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Harmonic oscillator

by asdf1
Tags: harmonic, oscillator
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asdf1
#1
Dec4-05, 02:25 AM
P: 736
why is the lowest allowed energy not E=0 but some definite minimum E=E0?
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siddharth
#2
Dec4-05, 02:43 AM
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If you solve the Time Independent Schrodinger equation for the Harmonic Oscillator, that is
[tex] -\frac{\hbar^2}{2m} \frac{d^2\Psi}{dx^2} + \frac{1}{2}kx^2 \Psi = E \Psi [/tex]

The quantization of energy comes from the boundary conditions (ie, [itex] \Psi = 0 [/itex] when [itex] x= \infty [/itex] or [itex] x = -\infty [/itex]).

The permitted energy levels will be

[tex] E_n = (n+\frac{1}{2}) \hbar \omega [/tex]

So the lowest Energy is not E=0.
Galileo
#3
Dec4-05, 04:52 AM
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I could give a hand-wave argument. We have E=1/2mv^2+1/2kx^2.
If E=0 both x and v are zero, which contradicts Heisenberg.

asdf1
#4
Dec4-05, 07:13 AM
P: 736
Harmonic oscillator

thank you very much!!! :)


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