1. Nov 11, 2015

### Harel

If we have 2 particles A and B with entangled spins like in the EPR case. we know by the pauli matrices that there is an uncertainty relation between the x and z spin direction for example so what happens if we Alice measures the spin in the x direction while simultaneously Bob measures the spin in the z direction. Alice measurement should define a defnite spin in the x direction in bob's particle while his own measurement define a spin in the z direction, but it's violite the uncertainty relation. I'm sure that there is something wrong with my understanding but I don't know what.

2. Nov 11, 2015

### StevieTNZ

When Bob measures his particle along spin z axis, then that particle does not have a defined spin x, despite Alice measuring her particle along the x axis.

3. Nov 11, 2015

### DrChinese

StevieTNZ is correct, and in a way you are too.

1. Alice performs a measurement in x; Bob's x is now defined (and known). But Bob's z is completely undefined if his x is defined, because of said uncertainty relation.
2. Bob later* performs a measurement on z; Bob's z is now defined. But Bob's x is completely undefined if his z is defined, because of said uncertainty relation.
3. There is no correlation between Alice's z and Bob's z after 1. is performed. They are no longer entangled.
4. There is no correlation between Alice's x and Bob's x after 2. is performed. They are completely independent at this point.
5. There is never a time at which you know both Bob's x and z.

*You can perform a "later" measurement in all reference frames, so as to remove issues of relativity.

4. Nov 11, 2015

### Strilanc

If you work out the math of what happens, you'll find that effects of Alice's operations commute with the effects of Bob's operations (including measurement). Actually it's a bit stronger than commutativity, because they can apply the operations simultaneously.The expected outcome for the simultaneous case is equivalent to the Bob-first and Alice-first cases.

More concretely, $(A \otimes I) \cdot (I \otimes B) = A \otimes B = (I \otimes B) \cdot (A \otimes I)$. For reference: $A$ is the matrix representation of Alice's operations on her local state, $B$ is Bob's operations on his local state, $I$ is the "doing nothing" identity matrix (appropriately sized), $\otimes$ is the tensor or kronecker product, and $\cdot$ is the usual matrix product.

5. Nov 11, 2015

### Simon Phoenix

I find that a convenient way to think of this is to consider a measurement in QM as a projection from one state to another.

So let's just consider a single spin-1/2 particle. We'll prepare it in the state |+>z in the spin-z direction - so it has a definite value of the spin when we make a measurement of spin-z. OK so if we do this experiment we'll get the answer + with unit probability.

Let's take the same state and make a measurement of spin-x. In the spin-x basis we can write our state |+>z as |+>z = ( |+>x + |->x )/√2

where |+(-)>x is the spin 'up' ('down') eigenstate of the spin-x operator.

So when we make a measurement of spin-x on our state, initially prepared as |+>z the quantum rules tell us we'll get the result + with probability of 1/2 and the result - with probability of 1/2

Furthermore, after the measurement the spin-1/2 particle is in the state |+>x if we found the result + and in the state |->x if we found the result -

Now let's see what happens with the entangled state ( |+,+>z + |-,->z )/√2
where |+,+>z means that particle 1 is in the state |+>z and particle 2 is in the state |+>z, in their respective spaces.

Now let's suppose Bob measures spin in the z-direction. What is the result of this measurement? Well Bob is only acting on one of the particles (let's suppose it's particle 2). The quantum rules tell us that he will obtain the result + or - with 1/2 probability.

If Bob finds the spin to be + then the state of both particles after measurement is now |+,+>z. If Bob finds the result - then the state of both particles after the measurement is |-,->z

If Alice measures spin-z (she's performing this measurement on particle 1) her result will be identical to Bob's and we'll get perfect correlation between the results.

If, however, Alice measures spin-x, then if Bob's result was + she's measuring a particle (particle 1) that is in the state |+>z. But we know that this gives either + or - with probability 1/2. So Alice's result is completely uncorrelated with Bob's result if Bob measures spin-z and Alice measures spin-x

So despite the fact that we have maximal entanglement - and the strongest 'correlation' that QM will allow between 2 spin-1/2 particles - we find that not all observable properties are correlated. Here is an example where the results of one measurement (spin-z) are completely uncorrelated with the results of another (spin-x) on the particles, respectively.

OK - there's a couple of technical/interpretation issues here. First off I'm talking about a rather specialized form of measurement known as a von Neumann measurement of type I - it's not the most general kind of measurement allowed by QM - although, in a sense, all measurements in QM boil down to measurements of this kind. Secondly, there are issues with thinking in terms of this 'state projection' picture - particularly when we consider the ideas of SR. If Alice's and Bob's measurements are spacelike separated then we can find a frame in which Alice's measurement occurs before Bob's - so the question of who is 'doing' the projection is somewhat awkward. Having said that, it's kind of OK because it doesn't matter, as far as experimental predictions go, whether we consider Bob to do the projecting or Alice to do the projecting.

Hope that helps. I find thinking of terms of projective measurements is the clearest - for me at any rate - but others may not.