This question is known as the Elitzur-Vaidman bomb-testing problem, and although one can arrive after reasonably little thought at the fairly obvious answer that no such ensemble is possible (as any direct observation using light or matter will detonate any working bombs), in actual fact such an ensemble is possible! How can this be the case? The short answer is: quantum effects. The long answer? Read on!
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| Figure 1: A Mach-Zehnder interferometer with faulty bomb $B$ in place and all branches labelled. Note that the $d$-branch is drawn only for illustrative purposes; the photon cannot be detected along the $d$-branch due to destructive interference (see equation \ref{eq:MZ}). (1) The single-photon source $S$. (2) One of the the two 50:50 beam-splitters which are both assumed to be lossless. (3) One of the two mirrors which are assumed to be perfectly reflective. (4) One of the two detectors which are assumed to be perfect detectors. |
The solution to this problem involves the use of a Mach-Zehnder interferometer (Fig. 1) with a single-photon source. To see how, let's consider the case of the interferometer without any bomb in place. We then have
\begin{align}\label{eq:MZ}
\left|s\right\rangle &\rightarrow \frac{i}{\sqrt{2}}\left|u\right\rangle + \frac{1}{\sqrt{2}}\left|v\right\rangle \nonumber \\
&\rightarrow \frac{i}{\sqrt{2}}\left(\frac{1}{\sqrt{2}}\left|c\right\rangle + \frac{i}{\sqrt{2}}\left|d\right\rangle\right) + \frac{1}{\sqrt{2}}\left(\frac{i}{\sqrt{2}}\left|c\right\rangle + \frac{1}{\sqrt{2}}\left|d\right\rangle\right) \nonumber \\
&= \frac{i}{2}\left|c\right\rangle + \frac{-1}{2}\left|d\right\rangle + \frac{i}{2}\left|c\right\rangle + \frac{1}{2}\left|d\right\rangle \nonumber \\
&= i\left|c\right\rangle,
\end{align}
where $\left|a\right\rangle$ represents the quantum state in the $a$-branch of the interferometer (as labelled in Fig. 1) and $i$ is the imaginary unit.$^{1}$ What the above calculation shows$^{2}$ is that (somewhat surprisingly) despite the branching at the second beam-splitter, destructive interference along $d$ and constructive interference along $c$ causes the photon to always be detected at $C$ and never at $D$ (for this alignment).
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| Figure 2: A Mach-Zehnder interferometer with working bomb $B$ in place and all branches labelled. Note that $B$ blocks the $u$-branch whether the photon interacts with the detector or not (the case of an interaction is not illustrated here as this would correspond to the detonation of the bomb). |
Now let's consider the same Mach-Zehnder interferometer but with a bomb placed such that the detector will be along the $u$-branch (as shown in Fig. 2). In this case we have
\begin{align}\label{eq:bomb}
\left|s\right\rangle\left|B_0\right\rangle &\rightarrow \frac{i}{\sqrt{2}}\left|u\right\rangle\left|B_0\right\rangle + \frac{1}{\sqrt{2}}\left|v\right\rangle\left|B_0\right\rangle \nonumber \\
&\rightarrow \frac{i}{\sqrt{2}}\left|X\right\rangle + \frac{1}{\sqrt{2}}\left|v\right\rangle\left|B_0\right\rangle \nonumber \\
&\rightarrow \frac{i}{\sqrt{2}}\left|X\right\rangle + \frac{1}{\sqrt{2}}\left(\frac{i}{\sqrt{2}}\left|c\right\rangle + \frac{1}{\sqrt{2}}\left|d\right\rangle\right)\left|B_0\right\rangle \nonumber \\
&= \frac{i}{\sqrt{2}}\left|X\right\rangle +\frac{i}{2}\left|c\right\rangle\left|B_0\right\rangle + \frac{1}{2}\left|d\right\rangle\left|B_0\right\rangle,
\end{align}
where $\left|B_0\right\rangle$ is the 'primed' or unexploded bomb, $\left|X\right\rangle$ represents the state where the bomb has been detonated$^3$ and $\left|a\right\rangle\left|b\right\rangle\equiv\left|a\right\rangle\otimes\left|b\right\rangle$. Note that for the purposes of this thought experiment we are assuming the detonator is a perfect detector, i.e., the photon wave cannot travel down $u$ without being absorbed.
As is clear from equation \ref{eq:bomb}, the inclusion of the detonator destroys the constructive/destructive interference that caused the simplification in equation \ref{eq:MZ}. Therefore, in the detonator case, rather than having every photon detected at $C$, we have the photon detected at $C$ with a probability of $1/4$, detected at $D$ with a probability of $1/4$ and the bomb detonated with a probability of $1/2$.$^4$
This is what makes it possible to assemble a set of functional bombs without detonating them—if a photon is detected by $D$ then the bomb must have a detonator attached and so we can set it aside knowing it works. If a photon is detected by $C$ then the functionality is indeterminate as we expect a detection at $C$ with non-zero probability in both detonator and no-detonator cases, but this is not a problem as we can simply emit another photon and re-run the test.
Note that while the probabilities above can be derived (in a fairly straightforward manner) from classical principles, we cannot apply a classical interpretation here as the quantum nature of the experiment is indispensable. In the classical (many-photon) run it is possible to both detonate a bomb and make a detection at $D$; this is precluded in the quantum case as the single photon cannot be absorbed by multiple objects. Furthermore, it is the wave-nature of the photon that permits the destructive interference at $D$ in the no-detonator case and thus provides 'detection by $D$' to signify the presence of the detonator and thus successfully make an 'interaction-free' measurement.
