Measuring Entangled Particles in two different Basis

[randomuser3210]

Consider two entangled spin half particles given by the generic form of Bell Equation in Z-axis:
$\psi = (a\uparrow \uparrow + b\downarrow \downarrow)$ where $a^2+b^2=1$

In a (2D) planer rotated (by an angle $\theta$) direction the new equation can be given by:

$|\psi \rangle = [\alpha \cos^2(\theta/2) + \beta \sin^2(\theta/2)] |\uparrow_a\uparrow \rangle +$
$[(\alpha \sin^2 \theta/2 + \beta \cos^2 \theta/2)] |\downarrow\downarrow \rangle$
$[(\beta-\alpha) \cos(\theta/2) \sin(\theta/2)] (|\uparrow\uparrow \rangle +|\uparrow\downarrow \rangle +$

Now Alice keeps one particle and sends the other to Bob. Here is the Q:

1. Suppose, Alice measures the particle in $theta$ direction. Then after that, Bob measures in the Z direction. What is the probability of Bob getting the particle $\uparrow$ and $\downarrow$ in Z direction.

I know that as soon as Alice measures the particle in $\theta$ direction, the entanglement collapses. So we can measure the probability by using wave function. But, can someone help with the cases for 'cross measurement' i.e. as above?

 

[vanhees71]

Let $|\sigma_{\theta} \rangle$ and $|\sigma_z \rangle$ be the eigenvectors of the spin component in $\theta$ and $z$ direction respectively. Then the probability that A meausures the value $\sigma_{\theta}$ on her spin and B $\sigma_z$ on his, is given by
$$P(\sigma_{\theta},\sigma_z)=|\langle \sigma_{\theta},\sigma_z|\psi \rangle|^2.$$
You only need to express everything in a common basis (I've given all this in a thread some days in this forums, so you can just copy it from there).

 

[randomuser3210]

Let $|\sigma_{\theta} \rangle$ and $|\sigma_z \rangle$ be the eigenvectors of the spin component in $\theta$ and $z$ direction respectively. Then the probability that A meausures the value $\sigma_{\theta}$ on her spin and B $\sigma_z$ on his, is given by
$$P(\sigma_{\theta},\sigma_z)=|\langle \sigma_{\theta},\sigma_z|\psi \rangle|^2.$$
You only need to express everything in a common basis (I've given all this in a thread some days in this forums, so you can just copy it from there).

Thank you. I don't have background in QM/Physics. So I am lost at this compact reply and notation. If possible at all, could you provide more explicit probabilities (in notations in OP). Or point to a resource that goes step by step..

 

[PeroK]

Now Alice keeps one particle and sends the other to Bob. Here is the Q:

1. Suppose, Alice measures the particle in $theta$ direction. Then after that, Bob measures in the Z direction. What is the probability of Bob getting the particle $\uparrow$ and $\downarrow$ in Z direction.

I know that as soon as Alice measures the particle in $\theta$ direction, the entanglement collapses. So we can measure the probability by using wave function. But, can someone help with the cases for 'cross measurement' i.e. as above?

It doesn't matter who measures first. Bob's measurement has a probability $|a|^2$ or $|b|^2$ to be up or down. If Bob's outcome is up (or down), then the probability that Alice's measurement is up is $\cos^2 \frac \theta 2$ (or $\sin^2 \frac \theta 2$). That gives the four probabilities:
$$p(\uparrow_A \uparrow_B) = |a|^2\cos^2 \frac \theta 2; \ \ p(\downarrow_A \uparrow_B) = |a|^2\sin^2 \frac \theta 2; \ \ p(\uparrow_A \downarrow_B) = |b|^2\sin^2 \frac \theta 2; \ \ p(\downarrow_A \downarrow_B) = |b|^2\cos^2 \frac \theta 2$$

 

[randomuser3210]

