Introduction to Quantum Mechanics I - Final Exam Solutions | PHY 4604, Exams of Physics

Download Introduction to Quantum Mechanics I - Final Exam Solutions | PHY 4604 and more Exams Physics in PDF only on Docsity! PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 1 of 25 PHY 4604 Final Exam Solutions Wednesday December 6, 2006 (Total Points = 100) Problem 1 (25 points): Consider the following one-dimensional potential:     ∞+ −= 0 0 )( VxV Lx Lx Lx ≥ <≤ < 0 Where V0 is a positive (real) constant. (A) Suppose that particles with mass m and energy E > 0 enter from the left in region 1 and travel to the right and encounter the potential V(x). Let 22 2 2 ε mL E h= and E Vr 01+= . (1) (5 points) Calculate the probability that particles entering from the left in region 1 will be reflected back (i.e. calculate the ratio of the probability flux in region 1 traveling to the left to the incident flux in region 1 traveling to the right, 11 / jjPR rs = ). Express your answer in terms of ε and r. Answer: PR = 1. Solution: We look for solutions of the time-independent Schrödinger equation )()()()( 2 2 22 xExxV dx xd m ψψψ =+− h or )())(( 2)( 22 2 xxVEm dx xd ψψ −−= h with h/)(),( iEtextx −=Ψ ψ . In the region x < 0 (region 1) for E > 0 and V(x) = 0 we have )()(2)( 222 2 xkxmE dx xd ψψψ −=−= h with L mEk ε== 2 2 h and 22 222 22 ε mLm kE hh == The most general solution is ikxikx BeAex −+ +=)(1ψ , In the region 0 < x < L (region 2) we have iqxiqx DeCex −+ +=)(2ψ , with rkVEmq =+= 2 0 )(2 h , where E V k qr 01+== . In the region x > L (region 3) we have 0)(3 =xψ , The boundary conditions at x = L implies that 0)(2 =Lψ and hence 0=+ −+ iqLiqL DeCe or iqLCeD 2+−= . Thus, )()( 22 iqxiqLiqx eeeCx −++ −=ψ , At x = 0 the wave function must be continuous with a continuous derivative which yields x V(x) -V0 V = +infinity L Region 1 Region 2 Region 3 PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 2 of 25 )0()0( 21 ψψ = and )1( 2iqLeCBA +−=+ 0 2 0 1 )()( == = xx dx xd dx xd ψψ and )1( 2iqLeCiqikBikA ++=− Thus, )1( 2iqLeCBA +−=+ and )1( 2iqLeCrBA ++=− And ))1()1((2 2iqLerrCA +−++= and ))1()1((2 2iqLerrCB +++−−= Thus, iqLerr AC 2)1()1( 2 +−++ = and A err errB iqL iqL 2 2 )1()1( )1()1( + + −++ ++− −= . The relfection probability is given by 1 )1)(1()1)(1()1()1( )1)(1()1)(1()1()1( )1()1( )1()1( )1()1( )1()1( )1()1( )1()1( || || 2222 2222 2 2 2 22 2 2 2 2 = +−++−+−++ −++−++++− =       −++ ++−       −++ ++− = −++ ++− == −+ −+ − − + + + + iqLiqL iqLiqL iqL iqL iqL iqL iqL iqL m k m k R errerrrr errerrrr err err err err err err A B P h h (2) (5 points) Calculate the probability that particles traveling to the right in region 2 will be reflected back (i.e. calculate the ratio of the probability flux in region 2 traveling to the left to flux in region 2 