Problem Set 2 Solutions - Introduction to Quantum Mechanics I | PHY 4604, Assignments of Physics

Download Problem Set 2 Solutions - Introduction to Quantum Mechanics I | PHY 4604 and more Assignments Physics in PDF only on Docsity! PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 1 of 14 PHY 4604 Problem Set #2 Solutions Problem 1 (10 points): The time-dependent Schrödinger’s equation is ),()(),( 2 ),( 2 22 txxV x tx mt txi Ψ+ ∂ Ψ∂ −= ∂ Ψ∂ h h , where we take the potential energy, V, to be only a function of x (not of time). We look for solutions of the form )()(),( txtx Φ=Ψ ψ . The time-dependent equation separates into two equations (separation constant E). )()()()( 2 2 22 xExxV xd xd m ψψψ =+− h and )()( tE dt tdi Φ=Φh . Thus, h/)( iEtet −=Φ . (a) (3 points) For normalizable solutions, show that the separation constant E must be real. Solution: Following Griffith’s problem 2.1 we set Γ+= iEE r , where Er and Γ are real. Thus, hhh /// )()(),( tiEtiEt reexextx −Γ− ==Ψ ψψ and ∫∫ +∞ ∞− ∗Γ +∞ ∞− ∗ ==ΨΨ dxxxedxtxtx t )()(1),(),( /2 ψψh . The second term is independent of time and hence Γ = 0 and E is real. One can prove E is real by noting that if V(x) is real then ( )∫∫ +∞ ∞− ∗ +∞ ∞− ∗ = dxxxHdxxHx opop )()()()( ψψψψ , where )( 22 )( 2 222 xV xd d m V m p H op opx op +−=+= h . The time-independent Schrödinger equation is )()( xExHop ψψ = which yields ∫∫ +∞ ∞− ∗ +∞ ∞− ∗ = dxxxEdxxHx op )()()()( ψψψψ ( ) ( ) ∫∫∫ +∞ ∞− ∗∗ +∞ ∞− ∗ +∞ ∞− ∗ == dxxxEdxxxEdxxxHop )()()()()()( ψψψψψψ Thus, E = E* which means that E is real. (b) (3 points) Show that E corresponds to the total energy and that )()(),( txtx Φ=Ψ ψ corresponds to a state with definite energy (i.e. ΔE = 0). Solution: EdxxxEdxxHxHE opop ==>=>=<< ∫∫ +∞ ∞− ∗ +∞ ∞− ∗ )()()()( ψψψψ PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 2 of 14 22 222 )()( )())()(()())(( EdxxxE dxxHHxdxxHxHE opopopop == =>=>=<< ∫ ∫∫ ∞+ ∞− ∗ +∞ ∞− ∗ +∞ ∞− ∗ ψψ ψψψψ Thus, (ΔE)2 = <E2> - <E>2 = 0. Also note that since h/)(),( iEtextx −=Ψ ψ we have EdxxxEdxtx t itxdxtxEtxEE opop ==Ψ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ Ψ=ΨΨ>=>=<< ∫∫∫ +∞ ∞− ∗ +∞ ∞− ∗ +∞ ∞− ∗ )()(),(),(),(),( ψψh . (c) (4 points) Show that the time-independent wave function, )(xψ , can always be taken to be a real function. (Hint: take V to be real and see Griffith’s problem 2.1) Solution: If )(xψ satisfies )()( xExHop ψψ = we can take the complex conjugate of both sides and get )()( xExHop ∗∗∗∗ = ψψ , but Hop* = Hop and E* = E and hence )()( xExHop ∗∗ = ψψ . Thus, ( ) ( ))()()()( xxExxHop ∗∗ +=+ ψψψψ and we see that ( ))()( xx ∗+ψψ is a solution to the time- independent equation and ( ))()( xx ∗+ψψ is real. Problem 2 (50 points): Consider an infinite square well defined by V(x) = 0 for 0 < x < a, and V(x) = ∞ otherwise. (a) (4 points) Show that the stationary states are given by h/)(),( tiEnn nextx −=Ψ ψ with )/sin(2)( axn a xn πψ = and 2 222 2ma nEn hπ = , and n is a positive integer. Solution: For the region outside of 0 < x < a 0)( =xψ and inside the region )()( 2 2 22 xE xd xd m ψψ =− h or )()( 22 2 xk xd xd ψψ −= with m kE 2 22h = The most general solution is of the form )cos()sin()( kxBkxAx +=ψ . The boundary condition at x = 0 gives 0)0( == Bψ and the boundary condition at