Why does multiplying a wave function by a constant still satisfy the Schrödinger equation?

[ohhhnooo]

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

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

 

[dextercioby]

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]

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]

How about
$$\frac{d|\Psi\rangle}{dt}=\frac{1}{i\hbar}\hat{H}|\Psi\rangle$$

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

Daniel.

 

[masudr]

If you look at the equation dextercioby has put up (that is the most general form of Schrödinger'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 Schrödinger equation, if you replace $$|\psi\rangle$$ by $$k|\psi\rangle$$ you can take all the $$k$$s out and so cancel them.

 

[dextercioby]

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.

 

[selfAdjoint]

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]

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

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...

Daniel.

 

[masudr]

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