traveling to the right, 22 / jjR rs = ). Express your answer in terms of ε and r. Answer: R = 1. Solution: We see that 1 || || 22 2 2 === + iqL m q m q e C D R h h . (3) (5 points) Calculate the ratio of the probability flux in region 2 traveling to the right to the incident flux in region 1 traveling to the right, 12 / jjP rr = ). Express your answer in terms of ε and r. What are the maximum and minimum values of P(ε) (express your answer in terms of r)? Evaluate the maximum and minimum values of P(ε) for the case V0 = 3E. Explain what is going on. Answer: )2cos()1(1 2 22 εrrr rP −++ = , 2 11 2min →= =rr P , 22max →= =rrP Solution: We see that PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 5 of 25 Solution: There are three bound states. 202 2 0 2 ε mL E h−= with ε0 ≈ 8.17. Solution: For α = 11π/4 we have three bound states. The ground state has the smallest y, which is y ≈ 2.81 and hence 74.66 2 )17.8( 2 ))81.2()4/11(( 2 )( 2 2 2 2 2 2 22 2 2 2 min 2 2 2 0 mLmLmL y mL E hhhh −=−≈−−≈−−= πα -1 0 1 2 3 4 5 0.0 1.6 3.1 4.7 6.3 7.9 9.4 f(y)g(y) y π/2 π 3π/2 2π α = 11π/4 Problem 2 (25 points): The Hamiltonian of a charged particle with intrinsic spin S r at rest in a magnetic field is BSBH rrrr ⋅−=⋅−= γµ , where m q =γ (q is the electric charge, m is the mass). For spin 1 we have zSySxSS zyx ˆˆˆ ++= r with           = 010 101 010 2 h xS           − − = 010 101 010 2 hiSy           − = 100 000 001 hzS (A) (2 points) Show that           =++= 100 010 001 2 22222 hzyx SSSS . Solution: We see that           =                     = 101 020 101 2 010 101 010 010 101 010 2 22 2 hh xS           − − =           − −           − − − = 101 020 101 2 010 101 010 010 101 010 2 22 2 hh yS           =           −          − = 100 000 001 100 000 001 100 000 001 222 hhzS Hence, PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 6 of 25           =++= 100 010 001 2 22222 hzyx SSSS . (B) (6 points) The W+ boson is a spin 1 elementary particle with charge q = +e and mass MW. Suppose a W+ is at rest in a uniform magnetic field which points in the z-direction, zBB ˆ0= r . What are the energy levels and the corresponding eigenkets of the system? What is the ground state energy and ground state eigenket? Express your answers in terms of 0Bγω = . Answer: The ground state (i.e. lowest) energy is ωh−=0E with eigenket           >=>= 0 0 1 11|| 0χ . The 1st excited state has energy 01 =E with eigenket           >=>= 0 1 0 10|| 1χ . The 2nd excited state has energy ωh+=2E with eigenket           >=−>= 1 0 0 11|| 2χ . Solution: The eigenvalues of Sz are determined from 0 00 00 00 = −− − − λ λ λ h h and hence 0)()( =+− λλλ hh Thus, there are three eigenvalues hh −= ,0,λ . The eigenket of Sz corresponding to λ = 0 is determined from           =           − =                     − 0 0 0 0 100 000 001 c a c b a hh which implies that a = c = 0 and           >= 0 1 0 10| The eigenkets of Sz corresponding to h±=λ is determined from           ±=           − =                     − c b a c a c b a hhh 0 100 000 001 which implies that a = ±a and -c = ±c and hence           >= 0 0 1 11| and           >=− 1 0 0 11| and where >>= mmmSz 1|1| h . The Hamiltonian is given by zz SSBBSH ωγγ −=−=⋅−= 0 rr . Hence there are three energy levels ωh−=0E , 01 =E , ωh+=2E . The ground state (i.e. lowest) energy is ωh−=0E with eigenket PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 7 of 25           >=>= 0 0 1 11|| 0χ . The 1st excited state has energy 01 =E with eigenket           >=>= 0 1 0 10|| 1χ . The 2nd excited state has energy ωh+=2E with eigenket           >=−>= 1 0 0 11|| 2χ . (C) Suppose that at t = 0 the W+ boson is in the state           >= 1 1 1 3 1)0(| χ . (1) (2 points) What is >)(| tχ ? Express your answer in terms of 0Bγω = . Answer:           >= − + ti ti e e t ω ω χ 1 3 1)(| Solution: We know that hhh / 22 / 11 / 00 210 |||)(| tiEtiEtiE ececect −−− >+>+>>= χχχχ where 3 1 1 1 1 3 1)001()0(|00 =           >==< χχc 3 1 1 1 1 3 1)010()0(|11 =           >==< χχc 3 1 1 1 1 3 1)100()0(|22 =           >==< χχc Hence, PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 10 of 25 ( ) ( ) ( ) 011 3 01 3 1 100 000 001 1 3 )(||)()( 0 0 =−=           − =                     − >=>=<< − + +− − + +− hh h ti ti titi ti ti titi zz e e ee e e eetSttS ω ω ωω ω ω ωωχχ (E) (5 points) What is ∆Sx(t), ∆Sy(t), and ∆Sz(t) in the state >)(| tχ ? Express your answer in terms of 0Bγω = . Answer: )(cos23 3 )( 2 ttSx ω−=∆ h , )(sin23 3 )( 2 ttSy ω−=∆ h , h 3 2)( =∆ tSz . Solution: We see that ( ) ( ) ( ) ( ) ( )1)(cos2 3 1)cos()( 3 )cos(22)cos(2 6 21 6 1 101 020 101 1 6 )(||)()( 2 22 22 2 22 0 +=++= ++=           + + =                     >=>=<< −+ +− −+ −+ +− − + +− ttee tete ee ee ee e e eetSttS titi titi titi titi titi ti ti titi xx ωω ωω χχ ωω ωω ωω ωω ωω ω ω ωω hh hh h ( ) ( ) ( ))(cos23 9 )(cos1 3 )(cos1)(cos2 3 )()())(( 2 2 2 3 2 2 2 3 82 2 222 tt tttStStS xxx ωω ωω −=−= −+=><−>=<∆ hh h and hence )2cos(2 3 )(sin21 3 )(cos23 3 )( 22 ttttSx ωωω −=+=−=∆ hhh . Also, ( ) ( ) ( ) ( ) ( )1)(sin2 3 1)sin()( 3 )sin(22)sin(2 6 21 6 1 101 020 101 1 6 )(||)()( 2 22 22 2 22 0 +=+−−= −+=           +− − =                     − − >=>=<< −+ +− −+ −+ +− − + +− tteei tietie ee ee ee e e eetSttS titi titi titi titi titi ti ti titi yy ωω ωω χχ ωω ωω ωω ωω ωω ω ω ωω hh hh h PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 11 of 25 ( ) ( ) ( ))(sin23 9 )(sin1 3 )(sin1)(sin2 3 )()())(( 2 2 2 3 2 2 2 3 82 2 222 tt tttStStS yyy ωω ωω −=−= −+=><−>=<∆ hh h and hence )2cos(2 3 )(cos21 3 )(sin23 3 )( 22 ttttSy ωωω +=+=−=∆ hhh . Also, ( ) ( ) ( ) 3 211 3 01 3 1 100 000 001 1 3 )(||)()( 222 2 22 0 hhh h =+=           =                     >=>=<< − + +− − + +− ti ti titi ti ti titi zz e e ee e e eetSttS ω ω ωω ω ω ωωχχ 3 2)()())(( 2 222 h=><−>=<∆ tStStS zzz and hence h 3 2)( =∆ tSz . Problem 3 (25 points): Suppose we have two vector operators opJ )( 1 r and opJ )( 2 r with 0])(,)[( 21 =opop JJ rr and each of the vectors obey the same SU(2) “lie algebra”: opkijkopjopi JiJJ )(])(,)[( 111 ε= and opkijkopjopi JiJJ )(])(,)[( 222 ε= . The states |j1m1> are the eigenkets of opJ )( 2 1 and opzJ )( 1 and the states |j2m2> are the eigenkets of opJ )( 2 2 and opzJ )( 2 as follows: >>= >+>= 111111 111111 2 1 ||)( |)1(|)( mjmmjJ mjjjmjJ opz op >>= >+>= 222222 222222 2 2 ||)( |)1(|)( mjmmjJ mjjjmjJ opz op Also we know that >±±−+>= >±±−+>= ± ± 1|)1()1(|)( 1|)1()1(|)( 222222222 111111111 mjmmjjmjJ mjmmjjmjJ op op where opyopxop JiJJ )()()( 111 ±= ± and opyopxop JiJJ )()()( 222 ±= ± . Now consider the vector sum of the two operators, opopop JJJ )()()( 21 rrr += or opiopiopi JJJ )()()( 21 += for i = 1,2, 3. PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 12 of 25 (A) (1 point): Show that opzopzopopopopopop opopopopopopopopop JJJJJJJJ JJJJJJJJJJJ )()(2)()()()()()( )()(2)()()()()()()( 212121 2 2 2 1 21 2 2 2 12121 2 ++++= ⋅++=+⋅+=⋅= +−−+ rrrrrrrr Solution: First we note that )( )( 112 1 1 112 1 1 −+ −+ −= += JJJ JJJ iy x and )( )( 222 1 2 222 1 2 −+ −+ −= += JJJ JJJ iy x Hence, zz zz zzyyxx JJJJJJJJ JJJJJJJJJJJJ JJJJJJJJJJJJJ 212121 2 2 2 1 2122112 1 22112 12 2 2 1 212121 2 2 2 121 2 2 2 1 2 2 2))(())(( 2222 ++++= +−−−++++= ++++=⋅++= +−−+ −+−+−+−+ rr (B) Now consider the case where j1 = 1 and j2 = 1 (i.e. 3 × 3) and define the states as follows: 1111 1110 1111 11|| 10|| 11|| >−=> >=> >=> −Y Y Y and 2211 2210 2211 11|| 10|| 11|| >−=> >=> >=> −Y Y Y Consider the three superposition states 211111210110211111 ||3 1|| 3 1|| 3 1| >>+>>−>>>≡ −− YYYYYYaψ 211111211111 ||2 1|| 2 1| >>−>>>≡ −− YYYYbψ 211111210110211111 ||6 1|| 3 2|| 6 1| >>+>>+>>>≡ −− YYYYYYcψ Calculate the following: (1) (1 point) >=< aa ψψ | (2) (1 point) >=< bb ψψ | (3) (1 point) >=< cc ψψ | (4) (1 point) >=< ba ψψ | (5) (1 point) >=< ca ψψ | (6) (1 point) >=< cb ψψ | Calculate the following and express your answer in terms of |ψa>, |ψb>, and |ψc> : (1) (1 point) >=+ aJJ ψ|)( 2 2 2 1 (2) (1 point) >=+ bJJ ψ|)( 2 2 2 1 (3) (1 point) >=+ cJJ ψ|)( 2 2 2 1 (4) (1 point) >=azJ ψ| PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 15 of 25 >−= >>−−−+>>++>−>−−= >>−−+>>−>−>−+ >>−−>>+ >>+>−>−=       >>−+>>−>−>+       >>−+>>−>−>+       >>−+>>−>−>=       >>−+>>−>−>++= >++ +− −+ +−−+ +−−+ a zz zz azz JJ JJ JJ JJJJJJ JJJJJJ ψ ψ |4 11|11| 3 1)22(10|10| 3 1)22(11|11| 3 1)22( 11|11| 3 1)1)(1)(2(10|10| 3 1)0)(0)(2(11|11| 