Fund Quantum Mechanics Lect & HW Solutions 51

Fund Quantum - the same from any angle 3.6.4.2 Solution harmd-b Question Show that the ground state wave function is maximal at the origin and like

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3.6. THE HARMONIC OSCILLATOR 33 3.6.4.1 Solution harmd-a Question: Write out the ground state wave function and show that it is indeed spherically symmetric. Answer: Repeating an earlier exercise, taking the generic expression ψ n x n y n z = h n x ( x ) h n y ( y ) h n z ( z ) and substituting n x = n y = n z = 0, you get the ground state eigenfunction ψ 000 = h 0 ( x ) h 0 ( y ) h 0 ( z ) . Now substitute for h 0 from table 3.1: ψ 000 = 1 ( πℓ 2 ) 3 / 4 e x 2 / 2 2 e y 2 / 2 2 e z 2 / 2 2 where the constant is as given in table 3.1. You can multiply out the exponentials: ψ 000 = 1 ( πℓ 2 ) 3 / 4 e ( x 2 + y 2 + z 2 ) / 2 2 . According to the Pythagorean theorem, x 2 + y 2 + z 2 is the distance from the origin r , so ψ 000 = 1 ( πℓ 2 ) 3 / 4 e r 2 / 2 2 . It follows that the wave function only depends on the distance r from the origin, not on the angular orientation compared to it. That is the deFnition of spherically symmetric: it looks
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Unformatted text preview: the same from any angle. 3.6.4.2 Solution harmd-b Question: Show that the ground state wave function is maximal at the origin and, like all the other energy eigenfunctions, becomes zero at large distances from the origin. Answer: According to the answer to the previous question, the ground state is ψ 000 = 1 ( πℓ 2 ) 3 / 4 e − r 2 / 2 ℓ 2 . where r is the distance from the origin. Now according to the qualitative properties of expo-nentials, an exponential is one when its argument is zero, and becomes less than one when its argument becomes negative. So the maximum is at the origin r = 0....
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This note was uploaded on 01/06/2012 for the course PHY 3604 taught by Professor Dr.danielarenas during the Fall '11 term at UNF.

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