A question on Schroedinguer equation

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The discussion centers on the Schrödinger equation in the form i\bar\frac{d\psi}{dt}=-\frac{\hbar^{2}}{2m}D^{2}\psi+V(x)\psi+NV_{0}\psi, where N is a significantly large number (N>>1). It concludes that physical probabilities remain independent of N and V_0, with the average total energy being the only aspect influenced by these variables. The mathematical proof involves substituting the wave function with a phase factor, demonstrating that the modified wave function produces equivalent physical outcomes.

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eljose
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let be the SE equation in the form:

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

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 Schrödinger 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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