Elitzur–Vaidman bomb tester in multi-world interpretation

[Spathi]

TL;DR Summary

When a photon passes through the first semi-transparent mirror, it turns into a superposition of photons traveling along the upper and lower paths. In MMI this means that two "parallel worlds" appear...

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

In the Elitztur-Weidman experiment there are bombs that are triggered by a single photon hitting their detector, and with some probability you can find out that the bomb is working, despite the fact that it did not explode.

I'm trying to formulate how this experiment can be described in the many-worlds interpretation. When a photon passes through the first semi-transparent mirror, it turns into a superposition of photons traveling along the upper and lower paths. In MMI this means that two parallel worlds appear (I do not like the word "parallel" but there is no other yet), in the first one the photon goes along the upper path, in the second - along the lower one. In the first world the photon hits detector D or C, in the second it is absorbed by the bomb and the detectors see nothing. Yet it didn't become clearer; probably, in order to move forward, you need to figure out what happens to the photon or superposition on the second semitransparent mirror. It is written that the superposition interferes with itself and is detected in C; what is the correct name for the reverse transformation of a superposition into a single reality - decoherence or destructive interference?

In the Copenhagen interpretation, the experiment can be stated as follows. The bomb is an observer, the hit of a photon on its detector is a measurement process, i.e. wave function collapse. As a result of the collapse, the superposition turns into a single photon flying either along the lower path or along the upper one. If on the top - it is detected on D and so we find out that the bomb worked. If I'm not confused, Wikipedia says that if a superposition of photons hits the second mirror, then it decoheres and a photon is always detected at C; if a photon flying along the upper path hits, then it is reflected and detected either on C or on D. I did not find in the article what will happen if the photon passes only through the lower path. Maybe we are talking about the fact that this is impossible - if the bomb is faulty, then there will always be only a superposition?

It is still unclear to me how such an experiment is possible from a technical point of view. In order for the superposition of two photons to merge again into one photon, is it necessary that the first and second components of the superposition meet exactly at a single point in space? I mean that if, for example, in this setting, the length of the horizontal path is increased by one and a half times, and the length of the vertical path is not changed, then the superposition will not merge on the second translucent mirror, because the top and bottom photon will not meet in the same place? Then it turns out that for a correct experiment it is necessary to adjust the geometrical parameters of the installation to within an atom, very strange.

Just in case, I will state, as it is written in Wikipedia: if the bomb is in good condition, with a probability of 50% the photon will not be detected (the bomb exploded), with a probability of 25% it will be detected on C (as I understand it, the bomb also exploded) and with a probability of 25% it will be detected on D (the bomb is working but did not explode).

 

[PeroK]

There are a number of misconceptions in your post. The main one is that there is only every one photon and never two photons. A superposition does not increase the number of photons, but describes a spatially more complicated state.

In general, photons don't behave the way you imagine. You might want to start by reading Feynman's introduction to QED:

https://www.goodreads.com/book/show/5552.QED

 

[PeterDonis]

I'm trying to formulate how this experiment can be described in the many-worlds interpretation.

Simple: there is some fraction of worlds in which you know that the bomb is working, even though it didn't explode.

 

[Spathi]

Simple: there is some fraction of worlds in which you know that the bomb is working, even though it didn't explode.

You mean - in these worlds the bomb didn't explode but the observers know that it exploded in other worlds?

 

[PeterDonis]

You mean - in these worlds the bomb didn't explode

The bomb didn't explode and a photon was detected at detector D, which can only happen if the bomb is working (not a dud).

but the observers know that it exploded in other worlds?

No, they just know that what they observed in their world (see above) means that the bomb is working but did not explode.

 

[Spathi]

As I understand it, if there is an ordinary mirror instead of the first translucent mirror, and the photon always goes along the upper path, after the second mirror it is detected with a 50% probability at C and with a 50% probability it is detected at D. If the photon goes only along the lower path and there is no bomb, the ratio will be similar?

After passing the first mirror, two worlds are formed, in the first one the photon goes along the upper path, in the second one along the lower one. After the second mirror, these two worlds split again and 4 worlds are formed (more precisely - two identical pairs of worlds). And the two worlds in which the photon goes to detector C are somehow annihilated, leaving only the worlds with detector D. Did I formulate this correctly? Sounds pretty weird.

