 2 , B A .......... . 2! 3! A where A, B I , A 2 , B 2 A and A3 , B 3 A 2 . Two identical bosons of mass m are placed in a one-dimensional potential V x 1 m 2 x 2 . The bosons interact via a weak potential, 2 V12 V0 exp m x1 x2 / 4 2 where x1 and x 2 denote coordinates of the particles. Given that the ground state 1 m 4 wavefunction of the harmonic oscillator is e 0 x m x 2 2 . The ground state energy of the two-boson system, to the first order in V 0 , is (a) 2V0 (b) (c) V0 1 2 Ans: 1 2 V0 (d) V0 1 (c) Solution: There is two bosons trapped in harmonic oscillator so energy for ground state without perturbation is 2. If perturbation . 2 is introduced we have to calculate V1,2 where 2 V12 V0 exp m x1 x2 / 4 . 1 2 2 m x1 m x2 2 m e 2 e 2 But calculating V1,2 on state 0 x is very tedious task. So lets use a trick i.e perturbation is nothing but approximation used in Taylor series. So just expand V12 V0 exp m x1 x2 / 4 and take average value of first term 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 21 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES V12 V0 exp m x1 x2 / 4 = V0 (1 V12 V0 (1 2 2 m x12 x2 2 x1.x2 4 V1,2 = V0 (1 x1 x2 4 2 ...) ...) 2 m x12 x2 2 x1 . x2 ...) = V (1 m ( o 4 0) 2m 2m )... 4 1 2 V12 Vo (1 ) V0 1 . 4 2 1 2 So E V0 1 . 2 NET/JRF (DEC-2013) Q36. A spin - 1 1 1 i 2 particle is in the state 3 in the eigenbasis of S and S z . If we 2 11 measure S z , the probabilities of getting (a) Ans: 1 1 and 2 2 (b) h h and , respectively are 2 2 2 9 and 11 11 (c) 0 and 1 (d) 1 3 and 11 11 (b) Solution: P 2 P 2 2 1 i 1 10 1 2 2 3 2 11 11 1 2 1 i 1 9 01 11 3 11 i.e. probability of S z getting and 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 22 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q37. The motion of a particle of mass m in one dimension is described by the Hamiltonian H p2 1 m 2 x 2 x . What is the difference between the (quantized) 2m 2 energies of the first two levels? (In the following, x is the expectation value of x in the ground state.) (a) x Ans: (b) x (c) 2 2m 2 (d) (d) Solution: H p2 1 1 m 2 x 2 x V x m 2 x 2 x 2m 2 2 V x 1 2 2 2 1 2 2 m 2 x 2 x m x 2 x 2 m 2 m 2 m 2 4 m 2 4 2 V x 1 2 m 2 x 2 m 2 2m 2 2 1 2 3 1 En n E1 E0 2 2 2m 2 2 Q38. Let nlm denote the eigenfunctions of a Hamiltonian for a spherically symmetric potential V r . The expectation value of L z in the state 1 6 200 5 210 10 21 1 20 211 is (a) Ans: 5 18 5 6 (c) (d) 5 18 (d) Solution: Lz Lz = Q39. (b) 1 5 10 20 10 5 0 0 ( 1 ) (1 ) = 36 36 36 36 36 18 1 If x A exp x 4 is the eigenfunction of a one dimensional Hamiltonian with eiggenvalue E 0 , the potential V x (in units where 2m 1 ) is (a) 12x 2 (b) 16x 6 (c) 16 x 6 12 x 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com (d) 16x 6 12x 2 Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 23 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (d) Solution: Schrodinger equation 2 V 0 (where 2m 1 and E 0 ) 4 4 4 4 2 Ae x VAe x 0 e x 4 x3 Ve x 0 2 x x 4 3x 2 e x x 3 4x 3e x 4 4 Ve x4 0 12 