Imagine two entangled photons came into existence and one of them is being measured. We have altered one of the quantum states of the photon and caused its wavefunction to collapse and then we measured it's is spin up. Since we know the other pair must be spin down because both the photons are correlated mathematically speaking I am not sure if any physical exchange is going on but right now I trust SR no information can beat the cosmic speed limit.

Now I want to know, what happens to the other photon, did its wavefunction collapse too because we knew it is spin down although we never directly interacted with it?

  • $\begingroup$ Two photons can be physically correlated as with polarization but can you explain how they are entangled? There is no reason to believe measuring one will effects the other. $\endgroup$ – Bill Alsept Jan 18 '20 at 5:54

This is a very interesting question, and to answer it I am going to consider the following example. Suppose, as you said, that we have a source that, somehow (there are different techniques which allow us to obtain entangled photons such as employing nonlinear cristals) generates a pair of photons that are in the following global state \begin{equation} |\Phi\rangle = \dfrac{1}{\sqrt{2}}\big[|0_A1_B\rangle + |1_A0_B\rangle\big], \end{equation} where the $0$ represents vertical polarization and $1$ horizontal polarization, for instance. This is an example of maximally entangled state, which belong to the basis of Bell states (see the book Quantum Information and Quantum COmputation by Nielsen and Chuang, for example). Each of the photons are sent to system $A$ (usually known as Alice) and system $B$ (usually known as Bob), which explains the presence of the subscripts that I have considered above. Also, I assume that Alice and Bob are far away from each other and that they can only communicate classically (via telephone, for example).

Now, as you said, Alice performs a random polarization measurement so with probability $p_A = \frac{1}{2}$ she can obtain either horizontal or vertical polarization. So, if she obtains after the measurement horizontal polarization, i.e., a $1$, the global state of the system collapses to \begin{equation} |\Phi'\rangle = |1_A 0_B\rangle. \end{equation}

But, what about Bob? He doesn't know what the result of Alice measure was, so if we want to put ourselves in his position, we have to eliminate somehow Alice degrees of freedom. This is done in quantum information via the density matrix representation, which in our case is given by \begin{equation} \rho = |\Phi\rangle \langle \Phi | = \dfrac{1}{2} \Big[ |1_A0_B\rangle \langle 1_A 0_B| + |0_A1_B\rangle \langle 0_A1_B| + |1_A0_B\rangle\langle 0_A1_B| + |0_A1_B\rangle \langle 1_A 0_B|\Big], \end{equation} and taking the partial trace with respect to $A$, that is, to sum over Alice degrees of freedom, we get that Bob sees the state \begin{equation} \rho_B = \text{tr}_A \rho = \sum_{i=0,1} \langle i |\rho|i \rangle = \dfrac{1}{2}|0\rangle \langle 0 | + |1\rangle \langle 1 | = \dfrac{\boldsymbol{1}}{2}, \end{equation} where $\boldsymbol{1}$ represents the identity. Hence, no matter what Alice does on her state that, if Bob does not know what the result o her measurement was, every horizontal or vertical polarization measurement that he does on his state will have an outcome probability equal to $\frac{1}{2}$.

Nevertheless, if after the measurement Alice tells Bob her result (in our case horizontal polarization), then the density matrix of the global system will be \begin{equation} \rho = |1_A 0_B\rangle \langle 1_A 0_B|, \end{equation} and then Bob gets with probability $1$ vertical polarization.

Hope this answers your question satisfactorilly!

  • $\begingroup$ "every horizontal or vertical polarization measurement that he does on his state will have an outcome with probability equal to 1/2" So in other words, Bob can see $|0_A 1_B>$ as well? But if he ended up seeing this state, what will happen if they compare their results? A photon surely can't have two polarization at the same time after measurement, can it? $\endgroup$ – Paradoxy Jan 18 '20 at 21:30
  • $\begingroup$ What the previous result says is that Bob, on his own and without communicating with Alice, cannot say which state he has. So, if he measures randomly a different polarization as Alice, for example left or right circular polarization, the results will have an outcome probability equal to $\frac{1}{2}$. Simplifying, Bob cannot see the state $|0_A 1_B\langle$ unless he checks his results with Alice (and after repeating the measurements a large number of times). Hope this helps to solve your question! $\endgroup$ – Khal_Rhaegarys Jan 19 '20 at 16:58

So the question is basically about how entanglement works. There's probably a gazillion answers for that question by now, but there's no harm done in addressing it again in the context of the current question.

First, let's clear up some possible misconceptions that may be implied in the question. Photons carry many degrees of freedom. Therefore, they could also be entangle in other degrees of freedom, not only polarization. But for the sake of the question, let's assume they are entangled in polarization.

There are different ways in which polarization of two photons can be entangled. For maximally entangled states, these different ways are represented by the Bell states. The one case that you represent here is where their spin is anti-correlated, but it could just as well have been correlated.

