Physically acceptable solutions to the Schroedinger equation?

[etotheipi]

I apologise in advance if this is a silly question! We are aware that the wave function $\Psi(x, t)$ of a particle moving along one dimension will satisfy the differential equation$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi$$We are told that if we assume a general form $\Psi = Ae^{i(kx - \omega t)}$ (Edit, cred. PeroK: For $V(x) = 0$) then the equation will spit out physically correct results. But what about all of the other possible forms for $\Psi$, do we just discard them? There are some examples that we can discard immediately, e.g. something like $\Psi = Ae^{i(x - vt)}$ that is not dimensionally homogenous. More generally if we run other forms than $Ae^{i(kx-\omega t)}$ through the equation then whilst we can find mathematically acceptable solutions, the results are silly physically (for starters, they don't agree with with the De Broglie relations).

Are there any other forms for $\Psi$ that give valid results?

 

[PeroK]

I apologise in advance if this is a silly question! We are aware that the wave function $\Psi(x, t)$ of a particle moving along one dimension will satisfy the differential equation$$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V(x)\Psi$$We are told that if we assume a general form $\Psi = Ae^{i(kx - \omega t)}$, then the equation will spit out physically correct results. But what about all of the other possible forms for $\Psi$, do we just discard them? There are some examples that we can discard immediately, e.g. something like $\Psi = Ae^{i(x - vt)}$ that is not dimensionally homogenous. More generally if we run other forms than $Ae^{i(kx-\omega t)}$ through the equation then whilst we can find mathematically acceptable solutions, the results are silly physically (for starters, they don't agree with with the De Broglie relations).

Are there any other forms for $\Psi$ that give valid results?

Are you talking about solutions for $V(x) = 0$? Or, are you talking about separation of variables for a general time-independent potential $V(x)$?

 

[etotheipi]

Are you talking about solutions for $V(x) = 0$? Or, are you talking about separation of variables for a general time-independent potential $V(x)$?

I was just thinking about general solutions for a particle in any old arbitrary potential. Is that a bit too vague?

 

[PeroK]

I was just thinking about general solutions for a particle in any old arbitrary potential. Is that a bit too vague?

No, but they won't have solutions of the form $Ae^{i(kx - \omega t)}$.

Those solutions are for the free particle where $V(x) = 0$.

 

[etotheipi]

Oh hang on, yes you're right, I'm being thick . So yes, I guess I am restricting this to $V(x) = 0$.

 

[PeroK]

Oh hang on, yes you're right, I'm being thick . So yes, I guess I am restricting this to $V(x) = 0$.

So, the question is whether all solutions must be of this form?

 

[etotheipi]

Actually I realize now that the question is flawed. Of course all solutions must be like that. Don't know what I was thinking. Sorry to be a pain!

 

[PeroK]

Yes. Sorry

If you tackle the general equation (with $V(x)$) using separation of variables, then you get the common form of the time-dependent function: $e^{-iwt}$ or $e^{-iEt/\hbar}$. And, you get the TISE (time-independent Schroedinger equation):
$$-\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{d x^2} + V(x)\psi(x) = E\psi(x)$$
In the special case where $V(x) = 0$, you have solutions:
$$\psi(x) = Ae^{ikx} + Be^{-ikx}$$
Where $k = \frac{\sqrt{2mE}}{\hbar}$.

And that is more or less it. You need to appeal to the theory of differential equations to tell you that any other other solutions must be linear combinations of these basis solutions. Note that in all cases, as the SDE is linear, you can superimpose solutions with different energies and you still have a solution. Another mathematical technique (non separation of variables) may yield a set of solutions that look different, but ultimately they must be linear combinations of the basis solutions we found by separation of variables.

Note that the physical importance of these particular solutions is that they are energy eigenstates.

 

[PeroK]

Actually I realize now that the question is flawed. Of course all solutions must be like that. Don't know what I was thinking. Sorry to be a pain!

See above, from the linearity of the SDE you could find a set of solutions that look very different, but must be a linear combination of the energy eigenstates.

 

[etotheipi]

You're right about the differential equations, yes. I was getting muddled about why the coefficients always magically come out to be $k$ and $\omega$ with their usual meanings, but this is of course the De Broglie hypothesis and the SE was itself built to be consistent with this in the first place. I think I must get more sleep, thanks for your patience