Exploring Entangled States: Z-Dir & N-Dir Spins

[JaneHall89]

The first spin analyser is orientated on the z-axis and the second is in a arbitrary N-direction.

My thought process I can't get past is below

• If I find particle 1 to have spin up in Z-direction, the state collapses and the second must be spin down in the Z-direction for the particle at the second analyser, But if the second analyser is orientated in the arbitrary N-direction we can't get spin down in the Z-direction for the second particle, we get either spin up or spin down in the N-direction.

If I orientate the first spin analyser to the Z-direction but now the second to the Y-direction.

• Particle 1 will be either spin up or down in the Z-direction, we then measure the particle 2 in the Y-direction which is either spin up or down. I now know particle 1 must have the opposite Y-direction to particle 2 and particle 2 must have opposite Z-direction to particle 1. Thus my thought process is we have Z and Y spins for both particles by the process of elimination and measurement.

Can someone please explain why I'm stupid?

 

[drvrm]

If I find particle 1 to have spin up in Z-direction, the state collapses and the second must be spin down in the Z-direction for the particle at the second analyser, But if the second analyser is orientated in the arbitrary N-direction we can't get spin down in the Z-direction for the second particle, we get either spin up or spin down in the N-direction.

please clarify your thought experiment -so that one can understand the physical part.
are you having only two particle with fixed spin orientations (specially prepared or two particles with random spin directions)

 

[DrChinese]

If I orientate the first spin analyser to the Z-direction but now the second to the Y-direction.

• Particle 1 will be either spin up or down in the Z-direction, we then measure the particle 2 in the Y-direction which is either spin up or down. I now know particle 1 must have the opposite Y-direction to particle 2 and particle 2 must have opposite Z-direction to particle 1. Thus my thought process is we have Z and Y spins for both particles by the process of elimination and measurement.

Welcome to PhysicsForums, JaneHall89!

This is the EPR Paradox. There is nothing wrong with you, this has puzzled people since at least 1935.

You can perform an experiment on an entangled pair as you describe. The questionable part is the conclusion, in which you "believe" you reasonably (and simultaneously) know the non-commuting Y and Z spin components of the particles (let's call them Alice and Bob).

You don't really know both! You can confirm this by simply examining Alice's spin component in the same basis as you checked Bob. Alice's result will match Bob's only half the time. That's because it is randomly oriented on that basis. And vice versa with Bob. On the other hand, repeated tests of Alice's orientation at the same basis as she was originally tested will reveal the same (original) value.

The point is that you don't actually learn more information about the pair than the Heisenberg Uncertainty Principle allows. Any inference otherwise is invalid.

 

[JaneHall89]

Thanks. But I am confused to how if I measure spin up in the Z direction for particle 1 that it is also possible to have spin up in an N-arbitrary direction. I thought that we cannot measure the second particle in a N-arbitrary direction after measuring particle 1 in Z direction because the state has collapsed to Z-direction spins and particle 2 now must return a spin down in Z direction and not a spin up or spin down in N direction

My textbook has really confused me

 

[DrChinese]

Thanks. But I am confused to how if I measure spin up in the Z direction for particle 1 that it is also possible to have spin up in an N-arbitrary direction. I thought that we cannot measure the second particle in a N-arbitrary direction after measuring particle 1 in Z direction because the state has collapsed to Z-direction spins and particle 2 now must return a spin down in Z direction and not a spin up or spin down in N direction

Can anyone recommend text to read on it? My textbook has really confused me

A measurement of Z on particle 1 has the same effect as measuring Z on entangled particle 2. If you then measure on N basis on particle 2, the result is no different than if you had previously measured on the Z basis on particle 2.

And vice versa. Time ordering (of the measurement of particle 1 and particle 2) does not change the observed in any perceptible manner.

 

[JaneHall89]

A measurement of Z on particle 1 has the same effect as measuring Z on entangled particle 2. If you then measure on N basis on particle 2, the result is no different than if you had previously measured on the Z basis on particle 2.

