PS 4 Sol - CHEM 120A Problem Set 4 Solutions David Hoffman...

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Unformatted text preview: CHEM 120A Problem Set 4 Solutions David Hoffman , Tara Yacovitch , Doran Bennett 2/22/2010 1. The solutions can be found in the McQuarrie Solutions manual. 2. Here we follow the same procedure that we saw in class. Well assume that our wave function of three variables can be expressed as a product of three wave functions of a single variable. What we get is,- ~ 2 2 m 2 + k ( x 2 + y 2 + z 2 ) / 2 x ( x ) y ( y ) z ( z ) = E x ( x ) y ( y ) z ( z ) (1a) Because E is a constant we can conveniently express it as the sum of three constants, E = E x + E y + E z . Using this we can re-express equation ( 1a ) as, 1 x ( x ) 2 x ( x ) x 2- mkx 2 ~ 2- 2 mE x ~ 2 + 1 y ( y ) 2 y ( y ) y 2- mky 2 ~ 2- 2 mE y ~ 2 + 1 z ( z ) 2 z ( z ) z 2- mkz 2 ~ 2- 2 mE z ~ 2 = 0 (1b) We can see that each value in brackets depends on only one variable. In fact, each is identical to the one dimensional harmonic oscillator. Therefore each of our one dimensional wave functions x ( x ), y ( y ), and z ( z ) are 1D harmonic oscillator wave functions. This means that our energy is the sum of three 1D harmonic oscillator energies, i.e., E n x n y n z = n x + n y + n z + 3 2 ~ (1c) Where = p k/m . See table 1 for a list of the first five energy levels and the accom- panying configurations. Note that the degeneracy of the n t h level is, g = ( n + 1)( n + 2) 2 (2) 3. Before we begin it should be noted that the given wave function is already normalized, you can verify that for yourself. 1 Table 1: First 5 Energy Levels of a 3D Harmonic Oscillator E/ ~ g ( n x ,n y ,n z ) 3 / 2 1 (0,0,0) 5 / 2 3 (1,0,0),(0,1,0),(0,0,1) 7 / 2 6 (2,0,0),(0,2,0),(0,0,2),(1,1,0),(0,1,1),(1,0,1) 9 / 2 10 (3,0,0),(0,3,0),(0,0,3),(2,1,0),(2,0,1),(1,2,0),(0,2,1),(1,0,2),(0,1,2),(1,1,1)...
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PS 4 Sol - CHEM 120A Problem Set 4 Solutions David Hoffman...

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