I would like to know how this "annihilation" can be described mathematically.

https://en.wikipedia.org/wiki/Mach–Zehnder_interferometer

It is unclear for me, what the psi vector in the formulas from Wikipedia mean – two components of the polarization of photon? Do they change with time or space?

 

[Demystifier]

When a photon passes through the first semi-transparent mirror, it turns into a superposition of photons traveling along the upper and lower paths. In MMI this means that two parallel worlds appear

No, that's not sufficient for appearance of two worlds. Two worlds appear when you perform a measurement to determine which path the particle took, or more generally, when the photon interacts with something macroscopic (the notion of "world" is macroscopic) so that the photon suffers a decoherence.

 

[Demystifier]

what is the correct name for the reverse transformation of a superposition into a single reality - decoherence or destructive interference?

Neither. Photon in a superposition in two arms already is a "single reality", because it's not measured. The two realities appear upon measurement, i.e. decoherence. A reverse of that is a reverse of decoherence, so can be called recoherence.

 

[Nugatory]

Then it turns out that for a correct experiment it is necessary to adjust the geometrical parameters of the installation to within an atom, very strange.

Not to “within an atom” but to within the coherence length of the photons generated by our source.

 

[martinbn]

No, that's not sufficient for appearance of two worlds. Two worlds appear when you perform a measurement to determine which path the particle took, or more generally, when the photon interacts with something macroscopic (the notion of "world" is macroscopic) so that the photon suffers a decoherence.

I have seen David Deutch describe it exactly like this. His many world view might be different from the standard.

 

[Demystifier]

I have seen David Deutch describe it exactly like this. His many world view might be different from the standard.

That's true, for him quantum computer with a single qubit is already an evidence of many worlds.

 

[Spathi]

Sorry, is there really no one on this forum who can answer my questions? Let me rephrase them:

1) Do I understand correctly, that if there is no first mirror and bomb, and the photon always follows the lower path, then detector C is triggered with a probability of 50%, detector B with a probability of 50%? And the same if the photon always follows the upper path?

2) If there is no bomb, after the second mirror, two “parallel worlds”, in which the photon goes towards D, “annihilate”?

3) What formulas describe this annihilation?

 

[Nugatory]

1) Do I understand correctly, that if there is no first mirror and bomb, and the photon always follows the lower path, then detector C is triggered with a probability of 50%, detector B with a probability of 50%?

Yes, if you meant “D” instead of “B”. If you really do mean “B”, then no - B will always trigger, and the photon will never reach C or D.

And the same if the photon always follows the upper path?

That will depend on how we arranged to always send the photon on the upper path, but if we did so by replacing the bottom left half-silvered mirror with an ordinary full mirror like at the top left and bottom right ones…. C is triggered with 50% probability, D is triggered with 50% probability.

2) If there is no bomb, after the second mirror, two “parallel worlds”, in which the photon goes towards D, “annihilate”?

No, we just have a quantum superposition in which the amplitudes for the path to one detector interfere destructively. We don’t get multiple worlds until we have an irreversible outcome. So a photon enters the apparatus and soon after we have three worlds, one for each possible outcome: the bomb has exploded (if it is there and not a dud), detector C has triggered, or detector D has triggered.

3) What formulas describe this annihilation?

You’ve already linked the Wikipedia article on the Mach-Zehnder interferometer - it has the the math that describes the superposition and how to calculate the probabilities of the different possible outcomes.

 

[Spathi]

Yes, if you meant “D” instead of “B”. If you really do mean “B”, then no - B will always trigger, and the photon will never reach C or D.

Sorry, of course I meant detectors C and D.

You’ve already linked the Wikipedia article on the Mach-Zehnder interferometer - it has the the math that describes the superposition and how to calculate the probabilities of the different possible outcomes.

From this article it is unclear for me, what means the ψ in this formula (which is multiplied by the P*B*P matrix). Mayve it is a vector of electric intensity of a photon?

 

[Nugatory]

From this article it is unclear for me, what means the ψ in this formula (which is multiplied by the P*B*P matrix). Mayve it is a vector of electric intensity of a photon?