x 2 e x 16 x 6 e x Ve x 0 4 4 4 V 16x 6 12x 2 Q40. A particle is in the ground state of an infinite square well potential given by, 0 V x for a x a otherwise The probability to find the particle in the interval between 1 2 (a) Ans: (b) 1 1 2 (c) a a and is 2 2 1 1 2 (d) 1 (b) Solution: The probability to find the particle in the interval between a/2 a / 2 a a and is 2 2 a/2 a/2 2 2 x x 1 x 1 1 2 x dx 1 cos cos cos dx cos2 dx a 2a a 2 a / 2 2a 2a 2a 2a 2a a / 2 1 a a a 1 2a 1 1 1 a x 1 1 a x sin 2 2a a a / 2 2a 2 2 2a a/2 Q41. The expectation value of the x - component of the orbital angular momentum L x in the state 1 3 2,1, 1 5 2,1,0 11 2,1, 1 5 (where nlm are the eigenfunctions in usual notation), is (a) 10 25 11 3 (b) 0 (c) 10 25 11 3 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com (d) 2 Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 24 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (a) Solution: L l , m l l 1 m m 1 l , m 1 and L l , m l l 1 m m 1 l , m 1 Lx L L 2 Lx L L 2 1 L 3 2 210 5 2 211 5 L 1 1 1 .3 10 110 10(3 11) 25 25 25 1 L 2 5 21 1 2 11 210 5 L Lx L L 2 Lx Q42. 1 1 .3 10 10 11 25 25 = 1 10(3 11) 25 1 1 10 .3 10 10 11 = 25 25 25 11 3 A particle is prepared in a simultaneous eigenstate of L2 and Lz . If l 1 2 and m are respectively the eigenvalues of L2 and Lz , then the expectation value L2 of the particle x in this state satisfies (a) L2 x 0 (c) 0 L 2 x Ans: 2 2 (b) 0 L2 x 1 2 2 2 Lx (d) 2 2 1 2 2 (d) Solution: L2 x 1 l l 1 2 m2 2 2 For max value m 0 and for min m l l 2 l l 1 2 L2 x 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 25 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES A, B, C are Non zero Hermitian operator. A, B C AB BA AB Ab 0 C but C 0 if AB BA ie A, B C 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 E 5 2 2 is 2ma 2 (a) (b) 2 x1 2 x2 2 x1 x2 sin sin sin sin a a a a a (c) 2 x1 3 x2 3 x1 x2 sin sin sin sin a a 2a 2a a (d) Ans: 2 x1 2 x2 2 x1 x2 sin sin sin sin a a a a a 2 x1 x2 x x sin cos sin 2 cos 2 a a a a a (a) Solution: Fermions have antisymmetric wave function 2 x 2 x2 2 x1 x2 x1 x2 sin 1 sin sin sin a a a a a En 5 2 2 nx1 1, nx2 2 2ma 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 26 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Q44. A particle of mass m in the potential V x, y energy E 1 m 2 4 x 2 y 2 , is in an eigenstate of 2 5 . The corresponding un-normalized eigen function is 2 m (a) y exp 2 x 2 y 2 2 m 2 (c) y exp x y 2 2 Ans: m (b) x exp 2 x 2 y 2 2 m 2 (d) xy exp x y 2 2 (a) Solution: V x, y 1 5 m 2 4 x 2 y 2 , E 2 2 V x, y 1 1 2 m 2 x 2 m 2 y 2 2 2 1 1 1 1 Now, E n n x x n y y n x 2 n y 2 2 2 2 3 En 2nx n y 2 En Q45. 5 2 when n x 0 and n y 1 . A particle of mass m in three dimensions is in the potential 0 V r Its ground state energy is (a) Ans: 2 2 2ma 2 2 2 (b) ma 2 r a r a 3 2 2 (c) 2 ma 2 9 2 2 (d) 2ma 2 (a) 2 2 d u r l l 1 V r u r Eu r Solution: 2 2mr 2 2m dr d 2u r dr 2 Ku r K 2mE , l 0, V r 0 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 27 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES u r A sin Kr B cos Kr Using boundary condition, B 0, u r A sin Kr , r a, u r 0 sin Ka 0 Ka n E Q46. 