Momentum eigenfunctions proof and Fourier Transform question

[Joker93]

I have the following problem:

 

[DrDu]

How do you define the operator $\Psi$?

 

[Joker93]

How do you define the operator $\Psi$?

the equations are from this link

https://en.wikipedia.org/wiki/Position_and_momentum_spaceBut, is it a matter of how you define things? Isn't there a right way to do it that does not depend on how you define the operator? Excuse me if these questions are moronic, but I haven't been far into quantum mechanics yet(I am a self learner at this point).

 

[blue_leaf77]

I think you misunderstood the terms here. The first expression is the eigenfunction of momentum operator in position representation, and the second expression is the eigenfunction of position in momentum representation. They are not related by Fourier transformation.

 

[DrDu]

But, is it a matter of how you define things? Isn't there a right way to do it that does not depend on how you define the operator? Excuse me if these questions are moronic, but I haven't been far into quantum mechanics yet(I am a self learner at this point).

All I know from your post is that $\Psi$ apparently is an operator which has some eigenfunctions. So it seems appropriate to as for some more details before answering your questions.

 

[Joker93]

I think you misunderstood the terms here. The first expression is the eigenfunction of momentum operator in position representation, and the second expression is the eigenfunction of position in momentum representation. They are not related by Fourier transformation.

check this link by Wikipedia please

https://en.wikipedia.org/wiki/Position_and_momentum_space

 

[Joker93]

All I know from your post is that $\Psi$ apparently is an operator which has some eigenfunctions. So it seems appropriate to as for some more details before answering your questions.

its all in the Wikipedia link that I have provided you with.

 

[DrDu]

its all in the Wikipedia link that I have provided you with.

I can't find any operator $\Psi$ in this link.

 

[Joker93]

I can't find any operator $\Psi$ in this link.

I think its r(hat)=i*d/dk
its in the link under "Functions and operators in momentum space"

 

[DrDu]

You think so?!? So basically, we shall not only answer your question, but also guess what your question might be?

 

[Joker93]

You think so?!? So basically, we shall not only answer your question, but also guess what your question might be?

I found it in the link(by Wikipedia) and I also have given you the link and told you where to look for it. I "think so" because I have not seen operators in my self-stude of quantum mechanics yet, I am a beginner as I warned you.

 

[Joker93]

You think so?!? So basically, we shall not only answer your question, but also guess what your question might be?

in the link, it writes: "position operator r(hat)".

 

[blue_leaf77]

Note the presence of subscripts in $\psi_{\mathbf{k}}(\mathbf{r})$ and $\phi_{\mathbf{r}}(\mathbf{k})$, their presence give them different meaning from $\psi$ and $\phi$ without subscripts. It's true that $\psi(\mathbf{r})$ and $\phi(\mathbf{k})$ are related through Fourier transform, because both functions actually represent the same state, they are just represented in different space. On the other hand, $\psi_{\mathbf{k}}(\mathbf{r})$ and $\phi_{\mathbf{r}}(\mathbf{k})$ are not related through Fourier transform as they correspond to different state (more precisely because they are eigenfunctions of different operators).

I am a beginner as I warned you

For beginners, Wikipedia and other online sources are often not very good resources to learn because they are generally not designed to initiate learning. It's advisable to start your adventure from standard textbooks.

 

[Joker93]

Note the presence of subscripts in $\psi_{\mathbf{k}}(\mathbf{r})$ and $\phi_{\mathbf{r}}(\mathbf{k})$, their presence give them different meaning from $\psi$ and $\phi$ without subscripts. It's true that $\psi(\mathbf{r})$ and $\phi(\mathbf{k})$ are related through Fourier transform, because both functions actually represent the same state, they are just represented in different space. On the other hand, $\psi_{\mathbf{k}}(\mathbf{r})$ and $\phi_{\mathbf{r}}(\mathbf{k})$ are not related through Fourier transform as they correspond to different state (more precisely because they are eigenfunctions of different operators).

yes, sorry for the misunderstanding.
But, why does Ψk have a plus sign on the exponent while Φr has a minus sign?

 

[blue_leaf77]

why does Ψk have a plus sign on the exponent while Φr has a minus sign?

$\psi_{\mathbf{k}}(\mathbf{r})$ is an eigenfunction of momentum operator $\hat{\mathbf{p}} = -i\hbar\frac{\partial}{\partial \mathbf{r}}$, so, to calculate $\psi_{\mathbf{k}}(\mathbf{r})$ you need to solve
$$
\mathbf{p} \psi_{\mathbf{k}}(\mathbf{r}) = -i\hbar\frac{\partial}{\partial \mathbf{r}} \psi_{\mathbf{k}}(\mathbf{r})
$$