# B Two Confusing Entanglement Scenarios

1. Dec 20, 2017

### Isaac0427

Hi,

There are two particular scenarios involving entanglement that are confusing me. For both of these, assume we have an electron-positron pair created from a spin-0 particle, in which all components of spin must be opposite in order to conserve angular momentum. The electron and positron are a considerable distance apart.

1. The x-component of the electron's spin is measured to be, say, up, which would mean the x-component of the positron's spin is down. At the exact same instance, the y-component of the positron's spin is measured to be down, which would mean that the y-component of the electron's spin is up. At that exact moment, why is the HUP not violated? If, shortly after, we measure either quantity again, will we be guaranteed to get the same result, or does the previous "double measurement" randomize the results?

2. The scientist with the positron asks the scientist with the electron a yes-or-no question. They then get their watches completely synchronized and each travel a lightyear in opposite directions with the same speeds and accelerations, so their watches are still synchronized. After they both simultaneously arrive at their final destinations, the scientist with the electron has exactly one hour to repeatedly measure x-spin and y-spin over and over again until he/she gets the desired result; y-spin up for yes, y-spin down for no (the x-spin is measured to randomize the y-spin results). After the hour, which, especially if the scientist can perform a spin measurement every few seconds, would be far more than enough time to surely get the desired answer, the scientist with the positron measures its y-spin and finds the answer to the question, at most an hour after the answer was sent. Is this not information traveling at the speed of light? Now, I do understand that if, say, an hour is enough time for 60 x-spin-y-spin measurements, the scientist with the positron has a 1/2^60 chance of not having the correct answer (that is the probability that all y-measurements will end up with the wrong answer), so definite information was not really sent. But, take the limit of this situation. Say the scientists are x light years apart, and the scientist has x/2 years to answer the question, and then let x tend towards infinity. Wouldn't an answer to the question that has a probability of being correct approaching 100% travel at double the speed of light?

2. Dec 20, 2017

### DrChinese

Whew!

1. If you measure any particle's X-spin, and then its Y-spin, have you violated the HUP? Of course not. No different in your scenario. In fact, the one you give is almost identical to the 1935 EPR paper. The particles are entangled, true, but you have no way to make the experimentally verifiable deduction that the HUP is violated.

2. Not sure where you are going with this. Once any measurement is made on either particle, the entangled state ends on that measurement basis. So subsequent measurements will not reflect a state of entanglement.

3. Dec 20, 2017

### Staff: Mentor

#2 is easier, so we'll do it first:
The first measurement of either member of the entangled pair breaks the entanglement, so there's no possibility of repeating measurements for an hour until you get what you want. You measure the electron once, you get what you get, you know that if the other scientist measures the positron on the same axis they'll get the opposite of what you got, and that's it. Any subsequent measurements you make will not be correlated with the remote particle in any way.

Your #1 is David Bohm's version of the original EPR argument, using spin instead of position and momentum. However, there is a subtle difference between what you (and EPR) are assuming about how the world works and what QM says. You are saying that when we measure the electron's spin on the x-axis and get spin-up, then we know that the positron is spin-down. QM says something different: we know that if the positron is measured on the x-axis then the result will be spin-down. But here we aren't measuring the positron along the x-axis, we're measuring it along the y-axis, so any statements about what we would have gotten if we had measured on the x-axis are just irrelevant hypotheticals.

In mathematical terms, when we're done with both measurements the electron is in an eigenstate of $S_x$ (a superposition of the two possible $S_y$ states), the positron is in an eigenstate of $S_y$ (a superposition of the two possible $S_x$ states), and neither has both definite $S_x$ and $S_y$ values.

One of the most amazing discoveries of the last century is that there are statistical differences between "the positron is spin-down on the x-axis" and "the positron would be spin-down on the x-axis if we measured it on that axis". Although subtle, these differences are experimentally detectable and have been observed - and it's the quantum mechanical view that is supported by these experiments. Google for "Bell's Theorem Dr Chinese" to find the web site maintained by our own @DrChinese

[edit: who posted while I was composing my reply :)]

Last edited: Dec 21, 2017
4. Dec 21, 2017

### mikeyork

Not really wanting to be pedantic, but you don't know that they are entangled with opposite spin, because they could have orbital angular momentum too. All you know from angular momentum conservation is that J = L+S = 0.