The notion of collapse is incorporated into the Copenhagen interpretation of quantum mechanics. As such it is not a scientific establish fact. In other words, nobody knows whether there really is such a thing as quantum collapse. Another interpretation is the many worlds interpretation. As far as our experimental observations are concerned all the different interpretations are equivalent and there does not currently seem to be much hope that we'll ever be able to select the correct one in a scientific way, in other words, based on experimental observations.

So if we cannot know which interpretation is the correct one, then we can just as well pick the simplest one and use it to figure out what is happening. For me, the simplest interpretation is the many worlds interpretation. I'm not saying this is how it really works, but it helps me to know what to expect from observations.

According to the many worlds interpretation, the two terms in the Bell state represent two different realities. When we perform an experiment to observe one of the photons then in one reality we'll see an up-spin and then the corresponding spin of the other photon in the same reality is down. In the other reality it is the other way around. Here we don't need the complicated mind-twisting idea of collapse. That makes it simpler to understand.

Hope this helped.

  • $\begingroup$ I like the fact that you mentioned “photons carry many degrees of freedom”. Unfortunately most people can only imagine photons as billiard balls. When you take photons seriously and give them at least two degrees of freedom as with any other particle, you can derive any phenomenon having to do with light. Why do we need to fudge things with entanglement and wave collapse when ordinary correlation easily answers everything. $\endgroup$ – Bill Alsept Jan 19 '20 at 17:08

Now I want to know what happens to the other photon did it's wavefunction collapsed too because we knew it is spin down although we never directly interact with it?

yup, that's how it works.


Entanglement is a tricky beast. It is very important to understand that as soon as the entangled photons are created (usually at the same point in space), like for example with SPDC crytals, the entanglement is there, and the two photons create a single QM system.

Thus they do have a single common wavefunction. This wavefunction from now on not only describes the probability distribution for one particle, but for both. More precisely in the case of entanglement, you cannot separate the wavefunction into two separate parts for the two particles, they become indistinguishable.

Any physical system has a wavefunction that describes the whole system. So even when we have two particles that are not entangled the two of them are described by some single wavefunction Ψ. However when we say the particles are not entangled we mean the total wavefunction can be separated into different parts for each particle. When the two particles become entangled the means the total wavefunction Ψ can no longer be separated into an A part and a B part, so we can no longer write down an equation like equation (1). In effect the two particles have become mixed up and can no longer be distinguished from each other.

Does Quantum Entanglement happen between two wavefunctions?

When one of the photon's (wavefunction collapses in the Coppenhagen interpretation), its superposition decohers to an eigenstate (because of decoherence with the measuring device, information leaks to the environment), this eigenstate will represent both particles' state. This means, that you will now have an eigenvalue pair, and not just an eigenvalue for one of the photons, but for both.

This answers your question on whether the other photon's wavefunction collapses too. Since there is only one single common wavefunction, whenever it collapses, that means it collapses for both particles, that is, the superposition reduces into an eigenstate and an eigenvalue pair that represents the state for both particles. Thus, there is no need for any information to propagate between the particles.

Since there is no need for any information to propagate between the two particles, they can be arbitrarily far separated in space, the entanglement remains. This is in accordance with SR and this is QM.

  • $\begingroup$ you talk about entangled photons but what does that really mean? In particular you mention crystals. As I understand it, this is one way to correlate the polarizations of two photons. This is basic correlation in the way you would correlate the motion of any two objects. That being said why even use the word entangled and why does it need to be a tricky beast? $\endgroup$ – Bill Alsept Jan 20 '20 at 17:57
  • $\begingroup$ @BillAlsept I was mentioning Spontaneous Parametric Down Conversion through crystals, that is a basic way to create two entangled photons. I tried to answer the question about entanglement, and the question was about why there is no information needed to propagate between the particles. $\endgroup$ – Árpád Szendrei Jan 20 '20 at 19:33
  • $\begingroup$ Yes but isn’t spontaneous parametric down conversion simply correlating the polarizations of the two photons. Shouldn’t we be saying correlated instead of entangled? $\endgroup$ – Bill Alsept Jan 20 '20 at 19:42
  • $\begingroup$ @BillAlsept "First, it's important to realise that entanglement is a type of quantum correlation between two distinct systems." physics.stackexchange.com/questions/57860/… $\endgroup$ – Árpád Szendrei Jan 20 '20 at 22:26
  • $\begingroup$ yes but that is convoluting the conversation. It still comes down to physically correlating the polarizations of the photons. In addition there will be linear dependency. What else is entanglement adding? I mean can you explain what entanglement is adding? When you measure the particles you are measuring for polarization and timing. Calling it entanglement instead of correlation gives you the impression they are somehow spookily connected. $\endgroup$ – Bill Alsept Jan 20 '20 at 22:41

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