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why is that multplying psi by a constant does not its value?

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why is that multplying psi by a constant does not its value?

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Daniel.

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jtbell

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[tex]- \frac {\hbar^2} {2 m} \frac {\partial^2 \Psi} {\partial x^2} + V \Psi = i \hbar \frac {\partial \Psi} {\partial t} [/tex]

All the terms contain either psi itself, or one of its derivatives, therefore we call this a

The fact that Schrödinger's equation is linear and homogeneous guarantees that if if a particular psi is a solution, then k*psi is also a solution.

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How about

[tex] \frac{d|\Psi\rangle}{dt}=\frac{1}{i\hbar}\hat{H}|\Psi\rangle [/tex]

,where [itex] \hat{H} [/itex] is a densly-defined,self adjoint LINEAR operator (in agreement with the second principle)...

Daniel.

[tex] \frac{d|\Psi\rangle}{dt}=\frac{1}{i\hbar}\hat{H}|\Psi\rangle [/tex]

,where [itex] \hat{H} [/itex] is a densly-defined,self adjoint LINEAR operator (in agreement with the second principle)...

Daniel.

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[tex]\hat{L}(\lambda\mathbf{a} + \mu\mathbf{b}) = \lambda\hat{L}\mathbf{a} + \mu\hat{L}\mathbf{b}[/tex]

What this means is that in the Schrodinger equation, if you replace [tex]|\psi\rangle[/tex] by [tex]k|\psi\rangle[/tex] you can take all the [tex]k[/tex]s out and so cancel them.

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Daniel.

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selfAdjoint

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Daniel, I certainly agree with your reasoning here, but I humbly suggest that in that case the thread be moved to Homework. I sort of like to see different posters working things out one by one down a thread on this QM thread. Just my prejudice though.dextercioby said:

Daniel.

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But,in fact,they are.Maybe my posts are too numerous,but i can't help it... :tongue2:selfAdjoint said:I sort of like to see different posters working things out one by one down a thread on this QM thread.

Daniel.

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