Momentum Conversion: Why p-hat^2/2m = p^2/2m?

[ehrenfest]

Why does
$$1/\sqrt{2\pi\hbar }\int p-hat/2m * \psi(x) exp(-ipx/\hbar)*dx = p^2/2m \phi(p)$$, where phi(p) is that wavefunction in momentum space
?

I understand why $$1/\sqrt{2\pi\hbar }\int \psi(x) exp(-ipx/\hbar)*dx = \phi(p)$$.

by the way, how do you make hats in latex

EDIT: It should be $$1/\sqrt{2\pi\hbar }\int p-hat^2/2m * \psi(x) exp(-ipx/\hbar)*dx = p^2/2m \phi(p)$$

 

[malawi_glenn]

try

$$\hat{p}$$

And have done the p operator on the wave function?

 

[ehrenfest]

And have done the p operator on the wave function?

So, I should apply the opetor then do IBT two times to get rid of the second derivative? Aren't you left with a psi' and a psi if you do that?

 

[ehrenfest]

If you do the p operator to the right:

$$\hat{p}(\psi(x) exp(-ipx/\hbar))/(2m)$$

what do you get?

$$-i\hbar \frac{d \psi}{dx} exp(-ipx/\hbar)/(2m)$$
and if you apply it twice you get

$$\frac{\hbar^2 d\psi} {dx^2} exp(-ipx/\hbar)/(2m)$$

 

[malawi_glenn]

why twice? :|

Consider this thread:
https://www.physicsforums.com/showthread.php?t=179887

 

[ehrenfest]

why twice? :|

Consider this thread:
https://www.physicsforums.com/showthread.php?t=179887

Because I stupidly left out the squared sign after the $$\hat{p}$$ !

 

[malawi_glenn]

Because I stupidly left out the squared sign after the $$\hat{p}$$ !

Ah, ok that explains why i got zero from my first calculation;)

Now check at the tips given in that thread I gave you, read all posts, you shall find how to apply the p-operator when it is "squared", and when it is applied to a function that is a product of two functions. Then after that, perform the Fourier-transformation.

 

[malawi_glenn]

I probably gave you the wrong thread for showing you how to apply the p-operator to an arbitrary function. The operator is always going as far to the right as possible.

So:

$$-i\hbar \frac{d }{dx} (\psi (x)exp(-ipx/\hbar))/(2m)$$

use product rule for differentiation.

 

[ehrenfest]

I probably gave you the wrong thread for showing you how to apply the p-operator to an arbitrary function. The operator is always going as far to the right as possible.

So:

$$-i\hbar \frac{d }{dx} (\psi (x)exp(-ipx/\hbar))/(2m)$$

use product rule for differentiation.

I am pretty sure it the operator only applies to psi here. Perhaps I made another mistake in the initial post but I am not sure. The original equation which I did not show was:

$$\hat{p}/2m \psi(x) - g\delta(x) \psi(x) = -|E|\psi(x)$$

my book then multiplies $$1/\sqrt(2\pi\hbar)exp( - ipx/\hbar)$$ to both sides and then integrates.

Since you usually multiply things on the left, I think that you would not apply the operator to the exponential. I rearranged it because I did not think it made a difference. I thought an operator took as its argument only the function immediately following it. Was that wrong?

 

[malawi_glenn]

Iam just stating what is usual in Quantum mechanics. But try both and see. Then write some clearer and nicer forumulas so can we see if it is good and maybe some more guys can come and help.

This operator is a differential operator, and therefore one must be careful, we do not multiply operators, we operate with them =P

 

[ehrenfest]

$$\frac{1}{\sqrt{ 2\pi\hbar}}\int \frac{\hat{p}}{2m} \psi(x) e^{-ipx/\hbar}dx = \frac{\hbar^2}{2m\sqrt{ 2\pi\hbar}} \int \frac {d^2}{dx^2}\ \psi(x) e^{-ipx/\hbar}dx$$

If I integrate this by parts twice I get an expression with a psi'(x) and a psi(x) which I do not know how to get rid of.

 

[olgranpappy]

$$-i\hbar \frac{d \psi}{dx} exp(-ipx/\hbar)/(2m)$$
and if you apply it twice you get

$$\frac{\hbar^2 d\psi} {dx^2} exp(-ipx/\hbar)/(2m)$$

ummm... no, on both counts. But anyways, this whole thread is confused, so let me just sort it out for you.

We can write psi (a function of x) as a Fourier transform like this:
$$\psi(x)=\int \frac{dp}{2\pi}e^{ipx}\phi(p)$$

Then

$$\frac{d}{dx}\psi(x)=\frac{d}{dx}\int \frac{dp}{2\pi}e^{ipx}\phi(p)<br /> =\int \frac{dp}{2\pi}\left(\frac{d}{dx}e^{ipx}\right)\phi(p)<br /> =\int \frac{dp}{2\pi}ipe^{ipx}\phi(p)$$

And

$$\frac{d^2}{dx^2}\psi(x)=\int \frac{dp}{2\pi}(ip)^2e^{ipx}\phi(p)\;.$$

 

[ehrenfest]

The only problem is that i^2.
Should the exponential be exp( -ipx)? We want to show that

$$\frac{1}{\sqrt{ 2\pi\hbar}}\int \frac{\hat{p}}{2m} \psi(x) e^{-ipx/\hbar}dx = p^2/2m$$

and there is no sign change there.

 

[olgranpappy]

The only problem is that i^2.

That is there because I acted with $$\frac{d}{dx}$$ and
not $$-i\frac{d}{dx}$$ which is the momentum operator in position space.

Should the exponential be exp( -ipx)?

This depends on your Fourier transform conventions and does not change the physics. Typically the Fourier transform conventions are specified in the way I specified them--A function of x (In this case \psi(x)) is written with the plus sign in the exponential of the Fourier kernel and a function of p is written with the minus sign in the exponential.

We want to show that

$$\frac{1}{\sqrt{ 2\pi\hbar}}\int \frac{\hat{p}}{2m} \psi(x) e^{-ipx/\hbar}dx = p^2/2m$$

and there is no sign change there.

I doubt very much that anyone in their right mind has asked you to show that the above is true... the above makes no sense.

I'm sorry, but it just doesn't make sense.

I have a feeling that the problem comes from how you are thinking about Fourier transforms. But, keep at it and you will understand eventually. Good luck.

 

[ehrenfest]

I see. Thanks.