And vice versa. Time ordering (of the measurement of particle 1 and particle 2) does not change the observed in any perceptible manner.

I can't see the proof of this in Griffiths intro to quantum mechanics and I've tried to search on google but its difficult going through all the information for this proof. Where can I find it with explanation?

 

[secur]

Of course DrChinese is right but since I think I see where the confusion is I'll throw in my 2 cents.

Consider your first post. Before any measurements are made, the particles are in unknown states, but at least we know they're opposite to each other.

As you say, after particle 1 is found to be spin-up in the Z-direction, the other must be down. So if you measured it in Z-direction, that's what you'd get: down.

Now let's ignore your first paragraph, using N-direction, and consider the Y-direction paragraph. There's no reason to repeat the thought process twice.

So, instead of measuring particle 2 in Z-direction, you measure particle 2 in Y-direction. Let's say we get spin up there also. So far so good.

Now this is where you go wrong:

JaneHall89: I now know particle 1 must have the opposite Y-direction to particle 2 and particle 2 must have opposite Z-direction to particle 1.

- No, that's wrong. The different measurements changed ("collapsed") each one differently. At this point, particle 1 is up in Z-direction, particle 2 is up in Y-direction, and there's no "entanglement" - no special relation - any more.

Evidently you're thinking that a particle can have two separate, independent spin orientations in the two different axes, Z and Y, but no, it has only one direction of spin. Once you get particle 1 up in the Z-direction, you don't know anything at all about its Y-direction spin. You could say it doesn't even have a Y-direction spin; not really kosher, but the right idea.

If you measure particle 1 in Y-direction now, it might be spin up, it might be spin down.

Similarly if you measure particle 2 in Z-direction, it could be up or down, no way to know until you do it. And, note, after you do it, you lose the information about the Z-direction; now its orientation is unknown.

I hope that helps?

 

[Strilanc]

I can't see the proof of [the time ordering of measurements doesn't matter] in Griffiths intro to quantum mechanics and I've tried to search on google but its difficult going through all the information for this proof. Where can I find it with explanation?

It's not too difficult to prove this for yourself.

To keep things simple, we'll restrict our focus to pure states. (Most properties that hold for pure states also hold for mixed states, because mixed states are mathematically equivalent to not knowing which pure state the system is in.) Also we'll restrict ourselves to talking about qubits, instead of continuous states.

The proof follows directly from two facts:

1. Measurement is observationally equivalent to a Controlled-Not operation with the qubit you wanted to measure as the control and an otherwise unused qubit in the ##|0\rangle## as the target. No experiment can distinguish between those two possibilities, unless quantum mechanics is wrong (or you break the rule about not otherwise operating on the target qubit). ( See http://www.scottaaronson.com/democritus/lec10.html )

2. Operations commute when they apply to disjoint subsets of qubits. A full proof requires a bit more work than the following, but here's the basic idea:

##\begin{align} [A \otimes I, I \otimes B] &= (A \otimes I_b) \cdot (I_a \otimes B) - (I_a \otimes B) \cdot (A \otimes I_b)\\&= (A \cdot I_a) \otimes (I_b \cdot B) - (I_a \cdot A) \otimes (B \cdot I_b)\\&= A \otimes B - A \otimes B \\&=0 \end{align}##

So suppose you have some situation where two parties, Alice and Bob, have some qubits and measure them. Suppose Alice measures slightly before Bob. Replace the measurements with controlled-nots without affecting the expected result (by 1), use the fact that Alice is applying operations to qubits disjoint from Bob's to re-order the controlled-nots without affecting the expected result (by 2), then un-replace the controlled-nots with the measurements without affecting the expected result (by 1 again). Therefore the expected result when Alice measures first is the same as the expected result when Bob measures first.

Not that this proof doesn't apply if Alice and Bob interact after Alice's measurement but before Bob's measurement (e.g. Alice says "hey I got +1, you should measure X instead of Y"). That's because the operations no longer apply to disjoint subsets of qubits: some of Bob's operations are de-facto controlled by some of Alice's qubits.