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Fund Quantum Mechanics Lect &amp; HW Solutions 49

# Fund Quantum Mechanics Lect &amp; HW Solutions 49 -...

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3.6. THE HARMONIC OSCILLATOR 31 3.6.3.1 Solution harmc-a Question: Verify that the sets of quantum numbers shown in the spectrum Fgure 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 ¯ or deFning N = n x + n y + n z , E n x n y n z = 2 N + 3 2 ¯ 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 ¯ . Similarly, in each of the three sets of the second energy level in Fgure 3.3, the three quantum numbers n x , n y , and n z add up to N = 1, giving this state energy 5 2 ¯ . ±or the third energy level, the three quantum numbers of each set add up to N = 2, giving energy 7 2 ¯ , and for the fourth set, the quantum numbers in each of the ten sets add up to N = 3 for an energy 9 2 ¯ . 3.6.3.2 Solution harmc-b Question: Verify that there are no sets of quantum numbers missing in the spectrum Fgure 3.3; the listed ones are the only ones that produce those energy levels.
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Unformatted text preview: 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 ¯ hω or deFning N = n x + n y + n z , E n x n y n z = 2 N + 3 2 ¯ hω 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. If any one of n x , n y , or n z would be positive, then so would be N . So there is only one state n x = n y = n z = 0. ±or the second energy level, N = 1. To get a nonzero sum N , you must have one of n x , n y , and n z to be nonzero, but not more than 1, or N would be more than 1 too. Also, if one of...
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