Schrodinger Equation in momentum space? ?

[0ddbio]

Schrödinger Equation in momentum space? ??

I don't know if this makes any sense at all, but I'm studying QM and just trying to generalize some things I'm learning. Please let me know where I go wrong..

Basically by my understanding the most general form of the Schrödinger Equation can be written as follows:
i\hbar\frac{\partial\Psi}{\partial t}=\hat{H}\Psi

Then we have that H is given by:
\hat{H}=\frac{\hat{p}^{2}}{2m}+V

Then we just need to know what the 'p' operator is. My book shows that in configuration space it is:
\hat{p}=-i\hbar\nabla (I generalized it a bit, the book just has an x derivative)
however, in momentum space it is just: \hat{p}=p

Now.. if we take the momentum operator in configuration space and plug that into H and then plug that into the form of the Schrödinger Equation I wrote at the top we get the familiar form:
i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\Psi+V\PsiOk... but what I am wondering now is what happens if we use the momentum operator in momentum space instead of configuration space, then plugging that into H and H into the first equation as before we would end up with:
i\hbar\frac{\partial\Psi}{\partial t}=\frac{p^{2}}{2m}\Psi+V\Psi

This seems very odd though, I think it must be wrong but I don't know why... perhaps it is right but only for a wavefunction expressed in momentum space... ??Can anyone help me understand what is going on here?? Perhaps there is no justification in plugging the momentum operator for momentum space into H like I did.. I just don't know?

Thanks for any insight.

 

[dextercioby]

Yes to the first question, in the <new> equation without the nabla squared, but with p^2 you have both the wavefunction in terms of p and the potential expressed in terms of p with the help of a Fourier transformation.

 

[Avodyne]

The potential term is

\int_{-\infty}^{+\infty}dp&#039;\;\widetilde V(p-p&#039;)\widetilde\Psi(p&#039;)

where \widetilde V(p) and \widetilde\Psi(p) are the Fourier transforms of V(x) and \Psi(x).

 

[0ddbio]

Awesome guys, thanks!

Perhaps you could answer another question so that I don't have to start a new thread.

The motivation for me starting this thread was that the book is going over eigenvalue problems and for instance this one:
\hat{H}\psi(x)=E\psi(x)
makes perfect sense, however the other ones like:
\hat{p}\psi_{p}(x)=p\psi_{p}(x)
I don't know why this works.. That is why I tried doing what I was describing in my previous post.
I was hoping that it would have separable solutions, one of which would appear as the above eigenvalue problem, this is the way the book shows it done for the energy. But it gives no justification for why we can write eigenvalue equations for other operators.

 

[osturk]

Hello. I had the QM course this year, so I'll try to help:

however, in momentum space it is just: \hat{p}=p

this is not actually true. We can not substitute operators with scalars, in general. An operator, operates on a function, and yields another function which is, in general, different than the initial function. However, if an operator acts on one of its eigenfunctions, it yields the same function, but multiplied with a scalar, which we call the eigenvalue. These are special functions and they have a core role in QM, as far as I understood. We can reveal those functions by eigenvalue analysis...

\hat{p}\psi_{p}(x)=p\psi_{p}(x)
I don't know why this works..

... We can do eigenfunction analysis for any operator to try to find its eingenfunctions and eigenvalues. So in the equation above, we are merely searching for eigenfunctions \psi_{p}(x) with eigenvalue p, of the operator\hat{p}. You could substitute p=-ih*d()/dx to do this analysis and find eigenfunction of the momentum operator in position space.. Now, returning to the first quote I made.. operator p, looks as though it is equal to the scalar p, only in the case it operates on its eigenfunctions, that is \hat{p}\psi_{p}(x)=p\psi_{p}(x). But even this time, this doesn't mean p equals p. It just means when p acts on one of its eigenfunctions, it yields the same function multiplied with scalar p, nothing more. So operator p should stay as operator p in the Schr. eqn. you wrote. But if you want to write the Schr. eqn. in the momentum space, then you could multiply both sides with \psi_{p}(x) and integrate, on the domain of the problem. This is the Fourier transform. It will transform \psi_{p}(x)'s into \Phi(p)'s, and this is the Schr. eqn. in the momentum space. \Phi(p) is actually the state of the particle, just like \psi_{p}(x). Think of it as the frequency spectrum of a soundwave. \psi_{p}(x) tells us the amplitudes of \psi_{p}(x) we should "mix" together (for varying p's), to yield the state of the particle. Writing the state of the particle in momentum space, allows us to directly calculate the probability of observing momentum p, which is \left\|\Phi(p)\right\|. Just as \left\|\psi_{p}(x)\right\| is the probability of observing the particle at position x.

Hope this helps..

Deniz

 

[0ddbio]

Ya that does help a lot thanks.

Although I did get that the momentum operator in momentum space is just p directly from the book.
It's just like in, say 1 dimension, the operator \hat{x}=x
like if you want to find the expectation value of the position you just put "x" as the operator which goes in between psi* and psi.

But in momentum space you just put "p" in between psi* and psi as the momentum operator and the "x" operator now becomes, instead of x, \hat{x}=i\hbar\frac{\partial}{\partial p} in momentum space.

But I think I understand what I was asking now.

 

[osturk]

ah, yes, it is true in momentum space. I missed that.