Measuring Wavelength of Photons: Precision & Variation

[liroj]

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?
Hope you understand what I mean because I'm not a physicist and my knowledge of physics is limited.

 

[ZapperZ]

The precision is limited by technology. As we improve our instruments and our techniques, such precision also improves.

BTW, it is "wavelength".

Zz.

 

[liroj]

Thanx. I know it's wavelength but was typing on my mobile phone.
Is there a theoretical limit to the value of the wavelength? "A quantum of it"... the energy a photon can absorb, which is quantized, should relate to it, but my math skills are not that great... :)

PS
What is the current limit to the measurement of the wavelength?

 

[Drakkith]

Is there a theoretical limit to the value of the wavelength? "A quantum of it"... the energy a photon can absorb, which is quantized, should relate to it, but my math skills are not that great... :)

There is no limit to how small a change in wavelength can be. In other words, wavelength is not quantized. The fact that doppler shift exists, which is a continuous process you get just by moving towards or away from a light source, guarantees this even if there was some underlying process that quantized the wavelength (there isn't).

 

[sophiecentaur]

How precise can a wavelength of photons be measured

The precision with which anything can be measured depends on the amount of Energy available to the measuring system compared with random effects. (i.e. Signal to Noise Ratio). The energy of a single photon is pretty hard to determine because it has only one chance to interact with a sensor of some kind. If you, for instance, used a spectrometer and looked at which of the arc of sensors (at different angles) reacted to it, you would get just one and the experiment would be over. Using more photons (enough to treat it as a wave measurement) you would get a response of several of those sensors, with a central peak, which you could identify with much more confidence. That single photon could have been anywhere on the 'bell shaped' distribution that your wave measurement gave you.
There are other measurement methods for measuring wavelength or frequency and the confidence in the result is always related to signal to noise, when you get down to it.
PS Wavelength changes are much more accurate to measure. You can take laser light with a pretty well defined frequency and by splitting the beam so that one path is direct and the other beam bounces off a very slowly moving mirror, you can re-combine the two beams and producing an interference 'beat' of a few Hz to show the frequency shift due to the Doppler effect. That's pretty good as a fraction of the (say) 600THz of the original light beam. But, again, that involves a wave with many photons.

 

[Michaela SJ]

The recent publishing of the Event Horizon Telescope's fantastic accomplishment vis-a-vis the imaging of a black hole sometimes refers to the capturing of 'photons'.
Is any emission in the electromagnetic spectrum considered a photon? As an amateur visual astronomer, I am familiar with the visible light wavelength of say green at ~510 nm. But the radio telescopes were capturing date in the millimeter range around 1.3 to 3 mm.

 

[sophiecentaur]

All EM interactions can be interpreted in terms of photons. Sometimes the photon approach is more useful than others.
Waves are wonderful models to use for the transfer of EM energy.

 

[f todd baker]

I cannot believe that nobody in this thread answered this question vis a vis the uncertainty principle. When the wavelength of a photon is specified, it is calculated from its energy via Einstein's photon energy E=hf=h(c/λ)⇒λ=hc/E. But you cannot know the energy exactly unless you observe it for an infinitely long time (ΔEΔt≤ħ). Therefore, no real photon has a wavelength, rather a distribution of wavelengths. Indeed, as other answers indicate, our technological limits on measurements in most cases limit how accurately we can actually determine λ, in other words uncertainties related to the actual measurements are greater than the ultimate uncertainty due to Heisenberg. Still, the essential physics here is that there is a limit regarding how accurately a photon energy, and therefore a photon wavelength, can be known.

 

[sophiecentaur]

I cannot believe that nobody in this thread answered this question vis a vis the uncertainty principle. When the wavelength of a photon is specified, it is calculated from its energy via Einstein's photon energy E=hf=h(c/λ)⇒λ=hc/E. But you cannot know the energy exactly unless you observe it for an infinitely long time (ΔEΔt≤ħ). Therefore, no real photon has a wavelength, rather a distribution of wavelengths. Indeed, as other answers indicate, our technological limits on measurements in most cases limit how accurately we can actually determine λ, in other words uncertainties related to the actual measurements are greater than the ultimate uncertainty due to Heisenberg. Still, the essential physics here is that there is a limit regarding how accurately a photon energy, and therefore a photon wavelength, can be known.

