Alice’s measurement

Alice

transmits

This happens with probability

Collapsed state

Bob applies ${\sigma }_x^{b_1}{\sigma }_z^{b_2}$

$|{\Phi }^{+}\rangle \to 00$

$b_2{b}_1=00$

${\Vert \frac{1}{2}{\sigma }_x^0{\sigma }_z^0|\psi \rangle \Vert }^2=\frac{1}{4}$

$|{\Phi }^{+}\rangle {\sigma }_x^0{\sigma }_z^0|\psi \rangle$

${\sigma }_x^0{\sigma }_z^0|\psi \rangle =|\psi \rangle$

$|{\Psi }^{+}\rangle \to 01$

$b_2{b}_1=01$

${\Vert \frac{1}{2}{\sigma }_x^1{\sigma }_z^0|\psi \rangle \Vert }^2=\frac{1}{4}$

$|{\Psi }^{+}\rangle {\sigma }_x^1{\sigma }_z^0|\psi \rangle$

${\sigma }_x^1{\sigma }_z^0|\psi \rangle ={\sigma }_x|\psi \rangle$

$|{\Phi }^{-}\rangle \to 10$

$b_2{b}_1=10$

${\Vert \frac{1}{2}{\sigma }_x^0{\sigma }_z^1|\psi \rangle \Vert }^2=\frac{1}{4}$

$|{\Phi }^{-}\rangle {\sigma }_x^0{\sigma }_z^1|\psi \rangle$

${\sigma }_x^0{\sigma }_z^1|\psi \rangle ={\sigma }_z|\psi \rangle$

$|{\Psi }^{-}\rangle \to 11$

$b_2{b}_1=11$

${\Vert \frac{1}{2}{\sigma }_x^1{\sigma }_z^1|\psi \rangle \Vert }^2=\frac{1}{4}$

$|{\Psi }^{-}\rangle {\sigma }_x^1{\sigma }_z^1|\psi \rangle$

${\sigma }_x^1{\sigma }_z^1|\psi \rangle ={\sigma }_x{\sigma }_z|\psi \rangle$