Problem Set 5 Solutions - Introduction to Quantum Mechanics I | PHY 4604, Assignments of Physics

Download Problem Set 5 Solutions - Introduction to Quantum Mechanics I | PHY 4604 and more Assignments Physics in PDF only on Docsity! PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 1 of 9 PHY 4604 Problem Set #5 Solutions Problem 1 (20 points): Use separation of variables in Cartesian coordinates to solve the infinite cubical well (or “particle in a box”): ⎩ ⎨ ⎧ ∞+ = 0 ),,( zyxV otherwise LzLyLx <<<<<< 0,0,0 (a) (10 points) Find the stationary states and the corresponding energies. Answer: )/sin()/sin()/sin(8),,( 3 LznLynLxnL zyx zyxnnn zyx πππψ = and )(2 222 2 22 zyx nnnmL E ++= πh Solution: In this case the V(x,y,z) = V(x)+V(y)+V(z) and we look for stationary state solutions of the form )()()(),,( zZyYxXzyx =ψ and Schrödinger’s equation becomes EZzV dz Zd mZ YyV dy Yd mY XxV dx Xd mX =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−+⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +−+⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− )( 2 1)( 2 1)( 2 1 2 22 2 22 2 22 hhh and hence xEXxVdx Xd mX =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− )( 2 1 2 22h yEYyVdy Yd mY =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− )( 2 1 2 22h zEZzVdz Zd mZ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− )( 2 1 2 22h where the Ei is constants and Ex +Ey +Ez = E. The problem breaks into three one dimensional infinite wells and we know that the solution to the one dimensional problem is )/sin(2)( Lxn L xX xnx π= with 2 2 22 2 xn n mL E x πh = where nx = 1, 2, 3, … )/sin(2)( Lyn L yY yn y π= with 2 2 22 2 yn n mL E y πh = where ny = 1, 2, 3, … )/sin(2)( Lzn L zZ znz π= with 2 2 22 2 zn n mL E z πh = where nz = 1, 2, 3, … Thus, )/sin()/sin()/sin(8)()()(),,( 3 LznLynLxnL zZyYxXzyx zyxnnnnnn zyxzyx πππψ == , and )( 2 222 2 22 zyxnnn nnnmL EEEE zyx ++=++= πh . (b) (5 points) Call the energies E1, E2, E3, …, in order of increasing energy. Find E1, E2, E3, E4, E5, and E6. Determine the degeneraries (i.e. the number of different states with the same energy). PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 2 of 9 Answer: Energy nx ny nz Number of States E1= )2/(3 222 mLπh 1 1 1 1 E2= )2/(6 222 mLπh 2 1 1 E2 1 2 1 E2 1 1 2 3 E3= )2/(9 222 mLπh 2 2 1 E3 2 1 2 E3 1 2 2 3 E4= )2/(11 222 mLπh 3 1 1 E4 1 3 1 E4 1 1 3 3 E5= )2/(12 222 mLπh 2 2 2 1 E6= )2/(14 222 mLπh 1 2 3 E6 1 3 2 E6 2 1 3 E6 2 3 1 E6 3 1 2 E6 3 2 1 6 (c) (5 points) What is the degeneracy of E14, and why is this case interesting? Answer: The degeneracy of E14 is 4. Solution: The combinations after E6 are E7(322), E8(411), E9(331), E10(421), E11(332), E12(422), E13(431), and then E14(333 and 511). Simple combinatorics account for the degeneracies of 1 (nx = ny = nz), 3 (two the same, one