PS4Solutions - CHEM 120A Problem Set 4 Solutions David...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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. We’ll 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 ψ x ( x ) ψ y ( y ) ψ z ( z ) = Eψ x ( x ) ψ y ( y ) ψ 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 ψ x ( x ) ∂ 2 ψ x ( x ) ∂x 2- mkx 2 ~ 2- 2 mE x ~ 2 + 1 ψ y ( y ) ∂ 2 ψ y ( y ) ∂y 2- mky 2 ~ 2- 2 mE y ~ 2 + 1 ψ z ( z ) ∂ 2 ψ 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 – ψ x ( x ), ψ y ( y ), and ψ 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 ~ ω (1c) Where ω = 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/ ~ ω 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)...
View Full Document

This note was uploaded on 03/16/2010 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at University of California, Berkeley.

Page1 / 6

PS4Solutions - CHEM 120A Problem Set 4 Solutions David...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online