5. Dec 21, 2017

### Isaac0427

So what I am getting here is that a measurement breaks the entanglement, i.e. the entanglement is only valid for one measurement. Is that correct? If so, and after measurement of the x-spin the y-spin becomes unentangled, why does that not violate the conservation of angular momentum (i.e. if you measure both with y-spin up after they are unentangled)? Additionally, is there a way to show mathematically why they become unentangled?

6. Dec 21, 2017

### DrChinese

Yes, good for 1 measurement per entangled basis (say spin).

If you measure x-spin on A, you know what the result of an x-spin on B if performed. And if performed, would show conservation as expected - on x-spin. But recall that there is not simultaneously well-defined x-spin *and* y-spin.

Now compare that to your question about "if you measure both with y-spin up after they are unentangled"? Why would you expect that 2 unentangled particles, at least one of which had interacted with a measurement apparatus since an earlier time/state, would continue to exhibit a conservation rule? The answer is that they will not.

7. Dec 21, 2017

### mikeyork

The measurement is an interaction involving the apparatus. There is no violation of conservation of angular momentum because the particle is only part of the interacting system. The difference is absorbed by the apparatus.

Without understanding the role of the apparatus, every quantum measurement could posssibly violate some conservation rule because of the basis projection.

8. Dec 21, 2017

### Staff: Mentor

Forget entanglement for a moment and just consider a particle prepared in the state "spin up in the horizontal direction"; we might have prepared this particle by passing a bunch of particles through a horizontally oriented Stern-Gerlach device and then selecting one of the particles that was deflected to the right. Suppose we measure the particle's spin along the vertical axis. We know that this measurement will leave the particle in either either the state "spin up in the vertical direction" or "spin down in the vertical direction". How is that not a violation of conservation of angular momentum? Take a look at @mikeyork's post #7 above.
An entangled state is a superposition, collapsed by the first measurement.

Let's use the notation $|H_+H_-\rangle$ to represent the state "the electron will be spin up when measured on the horizontal axis and the positron will be spin down when measured on the horizontal axis". The state "the electron will be spin down when measured on the horizontal axis and the positron will be spin up when measured on the horizontal axis" would be written $|H_-H_+\rangle$, $|V_-V_+\rangle$ would be the same thing for measurements on the vertical axis, and so forth.

When the pair is created its state is something that can be written in the form $\psi=\alpha|H_+H_-\rangle+\beta|H_-H_+\rangle$ so that angular momentum is conserved. We don't know the values of the complex constants $\alpha$ and $\beta$; we just know that $\alpha^2+\beta^2=1$. We could have written that state as $\psi=\alpha'|V_+V_-\rangle+\beta'|V_-V_+\rangle$ where $\alpha'$ and $\beta'$ are some other constants; it's just two different ways of writing the same vector as a sum of two other vectors, and I chose the first one because it's convenient for the next step.

So now I measure the spin of the electron on the horizontal axis. This first measurement collapses the wave function to either $|H_+H_-\rangle$ (the measured result was spin-up, this happens with probability $\alpha^2$) or $|H_-H_+\rangle$ (the measurement result was spin-down, this happens with probability $\beta^2$). Either way, I know what I will get when if I measure the positron on the same axis - the collapse eliminated the superposition for both the electron and the positron.

Let's say that the state collapsed to $\psi=|H_+H_-\rangle$, and now I decide to measure the position of the positron along the vertical axis. A bit of algebra tells me that $\psi=|H_+H_-\rangle=\frac{1}{\sqrt{2}}(|H_+V_+\rangle+|H_+V_-\rangle)$. I'll get vertical spin up or vertical spin down for the positron with equal probability, but either way the electron is unaffected.

Last edited: Dec 21, 2017
9. Dec 21, 2017

### PeroK

There's an easier way to lose conservation of spin in a given direction. Take an electron with spin up in the x direction. Measure spin in the y-direction. Measure spin in the x direction and you will get spin down 50% of the time.

The conservation laws in QM generally apply to the time evolution of expected values. You can think of this as a deterministic time evolution under the Schrödinger equation, which gets disrupted by measurement to give a random eigenvalue. In other words, a measurement need not, and in some cases cannot, return the expected value that is being conserved.

10. Dec 21, 2017

### Isaac0427

Ah, this is very helpful. Thank you to everyone who responded.