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... or more generally, how can one represent mathematically an obstacle in interaction-free measurements? Would it be reasonable to represent it as an "absorption" that transforms $\mid 1\rangle$ into $\mid 0\rangle$ while leaving $\mid 0\rangle$ unchanged? I.e., if the obstacle were an operator $\hat{O}$, it would act on an arbitrary qubit $\hat{\rho}$ as $\hat{O}\hat{\rho}\hat{O}^\dagger$ where

$\hat{O} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$?

Is this a valid quantum mechanical operator, though? Don't I need to use Kraus operators instead if I want to account for probabilities?

PS: I don't like the "scattered" ket operator used in the original paper since it's not mathematically transparent, that's why I'm looking for an explicit mathematical representation in the truncated Fock basis that would simulate an absorption-like operation.

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  • $\begingroup$ "is this a valid operator" yes; any matrix is an operator (see creation/annihilation operators; not all operators used are hermetian or unitary). Is it an observable (which must be hermetian), or a time evolution operator (which must be unitary), no. But I can't tell which of those you want it to be because $B$ isn't used in any of your other equations. You wrote $O\rho O^\dagger$ where $B$ is blah. But $B$ wasn't in that equation... so why do I get it's defintion and where do I plug that in? Note I'm not the downvoter, but this is a potential reason one might downvote. $\endgroup$
    – AXensen
    Oct 11, 2023 at 21:14
  • $\begingroup$ @AXensen So how would the obstacle (or bomb) be represented then? $\endgroup$
    – Tfovid
    Oct 11, 2023 at 21:18
  • $\begingroup$ Also, that was I typo which I now corrected. By $B$, I actually meant $\hat{O}$. Regardless of what I wrote, how is the photon in the obstacle's mode transformed upon interacting with it. $\endgroup$
    – Tfovid
    Oct 11, 2023 at 21:21
  • $\begingroup$ not sure why you got downvoted, but maybe some people might not like that you started your question with an elepsis $\endgroup$ Oct 11, 2023 at 23:19

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I asked a question about the math of this a few years ago.

If the question and answer are not clear to you:

The key to interaction-free measurements is understanding that you are doing a "post selection" measurement after wavefunction collapse has happened. That is, you are collapsing the wavefunction at a certain point in time, then letting the wavefunction continue, and then you in the end only consider probabilities of certain outcomes.

Wave-function collapse cannot be represented using the normal "schodinger equation" and instead must be modeled by using a density matrix, which allows for states to either become classical probabilities. Once a wavefunction has been measured, it no longer becomes a probability amplitude that can interfere, and instead becomes a "normal" probability.

So your desired operator needs to be represented as operating on a mixed state. In the experiment, two cases happen. In one option, you get normal quantum interference. In the other, you cause collapse, forcing your state to become a mixed state:

$$ I_2 = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix} $$

Where the photon has only traveled through one of the two paths. One of these outcomes causes the bomb to explode, the other lets the photon continue on.

In the continue on case, you could write it as:

$$ M = \frac{1}{2} \begin{bmatrix} 1 & 0 \\ 0 & 0 \\ \end{bmatrix} $$

Which just says you have a 50% chance of going along the left path. Finally, this state goes through another beamsplitter, causing quantum interference. This will look something like:

$$ M = \frac{1}{2} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \\ \end{bmatrix} $$

You keep the 50% loss, and your final state is in a superposition of the two different paths -- so you have a combination of a "classical mixture" and a quantum superpostion as your final mixed state.

Then this is how you see that 1/4 of the time, you will obtain an outcome that is "interaction free".

Now for specific questions:

how can one represent mathematically an obstacle in interaction-free measurements?

Look at the input and output mixed states. If you want an "operator" you must just write out the transformations that happen at each step.

Would it be reasonable to represent it as an "absorption" that transforms $\mid 1\rangle$ into $\mid 0\rangle$ while leaving $\mid > 0\rangle$ unchanged?

Can't use bra's and ket's here. You have to go to the mixed state picture. Normal quantum mechanics cannot deal with probabilitys after wavefunction collapse.

I.e., if the obstacle were an operator $\hat{O}$, it would act on an arbitrary qubit $\hat{\rho}$ as $\hat{O}\hat{\rho}\hat{O}^\dagger$ where $\hat{O} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$? Is this a valid quantum mechanical operator, though?

Okay now you are writing ${\rho}$ which is the symbol for a density matrix. If you work with density matrices, then what you are doing is sufficient.

Don't I need to use Kraus operators instead if I want to account for probabilities?

I don't know what Kraus operators are, and I can work out these things..so maybe?

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