3 1)1)(1)(2( 11|11| 3 12210|10| 3 122 10|10| 3 12211|11| 3 122 11|11| 3 110|10| 3 111|11| 3 1)2( 11|11| 3 110|10| 3 111|11| 3 1)( 11|11| 3 110|10| 3 111|11| 3 1)( 11|11| 3 110|10| 3 111|11| 3 1)2( |)2( 212121 212121 2121 2121 21212121 21212121 21212121 212121212121 212121 and hence >>=−>=++++>= +−−+ aaazza JJJJJJJJJ ψψψψ |0|)44(|)2(| 212121 2 2 2 1 2 . Thus, the state |ψa> is an eigenstate of J2 and Jz with j = 0 and mj = 0 (i.e. |ψa>=|00>). >−= >>−−−>>+−+>−>−= >>−−−>−>−+ >>+ >>−=       >>−−>−>+       >>−−>−>+       >>−−>−>=       >>−−>−>++= >++ +− −+ +−−+ +−−+ b zz zz bzz JJ JJ JJ JJJJJJ JJJJJJ ψ ψ |2 11|11| 3 1)2(10|10| 2 1)22(11|11| 2 1)2( 11|11| 2 1)1)(1)(2(11|11| 2 1)1)(1)(2( 10|10| 2 122 10|10| 2 122 11|11| 2 111|11| 2 1)2( 11|11| 2 111|11| 2 1)( 11|11| 2 111|11| 2 1)( 11|11| 2 111|11| 2 1)2( |)2( 212121 2121 21 21 212121 212121 212121 2121212121 212121 and hence >>=−>=++++>= +−−+ bbbzzb JJJJJJJJJ ψψψψ |2|)24(|)2(| 212121 2 2 2 1 2 . PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 16 of 25 Thus, the state |ψb> is an eigenstate of J2 and Jz with j = 1 and mj = 0 (i.e. |ψb>=|10>). >= >>−−+>>++>−>−= >>−−+>>+>−>−+ >>−+>>+ >>+>−>=       >>−+>>+>−>+       >>−+>>+>−>+       >>−+>>+>−>=       >>−+>>+>−>++= >++ +− −+ +−−+ +−−+ c zz zz czz JJ JJ JJ JJJJJJ JJJJJJ ψ ψ |2 11|11| 6 1)24(10|10| 3 2)11(11|11| 6 1)24( 11|11| 6 1)1)(1)(2(10|10| 3 2)0)(0)(2(11|11| 6 1)1)(1)(2( 11|11| 3 22210|10| 6 122 10|10| 6 12211|11| 3 222 11|11| 6 110|10| 3 211|11| 6 1)2( 11|11| 6 110|10| 3 211|11| 6 1)( 11|11| 6 110|10| 3 211|11| 6 1)( 11|11| 6 110|10| 3 211|11| 6 1)2( |)2( 212121 212121 2121 2121 21212121 21212121 21212121 212121212121 212121 and hence >>=+>=++++>= +−−+ ccczzc JJJJJJJJJ ψψψψ |6|)24(|)2(| 212121 2 2 2 1 2 . Thus, the state |ψc> is an eigenstate of J2 and Jz with j = 2 and mj = 0 (i.e. |ψc>=|20>). Problem 4 (25 points): Consider the case of two non-interacting particles both with mass m in a one-dimensional infinite square well given by V(x) = 0 for 0 < x < L, and V(x) = ∞. For one particle we know that the stationary states of Schrödinger’s equation are given by h/)(),( tiEnn nextx −=Ψ ψ and 0 2EnEn = , and n is a positive integer and 2 22 0 2mL E hπ= and where )/sin(2)( Lxn L xn πψ = . For two (non-interacting) particles we look for a solution of the form hh / 21 / 2121 )()(),(),,( iEtiEt exxexxtxx −− ==Ψ ψψψ with E m p m p xx =+ 2 )( 2 )( 22 2 1 . (A) (1 point): Show that )()(),( 2121 xxxx βααβ ψψψ = , is a solution to the two particle non- interaction Schrödinger equations, where α and β are positive integers and ψα(x) and ψβ(x) are the one particle stationary state solutions. Show that the allowed energy levels are given by 0 22 )( EE βααβ += . Solution: We see that 0 L Two Particles in a Box ψ(x1) ψ(x2) PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 