Module 4 Physics-I PDF

PHYSICSI QUANTUM MECHANICS Chapter at a Glance =. he probability density P(x./) is defined as the probability. per unit length of tne...

PHYSICSI QUANTUM MECHANICS Chapter at a Glance =. he probability density P(x./) is defined as the probability. per unit length of tne s of finding the particle near the coordinate r at the time t. It can be expressed as P(x.) = w'(x./w(x./ukr =r(x./)f d 2. Position operulor: Operator represents the space cvordinate as X.y 3. Iinear momentum operator: p, -i,=-iand ii In general in three dimension momentum operator p = -ihV 4. Potentical energy operator: V(T) 5. 6. Total energy operator: E Kinetic energy operator: , = ih = 2 PL=-)(-)- 2 2m 7. Angular momentum operator: From elassical méchanics, we know L=fxp. Hence the components along three directions will be Similarly L, ž9-ih -and L,h, -h, -4, -n 8. Schrödinger equation in one dimension, - +P(x)= Ey(*) 2m) 9Schrödinger equation in three dimension, V w(r)+{E-V(r)}w(r)=0 10. A free particle is moving with a conslant momentunm and energy. lts wave function is W= Ael I1. Two operators f and are said to commule when their commutator vanishes i.e. i.=0 are lincar. An operator a is known as linear if it 12. The operators in quantum mechanics satisfies the following two conditions () &(y, +w) = cay, + ay, and (2) (w)= Cay, where C is a constant. PH-187 POPULARPUBLICATIONS ) Expectation value for position = w' (x,/) & V(x./) dk = J,V b) Expectation value for the momentum < pP, >= f'(-ih d,= -il of momentum] for x component ANS 4. Time Dependent Schrödinger's Equation: (lin One Dimension For a free particle i.c., V(x) =0 i) in +)2m 15. Time Dependent Schrödinger's Equation: (In threc Dimension) i)ih (For a free particle) 211 16. Time independent Schrödinger's Equation: (ln One Dimension) For a free. particle i.e, )0 i) ih Oy i) in + V Fo Po)+0] 17. Time independent Schrödinger's Equation: (In one Dimension) i)- Ey (For a ree particle) i) (E-VW) =0 [For Vr)0] 18. Time independent Schrodinger Equation (In three dimension) Ey (For a free particle) 19. Time independent Schrodinger Equation (ln three dimension) +V)|= Ey For V) 0 2m box of length I: For one dimension 20. Free particle in a wávefunctíon: w(r)= sin Energy eigenfunction: E= "r 2m PH-188  PHYSICS- 21. Free particle in, a box of dimensions a,b.c: For three dimension Wavefunction: (x.y.=) = sin,7* sin "sin Vube Energy eigenfunction: ANS E, =* 22. For Cubical box of length L Wave function w(x, y,.:)= sin sin L Energy Eigenvalue: E, 23. i) The energy eigen value 8mL +n +n corresponding to more than one wave.functions (independent quantum states) remains same is termed as degeneracy. i) The states for which same eigen value we are getting are called degenerate states. 11) The number of independent states corresponds to sáme energy eigen state is called degree of degeneracy. 24. Equation where of continuity: - u is termed.as probability current density or -v and 7 Wy+v) simple current. density. Multiple Choice Type guestions 1.1. The wave the correct option. function y,,(x) and , (x) are orthogonal to each other then find [WBUT 2018(ODD)] a) Jv.v.dr=lb).dk=0 d)Jvv.dr=1 Answer: (b) 1.2. Heisenberg's uncertainty relation is [WBUT 2018(ODD) a) ArA2 b). AsAp,a c) AxAp, Sh d) ArAp, =l Answer: (b) 1.3. In the limit of high temperature and large wavelength, Planck's law of radiation reducesto [WBUT 2018(ODD)]] a) Rayleigh-Jeans Law b) Wien's displacement Law c) Wien's Law d) Stephan's Law Answer: (a) 14.Fhe value of [, P,]: [WBUT2018(ODD)] a) ih/27 b) 0 c) Answer: (a) -ih/27 d)1 PH-189 POPULAR PUBLICATIONS 2007(EVEN), 2014(EVEN] 1.5. What is the value of WBUT d) 1 a)-1 b) 0 c)-2 Answer: (a) solution of Schrödinger Equation?.5, Which of the following wave function is the 2007(EVEN)] a) A sec ((x) b) A tan (x) c) A exp (-x*) d) A exp (x) Answer: (¢) oeeralized co-ordinate can be of any dimension. However WBUT 2007(ODD)] momentum is. a the