- #1

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[tex]i\bar\frac{d\psi}{dt}=-\frac{\hbar^{2}}{2m}D^{2}\psi+V(x)\psi+NV_{0}\psi [/tex]

where N is a big big number N>>1 then what would be the solution?..thanks.

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- Thread starter eljose
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- #1

- 492

- 0

[tex]i\bar\frac{d\psi}{dt}=-\frac{\hbar^{2}}{2m}D^{2}\psi+V(x)\psi+NV_{0}\psi [/tex]

where N is a big big number N>>1 then what would be the solution?..thanks.

- #2

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That depends on V(x) doesn't it??

- #3

Physics Monkey

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You will find that physical probabilities are independent of [tex] N [/tex] and [tex] V_0 [/tex]. The only thing that will depend on [tex] N V_0 [/tex] is the average total energy. This is a physical consequence of the fact that the reference point of potential energy is arbitrary in quantum mechanics.

This is mathematically evident in that any position independent potential can be absorbed into the wave function as a phase. You can check for yourself that the substitution [tex] \psi = e^{-i N V_0 \,t/\hbar} \psi' [/tex] yields a Schrodinger equation for [tex] \psi' [/tex] given by

[tex]

i \hbar \frac{\partial \psi'}{\partial t} = - \frac{\hbar^2}{2 m} \nabla^2 \psi ' + V(x) \psi'

[/tex]

However, since [tex] \psi [/tex] and [tex] \psi' [/tex] only differ by an overall phase, albeit a time dependent one, they produce the same physics.

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