# I If you change the spin of an entangled particle without knowing its original spin...

#### remormalise

If you change the spin of an entangled particle without knowing its original spin, what happens to the other entangled particle?

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#### Christopher Grayce

How do you know you've changed the spin if you don't know its original value?

#### remormalise

Assume you don't need to know its original state, can you impart up spin on all particles, would this break the entanglement or down spin all the entangled particles

#### Mentz114

Gold Member
If you change the spin of an entangled particle without knowing its original spin, what happens to the other entangled particle?
An entangled pair share one wave function. So they are in a way one thing. If either one is projected into a different spin state the partner will have the same/opposite spin ( the '/' depends on the preparation).

Oh, should add that in above model, it makes no difference if you know the initial spin. If you did know it, it must have been measured (in which case the entanglement is broken ) or the preparation procedure sets an initial state.

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#### PeterDonis

Mentor
can you impart up spin on all particles, would this break the entanglement
Yes, because this amounts to preparing a new state which is separable.

#### DrClaude

Mentor
If you change the spin of an entangled particle without knowing its original spin, what happens to the other entangled particle?
Nothing. You will change the total wave function of the two particles, but you can't say that something has happen to the first particle.

Consider the two-particle state
$$| \Psi \rangle = \frac{1}{\sqrt{2}} \left[ | \uparrow \downarrow \rangle + |\downarrow \uparrow \rangle \right]$$
Apply a rotation to the second particle such that
\begin{align*} | \uparrow \rangle &\rightarrow \frac{1}{\sqrt{2}} \left[ | \uparrow \rangle + |\downarrow \rangle \right] \\ | \downarrow \rangle &\rightarrow \frac{1}{\sqrt{2}} \left[ -| \uparrow \rangle + |\downarrow \rangle \right] \end{align*}
$$| \Psi \rangle = \frac{1}{2} \left[ -| \uparrow \uparrow \rangle + | \uparrow \downarrow \rangle + |\downarrow \uparrow \rangle + |\downarrow \downarrow \rangle \right]$$
which is still entangled.

Assume you don't need to know its original state, can you impart up spin on all particles, would this break the entanglement or down spin all the entangled particles
I don't understand here. If by "all particles" you mean the two you have, then putting them both with spin-up doesn't corresponds to an entangled state.

#### remormalise

Gentlemen, I am very grateful for your input to help me understand how entanglement works. I am not a physicist, but I have a deep and long-standing interest in how our universe works.

My apologies if my question was not precise or constrained enough, however, your answers have provided me with an increased understanding, which is all I am ever after.

Many thanks.

#### Abiologist1

I've recently seen the excellent youtube video on "Quantum Entanglement & Spooky Action at a Distance" by Veritasium - but either I'm not understanding something - or there seems to be something missing from the explanation.

It seems that the operator can choose the ANGLE of polarisation of both entangled photons (+_180 degrees) - even if one photon is very far away - is this correct ??

#### Nugatory

Mentor
I've recently seen the excellent youtube video on "Quantum Entanglement & Spooky Action at a Distance" by Veritasium - but either I'm not understanding something - or there seems to be something missing from the explanation.
You cannot learn quantum mechanics from videos on the internet. They're fun, they're interesting, some are better than others, but there is always something missing.
It seems that the operator can choose the ANGLE of polarisation of both entangled photons (+_180 degrees) - even if one photon is very far away - is this correct ??
The local operator can choose to measure the polarization on any axis they want; whatever result they get, a measurement of the other photon will produce the opposite result. Say the operator holds their polarizer at an angle of $\theta$ degrees. If their photon passes through the filter, they have just measured the photon to be polarized at that angle; if it doesn't pass they've just measured it to be polarized at 90 degrees to that angle. Either way, if we measure the other photon with a polarizer at the same angle, we will find the opposite polarization result.

However, that doesn't mean that the operator has any control over the angle of the remote photon. No matter what anyone does and no matter which angles are chosen for the two measurements, the remote observer will always find that half the photons clear his filter and half are absorbed. The remote observer has no way of knowing that the photons are entangled or even whether the operator came in to work that day until both sides get together after the fact and compare notes. Only then do they see that although their individual results are completely random, they always get opposite results when the polarizers are at the same angle.

#### Abiologist1

However, that doesn't mean that the operator has any control over the angle of the remote photon. No matter what anyone does and no matter which angles are chosen for the two measurements, the remote observer will always find that half the photons clear his filter and half are absorbed.
regardless of the percentage which clear the filter of the local operator ? - or do you mean only in the case that half the photons clear the local operator filter and half are absorbed ? Thankyou for your detailed first reply (!)

#### Nugatory

Mentor
regardless of the percentage which clear the filter of the local operator ? - or do you mean only in the case that half the photons clear the local operator filter and half are absorbed ?
The percentage that clears the local operator’s filter is always 50%, no matter what the angle of the filter, just as with the remote filter.

(And I should add that there are serious practical difficulties in actually performing this experiment in exactly this form - this is a thought experiment as described)

#### Abiologist1

(And I should add that there are serious practical difficulties in actually performing this experiment in exactly this form - this is a thought experiment as described)
OK - so what would be a better experiment ? I am aware that someone thinks that a hidden variable determining whether any photon is detected at all, or whether no photon is detected by both local and remote operators, could give results that allow the probabilities actually detected but still be consistent with non-quantum phenomena (see Explained & Debunked_ Quantum Entanglement & Bell Test Experiments - YouTube). Do you think this is a possibility ? Many thanks in advance.

"If you change the spin of an entangled particle without knowing its original spin..."

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