# Schrodinger equation

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?

## Answers and Replies

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dextercioby
Homework Helper
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.

jtbell
Mentor
Just to remind you, Schrödinger's equation in one spatial dimension looks like this:

$$- \frac {\hbar^2} {2 m} \frac {\partial^2 \Psi} {\partial x^2} + V \Psi = i \hbar \frac {\partial \Psi} {\partial t}$$

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.

dextercioby
Homework Helper
$$\frac{d|\Psi\rangle}{dt}=\frac{1}{i\hbar}\hat{H}|\Psi\rangle$$

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

Daniel.

Last edited:
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 $$|\psi\rangle$$ are linear. We have the result that, for linear operators

$$\hat{L}(\lambda\mathbf{a} + \mu\mathbf{b}) = \lambda\hat{L}\mathbf{a} + \mu\hat{L}\mathbf{b}$$

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

dextercioby
Homework Helper
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...

Daniel.

Staff Emeritus
Gold Member
Dearly Missed
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...

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.

dextercioby
Homework Helper
SA,if u want u can move it to HM section,no problem on my behalf.