Can someone explain why splitting light using a beam splitter is an example of entanglement?

I get the part where we cannot definitively tell which photos have gone in which direction, but i thought that entanglement meant that there would be a relationship or a connection between the two resulting beams other than just the fact that they have the same origin.

I may be confusing some different quantum concepts here but I was under the impression that if the photons of the resulting split beams are entangled then modifying one of them would affect the other?


Yes, the state of a photon in a superposition of two modes is an example of entanglement (1).

The reason why this statement may seem strange at first, is that studying basic quantum mechanics one often hears of entanglement as something that is shared between a number of particles. This is not the best way to think of entanglement, and not the way it is defined (see for example, Gühne and Toth (2008)). A state is entangled if it is not separable, end of the story.

You do not require to have many "particles" to talk about entanglement. You only need a state that lives in an Hilbert space of the form $\mathcal H_A\otimes\mathcal H_B$. Even when "two particles are entangled", it is actually more correct to talk of specific modes being entangled, rather than the particles themselves. In other words, the "entanglement" lies entirely in the way diffent measurement outcome are correlated with each other.

For example, in the typical case of the two "entangled photons" produced via parametric down-conversion, it is more correct to say that the polarization modes of the photons are entangled, not the photons themselves. Even the state of a single photon with polarization $\lvert\uparrow\rangle+\lvert\downarrow\rangle$ is, in some sense, entangled. It is not however very useful to talk of entanglement in such a case, because to access such entanglement one has to apply nonlinear operations conditioned to the polarization state, which is highly nontrivial (as in, I've no idea how you can do it directly). You can however always easily convert such entanglement between the polarization modes into entanglement between spatial modes, using a simple polarizing beam splitter. You can therefore see how if a superposition of two spatial modes is entangled, a superposition of two polarization modes must be as well.

That said, there is a tricky point in the question of whether a photon in a superposition of two modes represents an entangled state. In particular, to talk about entanglement, one first has to determine what are the Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$ of the state. In this case these are the Fock spaces for the two output modes of the beamsplitter. In other words, we write the output state of the photon as $(a_1^\dagger + a_2^\dagger)\lvert0\rangle\simeq\lvert1_1 0_2\rangle + \lvert0_1 1_2\rangle$, where in the ket notation "$0$" and "$1$" represent the state with $0$ or $1$ photons in the first or second mode. This is a maximally entangled state to all effects, where the information is encoded in the photon being present or not in one mode.

However, one may object, there is still only a single photon that carries the information. In particular, it may look like there is no way to act locally on one of the two "qubits" of the above state. How do you apply the transformation $X_1$, sending $\lvert10\rangle+\lvert01\rangle$ into $\lvert00\rangle+\lvert11\rangle$, for example? Note that such transformation effectively removes or add a photon in the first mode conditionally to the presence of a photon in the second mode... how can you do that?

The answer is that, yes, it is highly nontrivial to act locally when the single qubits are encoded in this way. In particular, it requires nonlinear iteractions. But this doesn't change the fact that the state is, to all effects, entangled. It is just entangled in a way that makes it harder experimentally to exploit said entanglement. Moreover, as discussed in S. J. van Enk (2005), one can imagine to make the photon interact with a pair of atoms, that get excited or not depending on whether they receive the photon or not. After such interaction one has a more standard form of entanglement, between the excited states of the atoms, which makes it easier to act locally on the two qubits.

(1) Canonical reference for this is probably S. J. van Enk (2005), which directly tackles the situation of a photon in a superposition of two modes. For an example of an experimental paper in which the authors demonstrate generation of a multipartite-entangled state (a W state) with a single photon, you can have a look at Gräfe et al. (2014). Many other papers use this definition of entanglement in one form or the other, as it is not so controversial.


The "text book" example of a photon through a beam splitter usually comes to demonstrate the principle of superposition, not entanglement. An (idealized) beam splitter does not by itself create entanglement. The entanglement enters when the photon interacts with the measuring device.

If, after passing through the beam splitter, the photon is in the state $\lvert \text{photon up}\rangle+\lvert \text{photon right}\rangle$ (a superposition of going up and going right), and it then comes into interaction with a measuring device (that measures the path the photon has taken, for example) with an initial state $\lvert\text{device idle}\rangle$, then the final state of combined system will be: $$\lvert \text{photon up}\rangle\lvert \text{device showing up}\rangle+\lvert \text{photon right}\rangle\lvert \text{device showing right}\rangle$$ So the photon and the measuring device are now entangled.

In a non-idealized environment, the photon might interact with other objects in its environment (including the beam splitter itself), which will cause immediate entanglement.

  • $\begingroup$ A measurement device, by its own definition, is a macroscopic system. You cannot so easily talk of "entanglement" between a macroscopic system and a quantum degree of freedom. In particular, in this case you cannot have the state you mention, because the outcome of the measurement is fixed, it's one or the other. You do not have a coherent superposition of measurement outcomes, and you cannot afterwards make different measurement outcomes "interfere" with each other. Also, once you have the outcome you do not have the photon anymore. $\endgroup$ – glS Dec 15 '17 at 10:02
  • $\begingroup$ You can make a somewhat similar argument talking about the photon being sent into atoms (or similar systems). Then it makes sense to say that the photon state gets transfered to an entangled state between the spin (or other degree of freedom) of the atoms. See van Enk 2005 for example. $\endgroup$ – glS Dec 15 '17 at 10:03

Splitting a beam with a beamsplitter is not an example of entanglement, it's an example of superposition. A single photon is in a superposition of being transmitted and reflected. However, it's not entangled with anything.

To get entangled photons, you need a two-photon source.

  • $\begingroup$ I see, so you need to have two source beams to produce two entangled beams. Can you explain what the entanglement means in practical terms? What is the relationship between the entangled beams and how that can be leveraged? Does manipulating one resulting beam have any effect on the other resuling beam? $\endgroup$ – td-lambda Mar 16 '16 at 20:38
  • $\begingroup$ Also, is this incorrect? "Figure 1: A single photon (filled circle) cannot divide into two when it hits a beam splitter. It must either pass through, or be reflected. According to quantum mechanics, both of these possibilities occur, producing an entangled state, in which a single photon is shared between the two beams after the beam splitter. Running this process in reverse (i.e. from right to left) provides a way to detect entanglement, since only an entangled state will always produce a single photon in the same place on the left." link $\endgroup$ – td-lambda Mar 16 '16 at 20:48
  • $\begingroup$ They should not be using the word "entangled" in that context. That's a misuse of vocabulary. $\endgroup$ – Jahan Claes Mar 16 '16 at 21:40
  • $\begingroup$ @td-lambda Entanglement just means that the result of measuring photon A depends on the result of measuring photon B. There's some statistical correlation between the state of photon A and the state of photon B. $\endgroup$ – Jahan Claes Mar 16 '16 at 21:42
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    $\begingroup$ @JahanClaes This is not (entirely) correct. E.g. you could imagine mapping the state of the photon (which is |0>|1>+|1>|0>) onto the internal state of two atoms, which is clearly entangled. Since this can be done at large distance, this means the photons were entangled. (There is some cheating involved, as this requires a phase reference. Such a reference can however be treated as classical, or it can be established by using a second such split photon. And if two copies of something are entangled, it's risky to say that a single copy wasn't.) See also arxiv.org/abs/quant-ph/0507189 $\endgroup$ – Norbert Schuch Mar 17 '16 at 9:24

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