Debunking Elitzur–Vaidman Bomb Tester

The Elitzur–Vaidman bomb tester thought experiment is often said to be the weirdest though experiment in all of quantum mechanics. Yet, the reasoning is fallacious and I will show you here why.

The Thought Experiment

First, consider the Mach–Zehnder interferometer.

If the “sample” does nothing to the light, then the light will interfere and only show up on one detector, let’s say, detector 2. If the sample inverts the phase of the light, then it will swap between destructive and constructive inference, causing it to only show up on detector 2. If the sample is instead a which-way detector which detects the photon to see if it went through that path, then the light will show up on both detectors.

Let’s ignore the second case and just focus on the cases where the light shows up only on detector 1, or on both detector 1 and 2. This is analogous to the double-slit experiment. If light is a simple particle, each photon should just bounce off of each beam splitter randomly, and so you should always get output on both detectors.

If light is a wave, then it should interfere and only ever get output on one detector. Placing a measuring device to see which way the photon gets seems to force it to behave like a particle. Note that the interference occurs even if you repeat this experiment by only firing one photon at a time. Some people claim that this is caused by the photon “taking all paths” and then “interfering with itself,” but I will show you later why this is wrong.

The bomb tester thought experiment is fairly simple. Imagine the sample is a bomb with a photon detector that, if it detects a photon, it will explode. Now, imagine you aren’t sure if the bomb is live or a dud, so for your sample you use this bomb.

If the bomb is a dud, it doesn’t have a real photon detector, so the photons should only ever come out of detector 2, implying the photons always take the top path. If the bomb is not a dud, however, the photon has a 50% chance of either taking the top path and going through the bomb and exploding it, or a 50% chance of taking the bottom path and showing up on detector 1.

The “weird” part is the fact that the last case only can happen if the bomb is live, yet the photon does not take the bomb’s path. This means you know for a fact the bomb is live, without the photon interacting with the bomb, and therefore, and thus without exploding the bomb. You can somehow measure things without interacting with them in quantum mechanics… or can you?

Fallacious Reasoning

The fallacious reasoning here comes down to an inconsistency in how the beam splitter is treated. Take a look at the diagram again.

Notice how in the second beam splitter, there are two inputs and two outputs, yet in the first beam splitter, there is only one input but two outputs. The beam splitters are literally the exact same physical object, you don’t use a different beam splitter for the second half of the experiment, so it is inconsistent to treat the beam splitters logically differently.

Furthermore, it also does not make sense in quantum mechanics as all logic gates in quantum mechanics are unitary. You cannot have a one-input, two-output logic gate. There must be an additional input, which we can add to the diagram.

Now, in this framing, we have something more consistent. Both beam splitters are treated as logic gates with two inputs and two outputs. If the most significant bit represents the top path and the least significant represents the bottom path, then the input will be |10⟩ since light is entering the top path but not the bottom path at the first beam splitter.

How does framing it this way help us? Let’s consider something different than photons. Let’s consider, for example, two electrons. Electron spins can only have two possible states, either up or down, so we can assign the up spin to |1⟩ and the down spin to |0⟩. We can then use two electrons to represent whether or not a photon will be in a particular path, where if the electron is an up-spin that means the photon is in that path, and if it is a down-spin that means the photon is not in that path.

First, we have to define some logic gate for the beam splitter. A simple logic gate can be shown with the truth table below.

• |00⟩ → |00⟩
• |01⟩ → 1/√2|01⟩ + 1/√2|10⟩
• |10⟩ → 1/√2|01⟩ - 1/√2|10⟩
• |11⟩ → |11⟩

If we apply this twice in a row, then this will carry out the operation where the bomb is a dud and does not act as a bomb detector.

• Initial state
• |10⟩
• Beam splitter
• 1/√2|01⟩ - 1/√2|10⟩
• Beam splitter x2
• 1/√2(1/√2|01⟩ + 1/√2|10⟩) - 1/√2(1/√2|01⟩ - 1/√2|10⟩)
• 1/2|01⟩ + 1/2|10⟩ - 1/2|01⟩ + 1√2|10⟩
• |10⟩
• Born rule
• [ 10: 100% ]

Hence, applying the beam splitter twice without a measurement just results in |10⟩. We can see what happens in the case where we measure |01⟩, meaning the path the bomb on is not taken.

