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Sabine Hossenfelder, a science YouTuber, talks about asking the following question to ChatGPT in one of her videos:

If you perform an operation which is not a measurement on one particle in a pair of entangled particles, does that affect the other particle?

(Source: https://youtu.be/cP5zGh2fui0?t=746)

More specifically, the question is whether flipping (without measuring) the spin of one of the particles in the EPR pair affects the spin of the other particle.

Sabine says that the answer is "No, you can't," and ChatGPT failed to generate the correct answer.

What I'd like to know is a mathematical description of this process of 'flipping.' What is the operator corresponding to flipping the spin of one of the particles of the EPR pair without measuring it? How do you show mathematically that such an operator does not affect the other particle if acted on one of the particles?

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  • $\begingroup$ No-communication theorem. Despite that, I couldn't see in the source anything regarding "switching" the spin... $\endgroup$ Mar 13, 2023 at 20:13
  • $\begingroup$ @TobiasFünke She talks about 'flipping the spin' around 13:20. $\endgroup$
    – Lory
    Mar 13, 2023 at 20:24
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    $\begingroup$ Could you please give the correct source/time stamps then?! Anyway, I guess she means a local unitary transformation. $\endgroup$ Mar 13, 2023 at 20:41
  • $\begingroup$ @TobiasFünke I've updated the source with a time stamp. $\endgroup$
    – Lory
    Mar 13, 2023 at 20:55
  • $\begingroup$ Wait the video seems to imply the opposite of what you are saying. The chatbot says that you CAN affect particle A by acting on particle B, while Sabine says the opposite, acting on A does not affect B. $\endgroup$
    – Mauricio
    Mar 13, 2023 at 21:04

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You might have gotten it wrong. Sabine says you can modify the state on one particle without modifying the state of the second particle. In the video ChatGPT says the opposite, affecting a single particle in the entangled pair affects the whole pair.

Both might be right in a sense, as both are speaking loosely. Let's write a Bell state for two particles A and B (or EPR pair as you call it):

$$|\psi\rangle=\frac{1}{\sqrt{2}}(|+1,-1\rangle+|-1,+1\rangle)$$ where I am using Sabine's notation. This state is a superposition between measuring particle A to be aligned with your measuring device (+1) and B to be anti-aligned (-1), and the state where A is anti-aligned and B is aligned. There is a 50% chance of measuring either.

There is always an operator $X$ such that acting on a single particle does the following $X|+1\rangle=|-1\rangle$ and $X|-1\rangle=|+1\rangle$. How do you create this operator physically? There are many ways, and example would be a magnetic field in a certain direction that makes the spin precess (note that this does not represent a measurement). In the language of spins it is the $\hat S_x$ operator or in the language of qubits it is the Pauli matrix X.

What happens if we apply $X$ only to the first particle (A) of $|\psi\rangle$?

$$X_A|\psi\rangle=\frac{1}{\sqrt{2}}(X_A|+1,-1\rangle+X_A|-1,+1\rangle)=\frac{1}{\sqrt{2}}(|-1,-1\rangle+|+1,+1\rangle)$$

Here you only acted on a single particle and still you have an entangled state: 50% of both being +1 or 50% -1. By flipping one of the spin, you perturb the system in the sense that you no longer have the same global state and thus different results. In another sense, you only acted on a particle you did not affect the state of the other (which was undefined previous to measurement anyway).

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  • $\begingroup$ This is very helpful, thanks. Is $X_{A}$ the same as $X \otimes I$? $\endgroup$
    – Lory
    Mar 13, 2023 at 22:17
  • $\begingroup$ @Triad yes, I omitted the $\otimes I_B$ to be exact. $\endgroup$
    – Mauricio
    Mar 13, 2023 at 22:25
  • $\begingroup$ ChatGPT says the opposite Please don’t use computer-generated text for questions or answers $\endgroup$
    – Ghoster
    May 22, 2023 at 17:35
  • $\begingroup$ @Ghoster I meant in the video not in my answer $\endgroup$
    – Mauricio
    May 22, 2023 at 19:25

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