What happens in the standard "interaction free measurement" when the detector connected to a bomb is replaced with attenuation (where light is lost through a semi-transparent medium)?

Consider the following experiment:

I know already that if I put a single photon through a lossy medium with absorption A and transmission T = (1-A) - that my photon will end up in a statistical mixture of photon and vacuum:

So what happens in the example experiment with a lossy medium in a MZI (Mach-Zender interferometer)?

I would think that since mixed states are classical probability states, I can break down this experiment into 2 separate cases: one in which the photon is absorbed by the lossy medium - and one in which the photon transmits through the lossy medium.

But if we can break things down into these two cases, then we see that these two cases are exactly the cases considered for the "interaction free measurement" experiment:

Now here's the part that I'm particularly not sure about. In the case where the photon is not absorbed - will it experience perfect Mach-Zender interference? Or will there be a reduction in the strength of the probability amplitude associated with the photon traveling through that arm.

I think it makes a lot of sense to use classical probability to separate the mixed state into two separate cases. So there would be a A% chance of the dud case, and a (1-A)% probability of the dud case.

But, I've struggled a bit to fully represent the entire situation in a full-density matrix framework.

(And if anyone is interested, I can edit in my attempt at writing this experiment mathematically in this picture and highlight the parts that are not clear)

Any ideas on the proper way to treat this case?


Here's an attempt at explaining my confusion in working out the math of this situation.

My state begins in a pure state, and I will expand it out as a density matrix so that I can include the part involving attenuation:

$$|\Psi\rangle = \frac{1}{\sqrt{2}}(|1_L\rangle + |1_R \rangle) $$ $$|\Psi\rangle \langle \Psi | = \frac{1}{2}(|1_R \rangle \langle 1_R|+ |1_R \rangle \langle 1_L|+ |1_L \rangle \langle 1_R \rangle + |1_L \rangle \langle 1_L \rangle)$$

Where $|1_L\rangle$ and $|1_R\rangle$ describe the states in which the reflects and transmits from the first beam splitter. Now mode in the "right path" $|1_R\rangle$ will experience the lossy medium and turn into a statistical mixture:

$|1_R\rangle \langle 1_R| \rightarrow A|0_R\rangle \langle 0_R|+T|1_R\rangle \langle 1_R|$

This is clear to me (single photons that become attenuated become mixed states, and I have confirmed this with a "quantum process operator" for attenuation).

But what happens the coherence between the $|1_L\rangle$ and $|1_R\rangle$ after attenuation?

$|1_L\rangle \langle 1_R| \rightarrow ?$

I don't know how to prove it, but I can guess that it becomes:

$|1_L\rangle \langle 1_R| \rightarrow (A|1_L\rangle \langle 0_R|+T|1_L\rangle \langle 1_R|)$

In this case if we plug this in our density matrix:

$$\frac{1}{2}(A |0_R\rangle \langle 0_R|+T|1_R\rangle \langle 1_R|+ A|0_R \rangle \langle 1_L| + T|1_R \rangle \langle 1_L|+ A|1_L \rangle \langle 0_R \rangle + T|1_L \rangle \langle 1_R \rangle + |1_L \rangle \langle 1_L \rangle)$$

$$\frac{1}{2}( A(|0_R\rangle \langle 0_R| +|0_R \rangle \langle 1_L|+|1_L \rangle \langle 0_R \rangle) + T(|1_R\rangle \langle 1_R|+ |1_R \rangle \langle 1_L| + |1_L \rangle \langle 1_R \rangle) + |1_L \rangle \langle 1_L \rangle)$$

Now I want to consider the case only when the photon is transmitted, I'm not sure exactly how to express this as an operator, but I can project my state to dump the $|0_R\rangle$ terms (only considering the transmitted photons). The density matrix reduces to:

$$\frac{1}{2}(T(|1_R\rangle \langle 1_R|+ |1_R \rangle \langle 1_L| + |1_L \rangle \langle 1_R \rangle) + |1_L \rangle \langle 1_L \rangle)$$

