How is the Matrix in Momentum Representation Derived?

[LCSphysicist]

Homework Statement

..

Relevant Equations

.

$$\langle p | W | p' \rangle = \int \langle p | x \rangle \langle x W | x' \rangle \langle x' p' \rangle dx dx'$$
$$\langle p | W | p' \rangle = \int \langle p | x \rangle \delta(x-x') W(x) \langle x' | p' \rangle dx dx'$$
$$\langle p | W | p' \rangle = \int \langle p | x' \rangle W(x') \langle x' | p' \rangle dx'$$
$$* \langle p | W | p' \rangle = W(x') \langle p |p' \rangle$$
$$\langle p | W | p' \rangle = \delta (p-p') W(x')$$

To get * from the previous equation, i assumed that $\int x'\rangle \langle x' dx' = 1$. But i am not sure if i can just take it out, because i don't really know if the W(x') does affect something in this integral.

Or is this right?

 

[vela]

No, you can’t do that. You’re integrating over $x’$, so it shouldn’t appear in the final result.

 

[vanhees71]

Hint: Think about what is $\langle x|p \rangle=\langle p|x \rangle^*$. Also it's clear that your result cannot depend on $x$, because $\hat{W}$ doesn't depend on $x$.

 

[LCSphysicist]

I could try to go on: $\langle x | p \rangle = c e^{ikx}$, with this the integral reduce to
$$c c^{*} \int W(x') e^{(ik'-ik)x'} dx'$$

Now, i need to review Fourier transform, but maybe i could say that $$ \langle x | p \rangle = c c^{*} \hat W (p)$$ ?

 

[vanhees71]

No, you first formula is right:
$$\langle x|p \rangle=u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
Now start again with your initial formula in #1.

 

[LCSphysicist]

No, you first formula is right:
$$\langle x|p \rangle=u_p(x)=\frac{1}{\sqrt{2 \pi \hbar}} \exp(\mathrm{i} p x/\hbar).$$
Now start again with your initial formula in #1.

I will try to go step by step.

$$\int \int \frac{dxdx'}{2 \pi \hbar} exp(i(p'x'-px)/\hbar) \delta(x-x') W(x)$$
$$ \frac{1}{2 \pi \hbar} \int exp( \frac{-2 \pi i (p-p')x'}{2 \pi \hbar}) W(x') dx' $$
$$ \frac{1}{2 \pi \hbar} \int exp( -2 \pi i f x' ) W(x') dx' = \frac{\hat W (t)}{2 \pi \hbar}$$
$$ \langle p | W | p' \rangle = \frac{\hat W (t)}{2 \pi \hbar} = \frac{K (\frac{p-p'}{2 \pi \hbar})}{2 \pi \hbar}$$

Where K is the spectrum of W
At least it is beautiful now