x = a gives 0)sin()( == kaAaψ which implies that πnka = with n = 1, 2, 3,… Thus, )/sin()( axnAxn πψ = with 2 222 2ma nEn hπ = . The normalization is arrived at by requiring that 24 )2sin( 2 )(sin)/(sin1)()( 2 0 2 0 2 2 0 22 aAyy n aAdyy n aAdxaxnAdxxx nna nn =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −==== ∫∫∫ +∞ ∞− ∗ ππ ππ πψψ Thus, aA /2= . These states are called stationary because the probability density and all the expectation values are independent of time. (b) (2 points) Show that the states, )(xnψ , form an orthonormal set. Namely, show that mnnm dxxx δψψ =∫ +∞ ∞− ∗ )()( . PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 5 of 14 where c1 = c2 = 2/1 which means that you get energy E1 with probability ½ and energy E2 with probability ½. The expectation value of E is 2 22 2 22 2 22 22 1 12 1 4 5 2 4 22 1 mamama EEE hhh πππ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +=+>=< (f) (10 points) Suppose that particle in this infinite square well has the initial wave function ⎩ ⎨ ⎧ − =Ψ )( )0,( xaA Ax x axa ax ≤≤ ≤≤ 2/ 2/0 Sketch )0,(xΨ and determine the normalization A. Find ),( txΨ . What is the probability that a measurement of the energy of this state will yield the ground state energy E1? What is the expectation value of the energy for the state ),( txΨ ? Answer: 3/32 aA = , ∑ ∞ = −−−=Ψ L h ,5,3,1 / 2 2/)1( 2 )/sin( 1)1(264),( n tiEn neaxn na tx π π , the probability of measuring E1 is 9855.0 96 41 ≈= π P , 2 26 ma E h>=< . Ψ(x,0) 0 1 2 0.0 0.5 1.0 x/a Solution: First we normalize the distribution as follows: 2 32/ 0 22 2/ 22 2/ 0 22 12 2 )()0,()0,()(1 AadxxA dxxaAdxxAdxxxdxx a a a a == −+=ΨΨ== ∫ ∫∫∫∫ +∞ ∞− ∗ +∞ ∞− ρ Thus, 33 /32/12 aaA == . We know that ∑ ∞ = −=Ψ 1 /)(),( n tiE nn nexctx hψ with ∫ +∞ ∞− ∗ Ψ= dxxxc nn )0,()(ψ . Thus, PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 6 of 14 [ ] [ ] [ ] ( ))2/cos()2/()2/sin(2)1(1 )sin(2)1(1)/sin(2)1(1 )/sin(2)/sin(2 )/sin()(2)/sin(2 2 2/ 0 22/ 0 2/ 0 2/ 0 2/ 2/ 0 πππ π π π πππ ππ π nnn an aA dyyy an aAdxaxnx a A dxaxnnx a Adxaxnx a A dxaxnxa a Adxaxnx a Ac n n n a n aa a a a n −⎟ ⎠ ⎞ ⎜ ⎝ ⎛−−= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛−−=−−= =−+= −+= ∫∫ ∫∫ ∫∫ For even n cn = 0 and for odd n we have 22 2/)1( 2 2/3 2/)1( 64)1(2322)1( ππ nan a a c nnn −− −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛−= and ∑ ∞ = −−−=Ψ L h ,5,3,1 / 2 2/)1( 2 )/sin( 1)1(264),( n tiEn neaxn na tx π π where 2 222 2ma nEn hπ = . The probability of measuring E1 is 9855.09664 4 2 2 2 1 ≈=⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == ππn cP . The expectation value of the energy is 2 2 22 2 222222 2 ,5,3,1 222 2 ,5,3,1 2222 22 1 2 6)2(48 7 1 5 1 3 1 1 148148 2 64 mamamanma m n n EcE n nn nn hh L hh h L L ==⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++++== ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =>=< ∑ ∑∑ ∞ = ∞ = ∞ = ξ πππ π π where I used 87 1 5 1 3 11 2 222 π =++++ L . (g) (12 points) Suppose a particle of mass m in this infinite square is in the ground state (i.e. n = 1). Now suppose that the well suddenly expands to twice its original size (the right wall moves from a to 2a) leaving the wave function (momentarily) undisturbed. If the energy of the particle is now measured, what is