Now if we incorporate the second beam splitter this density matrix becomes:

$$\frac{1}{2}\left( (\frac{1}{2}+\frac{3}{2}T)|D_1\rangle \langle D_1| + (\frac{1}{2}-\frac{1}{2}T)|D_2\rangle \langle D_2| + (\frac{1}{2} - \frac{1}{2}T)(|D_1\rangle \langle D_2|+|D_2\rangle \langle D_1|) \right)$$

Here I see when T = 1 I recover the normal MZI interference, but when T is not one, I see that I have imperfect MZI interference. I suspect that my intutition is correct, and that I not giving this system the proper mathematical treatment.

Maybe the way that I am projecting the state into $I_L \otimes |1_R\rangle \langle 1_R |$ is incorrect? Maybe I have to renormalize somewhere in there?

  • $\begingroup$ I don't fully understand what you are asking. If the state is $(\lvert0\rangle\langle0\rvert+\lvert1\rangle\langle1\rvert)/2$ that means that yes, you can describe the situation by analysing separately the cases in which there is or isn't a photon, and then average the results at the end $\endgroup$
    – glS
    Feb 15, 2019 at 15:47
  • $\begingroup$ Thanks for the reply. So there are two cases for the photon, absorption l0><0l and transmission l1><1l. I want to work out the detector-clicking probabilities for these two cases. In the transmission case, are you sure that the system experienced perfect MZI interference? I tried to work it out with density matrices and failed to show it explicitly. If you are interested I can edit in my math attempts in the question. $\endgroup$ Feb 15, 2019 at 17:11
  • $\begingroup$ I updated the question with my attempt at working things out with density matrices. Maybe that will make more clear what my confusion is. $\endgroup$ Feb 15, 2019 at 23:07

1 Answer 1


Mostly yes, you are correct.

There are several ways to do this calculation. Let me start with what I find to be the simplest one, and does not require the use of density matrices at all.

Without using density matrices

$\newcommand{\ket}[1]{\lvert #1\rangle}$The main idea is to notice that a "lossy beamsplitter" is nothing but a "normal" beamsplitter in which you are ignoring one of the output arms. With this in mind, let me denote with $\ket{1},\ket2,\ket3$ the states in which the photon is in the first ("left" in the image), second ("right" in the image), or additional "hidden" spatial mode.

The state after the first beamsplitter is $$\ket{\psi_1}=\frac{1}{\sqrt2}(\ket1+\ket2).$$ The "lossy" beamsplitter now evolves $\ket2$ into a balanced superposition of $\ket2$ and $\ket3$, leaving $\ket1$ untouched, so we get $$\ket{\psi_1}\mapsto\ket{\psi_2}\equiv\frac{1}{\sqrt2}\ket1+\frac{1}{\sqrt2}(t\ket2+a\ket3),$$ where $\lvert a\rvert^2=A, \lvert t\rvert^2=T$. The final beamsplitter acts by making $\ket1$ and $\ket2$ interfere but leaving $\ket3$ untouched, and therefore $$\ket{\psi_2}\mapsto\ket{\psi_3}\equiv\frac{1}{2}\left(1+t\right)\ket1 + \frac{1}{2}\left(1-t\right)\ket2+\frac{a}{\sqrt2}\ket3.$$ The probability of seeing the photon in the second mode, that is, of $D_2$ clicking, is finally given by $$p_2=\frac{(1-t)^2}{4}.$$ You can see that the photon does not interfere with itself as much as it would have done in a regular MZ interferometer, in which you would have had $p_2=0$.

With density matrix formalism

If we want to use density matrices for the calculation, then it is handier to use the notation $\ket{1_L},\ket{1_R}$ for when the photon is in the left/right mode, and $\ket0$ to denote the state in which no photon is in the interferometer. Also, let me use the shortand notation $\mathbb P(\ket\psi)\equiv\ket\psi\!\langle\psi\rvert$.

