## Monday, September 11, 2017 ... / / ### Why vanishing commutators imply there's no action at a distance

I believe that an extremely similar blog post has been written in the past but I can't find it or what I can find isn't quite the same so that's why I decided to write this one again.

Lots of people say that there must be non-local influences or some action at a distance in the real world, and this claim is implied by Bell's inequalities or something like that. This statement is completely wrong. Since the 1905 special theory of relativity, we have known that the non-local or superluminal influences would be equivalent – by the Lorentz transformation – to the influencing changing one's past, and those are logically inconsistent.

So why don't entanglement experiments imply any action at a distance?

In quantum mechanics, events are predicted probabilistically. Unless all the probabilities are calculated to be 100% or 0%, and they're usually in the middle, we can't say that the outcome will be something or something else with certainty. We can only say that the outcome will be something with some probability; and something else with some other probability.

In this setup, the action at a distance obviously means that the willful action at one place which we will call Alaska (A) will modify the probabilities of some properties of outcomes of measurements at another place which we will call Boston (B). OK, let's imagine we have an entangled pair of particles or other physical objects that were created as entangled somewhere in Texas, to make it general, but the subsystems have propagated to Alaska and Boston, respectively.

I chose Alaska and Boston for them to be on the left and on the right. Alabama's position didn't look convenient enough.

OK, we want to follow the probability that some measurement in Boston will have some property. It may be any property of any measurement of any quantity that is geometrically confined to Boston. Therefore, the corresponding quantity or operators could carry some extra indices that indicate that there are many options. But without a loss of generality, we will follow one particular property and it will be expressed by the projection operators $$P_B$$. It's some projection operator constructed from local fields in the Greater Boston area, if you use the framework of quantum field theory.

The initial (entangled) density matrix describing both subsystems will be called simply $$\rho$$ and we assume $${\rm Tr}(\rho)=1$$. I am using a density matrix in order to be really general. But the discussion below may of course be applied to the special case of the pure states when $$\rho=\ket\psi\bra\psi$$, too.

Fine, so what is the probability that the property in Boston will hold? It will be simply${\rm Tr} (P_B \rho P_B).$ I wrote the projection operators on both sides, in order to indicate that $$P_B$$ would act both on the ket vector $$\ket\psi$$ and on the bra vector $$\bra\psi$$ in the pure state case. But you can see that due to the cyclicity of the trace, one of the $$P_B$$ may be moved "around the world" through the parentheses next to the other one and we may use $$P_B^2 = P_B$$, a defining property of a projection operator. It follows that in the probability above, you could easily omit one of the two copies of $$P_B$$.

In the pure state case, the probability given by the trace is reduced to $$|\bra\psi P_B \ket\psi|^2$$, a classic example of Born's rule, but I will no longer write the special forms of the formulae for pure states below.

Fine. Note that in the case of a spin-1/2 particle, the projection operator saying that "the spin is up with respect to the axis $$\vec n$$" where $$|\vec n|=1$$ is simply$P_B = \frac{ 1 + \vec \sigma \cdot \vec n}{2}, \quad \vec\sigma \cdot \frac\hbar 2 \equiv \vec S.$ But again, the rest of our discussion holds not only for spins of the electrons but for any systems, and not only for the maximally entangled states, but for absolutely any states of any composite systems.

Now, the question is whether some willful action done on the other side of America, in Alaska, may change the probability${\rm Tr} (P_B \rho P_B)$ to something else. What are the willful actions? Well, there are people in Alaska. And one of them, Sarah Palin – the only name in Alaska I was able to memorize – may press buttons in her laboratory which determines what property of her subsystem is going to measured. For example, if the subsystem in Alaska is another spin-1/2 particle, Sarah Palin may press the button "x" or "y" or "z" which determines which polarization of the spin of her electron is gonna be measured.

The decision is somewhat random, mostly confined to her brain. It may be chaotic, it may depend on some objects in Russia that Sarah Palin has seen from her house, and so on. I mentioned the spin but any decision that a human makes may be parameterized by similar data as the choice of "x" or "y" or "z". The only inherently quantum impact of this decision is the choice of the basis of possible "after the measurement states" of her subsystem, e.g. the Alaskan spin-1/2 particle.

Again, such decisions may be arbitrary, influence any observable of the Alaskan system, so the operator $$P_{A,i}$$ could have extra indices. But again, our argument would work for all values of these extra indices, so without the loss of generality, we will omit these indices. So Sarah Palin decides to press a button, e.g. "y", and this choice singles out the projection operators$P_{A,i}$ that may act on her pure state or the density matrix. Here, $$i$$ is the index indicating the possible results. For example, in the case I mentioned, $$i=1,2$$ means $$\sigma_y=\pm 1/2$$, respectively. We obviously have$\sum_i P_{A,i}=1.$ Sarah Palin's measurement, when it's completed, induces a collapse of the wave function or density matrix, and it's this collapse that is said to "influence the situation in Boston" by the confused champions of non-locality. OK, did Sarah Palin influence Boston when she pressed a button and extracted a result of the measurement? Let's look carefully.

The probability that Sarah Palin gets the $$i$$-th result is${\rm Tr} (P_{A,i} \rho P_{A,i})$ and if she gets this $$i$$-th result, the density matrix for the composite system changes from $$\rho$$ to$\rho \to \frac{ P_{A,i} \rho P_{A,i} }{ {\rm Tr} (P_{A,i} \rho P_{A,i}) }.$ I have divided the density matrix by its trace in order to keep the trace of the ultimate density matrix equal to one. OK, at any rate, you can see that Sarah Palin has modified the density matrix – she has inserted her particular projection operator $$P_{A,i}$$ on both sides from the density matrix. Let me mention that in the numerator, there is no trace, so you cannot omit either copy of the two projection operators (with the omission, the density matrix wouldn't even be Hermitian). In the denominator, you can omit one for the reason I have already discussed back in Boston.

OK, this looks like a very different density matrix that depends on the $$i$$-th result of her Alaskan measurement. How does it affect Boston? How does it affect the probability that the Bostonian property we're discussing will be measured as "Yes"? Well, it's simple. In the formula for the probability of the chosen Bostonian property${\rm Tr} (P_B \rho P_B),$ we must simply replace $$\rho$$ with the new density matrix after the Alaskan measurement. It means that the probability of the Bostonian property will change to the conditional probability$\frac{{\rm Tr}(P_B P_{A,i} \rho P_{A,i} P_B) }{ {\rm Tr} (P_{A,i} \rho P_{A,i}) }$ in the case of the $$i$$-th result in Alaska. This looks complicated and dependent on some stuff in Alaska. So Sarah Palin has influenced Boston, hasn't she? Wait a minute.

No one in Boston could have learned about the value of the index $$i$$ indicating the outcome of the Alaskan measurement yet. Also and equally importantly, Sarah Palin couldn't "order" the result to be $$i$$ – the results are randomly chosen by Mother Nature, not by Sarah Palin. So the probability that the Bostonian property is obeyed must be calculated as the probabilistic weighting of the conditional probabilities above multiplied by the probabilities of the $$i$$-th outcomes in Alaska.

So after Sarah Palin picked her measurement type and measured the outcome chosen by Mother Nature, but before Boston learned about that outcome, the probability of the chosen Bostonian property is clearly$\sum_i \frac{{\rm Tr}(P_B P_{A,i} \rho P_{A,i} P_B) }{ {\rm Tr} (P_{A,i} \rho P_{A,i}) } {\rm Tr} (P_{A,i} \rho P_{A,i})=\dots$ I have copied and pasted the previous complicated formula, added the sum over $$i$$ on the left, and added the probability that she measured the corresponding $$i$$-th outcome. But as you can see, that probability is exactly the denominator that I previously inserted to normalize the density matrix, so those two cancel. The expression above is therefore equal to$\dots = \sum_i {\rm Tr}(P_B P_{A,i} \rho P_{A,i} P_B)=\dots$ That looks much simpler. Now, if the operators in Alaska and Boston commute with each other – and they commute with each other if the regions are spacelike-separated, we have$P_{A,i} P_B = P_B P_{A,i}.$ But if that's so, we can switch the order of the two adjacent operators (in Alaska vs Boston) at both places, to get$\dots = \sum_i {\rm Tr}(P_{A,i} P_B \rho P_B P_{A,i})=\dots$ Now it's simple. We use the cyclic property of the trace which allows us to move one of the operators $$P_{A,i}$$ next to the other one, to the opposite side of the parentheses. We use the property of the projection operator $$P_{A,i}^2 = P_{A,i}$$ and the expression above is therefore equal to$\dots = \sum_i {\rm Tr}(P_{A,i} P_B \rho P_B)=\dots$ which differs by the erasure of one of the two copies of the Alaskan projection operator. But by linearity, we may use the fact that the sum of the Alaskan projection operators over $$i$$ is simply equal to the identity operator $$1$$, so the expression above is simply equal to$\dots = {\rm Tr}(P_B \rho P_B).$ That's the same probability of the property in Boston that we had at the beginning. It follows that Sarah Palin's free decision to press some buttons that influence a part of the entangled subsystem and the subsequent collapse of the state vector or density matrix into one of the basis options (Palin could have affected the choice of the basis but not the particular vector, and folks in Boston couldn't know either) has not influenced the probability in Boston at all. Because $$P_B$$ as well as the set of $$P_{A,i}$$ was chosen arbitrarily, we may conclude that no decision made in Alaska influences any property (any probability of it) that may be measured in Boston. There is no action at a distance.

To make this discussion complete, one should revisit the corresponding technical section from basic quantum field theory, to explain why operators in spacelike-separated regions exactly commute with each other. And we could add a discussion how $$P_{A,i}$$ and $$P_B$$ are constructed as functionals of the quantum fields in Alaska or Boston, respectively, so they commute with each other, too. Also, to be safe, one should add a discussion explaining why the non-relativistic description of several particles may be embedded within a quantum field theory, so the commutators are obviously vanishing in the non-relativistic description of the subsystems, too (there are other arguments to demonstrate this simple fact).

At any rate, there exists a valid proof shown or sketched above that implies that the people's playing with the subsystems of entangled systems implies no action at a distance at all – the influence is exactly zero. All of the people who say that there is an action at a distance or non-local influence are just wrong. Most of the time, their demonstration of the wrong conclusion has nothing whatever to do with the proof above because it has nothing to do with quantum mechanics. These people aren't thinking quantum mechanically, their brains are confined somewhere in the 17th century.    