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Harmonic oscillator 
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#1
Dec405, 02:25 AM

P: 736

why is the lowest allowed energy not E=0 but some definite minimum E=E0?



#2
Dec405, 02:43 AM

HW Helper
PF Gold
P: 1,197

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. 


#3
Dec405, 04:52 AM

Sci Advisor
HW Helper
P: 2,002

I could give a handwave argument. We have E=1/2mv^2+1/2kx^2.
If E=0 both x and v are zero, which contradicts Heisenberg. 


#4
Dec405, 07:13 AM

P: 736

Harmonic oscillator
thank you very much!!! :)



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