product of generalized co-ordinate and generalized of dimension ml lt always D)dimension of the work done depends on the dimension of the generalized co-ordinate )the generalized momentum is always of the dimension mlli Answer: (a) (Given, mass = m, length , dthe generalized fore is always of the dimension mllr time = t) 1.8. Rigid body has constraints classifed ini which of the following group? a) Rheonomic and Holonomic WBUT 2007(ODD)1 b) Rheonomic and Non-holonomic c) Scleronomic and Holonomic d) Scieronomic and Non-holonomic Answer: (c) 1.9. If the quantum mechanical state of a particle is described by WBUT 2008(EVEN) (x) =Pe",for |x|sL and w(x)=0,otherwise then. a) the particle has a definite position bu uncertainty in momentum b) the particle has a definite momentum but uncertainty in position c the particle has uncertainties both in position and momentum d) the particle has no uncertainties in either momentum or in position Answer: (c) 1.10. If (x./) is a normalized one dimensional wave function, then [WBUT 2008(EVEN); 2014(0DD)J a) cs =1 b)w'wds= 2 c)v'wd=0 d) v'v=d Answer: (a) PH-190  PHYSICS-I 1.11. Which one of the following functions is an eigen function of the operator [WBUT 2008(EVEN)] a) X b) Answer: (d) c)e d) cos x 1.12. A system consists of three point masses is fixed with time point masses. If the mutual then the degree of freedom of distance between tne such system i a) zero b) three [WBUT 2008(ODD) Answer: (d) c) nine d) six 1.13. The Lagrangian of a system is given in terms of generalized coordinates'r, 6, and associated velocities as L=0 The cyclic co-ordinate/s is I are ) WBUT 2008(ODD)] b) and c)rand 0 d) Answer: (d) 1 1.14. The Lagrangian of a harmonic oscillator is given by L=m 2 The Hamiltonian is given by. WBUT 2008(ODD)] 2 ,wh k, where p-OL RE 2m mi Answer: (b) 1.15. The eigenvalue of the eigenfunction e for the operator wBUT 2009(EVEN)] a) 1 b) 0 c) Answer: (C) system has f degrees of freedom the number of Hamilton's equations for 1.16. If a the system is WBUT 2009(ODD)]l a) 2 b)f c) 2f Answer: (b) 1.17. Which of the following a functions is acceptable as wave function of a one- [WBUT 2011(ODD), 2014(EVEN)] dimensional quantum mechanical system? a) b) =Ae c) w= Ax d) y=A secx Answer: (6) PH-191 POPULAR PUBLICATIONS 1.18. A system with time functionof this stat potentiai is in an energy state E. The wave independent pote endependent On'of state a) independent is [WBUT 2011(ODD)] of time an exponentially decaying function Periodic function of time with time ofperiod proportion tO 5 time rtodic function of time with time period inversely proportionai to Answer: (d). ree particle in a cubicalt box the degeneracy of the first excited state is a) two-fold WBUT 2011(ODD)] b) three-fold c) six-fold d) nine-fold Answer: (6) 1.20. Wave function of a certain particle is given py WBUT 2011(ODD)] V=AcOs x for |xs 0 otherwise Then the value of A is a)2 2 c) 1 d) zero Answer: (b) 1.21. Which of the following functions are eigen functions of the operator ? de a) yclogx b) vcx C). d) y ce where c and m are årbitrary, constants] WBUT 2012(EVEN), 2014(0DD)] Answer: (d) 1.22. The degrees of freedom for à system of N. particles with K constraint relation is given by [WBUT 2012(EVEN)] a) N-K b) N- 3K c) 3N-K Answer: (C) d 3 (N-K 1.23. The ignorable co-ordinate corresponding to the motion of a particle under central force is given by [WBUT 2012(EVEN) a) r b) 6 c)i. d) Answer: (6) 124, The expectation value of. the position.of a particle in otential box of length L(V(X)=G Oin the grouno siat b)USing the principle of operator correspondence write down tne op LL (components of angular momentum) in terms of position and " operators. Then show WBUT 200a(EVEN)] that| L,. L, ]= iht,. Ansiwer: 1) erein one stretched we assume the particle is moving randomly insíde the infinite potetia dimension within x=0 to x=/. So its whole energy is kinetie energyWe know that the total energy (E) of any particle is equal to sum of its K.E and P.E. So total energy E K.E since P.E = 0 Since P.E is independent of time so, Schrodinger's wave equation becomes )=Ey[symbols havetheirusual meaníing) = {stated earlier} V 0 :.5 Evy V 2mE ii) To solve the equation V y =- 2mE let we choose =k 1et dx Solution of this equation, y = Acoskx+ Bsin k (1). where A and B are the arbitrary constant.: boundary conditions To find the values of A and B we-may use the w(r)=0 when x=0. (2).