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

[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?

 

[bhobba]

##<p|[x,p]|p>=i\hbar##

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/dp/0521558905/?tag=pfamazon01-20

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

Thanks
Bill

 

[Pring]

er formal manipula

I just regard this term as normalization 1.

 

[bhobba]

I just regard this term as normalization 1.

<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

 

[Pring]

<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

Aha, you are right, thank you!

 

[bhobba]

Aha, you are right, thank you!

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

 

[Pring]

No problem.

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

That 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

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

 

[stevendaryl]

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

[itex]\hat{p} \phi_p = (p + 2 i \lambda x) \phi_p[/itex]

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

[itex]\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[/itex]
[itex]= p \langle \phi_p|x|\phi_p \rangle - 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

[itex]\langle \phi_p | x \hat{p} | \phi_p \rangle = \langle \phi_p|(p + 2i \lambda x)x|\phi_p\rangle[/itex]
[itex]= p \langle \phi_p|x|\phi_p \rangle + 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

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

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

[itex]\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \frac{1}{4 \lambda} = -i[/itex].

 

[Pring]

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

[itex]\hat{p} \phi_p = (p + 2 i \lambda x) \phi_p[/itex]

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

[itex]\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[/itex]
[itex]= p \langle \phi_p|x|\phi_p \rangle - 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

[itex]\langle \phi_p | x \hat{p} | \phi_p \rangle = \langle \phi_p|(p + 2i \lambda x)x|\phi_p\rangle[/itex]
[itex]= p \langle \phi_p|x|\phi_p \rangle + 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

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

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

[itex]\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \frac{1}{4 \lambda} = -i[/itex].

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>##?

 

[stevendaryl]

We know the momentum operater is a observable physical quantity. Why its eigenvalue is complex?

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

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

I'm going to think about that.

 

[Demystifier]

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

[itex]\hat{p} \phi_p = (p + 2 i \lambda x) \phi_p[/itex]

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

[itex]\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[/itex]
[itex]= p \langle \phi_p|x|\phi_p \rangle - 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

[itex]\langle \phi_p | x \hat{p} | \phi_p \rangle = \langle \phi_p|(p + 2i \lambda x)x|\phi_p\rangle[/itex]
[itex]= p \langle \phi_p|x|\phi_p \rangle + 2i \lambda \langle \phi_p |x^2|\phi_p\rangle[/itex]

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

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

[itex]\langle \phi_p | \hat{p} x - x \hat{p} | \phi_p \rangle = -4i \lambda \frac{1}{4 \lambda} = -i[/itex].

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

 

[Demystifier]

But you're right, it's kind of weird that the expectation value of p̂ is a complex number.

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

 

[Demystifier]

For a mathematically more rigorous treatment of such stuff see also
http://lanl.arxiv.org/abs/quant-ph/9907069

 

[Pring]

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

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

 

[Demystifier]

How to understand this complex number based on physical reasons?

It can be interpreted in terms of complex weak values. See e.g.
http://arxiv.org/abs/0706.4207
http://arxiv.org/abs/1112.3986

 

[Pring]

It can be interpreted in terms of complex weak values. See e.g.
http://arxiv.org/abs/0706.4207
http://arxiv.org/abs/1112.3986

Thank you. You help me a lot!

 

[stevendaryl]

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

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 [itex]\frac{1}{2\pi} \int_{-\infty}^{+\infty} e^{ikx} dx = \delta(k)[/itex], I always think of as:
[itex]\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}}[/itex]

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

 

[Demystifier]

Well, I did it off the top of my head, but I'm sure it's been done before.

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.