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Let's say I have the two-particle state

$$|\psi\rangle=\frac{|H\rangle_a|H\rangle_b+|V\rangle_a|V\rangle_b}{\sqrt{2}}$$

where $H$ is horizontally polarized and $V$ is vertically polarized.

And I want to calculate the probability of finding particle $a$ with polarization $\alpha$.

I know that, if I want to calculate the probability of finding particle $a$ in $\alpha$ and $b$ in $\beta$, then I get

$$P_{a,b}(\alpha,\beta)=|(\langle\alpha|_a\otimes \langle\beta|_b)|\psi\rangle|^2 = \frac{1}{2} \cos^2(\beta-\alpha)$$

But how do I get the marginal $P_a(\alpha)$?

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  • $\begingroup$ Are you familiar with the density matrix formulation? $\endgroup$ – Feynmans Out for Grumpy Cat Feb 28 at 2:45
  • $\begingroup$ A little bit, I still get confused. I have not read about partial traces, which is what a friend told me I should. $\endgroup$ – The Bosco Feb 28 at 2:51
  • $\begingroup$ This is just a probability problem; to get $P_{a}(\alpha)$ you just sum $P_{a,b}(\alpha,\beta)$ over all possible states of $b$; i.e. the $\beta$ index! $\endgroup$ – rnels12 Feb 28 at 11:19
  • $\begingroup$ I assumed it would be like that, but I was reading that, but there's an infinite number of possible states. Also, since this is a maximally entangled state, it is supposed to be $1/2$ since you lose all the information about the individual particles, but not the ensemble $\endgroup$ – The Bosco Mar 1 at 3:54

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