# I Why doesn't momentum transfer on mirrors reveal which path?

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1. Apr 30, 2017

### greypilgrim

Hi.

In a Mach-Zehnder interferometer, there are mirrors in both paths. When a photon gets reflected on a mirror, it transfers momentum. Also, mirrors are classical objects that are usually not assumed to exist in superposition states.
Shouldn't it (at least in principle) be possible to measure the forces on the mirrors, which would reveal the path taken and therefore destroy the superposition and interference on the screen?

2. Apr 30, 2017

### andrewkirk

That's a very interesting idea.

I think the Uncertainty Principle would prevent this. In quantum interferometer experiments like delayed-choice, the which-path information is recorded on a permanent record, via the photon detectors, so that it can be examined after the experiment. In contrast, to use momentum to measure which-path info, one would need to measure the momentum increments in real time, on a photon by photon basis, so that changes in screen momentum can be mapped to individual photons showing up on the screen.

Measuring momentum of the mirror with the extraordinarily high accuracy that demands would - courtesy of the HUP - massively increase the uncertainty of the position of the mirror. Since the mirror needs to be very precisely positioned in order to align the reflected and transmitted beams and stabilise path-length, that uncertainty would - I imagine - be enough to prevent any meaningful results to be obtained from the readings.

3. Apr 30, 2017

### Mentz114

Is this what you mean ?

https://en.wikipedia.org/wiki/Elitzur–Vaidman_bomb_tester

4. Apr 30, 2017

5. May 1, 2017

### Strilanc

I asked this same basic question on the physics stackexchange awhile ago.

Basically it comes down to the fact that the mirrors don't have one specific momentum; they have a superposition of momentums. The spread in momentums is wide enough that the trace distance (the effective difference) between the nudged and not-nudged states is nearly zero. The overlap is nearly perfect. And the math doesn't care about nudge-vs-not-nudge, it cares about trace distance.

In other words, yes hitting the mirrors does cause decoherence. But only a negligible amount.

6. Jun 25, 2017

### injinear

so, change the mirror to a molecule that scatters the light. the molecule momentum could be known fairly accurately by cooling. its momentum could subsequently be measured with Doppler. seems like there is no way around the problem by claiming the effect is "small".

7. Jun 25, 2017

### Strilanc

Well obviously if you're only using a single molecule the approximation of the mirror as not being affected is going to stop working. The spread in total momentum is much sharper with a single molecule than with an object made up of over $10^{20}$ molecules. The trace distance between the two cases is going to jump from nearly-0 to nearly-1.

8. Jun 25, 2017

### injinear

Right. So the question stands in principle. The photon could be detected years later. I calculate for a hydrogen molecule and a 3 volt photon we are looking at about 1 meter per second. Piece of cake to detect.

9. Jun 25, 2017

### Strilanc

No, waiting years doesn't help. The uncertainty in the momentum of the mirror will swamp out the signal you want to extract.

The mirror's predicted position years later is not a fixed position, but a distribution of positions. The mirror is macroscopic, so that distribution has a standard deviation which is tiny. But the deviation is still significantly larger than the effect of a single photon bouncing off of the mirror. You can't tell the difference between "Oh it drifted a bit further than I expected because of inherent uncertainty in the momentum/position." and "Oh it drifted a bit further than I expected because the photon hit it.". Waiting longer increases the standard deviation of the expected position in tandem with the effect of the single photon, so your accuracy stays quantum-limited.

10. Jun 27, 2017

### Demystifier

This question is answered in detail in the book
https://www.amazon.com/Quantum-Processes-Information-Benjamin-Schumacher/dp/052187534X
Sec. 10.4.

Here is a short answer. The mirror is a macroscopic object with very small position uncertainty $\Delta x$. The mirror momentum uncertainty $\Delta p$ is at least $\hbar/2 \Delta x$ or larger. If the momentum transferred by photon is much smaller than $\Delta p$, then the mirror state after the reflection is not much different from the mirror state before the reflection, so the scalar product of those two states is close to 1. Consequently, the mirror cannot distinguish the cases with and without reflection, so it cannot perform the measurement.

EDIT: Now I saw that @Strilanc already said essentially the same.

Last edited: Jun 27, 2017