# B Beam splitter, one photon, one detector?

1. Nov 21, 2017

### Idunno

I would like to know the results of a few scenarios with a beam splitter.

(1) You send a single photon through a half silvered mirror with a reflector at either side, as above, but instead of having two detectors, and a 50% chance of either going off, you just have one detector, and where the other detector should be, there is just empty space for a long distance. What is the chance of the one detector going off? 100%? 50%?

(2) Same as before, but with a laser beam, not just one photon at a time, any difference?

(3) Two detectors, one photon at a time, but detectors are not equidistant, one further than the other.

Thanks, I am curious as to the results of this, but can't find an easy source.

2. Nov 22, 2017

### BvU

Well:

1) 50 percent
2) 50 percent
3) 50 percent

any objections ?

3. Nov 22, 2017

### Staff: Mentor

As I understand the problem, in case #2 both detectors will be continuously triggered (if their construction allows it) as they're both continuously illuminated.

4. Nov 22, 2017

### Idunno

Follow up question if you don't mind: What if, in a double slit experiment, you rotate the detector screen by 45 degrees or so? Usually the detector screen is parallel to the wall with the slits in it, but now it's at an angle. This make a difference? More intense signal at the edge closest to the slits perhaps?

The issue in my mind is: what if some of the wavefunction encounters a detector before the rest of the wavefunction encounters anything at all? In the case of the beamsplitter and one detector, "half" of the wavefunction encounters a detector, and the other half encounters nothing, can this make a difference to how collapse happens? Same deal with the rotated screen?

Last edited: Nov 22, 2017
5. Nov 22, 2017

### Staff: Mentor

This question is much easier to handle if you don't use a collapse interpretation. You just calculate the amplitude at each point on the screen - it's the sum of the amplitudes through each slit, and the only difference is that the distance from the slits and hence the phase at any given point is different when the screen is tilted. Square the amplitude and you'll have your probability at each point.

Last edited: Nov 22, 2017
6. Nov 22, 2017

### Idunno

Ah, Thank you. :)

7. Nov 24, 2017

### bahamagreen

Is there an asymmetry in the path without the detector when time reversed - the photon path apparently originating from "empty space for a long distance"?

Or, if the whole apparatus were enclosed within a box, that the path without the detector when time reversed has the photon path emission event located a little too conveniently at just the right place on the inner wall of the box?

These non-detector path time reverse emission events seem kind of fortuitously spontaneous... how are they understood?

8. Nov 26, 2017

### BvU

What is it you expect from bringing in time reversal ? (The wave equation is first order in time, there is no symmetry to be exploited)

9. Nov 26, 2017

### vanhees71

That's of course wrong. As long as the Hamiltonian is time-reversal invariant (a symmetry is always a property of the Hamiltonian in QT), the Schrödinger equation is time-reversal invariant. You must not forget that time-reversal invariance is realized by an anti-unitary operator!

10. Nov 26, 2017

### BvU

Oops... my bad. I still have a hard time imagining what you can achieve with a time reversal thought experiment in this setup

11. Nov 26, 2017

### vanhees71

Well, I can't either...

12. Nov 26, 2017

### bahamagreen

I can't tell if my question was answered or deflected, or if you think my question is wrong, or "not even wrong".