$\psi$ is a vector (in the general sense of being an element of an abstract vector space - not the “magnitude and direction” thing from your first year or so of physics) that mathematically represents the state of the system. We use the rules of quantum mechanics to calculate the probabilities of the various outcomes from it.

Unfortunately there is no way of doing QM without the math - it’s not like classical physics where we can form intuitive pictures of what’s going on without the equations. So if you haven’t acquired the math needed to follow the Wikipedia explanation the only answer to your original question is the one in post #3 of this thread.

 

[Spathi]

I have written the formulas, please look and say whether they are correct:

Here Ψl is the wavefunction of the electron moving at the lower path, Ψh correspondingly the higher path, Ψs is the superposition after the first beam splitter, Ψf the superposition after the second beam splitter, Ψsb the Ψs after detonation of the bomb, Ψfb the superposition after the bonb and the second beam splitter.

 

[Demystifier]

I have written the formulas, please look and say whether they are correct:

They are not even wrong, because you don't explain what those formulas describe. Besides, why do you denote the product by a star, are you a programmer or something?

 

[Spathi]

They are not even wrong, because you don't explain what those formulas describe. Besides, why do you denote the product by a star, are you a programmer or something?

I am a programmer. How the product should be denoted?
Ψl is the wavefunction of the electron moving at the lower path, Ψh correspondingly the higher path. When the electron passes the first beam splitter, Ψl is multiplied by B and the superposition Ψs appears. When this superposition passes the second beam splitter, it is multiplied by B again, and it is shown mathematically how the Ψl component (the higher digit in the vector) dissapears (this means that the D detector does not receive photons). Ψsb is Ψs where the Ψl is made zero because of the detonation of the bomb. When Ψsb passes the second beam splitter, it is multiplied by B and Ψfb appears (and since Ψfb contains the Ψl component, the D detector can receive photons).

 

[Demystifier]

I am a programmer. How the product should be denoted?

A product of x and y is denoted as xy.

 

[Spathi]

I have read the article of Elitzur, and one moment should be clarified. The authors describe the procedure for improving the efficiency of the experiment (to get a better percentage of good unexplided bombs):

But if the first mirror is almost transparent, this means that most of the photons will follow the lower path and accordingly detonate the bomb? Maybe the authors got it slightly wrong?

 

[PeterDonis]

I have written the formulas

No, you haven't, you've posted them in an image. That is not allowed. Please use the PF LaTeX feature to post equations directly in the thread.

 

[Spathi]

Sorry. Here they are:
$\psi _{l}=\begin{pmatrix}1\\0\end{pmatrix}$ $\psi _{h}=\begin{pmatrix}0\\1\end{pmatrix}$ $\psi _{sb}=\begin{pmatrix}0\\i\end{pmatrix}$
$$B=\frac{1}{\sqrt{2}}*\begin{pmatrix}1 & i\\i & 1\end{pmatrix}$$
$$
\psi_{s}=B*\psi_{l}=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

1 & i\\

i & 1

\end{pmatrix}*\begin{pmatrix}1

\\

0

\end{pmatrix}

=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

1*1+i*0 \\

i*1+1*0

\end{pmatrix}=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

1\\

i

\end{pmatrix}$$
$$
\psi_{f}=B*\psi_{s}=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

1 & i\\

i & 1

\end{pmatrix}*

\frac{1}{\sqrt{2}}*

\begin{pmatrix}1

\\

i

\end{pmatrix}

=

\frac{1}{2}*\begin{pmatrix}

1*1+i*i \\

i*1+1*i

\end{pmatrix}=

\frac{1}{2}*\begin{pmatrix}

1-1\\

i+i

\end{pmatrix}

=

\begin{pmatrix}

0\\

i

\end{pmatrix}
$$
$$
\psi_{fb}= B*\psi_{sb}=

=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

1 & i\\

i & 1

\end{pmatrix}*

\begin{pmatrix}0

\\

i

\end{pmatrix}

=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

1*0+i*i \\

i*0+1*i

\end{pmatrix}=

\frac{1}{\sqrt{2}}*\begin{pmatrix}

-1\\

i

\end{pmatrix}$$