# How can quantum entanglement not be non-local?

I know this kind of question has been brought up many times.I have read many posts here regarding this but I still have a problem with a certain aspect of it so please bear with me.

Lets consider the standard case where one of two singlet state electrons is send to Alice and the other is send to spacelike separated Bob.This is repeated many times.

I know that its impossible for Alice to transmit information by measuring her electrons.This is because the density operator for the state of Bob is invariant under a rotaion of Alices axes.To show that, we note that the state prior to Alices measurement is also rotation invariant (using the well known rotation matrices for spin 1/2 objects):

$$\frac{|\uparrow_z\downarrow_z\rangle-|\downarrow_z\uparrow_z\rangle}{\sqrt{2}}=\frac{|\uparrow_n\downarrow_n\rangle-|\downarrow_n\uparrow_n\rangle}{\sqrt{2}}$$

The state has the same components, no matter in which base it is expressed.That means, that no matter which axis Alice chooses, she will always get half of her results spin up and the other half spin down. Now if we use the "model" that Alices measurement collapses the composite state wavefunction, it follows that the particles Bob recieves will be a statistical mixture of one half spin up, one half spin down.This state is also independent of the basis,as can be shown using the same rotation matricies. $$\rho=1/2*|\uparrow_z\rangle\langle\uparrow_z|+1/2*|\downarrow_z\rangle\langle\downarrow_z|=1/2*|\uparrow_m\rangle\langle\uparrow_m|+1/2*|\downarrow_m\rangle\langle\downarrow_m|$$

This means that the statistical distribution of ups and downs that bob measures is completely independent of Alices choice of axis, or even her choice to measure her electron at all. Bob will statistically always get a fifty fifty result.

But nevertheless, and this is my question, Alices measurement makes Bobs wavefunction collapse to a pure state, and it seems to me this will leave an imprint on Bobs measurement in a non local fashion. Lets say Alice only measures along the z axis, and Bob chooses a random axis for every measurment.If Alice tells Bob the outcome of her measurements, Bob will be able to divide his outcomes in two groups: One where the corresponding spin of Alice was measured up, and the other one for the case where Alice has gotten down.And these two groups will show a perfect correlation between Bobs axis and Alices z-axis.Everytime Alice had found spin up, and Bob has chosen the z-axis aswell, he will find that his measurement was down, and the probability distribution will depend on the angle of his axis relative to Alices z-axis by the well known formula

$$|\langle \chi_z|\chi_n\rangle|^2=\cos\left(\frac{\theta}{2}\right)^2$$

Where theta is the angle that Bob measured the spin along, relative to the z-axis.And Alice can change the axis that Bobs measurements are biased to.If she chooses another axis, then Bobs outcomes will show the correlation with respect to this new axis.This correlation will be "burned" into Bobs list of outcomes one by one, everytime Alice measures an electron.

Of course, Bob will only see this correlation when he recieves Alices results, which can only be transmitted via a classical channel. But I mean, "something" has to change in the very moment Alice makes her measurement, because the correlation between the spins is already there, even before Alices information has reached Bob.

How can we rescue locality here?

(I guess this is deeply connected to the question whether the wavefunction actually collapses or something else is going on.But how can we say that there is not actually a collapse if the model of the collapse gives us the right probability distribution for the angles?Maybe you can adress this)

Edit:

I think these answers don't answer my question because I explained that I know that Alice cant send information to Bob this way.Nevertheless it seems to me there is some kind of non locality involved.Like a nonlocal event thats just useless for sending information.I wonder whether there is a way of looking at it that doesnt involve any nonlocal event.

• Possible duplicate of Does entanglement not immediately contradict the theory of special relativity? – Alex Robinson Oct 15 '17 at 16:37
• Yes, entanglement is (or can be construed to be) nonlocal, but what's wrong with that? You only require nature to be causal and that's not at risk here. – Emilio Pisanty Oct 15 '17 at 17:32
• I guess my concern still arises because I understand entanglement in the way that Alices measurement "does" something to Bobs electron.It actually changes to state of the electron.My understanding is that an electron actually has a definite state (in a quantum sense of course), some pure state or even some mixed state.And Alices measurement makes the electron change its state into something else.(That looks exactly the same to Bob, but only because hes missing Information). Is this approach flawed? – curio Oct 15 '17 at 17:53
• If the electron actually changed its state there would have to be some kind of interaction which would of course bring many problems wouldn't it? – curio Oct 15 '17 at 17:58
• Argument over this contradiction goes back all the way Einstein vs. Bohr. See Einstein/Podolsky/Rosen (en.wikipedia.org/wiki/EPR_paradox) and Bell's Theorem plato.stanford.edu/entries/bell-theorem. Bell's theorem is subject to experimental verification (the original EPR isn't). The experimental evidence at this time is quite clear: Bohr was right and Einstein was wrong: the universe appears to be non-local. – Hilmar Oct 15 '17 at 18:27