It doesn't matter who measures first. Bob's measurement has a probability $|a|^2$ or $|b|^2$ to be up or down. If Bob's outcome is up (or down), then the probability that Alice's measurement is up is $\cos^2 \frac \theta 2$ (or $\sin^2 \frac \theta 2$). That gives the four probabilities:
$$p(\uparrow_A \uparrow_B) = |a|^2\cos^2 \frac \theta 2; \ \ p(\downarrow_A \uparrow_B) = |a|^2\sin^2 \frac \theta 2; \ \ p(\uparrow_A \downarrow_B) = |b|^2\sin^2 \frac \theta 2; \ \ p(\downarrow_A \downarrow_B) = |b|^2\cos^2 \frac \theta 2$$

So if I understand correctly, no matter who measures in what direction first (in any of the 4 General Bell States, not just this one), the end result of the measurement of Bob (and the probabilities of $\downarrow_B or \uparrow_B$ only depends on the direction Bob measures in.

 

[vanhees71]

The arguments of the cos and sin factors must be $\theta/2$:

Let's calculate everything in the $\sigma_z$-eigenbasis, $|\pm 1/2 \rangle$. For the single spins you have the eigenvectors for the spin components in $\theta$-direction
$$|1/2_{\theta} \rangle=\cos(\theta/2) |1/2 \rangle + \sin(\theta/2)|-1/2 \rangle$$
$$|-1/2_{\theta} \rangle=-\sin(\theta/2) |1/2 \rangle +\cos(\theta/2) |-1/2 \rangle.$$
With
$$|\Psi \rangle=\alpha |1/2,1/2 \rangle + \beta|-1/2,-1/2 \rangle$$
we need the following scalar products
$$\langle 1/2_{\theta},1/2|\Psi \rangle=\alpha \cos(\theta/2), \quad \langle 1/2_{\theta},-1/2|\Psi \rangle = \beta \sin(\theta/2), \quad \langle -1/2_{\theta},1/2|\Psi \rangle=-\alpha \sin(\theta/2), \quad \langle -1/2_{\theta},-1/2|\Psi \rangle = \beta \cos(\theta/2).$$
The probabilities are the moduli squared:
$$P(1/2_{\theta},1/2)=|\alpha|^2 \cos^2(\theta/2), \quad P(1/2_{\theta},-1/2)=|\beta|^2 \sin^2(\theta/2), \quad P(-1/2_{\theta},1/2)=|\alpha|^2 \sin^2(\theta/2), \quad P(-1/2_{\theta},-1/2)=|\beta^2|\cos^2(\theta/2).$$

 

[PeroK]

So if I understand correctly, no matter who measures in what direction first (in any of the 4 General Bell States, not just this one), the end result of the measurement of Bob (and the probabilities of $\downarrow_B or \uparrow_B$ only depends on the direction Bob measures in.

Yes. You can always think of it as a simultaneous measurement. Bob's results are the same statistically whatever Alice does; and vice versa.

 

[PeroK]

PS another way to see this is to express the state as a tensor product, using different bases for the constituent particles:
$$\psi = a \uparrow \uparrow + b \downarrow \downarrow = a \uparrow \otimes (\cos \frac \theta 2 \uparrow_{\theta} - \sin \frac \theta 2 \downarrow_{\theta}) + b \downarrow \otimes (\sin \frac \theta 2 \uparrow_{\theta} + \cos \frac \theta 2 \downarrow_{\theta})$$ $$ = a \cos \frac \theta 2 (\uparrow \otimes \uparrow_{\theta}) - a \sin \frac \theta 2 (\uparrow \otimes \downarrow_{\theta}) + b\sin \frac \theta 2 ( \downarrow \otimes \uparrow_{\theta}) + b \cos \frac \theta 2 (\downarrow \otimes \downarrow_{\theta})$$
And then you can just read off the amplitudes.

 

[vanhees71]

Remark: Of course my notation $|a,b \rangle \equiv |a \rangle \oplus |b \rangle$. It's just laziness of physicists' notation not to write out the tensor (Kronecker) product symbols ;-)).