# I Wavelength of photons

#### sophiecentaur

Gold Member
As has been discussed many times on this forum (see, eg, https://www.physicsforums.com/threads/why-no-position-operator-for-photon.906932/), photons never have a well defined position because you cannot construct a position operator for photons.
I agree. But can't the photon state be looked of as being at the far end of the HUP conditions and that implies its Energy can be specified arbitrarily accurately.
I remember the 'photon in a box' thought experiment where its 'location' can be restricted to within the dimensions of a reflecting box. In that model, the Δx would seem to have a meaning. Or is that too Classical?

#### sophiecentaur

Gold Member
Is the similar interpretation of electrons as particles similarly wrong?
Not at all. A free electron's position can be described with a given accuracy and that limits the accuracy of its momentum. That's the basis of HUP and it can be seen to apply in practice to electrons.
If you're trying to compare and contrast electrons and photons, the most relevant differences are speed and mass, which will make a difference to how HUP can be applied.

Staff Emeritus
The HUP is a statement about identically prepared states. A photon with a definite position, if that state exists at all (it does not) must exist in an eigenstate of photon number (n=1). That means it's not in an eigenstate of anything that does not commute with the number operator.

An electron, on the other hand, has every operator commute with the number operator, because charge is conserved.

#### sophiecentaur

Gold Member
That means it's not in an eigenstate of anything that does not commute with the number operator.
I feel that could be expanded on a bit to make it suitable for an I level thread.

#### hutchphd

As has been discussed many times on this forum (see, eg, https://www.physicsforums.com/threads/why-no-position-operator-for-photon.906932/), photons never have a well defined position because you cannot construct a position operator for photons.
I believe that what you say is true. But then what is it that is being measured when an individual pixel (out of 10 million possible pixels) on a CCD indicates a photon mediated event in a short time interval. Does this not indicate a position?? If not what should we call it?
I'm not trying to be argumentative here I am truly mystified.

#### sophiecentaur

Gold Member
Does this not indicate a position??
It indicates the position at the time of the interaction - it is no longer a photon, once it has done its job of releasing a photoelectron (or whatever)

#### davenn

Gold Member
on a CCD indicates a photon mediated event in a short time interval. Does this not indicate a position??
as has been said earlier in the thread and again in post #26 above this one ....

That is the time when a position can be indicated... abit for a very brief moment
the photon no longer exists after that time. The act of detecting the photon, destroys it. by the act of absorption.
You cannot measure the position of a photon at some point along a "light" beam and then have it continue along in it's path

#### sophiecentaur

Gold Member
The act of detecting the photon, destroys it.
The 'destuction' of a photon is very undramatic compared with destroying a particle with mass. On the one hand we have hf 'released' and on the other hand we have mc2

#### Orodruin

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On the one hand we have hf 'released' and on the other hand we have mc2
Gamma ray photons have a higher hf than an electron has mc2, so it kind of depends on the photon.

on the other hand we have mc2
Tell that to an LHC proton ...

#### hutchphd

You cannot measure the position of a photon at some point along a "light" beam and then have it continue along in it's path
There are some folks who claim otherwise:

https://physics.aps.org/articles/v11/38

Also I don't believe the original question would be invalidated by absorption. Whether one can do non-destructive testing on the photon takes us rather far afield from the original question, parts of which remain unanswered:

How precise can a wavelength of photons be measured and how much can it vary?
For example, 300nm, 300.1nm, 300,11nm, 300.111 etc...
What is the limit up to which we can measure it or is there a point where there is no variation anymore - something like a "quantum" of wavelength?
I propose the following to make the measurement: Look for reflection from a fiber Bragg grating from an attenuated source using a cascade photo multiplier. Individual quantum events (photons??) can be differentiated. I believe the reflection wavelength FWHM for visible is around 1 ppm for a good long grating. Is this not the (albeit destructive) measurement of a photon wavelength?

#### f95toli

It also depends on what type of measurement you are happy with. In a Cs atomic clock we know the frequency (and since we are in a high vacuum the wavelength) to within 1 part in 10^15 and in a good optical clock (using photons of e.g. 700 nm) 1 part in in 10^18.

This is of course an "indirect" measurement of the wavelength

#### votingmachine

I feel that the original question had two components and the arguments here tend to mix them up:

How precise can a wavelength of photons be measured
and how much can it vary?
... is there a point where there is no variation anymore - something like a "quantum" of wavelength?
The answer to the question of whether photon wavelengths can be infinitessimally different in wavelengths or are discretely quantized is that they are NOT quantized in wavelengths, but can be ANY wavelength. The argument from the doppler shift is particularly compelling IMO.

The answer to how precisely wavelength can be measured slips into an uncertainty problem. And a technological problem.

But I think the question was primarily if wavelength is quantized, and the measurement issues from uncertainty were a tangential matter.

On that tangential issue, since a photon has momentum, and an atom is generally 0.1 nm or larger, doesn't that imply the wavelength certainty could never be measured beyond about 0.01 nm? (I'm reasoning that p=h/lambda and (delta-p)*(delta-x)=h/4pi ... so if delta-x is the width of an atom, delta-p leads to a delta-lambda of 1/4pi).

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#### hutchphd

They did go to a fair amount of effort. I also thought it interesting and couldn't find any flaw in their setup......but this is tricky stuff, mostly beyond my rapidly decreasing attention span.

#### sophiecentaur

Gold Member
There are some folks who claim otherwise:
I find that sort of article very disturbing because it suggests that you could take the idea much further and that could upset a lot of basic principles. Another of those trapdoors in Physics.

#### almostvoid

Gold Member
Potentially. There's something called a Planck length that could be the limiting lower size of anything in the Universe. The problem is it's not something we're near being able to test, so it's very hypothetical. You can read more about it and what the consequences of it could be here https://newt.phys.unsw.edu.au/einsteinlight/jw/module6_Planck.htm

It also covers a bit if the other part of your question, but many before me have already addressed that.

Hope that helps
It helped me---I write sci fi and this site -above was excellently explained. Thank you

"Wavelength of photons"

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