How can laser photons have the same precise energy?

[LarryS]

The photons generated by a conventional quantum laser are all in the same quantum state. Doesn't that mean that they all have the same exact energy?

Yet, because of energy-time uncertainty, the exact energy of any particle can never be measured. Also, the Copenhagen Interpretation says that it is meaningless to talk about qualities of quantum objects that can never be observed.

Comments?

Thanks in advance.

 

[Charles Link]

Photons are bosons, so this allows more than one photon in the same state. In all lasers, there is still some bandwidth, and you will not find all of the photons in the same boson state.

 

[Cryo]

The photons generated by a conventional quantum laser are all in the same quantum state. Doesn't that mean that they all have the same exact energy?

Thanks in advance.

Yes. No.

In an idealized treatment, the output of the laser is the coherent state which does not correpond to photons with the same energy.

 

[Vanadium 50]

Doesn't that mean that they all have the same exact energy?

Do you mean is the energy distribution narrower than the same line when its not lasing? No. Why should it be?

 

[PeterDonis]

The photons generated by a conventional quantum laser are all in the same quantum state. Doesn't that mean that they all have the same exact energy?

No, because the state that all of the photons generated by the laser are in is not an eigenstate of energy.

 

[LarryS]

Yes. No.
In an idealized treatment, the output of the laser is the coherent state which does not correpond to photons with the same energy.

No, because the state that all of the photons generated by the laser are in is not an eigenstate of energy.

Oh, okay. I just recently learned about the existence of these "Coherent States". Very interesting to me that eigenstates of non-Hermitian operators should play such an important role. Thanks.

 

[Paul Colby]

All oscillators have some line width (energy spread). Lasers are no different. Laser line widths are measurable

http://www.eblanaphotonics.com/downloads/Linewidth.pdf

 

[PeterDonis]

All oscillators have some line width (energy spread).

This is a separate issue from the issue raised in posts #3 and #5. Line width means that the photons emitted by any real laser are not perfectly in phase. But the laser photons being in a coherent state means that, even for an idealized laser that emits all its photons exactly in phase, the photons are not in an eigenstate of energy and so cannot be said to all have the same energy. All that can be said is that they all have exactly the same phase (in the idealized case).

 

[Charles Link]

Spectral line width indicates the laser is composed of photons of different frequencies.

 

[Paul Colby]

Yet, because of energy-time uncertainty, the exact energy of any particle can never be measured. Also, the Copenhagen Interpretation says that it is meaningless to talk about qualities of quantum objects that can never be observed.

It may be observed, just never without some error or limit to measurement. All devices and all measurements will have noise sources.

 

[Swamp Thing]

Since the photon arrival rate has a Poisson fluctuation, even an ideal laser is amplitude-modulated with noise, which might account for some of the line width.

But is there necessarily some phase / frequency fluctuation as well, in a coherent state?

 

[PeterDonis]

Since the photon arrival rate has a Poisson fluctuation, even an ideal laser is amplitude-modulated with noise

No, an ideal laser does not have a "photon arrival rate". A particular measuring device that detects photons might show a Poisson fluctuation in the rate of detections if the photons are coming from a laser. But the coherent state of the ideal laser itself does not fluctuate.

is there necessarily some phase / frequency fluctuation as well, in a coherent state?

Not for an ideal laser, no. For any real laser, there is some, but that's because no real laser produces an exact ideal coherent state.

 

[Swamp Thing]

If this page is not over-simplifying or just plain wrong, the Poisson thing is baked into the definition of the coherent state, and not an artifact of measurement.

It's a company site, so maybe not very rigorous, but... https://www.rp-photonics.com/coherent_states.html

I do seem to remember that the explanation in e.g. Saleh and Teich was similar but I didn't really understand it too well at the time.

 

[PeterDonis]

If this page is not over-simplifying or just plain wrong, the Poisson thing is baked into the definition of the coherent state

The Poisson distribution shown on that page is the "distribution" of the different photon number eigenstates (Fock states) in a coherent state; in other words, it's a distribution of the coefficients of the photon number eigenstates $|n\rangle$ as a function of the photon number $n$. But that has nothing to do with the distribution of detections of photons by a measuring device.

 

[Swamp Thing]

But don't the coefficients for each eigenstate $|n\rangle$ translate into measurement probabilities to see n photons in an ideal detector?

Edit: In an ideal photon counter ?

 

[PeterDonis]

don't the coefficients for each eigenstate $|n\rangle$ translate into measurement probabilities to see n photons in an ideal detector?

In an ideal detector that can detect any number $n$ of photons all at once, yes. But the Poisson distribution you were talking about earlier is a Poisson distribution of arrival times, i.e., one photon each detection, but random intervals between detections. The coefficients of the different eigenstates in the coherent state don't tell you about that.