2 2 n 1 2ma 2 1 r i , the uncertainty p r in the ground state. Given that p r r 0 r 1 a 3 0 e r / a0 of the hydrogen atom is (a) Ans: a0 (b) 2 a0 2a 0 (c) (d) 2 a0 (a) 1 1 r i , 0 r Solution: p e r / a0 3 r r a0 Pr Pr2 Pr Now Pr 0 2 r / a 1 e 0 e r / a0 i 3 3 r r a0 a0 1 4 i r / a0 1 r / a0 e e 3 a 0 0 a0 2 4 r dr 1 r / a0 2 r dr r e 4 i 1 2 r / a0 2 2 r / a0 e r dr re dr 3 a0 0 a0 0 4 i 1 2 ! 1 ! 3 a0 a0 2 / a0 3 2 / a0 2 2 2 a0 4 i a0 0 3 a0 4 4 Pr 0 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 28 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 2 r P 2 2 2 r / a0 1 r / a0 2 3 e e 4 r dr 2 r r a0 0 r 4 2 r / a0 r / a0 1 2 1 r / a0 2 e e 3 2 e r dr a0 a0 r a0 0 4 2 1 2 2 r / a0 2 4 2 2 r / a0 r e dr re dr 3 2 3 a0 a0 a0 0 0 a0 3 2 4 2 2 ! a0 2 a0 4 2 3 2 3 8 a0 4 a0 a0 a0 P Q47. Pr2 Pr 2 1 2 ! 2 1 ! 2 a0 2 / a0 3 a0 2 / a0 2 a0 a0 4 2 a0 2 3 2 4 2 a0 4 a0 2 0 2 a0 a0 The ground state eigenfunction for the potential V x x where x is the delta function, is given by x Ae x , where A and 0 are constants. If a perturbation H bx 2 is applied, the first order correction to the energy of the ground state will be (a) Ans: b 2 (b) 2 b (c) 2 2b (d) 2 b 2 2 (d) Solution: V x x , x Ae x 1 x e x E 1 H 1 1 1 e x bx 2 e x dx 0 0 0 2 x 2 x 2 2 2 x 2 2 x 2 2 x 2 e bx dx b e x dx b x e dx x e dx b 2 x e dx 2! 2! b 2 x 2 2 b 2 b 3 e bx dx 3 8 2 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 29 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 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 Ans: (b) 0.90 (c) 0.16 (d) 0.32 (d) 2 a0 Solution: Probability: 0 1 a 3 0 e r / a0 2 a0 1 3 a0 0 e r / a0 4 4 r dr 3 a0 2 4 4 r dr 2 ao 2 2 2 r / a0 r e a0 r e 2 2 r / a0 dr 0 a0 2e 2 r / a0 4 3 a0 4 3 a0 2 2aa0 a0 e 0 a0 2 2 a0 2 a 0 4 a0 a0 a0 a0 2 r / a0 2 2r e 2 2 0 0 a0 a0 a0 a0 2 2 2 0 3 3 2 a0 / a0 a0 2 a0 / a0 0 a0 e e 2 e 4 8 3 3 3 3 a0 a0 a0 1 a0 1 1 1 5 4 2 5 2 1 2 2 2 4 2 e 4 e 4e 4e 2 e 5 0.137 1 0.685 1 0.32 Q49. A particle in the infinite square well 0 x a 0 V x otherwise is prepared in a state with the wavefunction 3 x A sin x a 0 0 x a otherwise The expectation value of the energy of the particle is (a) Ans: 5 2 2 2ma 2 (b) 9 2 2 2ma 2 (c) 9 2 2 10 ma 2 (d) 2 2 2ma 2 (c) Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 30 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 3 x A sin x a otherwise, 0 0 x a otherwise 0 x a 0 Solution: V x 3 x 1 3 x x A sin sin 3 A 3sin A 4sin 3 A A sin 4 a 4 a a x A sin 3 x a 2 3 x 3 x 1 3 x A a 2 3sin sin A sin A sin 4 2 a a 2 a a 4 a 4 a x 1 1 4 9 x 3. x A a a 3 x 3 1 x 4 2 2 a 32 a 32 1 x 3 x 2 10a 2 10a 3 1 1 x 3 x 10 10 Now, E1 2 2 2ma 2 Probably P E1 E 10a 2 32 a 2 a 2 A 1 A A A 1 32 10a 32 32 , E2 1 2 9 2 2 E an P an 2ma 2 2 9 , P E2 10 2 1 10 9 2 2 1 9 2 2 9 2 2 E 10 2ma 2 10 2ma 2 10ma 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 31 