the most probable value and what is the probability of getting this value? What is the next most probable value, and what is its probability? Answer: Most probable energy is 2 22 2 2ma E hπ= with probability P2 = ½. The next most probable energy is 2 22 1 8ma E hπ= , with probability 36025.0 9 32 21 ≈= π P . Solution: The new allowed energies are 2 222 )2(2 am nEn hπ = and the new wave functions are PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 7 of 14 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= x a n a xn 2 sin 2 2)( πψ with ⎟ ⎠ ⎞ ⎜ ⎝ ⎛=Ψ x aa x πsin2)0,( We know that ∑ ∞ = −=Ψ 1 /)(),( n tiE nn nexctx hψ with ∫ +∞ ∞− ∗ Ψ= dxxxc nn )0,()(ψ . For n ≠ 2 we get, ( )( ) 4 1sin24 1 1 1 11 2 sin 2 2 1 2 1 2 sin 1 2 1 2 sin 2 2 1 2 1 2 sin 1 2 1 2 sin 2 2 1 2 cos1 2 cos 2 2sin 2 sin2 2 2 2 1 2 1 0 00 − + =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − −⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ += ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +−⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= ∫∫ nnn n n n n n a n a xn a n a xn a dx a xn a xn a dxx a x a n a c n a aa n π π π π π π π π π π π π ππππ For n = 2 we get, 2 2 4 )2sin( 2 2)(sin2)/(sin2 00 2 0 2 2 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=== ∫∫ ππ ππ π yydyydxax a c a . Thus, ⎪ ⎩ ⎪ ⎨ ⎧ = = = == − L L ,8,6,4 ,5,3,1 2 0 222 )4( 32 2 1 2 n n n cP nnn π The most probable energy is 2 22 2 2ma E hπ= (same as before the change) with probability P2 = ½. The next most probable energy is 2 22 1 8ma E hπ= , with probability 36025.0 9 32 2)4( 322 11 222 ≈=== − ππ n cP . Problem 3 (50 points): Consider the one dimensional harmonic oscillator with potential energy 2 2 1)( kxxV = . The Hamiltonian is given by ( )22 2 )( 2 1 2 xmp m V m p H x x ω+=+= where mk /=ω . We express the Hamiltonian in terms of the operators (px)op and (x)op, where x ip opx ∂ ∂ −= h)( and xx op =)( and 2 2 22 )()()( x ppp opxopxopx ∂ ∂ −== h . Namely, PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 10 of 14 ( ) ωψψω ψψωψψ h h h 2 1 00 02 1 000 )()( 2 )()()()()()( == +=>=>=<< ∫ ∫∫ ∞+ ∞− ∗ +∞ ∞− −+ ∗ +∞ ∞− ∗ dxxx dxxaaxdxxHxHE opopopop and also ( ) 000 2 1)( 2 10)( ψω ω ψω ω ψ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ∂ ∂ +=+==− x xm m pixm m a opxopop h hh Thus 0)(0 =xψ satisfies the following differential equation, )()( 00 xx m x x ψωψ h = ∂ ∂ , with solution 2 2 0 )( xm Aex h ω ψ − = . The normalization A is determined by requiring that ω πψψ ω m AdxeAdxxx xm h h 22 00 2 1)()( === ∫∫ +∞ ∞− − +∞ ∞− ∗ and hence 4/1 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= hπ ωmA . (h) (4 points) Prove that 11)( ++ += nnop na ψψ and 1)( −− = nnop na ψψ , were nψ is the n th excited state and show that 0)(! 1 ψψ nopn an + = . Solution: From (f) we have ( ) ∫∫ +∞ ∞− ∗ +∞ ∞− ∗ ± = dxxaxdxxxa nopmnmop )())(()()()( ψψψψ m From (g) we know that 1)( ++ = nnnop ca ψψ and 1)( −− = nnnop da ψψ where cn and dn are constants. In addition we know that ( ) nopopnnnnop aanEH ψωωψψψ 2121 )()()( +=+== −+hh and hence nnopop naa ψψ =−+ )()( . Similarly, ( ) nopopnnnnop aanEH ψωωψψψ 2121 )()()( −=+== +−hh and hence nnopop naa ψψ )1()()( +=+− . Thus, PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 11 of 14 ( ) )1()()()1()()())(( )()()()()()( 211 2 +=+== == ∫∫ ∫∫ ∞+ ∞− ∗ ∞+ ∞− +− ∗ +∞ ∞− + ∗ + +∞ ∞− + ∗ + ndxxxndxxaax cdxxxcdxxaxa nnnopopn nnnnnopnop ψψψψ ψψψψ so that 1+= ncn . Similarly, ( ) ndxxxndxxaax ddxxxddxxaxa nnnopopn nnnnnopnop === == ∫∫ ∫∫ ∞+ ∞− ∗ ∞+ ∞− −+ ∗ +∞ ∞− − ∗ − +∞ ∞− − ∗ − )()()()())(( )()()()()()( 211 2 ψψψψ ψψψψ so that ndn = . We see that 01 )( ψψ opa+= , 0 2 12 )(2 1)( 2 1 ψψψ opop aa ++ == , 0 3 23 )(123 1)( 3 1 ψψψ opop aa ++ ⋅⋅ == and 0)(! 