# B Uncertainty Principle and a photon

1. Nov 13, 2016

### backward

Let a photon of a definite wavelength (hence a definite momentum ) start it's journey at time 0. After 8.3 minutes it hits a detector on earth. So it's position is exactly known (in fact it can be predicted for any time less than 8.3 minutes). So we have particle with definitely known momentum and position at t=8.3 minutes in violation of the UP. Where have I committed a mistake.

2. Nov 13, 2016

### Staff: Mentor

By not doing the actual math. If you do, you will find that your ordinary language description is not correct, for, at a minimum, the following reasons:

...is not what you think it is. There actually is no such thing--there is no quantum state of the electromagnetic field that actually has this property.

This is not possible; there is no quantum state of the electromagnetic field that has this property either. (Strictly speaking, this is true for any particle, even one with nonzero rest mass; but the dodge that is used to get around that with massive particles, called Newton-Wigner localization, doesn't work with massless particles, so thinking of a "photon" as having a definite position is even more problematic than thinking that way about a massive particle.)

As should be obvious from the above, this is not correct.

And here, even if we disregard all of the above criticisms, you still have a problem: measuring a particle's position changes its state, so even if it was in a state of definitely known momentum before measurement, it will not be after measurement. So your scenario does not violate the HUP even if you formulate it for a quantum object that doesn't have all the issues that a "photon" has.

3. Nov 13, 2016

### Simon Bridge

The initial conditions violate HUP so of course the conclusion does also.
Consider - the only way you know that the photon takes 8.3 minutes to reach the detector from it's initial position is if you know it is 8.3 light minutes away from the detector at t=0 to some precision ... but any precision to the position measurement at t=0, with "definite wavelength" at t=0, violated HUP at t=0.
So even if it were possible to get those eigenstates IRL, which it isn't, the logic of the thought experiment fails.

Then there is the phenomenon that the subsequent measurement will disturb the photon ... so even if you had a very precise momentum to begin with, that precision changes with the position measurement. Strictly, by detecting the photon, you have destroyed it. So what is the momentum after detection?

4. Nov 14, 2016

### backward

Thanks.
I understand that the best way to describe QP is mathematical; but I was trying to get as much coceptual clarity as possible.
Does it imply that a travelling electromagnetic wave cannot be represented by a particle travelling with velocity c subject to HUP?
Of course I realise that as one measures a particles position accurately, it's momentum would become uncertain.

5. Nov 14, 2016

### houlahound

I swear this is not a joke question but wouldn't the destroyed photon have a precise momentum of zero?

6. Nov 14, 2016

### PeroK

One thing your are missing is that a measurement of a quantum object changes its state. Strictly speaking, this is not the HUP, but it is an equally important aspect of QM. In terms of a particle

1) You measure its momentum to a high degree of accuracy.

2) You measure its position to a high degree of accuracy.

3) If you now remeasure its momentum, then will get probabilistically a large range of values. Measurement 2 has effectively destroyed the state of "known" momentum. After measurement 2 you have uncertain momentum and no longer a known value.

If you repeat this experiment many times you will find that measurements 2 & 3 obey the HUP, which actually deals with the expected values of momentum and position over many identical cases. The HUP does not in fact say anything about any two specific measurements of position and momentum.

7. Nov 14, 2016

### backward

After destruction, the particle doesn't exist. So it's momentum has no value, which is different from having a value equal to 0.

8. Nov 14, 2016

### backward

Thanks a lot PeroK

9. Nov 14, 2016

### backward

Another way of posing the question which was in my mind is:
According to QP, a particle with precisely known momentum will have infinite uncertainty in its position. A monochromatic photon would then be impossible to locate anywhere. I do not know in what sense this could be true.
I know I am awfully vague.

10. Nov 14, 2016

### PeroK

It's not true at all, because you can never know the momentum precisely. The HUP, in fact, forbids you from ever knowing precisely the momentum or position of a particle. There is always a degree of uncertainty in both.

11. Nov 14, 2016

### backward

Thanks again.

12. Nov 14, 2016

### Staff: Mentor

In general, yes. A general traveling EM wave has too many degrees of freedom. In some special cases, we can use the "particle traveling at velocity c" description as a useful approximation.

13. Nov 14, 2016

### Staff: Mentor

There is nothing in the HUP that forbids knowing precicely either the position or momentum exactly.

Thee principle says suppose you have a large number of systems in the same state. Divide them into two large lots. Measure the position in one lot. You can measure exactly as you like - only practical considerations prevent exact values ie it would require a Dirac Delta function as a wave-function and such in practice is impossible. Do the same for momentum in the other lot. Again you can measure exactly as you like. Now compare the statistical spread of each set of measurements - it will be as per the HUP.

There is a bit of confusion around the HUP but Ballentine explains it precisely.

Thanks
Bill

14. Nov 14, 2016

### houlahound

I interpret (probably wrongly) that the HUP refers to making a successive measurement ie you measure the position accurately and only THEN you no longer know where the target thing is heading straight AFTER you just measured it's position?

15. Nov 14, 2016

### OCR

I'm not sure, but it would seem that an attempt to measure the one-way speed of light is also occuring ?

16. Nov 14, 2016

### Simon Bridge

The thought experiment can be set up within some convention for synchronising clocks that is independent of the experiment.
Some of this thought experiment is a little bogged down by the special status of light ... pick another particle.

That would be relevant to the thought experiment, but Bhobba has it strictly correct in post #13. The "measurement disturbs the system" idea is broadly correct, very commonly attributed to HUP, but not quite what HUP says.

Perhaps we should rephrase things so that we have a particle, other than a photon, produced with a well defined momentum, and then measure position with a detector that has a very small aperture.

The uncertainty in momentum has been determined by the apparatus of the source, while the uncertainty in position is determined by the detector ... these two devices will be functionally independent (technically they are linked by the particle - which is probably a clue right?) so how does HUP work in this case?

@backward : that about it? Intuitively the source and detector characteristics can have whatever relationship we like right?

17. Nov 14, 2016

### houlahound

could not the momentum and position be simultaneously defined with a set up with an infinitely small aperture at the detector as you mentioned to narrow down position but also a very long collimating tube/"lens"/physical channel because only the particles with the slightest deviation in the velocity will be going in the correct direction to make it to the detector.

might be hard to visualise without a diagram but I think my description is clear eg only a projectile perfectly aligned with say the axis of a barrel of a gun will be selected for the velocity that is in the precise direction of the axis of the barrel.

18. Nov 15, 2016

### PeroK

It depends on your definition of "know the position of a particle". Given the statistical nature of QM I would say that to know precisely the position of a particle in a given state, you would need to obtain precisely the same result on an ensemble of particles in that state. The variance of these measurements would be 0, violating the HUP.

If you do one measurement of position and declare the particle to be precisely at that position, then that is a different matter. I agree that does not violate the HUP.

I did already say in post #6 that the HUP has nothing to say about any specific measurement of position.

PS looking back, the purpose of my subsequent post #10 was to tackle the idea that the 0 uncertainty of position can be offset by an infinite uncertainty in momentum. Or, vice versa.

Last edited: Nov 15, 2016