(3) y(x)=0 when r=i Q=A putting (2) in equation (1) we have )=Bsin... (4) so kr have 0 = Bsin k putting (3) in equatíon (4) we B0 otherwise y(x) will not exist. So sin kl=0.kl = nz wheren=1,2, 3, ********************** k- wave function becomes y(x)= Bsinx1.e. the eigen function. PH-213 POPULAR PUBLICATIONS Determinution of the Constant A und B: ree parnicle is moving inside the hox nside the box is nmaximuim and it is cqual to only. probability of finding the pariicle I so wo can write. Bsinx F'sinr dt : or, or sin 2sin dr= dh=l =1 2 fa-cos) a ANS S 2 2e Normalised wave function becomes, y(r) =Sin ii) As we know the eigen function of then stateis.,(r) = So let us consider two different staies of n and n -2 having their eigen function as ,Cr)= 7sin and y()= si respectively. Now if we calculate. (x)w, ()dr we have or , cos coS (r)dr = 0 So in general we can say J(* W,XJLN=0 when *m lor the two orthogonal states 2 T sinand yv.,(x) = 7 sin* PH-214  PHYSICS iv). The expectation value of lhe component of momentum S P.V dx. Putting p, =-ih-and wave = 2 ve function (r) 7 si- have P,7sinxsin or, =-2in hr |sin cosr dx =0 SNVLLC S Now. or, b) Representation of the angular momentum operator L,, L, and L, are expressed from the vector expression = Fxp as l, = yP -=P,» L, =P; -P, and L, =»P, -w, p. =-ih Further the Iinear momentum operators P-i P, =-i oy and are substituted in these expression to get - and Nowoperator L,.,]=[L4-i,i. i, PH-215  POPULAR PUBLICATIONS Substituting the values of L,and L, we have [L.41-fe or4]-|-F 0-00-o-0-0 or.4,.4,]=n(-In) (Substituting the value of L,) ihl. Hence proved. 3.5. a) Write down Schrodinger's equation for one dimensional motion of a free particle in a one dimensional potentiäl box. Find its eigen function and eigen energy [WBUT 2009(EVEN), 2013(EVEN), 2013(ODD)] Answer: dy Ey =0 As is afunction ofx only, d This is one dimensional, time independent Schrodinger Equation for, free particle. ie. V=0 21 To solve this equation let we choose h -k*y dx Solution of this equation, w = Acoskxr+ Bsin kx. (1) where A and B are the arbitrary constant. To find the values of A and B we may use the boundary conditions yx) =0 when x= 0 (2) y(x) =0 when x=/ (3) putting (2) in equation (1) we have0 = A so, sO, r) = Bsinkr (4) putting (3) in equation (4) we have 0 = Bsin k otherwise y(*) will not exist. B#0. So.sind=0kl= nT where n = 1,2,3,. *** PH-216  PHYSICS- wave function becomes y(r)= x Bsin i.e., the eigen function. petermination of the Constant A and B: Since the Iree particle is moving inside the box only, probability of finding the parucic inside the box is maximum and it is equal to I so we can write, )vr =1 UTIA Bsin* sinr B sinXdr=1 dk=i UTIA or 2sin d1 fa-cos 1 2. Normalised wave function becomes. y(a)= sin17 So eigen function ,(x)= the probability density values sin. The value of the first three functions wvand and |. are shown in the fig ( ng with These wave functions iuuemble the vibrations of a string fixed at both ends because the wave function has the same form. Eigen values: = 21mE We have chosenk Again we have obtained k= Equating these above two we have 2mET cnece energy eigen value is|E ="Z 2m PH-217  POPULAR PUBLICATION D,ne wave function in a one-dimensional potential box with rigid walls is given by.(r)= sin s, = 0 otherwise Find the expectation value of i and p. WBUT 2011 (ODD), 2014(0DD Answer: *- » X=. dx=-u2du 20 fo du= 2 So, 22147Ya =1 A=F4= The normalised wave function y(x)=. e