17 of 25 hh / 21 / 2121 )()(),(),,( iEtiEt exxexxtxx −− ==Ψ ψψψ with E m p m p xx =+ 2 )( 2 )( 22 2 1 . Thus, ),(),( 2 ),( 2 2122 21 22 2 1 21 22 xxE dx xxd mdx xxd m ψ ψψ =−− hh )()()()( 2 )()( 2 2122 2 2 1 2 2 1 1 2 2 2 xxE dx xdx mdx xdx m ψψψψψψ =−− hh E dx xd xmdx xd xm =−− 2 2 2 2 2 2 2 1 1 2 1 2 )( )( 1 2 )( )( 1 2 ψ ψ ψ ψ hh Hence, E = E1 + E2 and )()( 2 1121 1 22 xE dx xd m ψψ =− h )( )( 2 2222 2 22 xE dx xd m ψ ψ =− h The total energy is therefore given by 2 2222 21 2 )()()( mL EEE βαπβααβ + =+= h 2 222 1 2 )( mL E απα h = 2 222 2 2 )( mL E βπβ h = where α = 1, 2, 3, ... and β = 1, 2, 3, ... and )/sin()/sin(2)()(),( 212121 LxLxL xxxx βπαπψψψ βααβ == )/(sin)/(sin4)()()()(),( 2 2 1 2 2212 * 1 * 21 LxLxL xxxxxx βπαπψψψψρ βαβααβ == This solution corresponds to the case where the two particles are distinguishable. Define the following three probabilities. Let LLPαβ be the probability of finding both particles in the left 1/3 of the box as follows: ∫ ∫= 3/ 0 3/ 0 21 2 21 |),(| L L LL dxdxxxP αβαβ ψ . Let LRPαβ be the probability of finding one particle in the left 1/3 of the box and one particle in the right 1/3 of the box as follows: ∫ ∫= 3/ 0 3/2 21 2 21 |),(| L L L LR dxdxxxP αβαβ ψ . Let CCPαβ be the probability of finding both particles in the center 1/3 of the box as follows: ∫ ∫= 3/2 3/ 3/2 3/ 21 2 21 |),(| L L L L CC dxdxxxP αβαβ ψ . Calculate the following observables for the ground state and the first excited state for the distinguishable case. (Show your work and fill in the table) Points Observable Ground State 1st Excited State 2 αβE 2 LLPαβ PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 20 of 25 ( ) ( ) ( ) πππ ππ ππ ππππ ππ 8 33 4 3 8 3 )sin( 2 1)sin( 4 1)/2cos(1)/4cos(1 )/2cos()/4cos(1)/3sin()/sin(2 3/2 0 3/4 0 3/ 0 3/ 0 3/ 0 3/ 0 13 −=−−= −=− =−== ∫∫ ∫∫ yydxLx L dxLx L dxLxLx L dxLxLx L I LL LL L ( ) ( ) ( ) πππππ ππ ππ ππππ π π π π 4 33 2 3 4 3 2 3 2 3 2 1 2 3 2 3 4 1 )sin( 2 1)sin( 4 1)/2cos(1)/4cos(1 )/2cos()/4cos(1)/3sin()/sin(2 3/4 3/2 3/8 3/4 3/2 3/ 3/2 3/ 3/2 3/ 3/2 3/ 13 =+=      −−−      += −=− =−== ∫∫ ∫∫ yydxLx L dxLx L dxLxLx L dxLxLx L I L L L L L L L L C ( ) ( ) ( ) πππ ππ ππ ππππ π π π π 8 33 4 3 8 3 )sin( 2 1)sin( 4 1)/2cos(1)/4cos(1 )/2cos()/4cos(1)/3sin()/sin(2 2 3/4 4 3/8 3/23/2 3/23/2 13 −=−−= −=− =−== ∫∫ ∫∫ yydxLx L dxLx L dxLxLx L dxLxLx L I L L L L L L L L R The results are summerized in the following table. 12 13 LI 2757.02 3 −≈− π 2067.0 8 33 −≈− π CI 0 4135.04 33 ≈+ π RI 2757.02 3 +≈+ π 2067.0 8 33 −≈− π sum 0 0 Now for distinguishable particles we have )/sin()/sin(2)()(),( 212121 LxLxL xxxx βπαπψψψ βααβ == )/(sin)/(sin4|),(|),( 2 2 1 2 2 2 2121 LxLxL xxxxD βπαπψρ αβαβ == The ground state is the state with α = 1 and β = 1, Thus, 00 2EE D = and PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 21 of 25 ( ) 2 2 1 3/ 0 22 2 3/ 0 11 2 3/ 0 3/ 0 212 2 1 2 211 4 3 3 1)/(sin2)/(sin2 )/(sin)/(sin4)(       −== = ∫∫ ∫ ∫ π ππ ππ L LL L L D LL PdxLx L dxLx L dxdxLxLx L P and ( ) 2 2 1 3/2 3/ 22 2 3/2 3/ 11 2 3/2 3/ 3/2 3/ 212 2 1 2 211 2 3 3 1)/(sin2)/(sin2 )/(sin)/(sin4)(       +== = ∫∫ ∫ ∫ π ππ ππ C L L L L L L L L D CC PdxLx L dxLx L dxdxLxLx L P and ( ) 2 2 111 3/2 22 2 3/2 0 11 2 3/2 0 3/2 212 2 1 2 211 4 3 3 1)/(sin2)/(sin2 )/(sin)/(sin4)(       −=== = ∫∫ ∫ ∫ π ππ ππ LRL L L L L L L D LR PPPdxLx L dxLx L dxdxLxLx L P The 1st excited state has α = 1 and β = 2, Thus, 01 5EE D = and ( )( )       +      −== = ∫∫ ∫ ∫ ππ ππ ππ 8 3 3 1 4 3 3 1)/2(sin2)/(sin2 )/2(sin)/(sin4)( 21 3/ 0 22 2 3/ 0 11 2 3/ 0 3/ 0 212 2 1 2 212 LL LL L L D LL PPdxLx L dxLx L dxdxLxLx L P and ( )( )       −      +== = ∫∫ ∫ ∫ ππ ππ ππ 4 3 3 1 2 3 3 1)/2(sin2)/(sin2 )/2(sin)/(sin4)( 21 3/2 3/ 22 2 3/2 3/ 11 2 3/2 3/ 3/2 3/ 212 2 1 2 212 CC L L L L L L L L D CC PPdxLx L dxLx L dxdxLxLx L P and       +      −== = ∫∫ ∫ ∫ ππ ππ ππ 8 3 3 1 4 3 3 1)/2(sin2)/(sin2 )/2(sin)/(sin4)( 21 3/2 22 2 3/2 0 11 2 3/2 0 3/2 212 2 1 2 212 RL L L L L L L D LR PPdxLx L dxLx L dxdxLxLx L P (B) For two identical bosons (i.e. particles with integral spins in the same spin state) we must use the symmetric wavefunction ( )),(),( 2 1),( 122121 xxxxxx S αβαβαβ ψψψ += (α ≠ β symmetric under 1↔2) PHY4604 Fall 2006 Final Exam Solutions Department of Physics Page 22 of 25 ( )),(),( 2 1),( 122121 xxxxxx S αααααα ψψψ += (α = β symmetric under 1↔2) Calculate the following observables for the ground state and the first excited state for the identical boson case. (Show your work and fill in the table) Points Observable Ground State 1st Excited State 2 αβE 2 LLPαβ 2 LRPαβ 2 CCPαβ Answer: Bosons Observable Ground State α = 1 β = 1 1st Excited State α = 1 β = 2 αβE 2E0 05E LLPαβ ( ) 0.03824 3 3 1 2 2 1 ≈      −= π LP ( )( ) ( ) 0.15460.07600.0786 2 3 8 3 3 1 4 3 3 1 2 2 1221 ≈+≈       +      +      −=+ πππ LLL IPP LRPαβ ( )( ) 0.03824 3 3 1 2 11 ≈      −= π RL PP ( )( ) ( )( ) 0.00260.0760-0.0786 2 3 8 3 3 1 4 3 3 1 2 121221 ≈≈       −      +      −=+ πππ RLRL IIPP CCPαβ ( ) 0.37092 3 3 1 2 2 1 ≈      += π CP ( )( ) ( ) 0.1191 4 3 3 1 2 3 3 12 1221 ≈       −      +=+ ππ CCC IPP Solution: In this case we have [ ])/sin()/sin()/sin()/sin(2),( 212121 LxLxLxLxLxx S απβπβπαπψαβ += (α ≠ β) )/sin()/sin(2),( 2121 LxLxL xxS απαπψαα = and ),(),(),( 21 int 2121 xxxxxx classicalBE αβαβαβ ρρρ += (α ≠ β) where ( )),(),(Re),( 212121int xxxxxx ∗≡ βααβαβ ψψρ (α ≠ β) and ),(),( 2121 xxxx classicalBE αααα ρρ = . where ( ))/(sin)/(sin)/(sin)/(sin2),( 22122212221 LxLxLxLxLxx classical απβπβπαπραβ += ( ))/sin()/sin()/sin()/sin(4),( 2211221int LxLxLxLxLxx βπαπβπαπραβ =