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES NET/JRF (DEC-2014) Q50. Suppose Hamiltonian of a conservative system in classical mechanics is H xp , where is a constant and x and p are the position and momentum respectively. The corresponding Hamiltonian in quantum mechanics, in the coordinate representation, is 1 (a) i x x 2 (c) i x Ans: 1 (b) i x x 2 x (d) i x 2 x (b) Solution: Classically H xp , quantum mechanically H must be Hermitian, So, H H H 2 xp px and H 2 xp px i x x x i i x 2 x x x x 2 i i 2x 2 x 1 2 x 2 x 1 H i x x 2 Q51. Let 1 and 2 denote the normalized eigenstates of a particle with energy eigenvalues E1 and E 2 respectively, with E 2 E1 . At time t 0 the particle is prepared in a state t 0 1 2 1 2 The shortest time T at which t T will be orthogonal to t 0 is (a) Ans: 2 E 2 E1 (b) E 2 E1 (c) 2 E 2 E1 (d) 4 E 2 E1 (b) Solution: t 0 1 2 1 2 and t T 1T 1 iE 1 iE 2T e 1 e 2 2 2 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 32 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 2 * 0 T dx 0 e iE1 T iE iE T 1T 2T i E2 E1 1 iE2 T e 0 e e e 1 2 T cos 1 1 Equate real part cos E2 E1 1 T E2 E1 E2 E1 Q52. Consider the normalized wavefunction a1 11 a 2 10 a3 1 1 where lm is a simultaneous normalized eigenfunction of the angular momentum operators L2 and Lz , with eigenvalues l l 1 2 and m respectively. If is an eigenfunction of the operator L x with eigenvalue , then (a) a1 a3 (c) a1 a3 Ans: 1 1 , a2 2 2 1 , 2 a2 (b) a1 a3 1 2 1 , 2 (d) a1 a 2 a3 a2 1 2 1 3 (b) Solution: Lx L L 2 L a1 11 a2 10 a3 1 1 a1 0 12 a2 2 11 a3 2 10 For L , a2 2 11 a3 2 10 For L , L a1 11 a2 10 a3 1 1 a1 2 10 a2 2 1 1 Given L L 2 L L 1 a2 2 11 a1 a3 2 10 a2 2 1 1 2 2 L L a1 11 a2 10 a3 1 1 (Given) 2 Thus a2 2 a1 a2 2a1 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 33 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES a1 a3 2 a2 a1 a3 a 2a1 2 a3 2 2 2 a12 a2 a1 a3 Q53. 2 a2 1 2 1 1 , a2 2 2 Let x and p denote, respectively, the coordinate and momentum operators satisfying the canonical commutation relation x, p i in natural units 1 . Then the commutator x, pe is p (b) i 1 p 2 e p (a) i 1 p e p Ans: (c) i 1 e p (d) ipe p (a) Solution: x, p i p 2 p3 p p p p x , p e p x , e ie p x ,1 p .... x , pe 2 3 p2 2ip 3ip 2 ie p p x,1 x, p x, .... ie p p 0 i ...... 2 3 2 x, pe ie p Q54. p p3 2 i p p ..... ie p ipe p i 1 p e p 2 Let 1 , 2 , 3 , where 1 , 2 , 3 are the Pauli matrices. If a and b are two arbitrary constant vectors in three dimensions, the commutator a , b is equal to (in the following I is the identity matrix) (a) a b 1 2 3 (c) a b I Ans: (b) 2i a b (d) a b I (b) , b bi a2 Solution: a a1i j a3 k 1 b2 j b3k , x i y j z k Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 34 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES a , b a1 x a2 y a3 z , b1 x b2 y b3 z a , b a1b1 x , x a1b2 x , y a1b3 x , z a2b1 y , x a2b2 y , y a2b3 y , z a3b1 z , x a3b2 z , y a3b3 z , z a1b1 0 a1b2 2i z 2ia1b3 y a2b1 2i z 0 a2b3 2i x a3b1 2i y a3b2 2i x 0 2i a b a , b Q55. The ground state energy of the attractive delta function potential V x b x , where b 0 , is calculated with the variational trial function x A cos , x 2a 0, a x a, otherwise, for is (a) Ans: mb 2 2 2 (b) 2mb 2 2 2 (c) mb 2 2 2 2 (d) mb 2 4 2 2 (b) Solution: V x b x ; Normalized T * a a b 0 and a 8ma 2 2a ; a x a 2 2 2 2 dx 2 m x 2 8ma 2 a E x 2 x cos 2a 2a V * b x dx 2 2 x A cos 2 b b 2a a b a E 2 2 2 2 2 2 2 b 2 0 b 0 a 4ma 4ma a 8ma 3 a Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 35 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Put the value of a in equation: Q56. 2 2 b 4mb b 4mb 2mb2 E 2 8ma 2 a 2 2 2 2 8m 2 2 2 2 2 2 c12 1 ) be a linear Let c0 0 c1 1 (where c 0 and c1 are constants with c 0 combination of the wavefunctions of the ground and first excited states of the onedimensional harmonic oscillator. For what value of c 0 is the expectation value x a maximum? (b) x (a) x (c) x Ans: , m 1 1 , c0 2 m 2 (d) x 1 , c0 2m 2 c0 , m 2 c0 1 2 (c) Solution: c0 0 c1 1 X X 2 2 2 X 2c0c1 0 X 1 c0 c12 c0 c1 0 X 1 1 c0 c1 0 X 1 For max X 2 2m Q57. X c0 c1 2 c0 c12 1 c0 1 2 1 1 0 X 1 0 X 1 2 2 0 a a 1 X 2m Consider a particle of mass m in the potential V x a x , a 0 . The energy eigenvalues E n n 0, 1, 2, .... , in the WKB approximation, are 3a 1 (a) n 2 4 2m 1/ 3 3a 1 (b) n 2 4 2m 3a 1 (d) n 2 4 2m 3a 1 (c) n 2 4 2m 2/3 4/3 Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 36 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES Ans: (b) Solution: V x a x , x 0 According to W.K.B., 2 1 E P2 a x P 2m E a x 2m E/ 0 E / E / 2 E / 0 1 pdq n where 1 and 2 are positive mid point 2 E / E / 1 2m E a x dx n 2 2m E ax dx E / 0 1 2m E ax dx n 2 1 2m E ax dx n 2 2m E ax t At x 0, t 2mE; x E / a, t 0 2madx dt 2ma 2 mE 0 4 ma t 3/2 3 2 mE 1 2 t1/ 2 dt n 2ma t 2 3 0 2 mE 0 1 n 2 1 4 1 3/ 2 n ma 2mE n 3 2 2 3a 4 1 1 23/ 2 am5 / 2 E 3/ 2 n E n 3 2 2 4 2m Q58. 2/3 The Hamiltonian H 0 for a three-state quantum system is given by the matrix 1 0 0 0 1 0 H 0 0 2 0 . When perturbed by H 1 0 1 where 1 , the resulting shift 0 0 2 0 1 0 in the energy eigenvalue E 0 2 is (a) , 2 Ans: (b) , 2 (c) (d) 2 None of the answer is correct. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 37 fiziks Institute for NET/JRF, GATE, IIT-JAM, JEST, TIFR and GRE in PHYSICAL SCIENCES 1 0 0 Solution: H 0 0 2 0 , 0 0 2 0 1 0 H 0 1 0 1 0 1 0 2 0 0 1 in H 0 is not 0 in H because H is not in block diagonal form. So we 0 2 1 0 must diagonalised whole H . The Eigen value at H 0, 2 0, 2 0. 0 After diagonalisation H 0 0 0 0 2 0 0 0 , 0 is correction for Eigenvalue at H 0 . 2 So 2 0 is the correction for eigenvalue of H 0 2 None of the answer is correct. Head office fiziks, H.No. 23, G.F, Jia Sarai, Near IIT, Hauz Khas, New Delhi-16 Phone: 011-26865455/+91-9871145498 Website: www.physicsbyfiziks.com Email: fiziks.physics@gmail.com Branch office Anand Institute of Mathematics, 28-B/6, Jia Sarai, Near IIT Hauz Khas, New Delhi-16 38

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