After the first beamsplitter (but before the lossy one), the state is $$\rho_1=\mathbb P\left[\frac{1}{\sqrt2}(\ket{1_L}+\ket{1_R})\right].$$ The lossy beamsplitter can now be modeled as a map of the form $$\mathcal E(\rho)=\operatorname{Tr}_2(\mathcal U\rho\mathcal U^\dagger),$$ and $\mathcal U$ is the Hamiltonian used also in the previous section: $$\mathcal U\ket{1_R}=t\ket{1_R}+a\ket{1_E}$$ for some extra mode $\ket{1_E}$. It follows that $$\mathcal E(\ket{1_L}\!\langle 1_R\rvert)=a\ket{1_L}\!\langle 0_R\rvert+t\ket{1_L}\!\langle 1_R\rvert,$$ and the rest of the calculation proceed like you showed.

An easier way to perform this calculation with this formalism is as follows: \begin{align} \rho_2 &=\operatorname{Tr}_E\left(\mathbb P\left[ \frac{1}{\sqrt2}(\ket{1_L}+t\ket{1_R}+r\ket{1_E}) \right]\right) \\ &= \left(\frac{1+T}{2}\right)\mathbb P\left[\frac{\ket{1_L}+t \ket{1_R}}{\sqrt{1+T}}\right] + \frac{R}{2}\mathbb P[\ket{\text{vac}}] \end{align} computing the last step now is easy because we can simply work separately on the pure states whose mixture makes up $\rho_2$: $$\rho_2\mapsto\rho_3=\left(\frac{1+T}{2}\right)\mathbb P\left[\frac{(1+t)\ket{1_L}+(1-t)\ket{1_R}}{\sqrt{2(1+T)}}\right] + \frac{R}{2}\mathbb P[\ket{\text{vac}}].$$ You can see in the expression for $\rho_2$ why you cannot simply consider the two different cases of "photon absorbed therefore zero interference" and "photon transmitted therefore full interference". What you actually have is a mixture of the photon being absorbed, and the photon being transmitted and therefore only partially interfering.

This is not surprising, because you cannot simply make the photon either stay there or "magically vanish". The only way to get something equivalent to the lossy beamsplitter is to have the photon "leak" away with some probability, but such an operation is fundamentally always unitary, and therefore evolves $\ket{1_R}$ into a superposition of $\ket{1_R}$ and something else. Because of this $\ket{1_L}$ cannot fully interfere with $\ket{1_R}$ anymore, because $\ket{1_R}$ is already interfering with something else (even though you are not looking at this something else). This is an effect analogous to the so-called monogamy of entanglement, but for coherences.

  • $\begingroup$ Thanks for the replay. Sorry if my question seems like a "moving target", but the question is in connection to the "interaction-free detection". I was thinking that detector 2 clicking could indicate that the photon must have traveled the left-path. And that the case where the photon is transmitted we would have 100% MZI interference. $\endgroup$ Feb 16, 2019 at 17:43
  • $\begingroup$ I certainly understand that the "partial trace" or average measurement will certainly experience lower visibility - but on a case-by-case basis, I would think we would be looking at 100% MZI interference (photon transmitted with probability T) + 0% MZI interference (photon absorbed with probability A). $\endgroup$ Feb 16, 2019 at 17:48
  • $\begingroup$ what are you asking then? Yes you can treat the problem from that point of view by considering the different cases, but isn't this what you already did in your post? $\endgroup$
    – glS
    Feb 16, 2019 at 17:58
  • $\begingroup$ Sorry, I should've made the motivation more clear from the beginning. I was thinking that you could use an "interaction-free measurement" to "filter out" the cases where the photon has been absorbed. Basically by ignoring cases where D2 clicks, do you have an enhanced "transmission" of clicks in D1 after this postselection? $\endgroup$ Feb 16, 2019 at 18:05
  • $\begingroup$ Basically, what are the statistics of detector 1 given that detector 2 has measured $|0\rangle$? $\endgroup$ Feb 16, 2019 at 18:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.