• Initial state
• |10⟩
• Beam splitter
• 1/√2|01⟩ — 1/√2|10⟩
• Measure
• |01⟩
• Beam splitter x2
• 1/√2|01⟩ + 1/√2|10⟩
• Born rule
• [ 01: 50%; 10: 50% ]

When a measurement is made due to a live bomb in place, there is only a 50% chance of that measurement being a |01⟩ where the bomb would read it as |0⟩. Then, after that, there is another 50% chance the output will be something |01⟩, which is what we need it to be so we can distinguish it from the first case.

That means there is a 25% chance we detect the bomb is live at the final detectors despite the bomb detecting a |0⟩, meaning it should not explode. This, however, clearly, is not an interaction-free measurement, because we said |0⟩ represents an electron with spin down, and so the bomb detected the electron with spin down, a real physical and tangible object. So clearly this is not an interaction-free measurement as the spin-down electron interacted with the bomb.

Now we see the solution to the puzzle. Have you ever noticed that a lot of the “weird” quantum effects are shown using photons? The Mach–Zehnder interferometer, the Elitzur–Vaidman bomb tester, the Wheeler’s delayed-choice experiment, etc. It’s always photons, photons. Why always photons? There is a simple reason.

The constant choice of photons is because the only way to use photons for computation is if |1⟩ represents the presence of a photon and |0⟩ represents the absence of a photon. If you try to measure the polarization of a photon in the same way as an electron spin, if the result is a |0⟩, then it blocks the photon.

Or at least, that’s what is usually assumed. In reality, photons can also be in the |0⟩ state and still carry information and interact with things like with the electron example. This is just the inevitable conclusion from treating the Mach–Zehnder interferometer consistently. You have no choice but to conclude that even when a measuring device measures |0⟩ for a photon, that this does not necessarily mean the absence of a photon, but would instead mean a photon in the |0⟩ state, we can call these void photons.

You might wonder how could a beam splitter create two discrete objects when only a single photon interacts with it. The key is to realize that fields permeate all of space already. When the photon enters the beam splitter, it already is intersecting with the electromagnetic field at that location, it just so happens that the electromagnetic field at that location is not excited so it does not produce a visible photon, but it also has an arbitrary phase.

The photon intersects with the state of the electromagnetic field in that particular location, producing an overall phase and excitation value, and these combined values then produce two new values as an output, an excited state of the electromagnetic field with a particular phase, and merely a phase without an excitation.

Even if no photons are entering the beam splitter, it still can be thought of as possessing two inputs, where these two inputs would be the summation of two possible states in the electromagnetic field which work out to the state of the field at the intersection of the beam splitter. Those two photons, would, of course, have no excitation, so the summation of those two states would be equivalent to two non-existent photons, so the beam splitter does not output any photons.

It is inconsistent to think about photons as simply particles. You have to consider the overall field, and this field can have phases, sort of like ripples in the field which can travel with an excited photon (which is just an excitation in the electromagnetic field) or on its own, independent of an excited photon. When you think of fields, then the inconsistency goes away.

This means that the intersection of the photon and the non-photon really are two discrete sets of numbers which lay on top of each other at the intersection of the beam splitter. The beam splitter is not creating two discrete objects, but just changing the numbers in the field that leave the beam splitter based on the numbers that intersect it at a particular point in time.

Once you understand the problem through this lens, then constructing quantum computers with photons makes a lot more sense and is easier to think about what is actually going on in the system. It’s not that the photon “takes all possible paths,” but that there are two inputs, and the logic gate, being the beam splitter, transforms those two inputs into two outputs, one being a photon (an excitation in the electromagnet field plus a phase) and the other one of these void photons (a phase on its own). It is this void photon that interacts with the bomb detector, and so it is not an interaction-free measurement, but the void photon is read by the bomb detector as a |0⟩ and so it does not go off.

There needs to be more investigation and consideration for the behavior of these void photons. They may come in other flavors for other particles, where similar kinds of interference-based experiments can be established. Any quantum state that can be expressed as |0⟩ and |1⟩ and could be used as a basis for quantum computation, both states should always be treated as discrete, tangible entities, because even if you think the |0⟩ state represents the nonexistence of that particle, it may be possible for the |0⟩ state to still carry a particular phase with it through the field associated with that particle.