I would think of HUP as implying that the position of a photon in a beam of well defined wavelength cannot be determined. This is why photons as little bullets going from A to B is such a 'wrong' notion as they can take any path of any length. You can only 'know' of its presence after the event of some interaction.

 

[hutchphd]

I would think of HUP as implying that the position of a photon in a beam of well defined wavelength cannot be determined. This is why photons as little bullets going from A to B is such a 'wrong' notion as they can take any path of any length. You can only 'know' of its presence after the event of some interaction.

Is the similar interpretation of electrons as particles similarly wrong?

 

[sophiecentaur]

Is the similar interpretation of electrons as particles similarly wrong?

No, but HUP applies differently because they are different particles and the 'numbers' are different. When you consider an electron in a bound state, its wavelength is well defined (solution to the wave equation) so you cannot determine where it actually is in its 'orbit' so it is no longer a defined particle under those conditions. The fuzziness is many nm in extent, despite its 'size' when free being around 3X10^-15m

 

[f todd baker]

I would think of HUP as implying that the position of a photon in a beam of well defined wavelength cannot be determined. This is why photons as little bullets going from A to B is such a 'wrong' notion as they can take any path of any length. You can only 'know' of its presence after the event of some interaction.

Whatever you are saying may be true, but HUP is not only delta x delta p. I do not see, however, what it has to do with the uncertainty of the energy of a photon.

 

[f todd baker]

No, but HUP applies differently because they are different particles and the 'numbers' are different. When you consider an electron in a bound state, its wavelength is well defined (solution to the wave equation) so you cannot determine where it actually is in its 'orbit' so it is no longer a defined particle under those conditions. The fuzziness is many nm in extent, despite its 'size' when free being around 3X10^-15m

Actually an electron state in an atom does not have a well defined wavelength because it does not have a well defined energy. Only the ground state has no width because it lasts forever. But thinking of an electron as having a wave length at all when in a bound state is very semi-classical. Are you thinking of "de Broglie standing waves" in the Bohr atom? That is a cute way to make semi-classical models but not really very meaningful.

 

[hutchphd]

I would think of HUP as implying that the position of a photon in a beam of well defined wavelength cannot be determined. This is why photons as little bullets going from A to B is such a 'wrong' notion as they can take any path of any length. You can only 'know' of its presence after the event of some interaction.

I do not understand. I know they are different objects, but what part of your statement is incorrect if you substitute the word "electron" for "photon" in the above ??

 

[EnSlavingBlair]

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?

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

 

[Nugatory]

There's something called a Planck length that could be the limiting lower size of anything in the Universe.

It could be, but...
This misunderstanding is so pervasive as to have inspired an Insights article here: https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/

 

[EnSlavingBlair]

It could be, but...
This misunderstanding is so pervasive as to have inspired an Insights article here: https://www.physicsforums.com/insights/hand-wavy-discussion-planck-length/

Thanks, that was very interesting and has helped me shift my understanding of how that bit if physics can be applied to the Universe.

 

[davenn]

I do not understand. I know they are different objects, but what part of your statement is incorrect if you substitute the word "electron" for "photon" in the above ??

Electrons are particles, photos are not

Photons are quantum packets of energy

 

[Nugatory]

Electrons are particles, photons are not

Both are particles in the quantum mechanical sense of the term, which is of course very different from the way the word is used in ordinary life. Neither can be properly modeled as little bullets, but that model is less likely to mislead when applied to electrons because...

Is the similar interpretation of electrons as particles similarly wrong?

Not nearly so much, because non-relativistic QM works for electrons and I'm not going to lose any sleep over the distinction between the expectation value of position as a function of time and a classical trajectory. But even then the "little bullet" model comes up short when we consider phenomena such as electron diffraction.

This thread is in some danger of becoming an argument about terminology and classification; the key point here is that photons don't work the way the original poster was thinking they do.

 

[Orodruin]

I would think of HUP as implying that the position of a photon in a beam of well defined wavelength cannot be determined

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.

 

[sophiecentaur]

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]

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.

 

[Vanadium 50]

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]

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]

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]

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]

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]

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).

 

[davenn]

There are some folks who claim otherwise:

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

Very interesting and very specialised but under normal conditions, my response still stands

 

[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]

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.