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

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- Thread starter asdf1
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- #1

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

- #2

siddharth

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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.

[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.

Last edited:

- #3

Galileo

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If E=0 both x and v are zero, which contradicts Heisenberg.

- #4

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thank you very much!!! :)

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