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Unformatted text preview: 3.6. THE HARMONIC OSCILLATOR 31 188.8.131.52 Solution harmc-a Question: Verify that the sets of quantum numbers shown in the spectrum figure 3.3 do indeed produce the indicated energy levels. Answer: The generic expression for the energy is E n x n y n z = 2 n x + 2 n y + 2 n z + 3 2 h or defining N = n x + n y + n z , E n x n y n z = 2 N + 3 2 h Now for the bottom level, n x = n y = n z = 0, so N = n x + n y + n z = 0, this state has energy 3 2 h . Similarly, in each of the three sets of the second energy level in figure 3.3, the three quantum numbers n x , n y , and n z add up to N = 1, giving this state energy 5 2 h . For the third energy level, the three quantum numbers of each set add up to N = 2, giving energy 7 2 h , and for the fourth set, the quantum numbers in each of the ten sets add up to N = 3 for an energy 9 2 h . 184.108.40.206 Solution harmc-b Question: Verify that there are no sets of quantum numbers missing in the spectrum figure 3.3; the listed ones are the only ones that produce those energy levels. Answer: The generic expression for the energy is E n x n y n z = 2 n x + 2 n y + 2 n z + 3 2 h or defining N = n x + n y + n z , E n x n y n z = 2 N + 3 2 h Now for the bottom level, N = 0, and since the three quantum numbers n x , n y , and n z cannot be negative, the only way that N = n x + n y + n z can be zero is if all three numbers are zero....
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This note was uploaded on 11/13/2011 for the course PHY 4458 taught by Professor Garvin during the Fall '11 term at University of Florida.
- Fall '11