Can HUP Explain the Instant Emission of Photoelectrons?

[lightarrow]

A sinusoidal EM wave which length l is very great, hits a photoelectric device.

According to HUP, a sinusoidal EM wave with finite length l has a spectrum of frequencies which broadness is inversely proportional to l. So, if l is very great, the energy uncertainty is very low, and this, still for HUP, should mean that the time required to make that energy measurement is high. Or not?

If this were true, why, instead,the photoelectrons are emitted almost instantly?

 

[DrChinese]

According to HUP, a sinusoidal EM wave with finite length l has a spectrum of frequencies which broadness is inversely proportional to l.

The HUP does *not* say that there is a connection between a photon's frequency "spectrum" and its wavelength. The HUP says there is a relationship between the frequency (actually energy) spectrum and the time spectrum.

\Delta E \Delta t & \ge \frac{\hbar}{2}

If you work through it, you will see that the wavelength itself is not a factor, even though wavelength and frequency themselves are inversely proportional. I.e. it is the deltas that are related by the HUP.

 

[lightarrow]

The HUP does *not* say that there is a connection between a photon's frequency "spectrum" and its wavelength. The HUP says there is a relationship between the frequency (actually energy) spectrum and the time spectrum.

\Delta E \Delta t & \ge \frac{\hbar}{2}

If you work through it, you will see that the wavelength itself is not a factor, even though wavelength and frequency themselves are inversely proportional. I.e. it is the deltas that are related by the HUP.

In my post I talked about the length l of the wave train, not the wavelenght. Probably I didn't explain well.

 

[RandallB]

If this were true, why, instead,the photoelectrons are emitted almost instantly?

And since as you put it the electron is produce "almost instantly" you must have some uncertainty about when it is, and I assume you have some uncertainty about where it or its energy is. Sounds consistent with the uncertainty defined with the originating photon, enough so as to not be able to question “why”.

 

[CarlB]

If this were true, why, instead,the photoelectrons are emitted almost instantly?

Yes, the length of the EM wave train is very long so delta E is very small. This means that the time at which the photon is absorbed (and the electron is emitted) is very imprecise. The HUP implication is not that the electron is spread out in time, but instead that the arrival time of the electron cannot be predicted in advance. And the energy of the electron is not ill defined, just the time that it arrives relative to the duration of the EM wave. If you run the experiment many times you will get many different electron arrival times. Let me explain this in more detail.

For example, suppose the wave is 500 feet long, and consists of a single photon. The wave takes about 500ns to pass a point in space, so you cannot expect the electron to appear except within a period of time of 500ns. If you run the experiment, you will find that the electron is emitted at various times, depending on how lucky you are.

Now anyone of those electrons will be measured to arrive at a very specific time. For exmple, one run might detect the electron at 139ns +- 1ns after the beginning of the wave train. But the +- 1ns is not the delta t of the HUP. The +- 1ns is the inaccuracy in your measurement apparatus. The delta t of the HUP is the fact that the photon is annihilated and the energetic electron appears at any time in that 500 ns.

Is that the explanation you were looking for?

 

[lightarrow]

Yes, the length of the EM wave train is very long so delta E is very small. This means that the time at which the photon is absorbed (and the electron is emitted) is very imprecise. The HUP implication is not that the electron is spread out in time, but instead that the arrival time of the electron cannot be predicted in advance. And the energy of the electron is not ill defined, just the time that it arrives relative to the duration of the EM wave. If you run the experiment many times you will get many different electron arrival times. Let me explain this in more detail.

For example, suppose the wave is 500 feet long, and consists of a single photon. The wave takes about 500ns to pass a point in space, so you cannot expect the electron to appear except within a period of time of 500ns. If you run the experiment, you will find that the electron is emitted at various times, depending on how lucky you are.

Now anyone of those electrons will be measured to arrive at a very specific time. For exmple, one run might detect the electron at 139ns +- 1ns after the beginning of the wave train. But the +- 1ns is not the delta t of the HUP. The +- 1ns is the inaccuracy in your measurement apparatus. The delta t of the HUP is the fact that the photon is annihilated and the energetic electron appears at any time in that 500 ns.

Is that the explanation you were looking for?

Yes, but not with a 500 feet long photon, with a 500 light seconds long one.

 

[CarlB]

Yes, but not with a 500 feet long photon, with a 500 light seconds long one.

500 feet is already very very long compared to the wavelength of light, and so the delta E of light is very accurately determined. (I.e. something like 1 part in 10^8).

If you make the light wavetrain instead be 500 seconds long, it just makes delta E that much smaller, and correspondingly increases the length of time over which the electron may arrive (i.e. it increases the delta t).

 

[reilly]

The standard way to do photoelectric assumes pure plane waves for the incident photon and the emittted electron. When you restrict the photon to a finite spatial range you get a term like sinkr1/kr1 - sinkr2/kr2 rather than the usual momentum delta function. r1-r2 is the length of the incident wave packet.(Don't worry, momentum is still conserved.Recall that the incident photon is in a state that encompasses all frequencies from 0 to infinity.So the term above needs to be integrated over the photon spectra) So the finite wavetrain will kick out an electron, with a probability dependent upon the momentum and the spatial characteristics of the incident wave train.

I'll leave it at that.

Re the HUP, first it is built into the computation of the matrix element. And, in fact, with a bit of effort, you can trace a Gaussian (or any other packet) to see how the uncertainties propagate through the system.

Regards,
Reilly Atkinson

 

[lightarrow]

The standard way to do photoelectric assumes pure plane waves for the incident photon and the emittted electron.

Regards,
Reilly Atkinson

So \Delta t should be infinite in that case?

 

[CarlB]

Yes, delta t is infinite.

 

[lightarrow]

Thanks to all for your answers.