1 ψψ nopn an + = . (i) (8 points) Find <x>, <x2>, <px>, and <px2> for the nth stationary state. Answer: 0=>< nx , ωm nx n h ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=>< 2 12 , 0=>< nxp , ωmnp nx h⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=>< 2 12 Solution: We see that ( )opopop aamx )()(2)( −+ += ω h ( )opopopx aamwip )()(2)( −+ −= h . Thus, ( ) 0)()( 2 )()(1 2 )()()()( 2 )())(( 11 =++= +==>< ∫∫ ∫∫ ∞+ ∞− ∗ − ∞+ ∞− + ∗ +∞ ∞− −+ ∗ +∞ ∞− ∗ dxxxn m dxxxn m dxxaax m dxxxxx nnnn nopopnnopnn ψψ ω ψψ ω ψψ ω ψψ hh h where I used 11)( ++ += nnop na ψψ and 1)( −− = nnop na ψψ and mnnm dxxx δψψ =∫ +∞ ∞− ∗ )()( . We can calculate <px>n from <x>n as follows: 0=><=>< dt xdmp nnx or we can calculate it the long way ( ) 0)()( 2 )()(1 2 )()()()( 2 )())(( 11 =−+= −==>< ∫∫ ∫∫ ∞+ ∞− ∗ − ∞+ ∞− + ∗ +∞ ∞− −+ ∗ +∞ ∞− ∗ dxxxnmidxxxnmi dxxaaxmidxxpxp nnnn nopopnnopxnnx ψψωψψω ψψωψψ hh h Now PHY4604 Fall 2007 Problem Set 2 Solutions Department of Physics Page 12 of 14 ( ) ( ) ω ψψψψ ω ψψ ω ψψ ω ψψ m ndxxxndxxxn m dxxaaaaaax m dxxaax m dxxxxx nnnn nopopopopopopn nopopnnopnn hh h h ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++++= +++= +==>< ∫∫ ∫ ∫∫ ∞+ ∞− ∗ ∞+ ∞− ∗ ∞+ ∞− −+−−++ ∗ +∞ ∞− −+ ∗ +∞ ∞− ∗ 2 10)()()1()()(0 2 )()()()()()()()( 2 )()()()( 2 )())(( 22 222 where I used ( ) 212 211)()( ++++ ++=+= nnopnop nnnaa ψψψ ( ) nnnopnopop nnnnaaa ψψψψ === −+−+ 1)()()( ( ) nnnopnopop nnnnaaa ψψψψ )1(111)()()( 1 +=++=+= +−+− ( ) 212 1)()( −−−− −== nnopnop nnnaa ψψψ Also, ( ) ( ) ωψψψψω ψψω ψψωψψ mndxxxndxxxnm dxxaaaaaaxm dxxaaxmdxxpxp nnnn nopopopopopopn nopopnnopxnnx h h h h ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ++++= +−− − = − − ==>< ∫∫ ∫ ∫∫ ∞+ ∞− ∗ ∞+ ∞− ∗ ∞+ ∞− −+−−++ ∗ +∞ ∞− −+ ∗ +∞ ∞− ∗ 2 10)()()1()()(0 2 )()()()()()()()( 2 )()()()( 2 )())(( 22 222 . (j) (4 points) Find Δx = σx and Δpx = xpσ for the n th stationary state and check that the uncertainty principle is satisfied. Answer: ωm nx n h ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=Δ 2 1)( , ωmnp nx h⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=Δ 2 1)( , 2 )12()( h+=ΔΔ npx nx Solution: From (h) we see that ωm nxxx nnn h ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=><−><=Δ 2 1)( 22 ωmnppp nxnxnx h⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=><−><=Δ 2 1)( 22 Thus, 22 )12( 2 1 2 1)( hhhh ≥+=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=ΔΔ nmn m npx nx ωω . (k) (4 point) Compute <T> and <V> for the nth stationary state, where T is the kinetic energy and V is the potential energy. What is sum <T> + <V>? Answer: ωh⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=>< 2 1 2 1 nT n , ωh⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=>< 2 1 2 1 nV n , nn EnVT =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +=><+>< ωh 2 1)(