How do wave-functions deal with energy (of the photon)?

[San K]

1. Are wave-functions, in some situations, also assumed to carry energy with them?
2. Can wave-functions be used to show/validate the law of conversation of momentum/energy?
3. In a single particle double slit/path experiments, if one path is blocked what happens to the wave-function ? and the energy?
4. Per the de-broglie bohm (DBB) interpretation:
In de Broglie–Bohm theory, the wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits.

If the wavefunction (per DBB is hypothesized) is to be traveling through both the slits, how is (the energy of) the photon supposed to be travelling?In single particle interference
(for example single particle double slit,
or
single particle through a Mach-Zehnder)

the particle interferes with "itself".

The path of the photon is different (when there is interference) than the path when there is no interference. Thus energy must be involved (at least per classical mechanics) during interference. ...to change the path of a photon

 

[The_Duck]

1. Are wave-functions, in some situations, also assumed to carry energy with them?

The energy of a particle is the frequency with which its wave function oscillates in time.

2. Can wave-functions be used to show/validate the law of conversation of momentum/energy?

Yes, it is easy to show that energy and momentum are conserved in quantum mechanics. But conservation of energy and momentum are more fundamental than quantum mechanics and arise directly from the invariance of the laws of physics under translations and time-translations. See http://en.wikipedia.org/wiki/Noether's_theorem

3. In a single particle double slit/path experiments, if one path is blocked what happens to the wave-function ? and the energy?

The part of the wave function that would have gone through the blocked slit is reflected or absorbed by whatever is blocking the slit. This prevents it from interfering with the part of the wave function that goes through the open slit.

4. Per the de-broglie bohm (DBB) interpretation:
In de Broglie–Bohm theory, the wavefunction travels through both slits, but each particle has a well-defined trajectory and passes through exactly one of the slits.

If the wavefunction (per DBB is hypothesized) is to be traveling through both the slits, how is (the energy of) the photon supposed to be travelling?

This wikipedia page shows the particle trajectories in the Bohmian interpretation of the double slit experiment: http://en.wikipedia.org/wiki/De_Broglie–Bohm_theory#Two-slit_experiment . The particle takes one of these paths. But since we don't know the initial position of the particle, we don't know which path it will take.

The path of the photon is different (when there is interference) than the path when there is no interference. Thus energy must be involved (at least per classical mechanics) during interference. ...to change the path of a photon

The path of a particle can be changed without changing its energy, in classical or in quantum mechanics. As the Moon orbits the Earth, its path is constantly deflected by the Earth's gravity, yet this process transfers no energy between the Earth and the Moon.

I think you should clearly distinguish between *forces*, which are what deflects particles' trajectories in classical mechanics, and *energy*, which is just a property of a particle's motion. The status of forces in quantum mechanics is actually quite interesting. Namely, force isn't a concept that really enters into quantum mechanics at all. Instead we deal with potential energy (force is the gradient of potential energy).

 

[San K]

The part of the wave function that would have gone through the blocked slit is reflected or absorbed by whatever is blocking the slit. This prevents it from interfering with the part of the wave function that goes through the open slit.

Thanks The Duck. Great answers.

So, in the above case:

Is the energy of the photon prior to encountering the slit different than later?

...because part of the energy has been absorbed (or reflected) by the obstacle...(?)

 

[The_Duck]

Is the energy of the photon prior to encountering the slit different than later?

...because part of the energy has been absorbed (or reflected) by the obstacle...(?)

No. At the end of the experiment we will find that either

1) the photon has passed through the open slit, maintaining the same energy it had before, or
2) the photon has been absorbed by the obstacle and destroyed, or
3) the photon has been reflected by the obstacle and has the same energy it had before.

 

[San K]

thanks The Duck

Re:The part of the wave-function that is absorbed (by the obstacle);

what happens to the energy carried/associated with it? ...say, per the wave-function treatment

 

[The_Duck]

It sounds like you are under the impression that the energy is spread out throughout the wave function in some fashion, so that we could say something like, "our photon has a total energy of 6 eV, 4 eV of which is carried by this part of the wave function and 2 eV of which is carried by this other part of the wave function." This is not the case.

When part of the wave function hits a barrier and is absorbed, and the other part doesn't, the interpretation is "with some probability the photon has been absorbed by the barrier and deposited all 6 eV of its energy, and with some probability the photon has missed the barrier, carrying all 6 eV of its energy with it and depositing none in the barrier."

As a somewhat glib and inaccurate summary: the wave function is made of probability, not energy.

 

[San K]

It sounds like you are under the impression that the energy is spread out throughout the wave function in some fashion, so that we could say something like, "our photon has a total energy of 6 eV, 4 eV of which is carried by this part of the wave function and 2 eV of which is carried by this other part of the wave function." This is not the case.

When part of the wave function hits a barrier and is absorbed, and the other part doesn't, the interpretation is "with some probability the photon has been absorbed by the barrier and deposited all 6 eV of its energy, and with some probability the photon has missed the barrier, carrying all 6 eV of its energy with it and depositing none in the barrier."

As a somewhat glib and inaccurate summary: the wave function is made of probability, not energy.

Thanks for clarifying The Duck.