Photons and HUP

[gort]

Quick question that probably has an obvious answer. I was reading Quantum Entanglement and the Loss of Reality by Marcella. Good read. But he insists that quantum particles (including photons) have no physical properties until they're measured. That said, photons, as quantum particles, should obey Heisenberg's Uncertainty Principle. For a single photon, we can compute its energy, hence, its momentum, quite easily. When it strikes a detector, we know exactly where it is. How does that agree with HUP - we seem to know momentum and position at the same time. What am I missing?

[George Jones]

How is this different from electrons that strike a detector?

Here is an interpretation of the HUP that I like. Prepare a large number, say 2N, of systems (e.g., electrons) that are in identical states. Measure the position of the electrons in half (i.e., N) of the systems. and measure the momenta of the electrons in the other N systems. Even though all the electrons are in identical states, the measured positions of the electrons will not all be same, i.e., there will be a statistical spread (standard deviation) of the measured positions. Similarly, there will be a statistical spread  for the measured momenta. The product of these statistical spreads will satisfy the HUP. Note that on any single system only one measurement is made. In this interpretation there is no system on which both position and momentum measurements are made.

[andrew s]

It's important to remember the statistical spread is not due to the "classical" measurement error but due to quantum nature of the microscopic world.

Regards Andrew 

[robin_astro]

22 hours ago, gort said:

For a single photon, we can compute its energy, hence, its momentum, quite easily.

You can compute the theoretical energy but if you measure it  you find a spread of values. This is seen for example  in the natural broadening of spectral lines. The lines from transitions with a short lifetime are broader in agreement with HUP

Cheers

Robin

Edited May 6, 2023 by robin_astro
typo

[vlaiv]

18 hours ago, gort said:

How does that agree with HUP - we seem to know momentum and position at the same time. What am I missing?

We don't.

Once you have measured photon's position - it no longer has energy it had to begin with.

Think of scattering event - say photon scattering of an electron. We know energy prior to scattering, we know where scattering took place - but we have no idea in which direction photon scattered and what frequency it has (until we do further measurements on electron).

[gort]

I'm still having a problem understanding. Please bear with me. I'm talking about a single photon. No spectrum spread. Consider an excited H atom and a transition from n=3 to n=2. That transition will emit a single photon in the red (656nm) and energy 1.89eV. Correct me if I/m wrong. That photon hits a detector. Please explain why we don't know p and x at impact time. Does it change p at the moment of impact?? Of course it imparts its p to an electron in the detector, but certainly there's conservation of momentum. Thanks!  

[robin_astro]

28 minutes ago, gort said:

 Consider an excited H atom and a transition from n=3 to n=2. That transition will emit a single photon in the red (656nm) and energy 1.89eV. 

The photons  have this energy on average but individual photons will have a spread of energy values even for photons repeatedly emitted from a single isolated atom, dependent on the lifetime of the excited state such that ∆E∆t ≈ h/2π. This is seen observationally as a broadening of the spectral line.  This is an example of the fundamental uncertainty inherent in our quantum world.

Robin