Schrodinger Equation in momentum space? ?

• 0ddbio
In summary, the book is showing how to do eigenvalue problems, but when p is substituted in for the operator, it doesn't actually equal p, but it behaves the same. So in the first equation, p^2 should be substituted in for operator p.
0ddbio
Schrodinger 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 Schrodinger 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 Schrodinger Equation I wrote at the top we get the familiar form:
$$i\hbar\frac{\partial\Psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\Psi+V\Psi$$Ok... 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.

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.

The potential term is

$$\int_{-\infty}^{+\infty}dp'\;\widetilde V(p-p')\widetilde\Psi(p')$$

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

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.

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

0ddbio said:
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...
0ddbio said:
$$\hat{p}\psi_{p}(x)=p\psi_{p}(x)$$
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

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.

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

1. What is the Schrodinger Equation in momentum space?

The Schrodinger Equation in momentum space is a mathematical equation used to describe the behavior of quantum particles in terms of their momentum. It is an alternative form of the Schrodinger Equation in position space, which describes the position of particles.

2. How is the Schrodinger Equation in momentum space different from the position space equation?

The position space equation describes the position and potential energy of a particle, while the momentum space equation describes the momentum and kinetic energy of a particle. The two equations are related through a mathematical transformation known as a Fourier transform.

3. What does the Schrodinger Equation in momentum space tell us about quantum particles?

The equation helps us understand how particles behave at the quantum level, including their wave-like properties and the probabilistic nature of their behavior. It also allows us to make predictions about the behavior of particles in different situations.

4. How is the Schrodinger Equation in momentum space used in practical applications?

The equation is used in a variety of fields, including quantum mechanics, chemistry, and materials science. It is used to study the behavior of particles in different systems and to make predictions about their properties and interactions.

5. Is the Schrodinger Equation in momentum space a fundamental principle of physics?

Yes, the Schrodinger Equation in momentum space is considered a fundamental principle of quantum mechanics. It is a cornerstone of our understanding of the behavior of particles at the microscopic level and has been confirmed through numerous experiments.

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