# The momentum representation of $x$ and $[x,p]$

Tags:
1. Oct 27, 2015

### Pring

To deduce the momentum representation of $[x,p]$, we can see one paradom
$<p|[x,p]|p>=i\hbar$
$<p|[x,p]|p>=<p|xp|p>-<p|px|p>=p<p|x|p>-p<p|x|p>=0$
Why? If we deduce the momentum representation of $x$, we obtain
$<p|x|p>=i\hbar \frac{\partial \delta (p'-p)}{\partial p'}|_{p'=p}$. This value is not definite. So, why two uncertain values can obtained a certain value $i\hbar$? In addition, the $x$ should be replace by $i\hbar \frac{\partial }{\partial p}$. Then the eigenvalue $p$ can't extract. However, if we consider $i\hbar \frac{\partial }{\partial p}$ to act on the bra, not the ket, then the eigenvalue $p$ can be extracted. Is anything wrong here?

2. Oct 27, 2015

### Staff: Mentor

What does <p|p> equal?

Remember formal manipulations involving infinity are rather dubious.

The real solution to this sort of stuff requires Rigged Hilbert Spaces which needs a very good background in analysis to understand. As a build up to it I suggest the following book which should be in armoury of any physicist nor even applied mathematician:
https://www.amazon.com/The-Theory-Distributions-Nontechnical-Introduction/dp/0521558905

It will not rigorously resolve what's going on, that requires considerable advanced analysis, but will allow an intuitive accommodation.

Thanks
Bill

Last edited: Oct 27, 2015
3. Oct 27, 2015

### Pring

I just regard this term as normalization 1.

4. Oct 27, 2015

### Staff: Mentor

<p|p'> = Dirac delta (p - p'). Substitute p = p' and you get Dirac delta (0) - which strictly speaking is undefined, but intuitively is taken as infinity.

Thanks
Bill

5. Oct 27, 2015

### Pring

Aha, you are right, thank you!

6. Oct 27, 2015

### Staff: Mentor

No problem.

This stuff can be tricky because its not really valid with the math you likely know at the beginner level.

That's why I STRONGLY suggest you get the book I mentioned. It will not resolve things completely, but things will be a LOT clearer.

Thanks
Bill

7. Oct 27, 2015

### Pring

Yeah, nowadays books about quantum physics are of many mathematical details. Thank you for your recommend.

8. Oct 28, 2015

### stevendaryl

Staff Emeritus
As Bill says, the problem is that $|p\rangle$ is not a normalizable state; there are no normalizable momentum eigenstates. However, you can get an approximate momentum eigenstate $\phi_p(x) = (\frac{2 \lambda}{\pi})^{\frac{1}{4}} e^{i p x - \lambda x^2}$. Then:

$\hat{p} \phi_p = (p + 2 i \lambda x) \phi_p$

which for small values of $x$ is approximately $p \phi_p$. So $\phi_p$ is an approximate momentum eigenstate. If you compute $\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle$ using the fact that $\hat{p}$ is Hermitian, you find:

$\langle \phi_p | \hat{p} x | \phi_p \rangle = \langle \hat{p} \phi_p | x | \phi_p \rangle = \langle (p + 2i \lambda x) \phi_p|x|\phi_p\rangle = \langle \phi_p|(p - 2i \lambda x)x|\phi_p\rangle$
$= p \langle \phi_p|x|\phi_p \rangle - 2i \lambda \langle \phi_p |x^2|\phi_p\rangle$

$\langle \phi_p | x \hat{p} | \phi_p \rangle = \langle \phi_p|(p + 2i \lambda x)x|\phi_p\rangle$
$= p \langle \phi_p|x|\phi_p \rangle + 2i \lambda \langle \phi_p |x^2|\phi_p\rangle$

So $\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \langle \phi_p |x^2|\phi_p\rangle$, not zero.

You might think that by choosing the convergence parameter $\lambda$ to be very small, you could ignore this term, but in fact, $\langle \phi_p |x^2|\phi_p\rangle = \frac{1}{4 \lambda}$, so regardless of the choice of $\lambda$,

$\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \frac{1}{4 \lambda} = -i$.

9. Oct 28, 2015

### Pring

Great! This is a specific handle,and you solve it in the coordinate representation. You miss the quantity $\hbar$, but I think it is just a unit assumed. I have one question. We know the momentum operater is a observable physical quantity. Why its eigenvalue is complex? It means that $<\phi_p|\hat p\hat x|\phi_p>-<\phi_p|\hat x\hat p|\phi_p>=<\phi_p|f^*(p,x)x|\phi_p>-<\phi_p|xf(p,x)|\phi_p>$?

10. Oct 28, 2015

### stevendaryl

Staff Emeritus
The state $|\phi_p\rangle$ isn't an eigenstate, and the quantity $p + i 2 \lambda x$ isn't an eigenvalue. Notice it depends on $x$, and eigenvalues must be constants.

But you're right, it's kind of weird that the expectation value of $\hat{p}$ is a complex number.

I'm going to think about that.

11. Oct 29, 2015

### Demystifier

This is really great! Did you devised it by yourself, or is it published somewhere else?

12. Oct 29, 2015

### Demystifier

Actually, it is a real number. We have
$\int dx \, \phi_p^* ix \phi_p =0$
so the imaginary part vanishes.

13. Oct 29, 2015

### Demystifier

14. Oct 29, 2015

### Pring

How to understand this complex number based on physical reasons? Thank you for your recommended paper.

15. Oct 29, 2015

### Demystifier

16. Oct 29, 2015

### Pring

17. Oct 31, 2015

### stevendaryl

Staff Emeritus
Well, I did it off the top of my head, but I'm sure it's been done before. I always try to make sense of dubious mathematics by inserting parameters to make everything converge and then see if the results make sense when I let the parameter go to zero (or infinity, or whatever).

So, for example, the dubious result that $\frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx} dx = \delta(k)$, I always think of as:
$\frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx - \lambda x^2} dx = \frac{1}{2 \sqrt{\pi \lambda}} e^{-\frac{k^2}{4\lambda}}$

Or alternatively, I think of it as:
$\frac{1}{2\pi} \int_{-L}^{+L} e^{ikx} dx = \dfrac{sin(kL)}{\pi k}$

18. Nov 2, 2015

### Demystifier

Perhaps, but it is certainly not well known. I believe you could publish it in a journal for pedagogic papers such as American Journal of Physics or European Journal of Physics.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook