The phenomenon of quantum entanglement in quantum physics at first glance seems to allow one to transmit information faster than the speed of light; however, one cannot transmit information in this way. In the following I will explain why with an example and then I'll discuss the general case at a more formal level. Here I am going to talk exclusively about non-relativistic quantum mechanics, though the same should be true for quantum field theory, but I have not done or seen that proof.

Now, one might wonder why the speed of light would come in as a limitation in non-relativistic quantum mechanics, and the answer is that it doesn't really. In fact, what I will illustrate is that entanglement alone can't be used to transmit any information at all, at any speed. Entanglement can be used together with a classical signal to transmit information, as in quantum teleportation, or it can be used together with an interaction between two systems, but by itself entanglement is useless for communication.

This material is related to the EPR paradox, and is somewhat in response to How to transmit information faster than light speed and to suplement How not to transmit information faster than light speed.

As a warning, this write-up might be more useful if you read something about entanglement and quantum superposition first.

Note: Some characters in formulae have been replaced, since they did not render correctly in some browsers. As a result, * will be used to represent the tensor product of two states or operators.

### An Example

Suppose that we have two spin 1/2 particles (if you don't know what that is, think of them as coins that have a "heads" side that can be either up or down). They are in an entangled state such that 50% of the time particle A is found to be spin up (heads up) and particle B is spin down (heads down, i.e. tails), and the other 50% of the time A is found to be spin down and B is spin up. This is called a maximally entangled bell state and can be written

|ψ> = 1/sqrt(2) (|up>A*|down>B + |down>A*|up>B)

in Dirac notation. Now, the important point about this entanglement is that it can never happen that A and B are found with both spins up or both spins down, so if we measure A and find that it is spin up, then we already know that B MUST be spin down, even before we measure it.

At this point you might object, "But you just said we know the state of particle B even before we measure it, so if it's like a million light years away, doesn't that mean we just got information faster than the speed of light?" The answer is no, not really. We only know what state B is in because we already knew the quantum state of the system made up of A and B at the beginning and we know the outcome of the measurement on A. The true test of whether information is sent is whether we're able to use this to send some sort of message to someone who is measuring particle B.

Ok, so let's say that Alice has particle A here on earth, and Cooper is light years away in deep space with particle B. They want to use their entangled pair to send a signal. Well, if Alice measures A, she can't control which outcome she gets, so she can't control which outcome Cooper gets. That makes it hard to send any message. Well, what if Alice and Cooper have agreed that at noon Alice will measure A if she want him to come back and she won't if she doesn't want him back. So instead of using the outcome of the measurement to carry a message, Alice would send a message based on whether or not she measured the state. Unfortunately, this doesn't work either, because if Alice measures, 50% of the time she gets up and 50% of the time she gets down. That means that when Cooper measures his afterward, 50% of the time he will get down (corresponding to Alice's up) and 50% of the time he will get up (corresponding to Alice's down). The problem is that those are the same probabilities as in the case that Alice hasn't measured at all. So it seems this scheme just isn't going to work.

Now you could get trickier than Alice and Cooper. For one thing you could first do things (introduce interactions) to particle A or particle B in order to change the entangled state and then measure them. In the end, though, this doesn't work either, but I'll leave that to my examination of the general case for those who are more expert in the field. Suffice it to say that no matter how tricky you try to get, the situation is always essentially the same as the preceding example. Now, it is true that something changed faster than light, the wave function (or state vector). The thing is that the wave function is not directly observable and this change has no consequence accessible to people observing either particle individually, which is why we couldn't use it to transmit information. The change in the state is only directly observable when you compare the two measurements, which can only be done at speeds less than or equal to light (unless you have some other faster than light communication technology).

### The General Case

This is the section for people who know some quantum mechanics. This would probably be clearer in terms of the density matrix, but I didn't want to limit my audience further. Here I will consider a system consisting of two entangled subsystems, because I believe it will be clearer, but I think the explanation is only trivially different when dealing with a system comprised of N entangled subsystems. We will work in the interaction picture here, meaning effectively that we will ignore the free evolution of the system and only worry about additional evolution introduced by the hypothetical communication process.

We consider, then, a system comprised of two subsystems A and B. There are two observers, who we will again call Alice and Cooper. Alice can only interact with subsystem A and can only measure subsystem A. Likewise Cooper can only interact with and measure subsystem B. I will show that no matter what interactions with A or measurements of A Alice makes, the measurements of B by Cooper will be unaffected.

When we say Cooper can only measure B this means simply that if subsystems A and B are in a separable (unentangled) state, then the value of a measurement by Cooper will be independent of the state of A. This implies the operator representing the dynamical quantity being measured must have the form

O2 = 1A*O2B

possibly apart from some multiplicative factor. Thus, if the system is the separable state

|ψ> = |φ>A*|θ>B

then the expectation value of the measurement is

<O2>ψ = <ψ|O2|ψ> = <θ|O2B|θ>B

satisfying our requirement that it is independent of the state of A. This tells us that Cooper is measuring B only. If this criterion were not met then he would not even need entanglement to transmit information.

We also require that Alice may only introduce interactions with subsystem A, meaning that she may only introduce evolution of the system with an operator of the form

U1 = U1A*1B

This is the class of evolutions that only effect subsystem A in a separable state. Likewise, Cooper can introduce interactions on subsystem B.

Once we have defined what it means for each observer to interact with and measure only his own subsystem, now we can examine whether they may use entanglement to transmit information. We now assume that the system begins in an arbitrary state, which may have any sort of entanglement.

|ψ> = Σj,k ajkj>A*k>B

We represent the state in terms of an orthonormal basis of eigenstates of the observables that Alice and Cooper will measure. First we need to show that any measurement on A will not effect the expectation value of a measurement on B. Ok, well, first we find an expression for the expectation value of Cooper's measurement of B for the general state.

<O2>ψ = Σj',k' Σj,k conj(aj'k')ajkj'|φj>Ak'|O2Bk>B_ = Σj,k,k'conj(ajk')ajkk'|O2Bk>B

If Alice measures the system A and finds it in the state |φl> then the total state of the system will be

|ψ>m(l) = 1/sqrt(P(l)) Σk alkl>A*k>B

where P(l) is the probability that the system will be found in the state |φl>. For this measured state, the expectation value of Cooper's measurement on B will be

<O2m(l) = <ψ|O2|ψ>m(l) = 1/P(l) Σk,k' conj(alk')alkll>Ak'|O2Bk>B = 1/P(l) Σk,k' conj(alk')alkk'|O2Bk>B

Now this isn't the same as what we had for the general state. That doesn't look good, right? Except, remember we're talking about just one measurement. We found the expectation value for a measurement of B provided that A is in the state |ψl> every time, but that won't happen, because as we discussed, Alice can't control what result she gets. Cooper can't tell anything from just one measurement (other than that the result is possible, which doesn't depend on Alice's measurement). He has to perform many measurements and look at the distribution before he can tell if Alice has done anything. If Alice and Cooper perform this process many times, Alice will get a different result each time with probability P(l). So, then the actual expectation value Cooper would find would be

<O2>ψm = Σl P(l)<O2>ψm(l) = Σl,k,k' conj(alk')alkk'|O2Bk>B

which is exactly what he would have found for the unmeasured state. Thus, Cooper cannot tell that Alice made any measurement.

Now you might get trickier and ask, "Well, what if Alice and Cooper also evolve their states as in quantum teleportation?" It turns out that doesn't help either. We can suppose that each observer performs a local evolution on his system before and after the measurement of A. The evolution could just be the identity, so we've also included the cases where one or both don't evolve their respective systems. Each performs a local evolution on his system at the outset, then the state of the total system becomes

|ψ'> = Σj,k ajkUA1j>A*UB1k>B

which gives an expectation value for the measurement of B

but

so

For simplicity, we may now write the state after the first evolution in our general form

|ψ'> = Σj,k bjkj>A*k>B

so the expectation value for measurement of B before the any measurement of A is

<O2>ψ' = Σj,k,k'conj(bjk')bjkk'|O2Bk>B

Now, we suppose Alice measures A and then both Alice and Cooper perform local evolutions on their systems. The measured and evolved state is then

U2|ψ'>m(l) = 1/sqrt(P(l)) Σk blkUA2l>A*UB2k>B

Again, we must take the expectation value for each possible measured state and add them all up with their probabilities. This yields

<O2>m = Σl P(l) <O2>m(l) = Σl,k',k conj(blk')blkl|adj(UA2)UA2l>Ak'|adj(UB2)O2BUB2k>B

If we consider what the expectation value would have been if we started with |ψ'> and only Cooper applied his evolution operator, we would have

|ψ''> = Σj,k bjkj>A*UB2k>B

<O2>ψ'' = Σj,k',k conj(bjk')bjkk'|adj(UB2)O2BUB2k>B

identical to the value with the measurements and evolution on A. So there you have it. No matter what Alice and Cooper do here, they cannot use the entanglement alone to transmit information. Now if Alice sends Cooper the result of her measurement via a classical channel and then Cooper chooses his evolution based on that result, they could teleport a quantum state, but then they would be constrained by the speed of their classical signal, which should presumably be subluminal.

As always please let me know if you find any errors or if anything is unclear, especially in the portion for non-physicists, because I'd especially like everyone to be able to understand that part. I may add an explanation of the general case in terms of density matricies in the future. If you're reading this and you'd like to see that, let me know. Also, please let me know if you have any suggestions to make the math more readable. Remember upvotes are nice, but feedback is nicer.

Sources:

• Modern Quantum Mechanics by J. J. Sakurai
• Introduction to Quantum Mechanics by D. J. Griffith

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