Classical vs quantum wave amplitudes?

[LarryS]

In classical mechanics and EM, the energy carried by a wave is the amplitude squared. In QM the (complex) amplitude squared of the position-space wave function is the position probability density. Do physicists regard this as anything more than just an interesting coincidence? Has anybody ever proposed a direct physical/mathematical relationship between the two (classical vs quantum) amplitudes squared?

Thank you in advance.

 

[davidge]

In QFT, consider for simplicity the Number Operator acting on a general-single-particle state $$N_a (k) \{ket \} \bigg(\sum_{k'}A_{k'} \exp(-ik'x) \frac{1}{V} \bigg) = |A_k|^2 \{ket \} \bigg(\sum_{k'}A_{k'} \exp(-ik'x) \frac{1}{V} \bigg)$$

Take the free Maxwell equations $\partial_\alpha \partial^{\alpha} A = 0$ for simplicity with $A^{0} = \varphi = 0$, (in which case $E = - \partial A/ \partial t$, with the general solution above, $$E = \pm i \omega A \exp{\pm i(\omega t - kx)}$$
Do you see the link between the two?

 

[PeterDonis]

Do physicists regard this as anything more than just an interesting coincidence?

Most definitely.

Has anybody ever proposed a direct physical/mathematical relationship between the two (classical vs quantum) amplitudes squared?

Yes. If you want to read an informal discussion of some of the issues involved, some time back there was an exchange on sci.physics.research, called the "photons, schmotons" thread, that got distilled into a long series of articles which you can find here:

http://math.ucr.edu/home/baez/photon/triv-ex.htm

The basic question being asked is the same as yours, except that instead of "energy" it talks about the photon number operator. But since the beam of light being considered is monochromatic, the energy operator is just the photon number operator times $\hbar \omega$, so they're really the same thing. I highly recommend reading the whole thing if you can take the time; it brings up a lot of issues that tend to be glossed over in quick discussions of quantum field theory.

 

[PeterDonis]

Do you see the link between the two?

The link is actually in the first equation you give, which basically says that the eigenvalues of the number operator are $| A_k |^2$. The question is how to actually demonstrate that.

I'm not sure about the second equation you give; it seems to be saying that energy is equal to $A$, not $A^2$. In particular, where are you getting $E = - \partial A / \partial t$ from?

 

[davidge]

That $E$ is the electric field, actually. I wanted to give it in a form given in Maxwell's theory for the OP to compare the two.

 

[PeterDonis]

That EE is the electric field, actually.

Ah, ok. I should have realized that from the reference to Maxwell's Equations.

 

[vanhees71]

Most definitely.
Yes. If you want to read an informal discussion of some of the issues involved, some time back there was an exchange on sci.physics.research, called the "photons, schmotons" thread, that got distilled into a long series of articles which you can find here:

http://math.ucr.edu/home/baez/photon/triv-ex.htm

The basic question being asked is the same as yours, except that instead of "energy" it talks about the photon number operator. But since the beam of light being considered is monochromatic, the energy operator is just the photon number operator times $\hbar \omega$, so they're really the same thing. I highly recommend reading the whole thing if you can take the time; it brings up a lot of issues that tend to be glossed over in quick discussions of quantum field theory.

Well, one should also be a bit more careful concerning the "photon number operator". Since photon number is not conserved, it's not so easy to define a Lorentz invariant quantity. That's why one usually plots the "invariant momentum spectrum" $\propto E \mathrm{d} N/\mathrm{d}^3 \vec{k}$.