If you're unconvinced of this argument because it is based on a purely theoretical consideration, consider that this thought experiment has (equivalently) been carried out in the real world (admittedly using an ordinary detector rather than a bomb) and in fact was first done about a year after this problem was first published. I can't speak to the practical applications, if any exist, but I love this problem regardless for the simple fact that the solution challenges your intuition but can be understood using reasonably straightforward quantum mechanical principles.
Notes
$1$. The inclusion of $i$ in these equations might seem unusual or arbitrary, so I will provide a derivation here that shows where it comes from. |
| Figure 3: A beam-splitter with two incoming beams ($\psi_1$ and $\psi_2$) and two outgoing beams ($\psi_3$ and $\psi_4$). The incoming and outgoing beams are related by the beam-splitter matrix for the beam-splitter in question, as shown in equation \ref{eq:BSM}. In the note below, the beam-splitter will be assumed to be 50:50 in accordance with the calculations in the main text. |
Consider a beam-splitter as shown in Fig. 3. This system can be represented by the matrix equation $\left|\psi_3,\psi_4\right\rangle = \hat{B}\left|\psi_1,\psi_2\right\rangle$, or explicitly,
\begin{equation}\label{eq:BSM}
\begin{pmatrix}
\psi_3 \\ \psi_4
\end{pmatrix}
=
\begin{pmatrix}
T & R \\ R & T
\end{pmatrix}
\begin{pmatrix}
\psi_1 \\ \psi_2
\end{pmatrix},
\end{equation}
where $T$ and $R$ are the transmission and reflection coefficients respectively. In the experiment we assume an ideal, lossless beam-splitter which demands that the beam-splitter matrix be unitary, i.e., $\hat{B}^{\dagger}\hat{B}=\hat{\mathbb{I}}$, or,
\begin{equation}\label{eq:unitary}
\begin{pmatrix}
T^{\ast} & R^{\ast} \\ R^{\ast} & T^{\ast}
\end{pmatrix}
\begin{pmatrix}
T & R \\ R & T
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\ 0 & 1
\end{pmatrix}.
\end{equation}
Equation \ref{eq:unitary} immediately implies the following relations:
\begin{equation}
|T|^2+|R|^2=1,
\end{equation}
\begin{equation}\label{eq:0}
T^{\ast}R+R^{\ast}T=0.
\end{equation}
As $T$ and $R$ are complex numbers, we can represent them in polar form as $T=|T|e^{i\theta_T}$ and $R=|R|e^{i\theta_R}$. For simplicity we choose $\theta_T=0$ and thus $T=|T|\implies T^{\ast}=T$ and so equation \ref{eq:0} becomes
\begin{align}\label{eq:0new}
T|R|e^{i\theta_R}+|R|e^{-i\theta_R}T&=0 \nonumber \\
2T|R|\cos{\left(\theta_R\right)}&=0
\end{align}
where we have made use of the identity $\cos{(\alpha)}=e^{i\alpha}/2+e^{-i\alpha}/2$. Equation \ref{eq:0new} is satisfied by $\theta_R=n\pi+\pi/2, n\in\mathbb{Z}$, but we will choose $n=0\implies\theta_R=\pi/2$ for simplicity, which in turn gives $R=|R|e^{i\pi/2}=i|R|$.
Finally, as the beam-splitter is 50:50 (50% transmission, 50% reflection) we demand $|T|=|R|=1/\sqrt{2}$ and so the beam-splitter matrix is given by
\begin{equation}\label{eq:B}
\hat{B}=\frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & i \\ i & 1
\end{pmatrix}.
\end{equation}
It should be clear that equation \ref{eq:B} is not a unique representation of $\hat{B}$; another choice of $\theta_T$ and/or $\theta_R$ would yield a different (unitary) matrix that would make no difference to the calculations shown in equations \ref{eq:MZ} and \ref{eq:bomb} (I leave proof of this as an exercise for the interested reader). With that said, the reason I like this representation is that it allows $i$ to function as a label for the states that result from a beam-splitter reflection, making it easier to write down interferometer equations directly from the diagram and keep track of where each term comes from. This is, of course, purely a matter of personal preference.
$2$. This equation is an example of quantum superposition in action. For example, the first line says that the photon exists in a superposition of the $\left|u\right\rangle$ and $\left|v\right\rangle$ states where the states are equally weighted (as we are assuming normalisation). Superposition is a fundamental aspect of quantum mechanics that follows from the linearity of the Schrödinger equation (linear combinations of solutions will themselves be solutions). In this case, the beam-splitter splits the photon probability wave along the two channels and so in some sense the photon travels along both branches, although no measurement can be made which will detect the photon in both channels at once—this is not a consequence of experimental limitations but is a restriction that is fundamental to quantum theory. The question of why this is the case is a deep and ongoing one, and I encourage the interested reader to investigate the literature on the philosophy (and especially interpretations) of quantum mechanics.
$3$. I have gone to some pains in this post to avoid using the term "wavefunction collapse" at any point, although for clarity I will say will say that in the Copenhagen interpretation, the case of the photon interacting with the detonator (or any of the detectors for that matter) is an example of wavefunction collapse.
$4$. So long as the beam-splitters are both 50:50, as we have assumed throughout this blog post. Naturally, other types of beam-splitters will yield different results, and in fact using a more sophisticated apparatus will permit a much better detection level (in theory, the detection fraction can be brought arbitrarily close to 1, although I cannot speak to the practicality of such an apparatus).
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