# Density matrix after measuring Bell state in CHSH game

In these notes, the author says the following about the CHSH game

Does Alice and Bob’s ability to succeed more than 75% of the time mean that they are communicating? Well, we know it’s not possible for either to send a signal to the other, by the No-Communication Theorem. But how can we reconcile that with their success in the CHSH game? One way to understand what’s going on, is to work out Alice and Bob’s density matrices explicitly. Bob’s initial density matrix is $$\begin{bmatrix}\frac{1}{2}&0\\0&\frac{1}{2}\end{bmatrix}$$ and after Alice measures it’s still $$\begin{bmatrix}\frac{1}{2}&0\\0&\frac{1}{2}\end{bmatrix}.$$

Before measurement, we have the Bell state $$\frac{1}{\sqrt{2}}(|00⟩+|11⟩)$$. Bob's bit (the second one) has the density matrix

$$\sum_1^np_i|ψ_i⟩⟨ψ_i|=\frac{1}{2}|0⟩⟨0|+\frac{1}{2}|1⟩⟨1|=\frac{1}{2}\begin{bmatrix}1&0\\0&0\end{bmatrix}+\frac{1}{2}\begin{bmatrix}0&0\\0&1\end{bmatrix}=\begin{bmatrix}\frac{1}{2}&0\\0&\frac{1}{2}\end{bmatrix}$$

(definition of density matrix from here).

However, after measurement, I don't understand how the density matrix is still the maximally mixed state. Once Alice measures her bit, shouldn't Bob's state be a pure state? For example, if Alice measures in the $$\{|0\rangle,|1\rangle\}$$ basis and gets $$|0\rangle$$, Bob's bit is $$|0\rangle$$, which is a pure state. How is this, after measurement, the maximally mixed state?

• My interpretation is that after measurement, there still hasn't been any communication between Alice and Bob. Thus, Bob's information about his bit is the same: 50% chance of being in either state. Jan 9 at 13:54

Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $$|\psi\rangle$$ with probability $$\frac12$$ and $$|\psi^\perp\rangle$$ with probability $$\frac12$$ for some orthonormal basis $$|\psi\rangle$$, $$|\psi^\perp\rangle$$. Therefore, his state is a mixture
$$\rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\frac12|\overline{\psi^\perp}\rangle\langle\overline{\psi^\perp}| = \frac{I}{2}.\tag1$$