# Calculating the probability of one particle being in a certain state in two-particle system

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 $$P_a(\alpha)$$?

• Are you familiar with the density matrix formulation? – Dvij Mankad Feb 28 at 2:45
• A little bit, I still get confused. I have not read about partial traces, which is what a friend told me I should. – The Bosco Feb 28 at 2:51
• 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! – rnels12 Feb 28 at 11:19
• 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 – The Bosco Mar 1 at 3:54