Schrodinger equation

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can you explain this statement "if psi is a solution of a schrodinger equation, then so is kpsi, where k is any constant".

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

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  • #2
dextercioby
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Try to see whether that's true or not by simply plugging the new state vector into Schroedinger's equation.It shouldn't be too hard...

Daniel.
 
  • #3
jtbell
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Just to remind you, Schrödinger's equation in one spatial dimension looks like this:

[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 homogeneous differential equation. Also, no term contains powers of psi or of one of its derivatives, or combinations of psi and its derivatives, therefore we call this a linear differential equation.

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.
 
  • #4
dextercioby
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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)... :wink:

Daniel.
 
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  • #5
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If you look at the equation dextercioby has put up (that is the most general form of Schrodinger's equation), you will see that both operators that act upon [tex]|\psi\rangle[/tex] are linear. We have the result that, for linear operators

[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.
 
  • #6
dextercioby
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I emphasized the word "linear" and left to the OP to see what implications that fact would have.After all,it's better for him to figure out things by himself,as i think that would give him a feeling of satisfaction,too... :wink:


Daniel.
 
  • #7
selfAdjoint
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dextercioby said:
I emphasized the word "linear" and left to the OP to see what implications that fact would have.After all,it's better for him to figure out things by himself,as i think that would give him a feeling of satisfaction,too... :wink:


Daniel.
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.
 
  • #8
dextercioby
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SA,if u want u can move it to HM section,no problem on my behalf. :smile:

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

Daniel.
 
  • #9
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Yes I do understand the need of the person to work it out for himself. But since we received no acknowledgement of his/her understanding the problem, I thought I should explain it in detail.
 

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