different), and 6 (all three different). But in the case of E14 there is a numerical “accident”: 32 + 32 +32 = 27 and 52 +12 +12 = 27, so the degeneracy is greater than combinatorial reasoning alone would suggest. Problem 2 (20 points): A particle of mass m and E < 0 is placed in a finite spherical well ⎩ ⎨ ⎧− = 0 )( 0 V rV Rr Rr > ≤ (a) (10 points) Find the ground state by solving the radial equation with l = 0. Solution: The radial equation is given by )()()()( 2 2 22 rEUrUrV dr rUd m eff =+− h where 2 2 2 )1()()( mr llrVrVeff h+ += , where U(r) = rR(r). In the region for r < L (region 1) with l = 0 we have )()()(2)( 22 0 2 2 rUkrUVEm dr rUd −= + −= h where 20 /)(2 hVEmk += and 0 22 2 V m kE −= h and )cos()sin()(1 krBkrArU += and rkrBrkrArR /)cos(/)sin()(1 += . The second term diverges as r → 0 and hence B = 0. In the region for r > L (region 2) with l = 0 we have PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 5 of 9 ( ) )/sin(2sin // 1 1 /)( 0 /)cos( h hh hhhh pr pr ee pr idede ipriprpripri =−== +− − −− ∫∫ ξθθ ξ π θ and ∫ ∫ ∞ − ∞ − = = 0 / 2/3 0 0 / 3 0 2/3 )/sin(2 )(2 1 )/sin(4 )2( 1)( 0 0 drprre pr drprre rp p rr rr h h h h h h r π π π π φ Now 22 0 3 0 222 0 0 2 0 2 0 0 2 00 )//1( 0 2 00 )//1( 0 )//1( 0 )//1( 0 // 0 // 0 / ))/(1( )/(4 ))/()/1(( )/(4 2 1 )//1( 1 )//1( 1 2 1 )//1( 1 )//1(2 1 )//1( 1 //12 1 2 1 2 1)/sin( 0 0 00 000 h h h h hh hh hh h h h hh hh pr pr pr rip iipripri ipripr re i ipripr re i drredrre i drreedrree i drprre ripr ripr riprripr iprrriprrrrr + = + =⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ + − +− − ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ +− − +− = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −= ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ −= ∞ +− ∞ +− ∞ +− ∞ +− ∞ −− ∞ − ∞ − ∫∫ ∫∫∫ and hence 22 0 2/3 0 22 0 3 0 2/3 0 ))/(1( 121 ))/(1( )/(42 )(2 1)( hhh hh h r pr r pr pr pr p + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= + = ππ φ . (b) (5 points) Show that )( prφ is normalized. Solution: We see that ∫∫∫ ∞∞ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛== 0 42 0 23 0 2 0 2232 ))/(1( 24||4|)(| dp pr prdpppdp hh r π πφπφ . From the integral tables I see that )/(tan 16 1 )(16)(24)(6)( 1 52242222322422 2 ax axaa x xaa x xa xdx xa x −+ + + + + + −= +∫ and hence 55 0 422 2 32216 1 )( aa dx xa x ππ == +∫ ∞ . Thus, 32)/( 1 )/(32)/( 1 ))/(()/( 1 ))/(1( 30 5 0 8 00 422 0 2 8 00 42 0 2 ππ hhhhhh rrr dp pr p r dp pr p == + = + ∫∫ ∞∞ PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 6 of 9 and 1 32)/( 132 ))/(1( 32|)(| 3 0 3 0 0 42 0 23 032 =⎟ ⎠ ⎞ ⎜ ⎝ ⎛= + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= ∫∫ ∞ π ππ φ hhhh r r rdp pr prpdp . (b) (5 points) Use )( prφ to calculate <p2>, in the ground state of hydrogen. Answer: 2 0 2 2 r p h>=< Solution: ∫∫∫ ∞∞ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛==>=< 0 42 0 43 0 0 423222 ))/(1( 32||4|)(| dp pr prdpppdppp hh r π φπφ . From the integral tables I see that )/(tan 16 1 )(16)(24 7 )(6)( 1 3222222322 2 422 4 ax axaa x xa x xa xadx xa x −+ + + + − + = +∫ and hence 33 0 422 4 32216 1 )( aa dx xa x ππ == +∫ ∞ . Thus, 2 0 2 3 0 8 0 3 0 0 422 0 4 8 0 3 0 0 42 0 43 02 )/(32)/( 132 ))/(()/( 132 ))/(1( 32 rrr r dp pr p r rdp pr prp h hhh hhhhh =⎟ ⎠ ⎞ ⎜ ⎝ ⎛= + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛= + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛>=< ∫∫ ∞∞ π π ππ Problem 5 (20 points): Angular momentum is the vector operator given by opzopyopx opop ppp zyx zyx prL )()()( ˆˆˆ =×= rrr Hence, opyopzopx pzpyL )()()( −= opzopxopy pxpzL )()()( −= opxopyopz pypxL )()()( −= . (a) (5 points) Starting with the canonical commutations for position and momentum, work out the following commutators: yixLz h=],[ , xiyLz h−=],[ , 0],[ =zLz , yxz pipL h=],[ , xyz pipL h−=],[ , 0],[ =zz pL . Solution: yixpyxypxypxxpxypxpxL xxxyxyz h=−=−=−=−= ],[],[],[],[],[],[ xiypxyxpyypyxpyypxpyL yyxyxyz h−===−=−= ],[],[],[],[],[],[ 0],[],[],[],[ =−=−= zypzxpzypxpzL xyxyz yyxxyxxxyxxyxz pippxpxppyppxppypxppL h===−=−= ],[],[],[],[],[],[ PHY4604 Fall 2007 Problem Set 5 Solutions Department of Physics Page 7 of 9 xxyyxyxyyyxyyz pippypyppyppxppypxppL h−=−=−=−=−= ],[],[],[],[],[],[ 0],[],[],[],[ =−=−= zxzyzxyzz pyppxppypxppL (b) (5 points) Use the results of (a) to prove that yxz LiLL h=],[ . Solution: yxz yzyzzzzz yzzzyzzxz Lizpixpi pzLpLzpyLpLy zpLypLzpypLLL hhh =−+−= =−−+ =−=−= 00 ],[],[],[],[ ],[],[],[],[ (c) (5 points) Evaluate the commutators ],[ 2rLz and ],[ 2pLz , where 2222 zyxr ++= and 2222 zyx pppp ++= . Answer: 0],[ 2 =rLz and 0],[ 2 =pLz Solution: 0 ],[],[],[],[],[],[],[],[ 222222 =−−+= +++=+=++= xyiyxiyxixyi yyLyLyxxLxLxyLxLzyxLrL zzzzzzzz hhhh 0 ],[],[],[],[],[],[],[],[ 222222 =−−+= +++=+=++= yxxyxyyx yyzyzyxxzxzxyzxzzyxzz ppippippippi ppLpLpppLpLppLpLpppLpL hhhh Note that one can show (in a similar way) that 0],[],[ 22 == pLrL ii for i = x, y, z. (d) (5 points) Show that the Hamiltonian V m pH += 2 2 commutes with all three components of L r provided that V depends only on 222 zyxr ++= . Solution: )](,[)](2/,[],[ 2 rVLrVmpLHL iii =+= where i = x, y, z and where I used the results from (c). Now 0)(coscotsin)](,[ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ = rVirVLx φ φθ θ φh 0)(sincotcos)](,[ =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ∂ ∂ + ∂ ∂ −= rVirVLy φ φθ θ φh 0)()](,[ = ∂ ∂ −= rVirVLz φ h Thus, 0],[ =HLi . Problem 6 (20 points): The three components of the vector hermitian operator zJyJxJJ zyx ˆˆˆ ++= r obey the “lie algebra”: [Ji,Jj] = iεijkJk, where J1 = Jx, J2 = Jy, and J3 = Jz. (a) (3 points) Prove that [J2,Jk] = 0 (k=1,2,3), where J2 = Jx2 + Jy2 + Jz2. Solution: