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Sakurai_Ch4_1_2_3_4_8_10_12

Homework 1: # 4.1, 4.2, 4.3, 4.4, 4.8, 4.10, 4.12 Michael Good Jan 20, 2005 Problem 4.1 Calculate the three lowest energy levels, together with their degeneracies, for the following systems (assume equal mass distinguishable particles): a. Three noninteracting spin 1 2 particles in a box of length L. b. Four noninteracting spin 1 2 particles in a box of length L. Solution: In a box we have three dimensions. So the energy for one particle is E 1 nx,ny,nz = (n 2
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  Homework 1: # 4.1, 4.2, 4.3, 4.4, 4.8, 4.10, 4.12 Michael GoodJan 20, 2005 Problem 4.1Calculate the three lowest energy levels, together with their degeneracies, forthe following systems (assume equal mass distinguishable particles):a. Three noninteracting spin 12 particles in a box of length L .b. Four noninteracting spin 12 particles in a box of length L .Solution:In a box we have three dimensions. So the energy for one particle is E  1 n x ,n y ,n z = ( n 2 x + n 2 y + n 2 z ) π 2 ¯ h 2 2 mL 2 If the particles are distinguishable, the composite wave functions are prod-ucts, as explained in an example by Griffiths, section 5.1. Therefore the energyof three particles are E  T  = E  1 + E  2 + E  3 This is E  T  = 3  i,j n 2 ij π 2 ¯ h 2 2 mL 2 where the index i represents the particle, and j represents the dimension. Notethat there are 9 distinct n ’s. The n ’s are non-zero integers, so the lowest energyis where all the n values are equal to 1. That is just 1 2 ∗ 9 = 9 E  1 =9 π 2 ¯ h 2 2 mL 2 The second lowest energy is where only one of the n values is 2 and the restare still 1. That is 2 2 + 8 = 12. So E  2 =12 π 2 ¯ h 2 2 mL 2 1  The third lowest energy is where two of the n ’s have value 2 and rest arestill 1. That is 2 2 + 2 2 + 7 = 15. So E  3 =15 π 2 ¯ h 2 2 mL 2 The degeneracy is found by multiplying the number of spatial functions bythe number of spin functions. There are 8 different spin functions. | − −− , | − − +  , | − + − , | + −− , | + + − , | − ++  , | + − +  , | + ++  or 2 3 = 8 combinations. The spatial functions are found by looking at thepossibilities for the n ’s. For E  1 there is only one possibility, that is, with all the n values equal to 1. For E  2 there are 9 possibilities, where each of the 9 valuesof  n could be 2 where the rest are 1. For E  3 its a little more difficult, becausetwo of the n ’s may have the value 2 so we use the combination formula ’9 choose2’. N  !( N  − n )! n !=9!(9 − 2)!2!= 36possibilities for E  3 . Multiplying the spatial possibilities times the spin pos-sibilities gives the degeneracy. E  1 → 1 ∗ 8 = 8 E  2 → 9 ∗ 8 = 72 E  3 → 36 ∗ 8 = 288For four particles we have 4 ∗ 3 = 12 distinct n ’s. So for the lowest energy,all of them are 1: E  1 =12 π 2 ¯ h 2 2 mL 2 For the second lowest energy we have 2 2 + 11 = 15: E  2 =15 π 2 ¯ h 2 2 mL 2 For the third lowest energy we have 2 2 + 2 2 + 10 = 18: E  3 =18 π 2 ¯ h 2 2 mL 2 There are 2 4 = 16 spin possibilities. There is only 1 spatial possibility for E  1 . There are 12 spatial possibilities for E  2 . And ’12 choose 2’ possibilities for E  3 12!(12 − 2)!2!= 662  The degeneracies are therefore E  1 → 1 ∗ 16 = 16 E  2 → 12 ∗ 16 = 192 E  3 → 66 ∗ 16 = 1056Problem 4.2Let T  d denote the translation operator (displacement vector  d ); D (ˆ n,φ ), therotation operator (ˆ n and φ are the axis and angle of rotation, respectively);and π the parity operator. Which, if any, of the following pairs commute? Why? ã T  d and T  d  (  d and  d  in different directions). ã D (ˆ n,φ ) and D (ˆ n  ,φ  ) (ˆ n and ˆ n  in different directions). ã T  d and π . ã D (ˆ n,φ ) and π .Solution:a. T  d and T  d  commute as can be seen graphically by translating a vector intwo directions in different orders or algebraically(where r and d are vectors) by T  d φ ( r ) = φ ( r + d ) → T  d  T  d φ ( r ) = φ ( r + d + d  ) T  d  φ ( r ) = φ ( r + d  ) → T  d T  d  φ ( r ) = φ ( r + d  + d )Therefore[ T  d , T  d  ] φ ( r ) = 0and[ T  d , T  d  ] = 0because φ ( r ) can be any function of vector r .b. D (ˆ n,φ ) and D (ˆ n  ,φ  ) do not commute. As we know that rotations alongdifferent axis do not commute and D acts on φ to cause a rotation. This can bedone by rotating a book around two different axis.c. T  d and π do not commute. This can be done graphically by translatinga vector, then mirroring it, and see if that point matches up if you mirror thevector then translate it. Algebraically,3  πφ ( r ) = φ ( − r ) → T  d πφ ( r ) = φ ( − r + d ) T  d φ ( r ) = φ ( r + d ) → π T  d φ ( r ) = φ ( − r − d )Therefore[ T  d ,π ] φ ( r )  = 0[ T  d ,π ]  = 0d. D (ˆ n,φ ) and π commute. Graphically by rotating then mirroring, andchecking to see if that matches up with mirroring then rotating. Algebraically, D φ ( r ) = φ ( x ) → π D φ ( r ) = φ ( − x ) πφ ( r ) = φ ( − r ) → D πφ ( r ) = D φ ( − r ) = φ ( − x )Therefore[ D ,π ] φ ( r ) = 0[ D ,π ] = 0Problem 4.3A quantum-mechanical state Ψ is known to be simultaneous eigen-state of twoHermitian operators A and B which anticommute, AB + BA = 0What can you say about the eigenvalues of  A and B for stat Ψ? Illustrate yourpoint using the parity operator (which can be chosen to satisfy π = π − 1 = π † )and the momentum operator.Solution:Taking a look at the eigenvalues we see Aψ = aψ Bψ = bψB ( Aψ ) = B ( aψ ) = abψA ( Bψ ) = A ( bψ ) = baψ And we have ABψ = − BAψ 4
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