# How do I measure a single qubit in a two-qubit state?

Let us suppose that I have the state $$\begin{equation} \frac{1}{\sqrt2}(\alpha|0\rangle|+\rangle+\beta|1\rangle|-\rangle) \end{equation}$$ and I choose to measure the first qubit in the basis $$\{(1/\sqrt{2})(|0\rangle\pm e^{i\phi}|1\rangle)\}$$. How can I perform this measurement and how can I find the related outcomes (i.e. the eigenvalues) provided only the basis in which I have to measure the system? Being kinda new to quantum mechanics I am really confused about the algebraic aspects of this problem, but I have an intuitive idea of what performing a measurement is.

Let $$P_{\pm}$$ be the projectors onto the two orthonormal basis states of the measurement. So, $$P_+=\frac{1}{2}(|0\rangle\langle 0|+|1\rangle\langle 1|+e^{-i\phi}|0\rangle\langle 1|+e^{i\phi}|1\rangle\langle 0|).$$ Also, let $$|\psi\rangle$$ be the state that you're measuring (note that this must be normalised. Yours might be depending on your constraints on $$\alpha$$ and $$\beta$$, but we wouldn't usually have a factor of $$1/\sqrt{2}$$ present).
Since you are measuring the first qubit of two, the actual projectors should be written as $$P_\pm\otimes I$$. Now, we get the + result with probability $$p_+=\langle\psi |P_+\otimes I|\psi\rangle,$$ and if the result is $$+$$, the state after measurement is $$(P_+\otimes I)|\psi\rangle/\sqrt{p_+}.$$ In this case, $$p_+=p_-=\frac12$$.
• Thank you, but I have one more question: how can I know that the eigenvalue is +1, in the case of $P_{+}$ (or -1 for the other projector)? I mean, which matrix do I have to diagonalize? – Hub One Jul 22 at 9:47
• It's not about diagonalising matrices (unless your measurement is specified as a Hermitian operator, which it isn't here). Instead, you're given a state $|\phi\rangle$ to project onto, so you construct the projector $P=|\phi\rangle\langle\phi|$. By construction this has one eigenvalue 1 and the rest 0. – DaftWullie Jul 22 at 9:49
• So it's just a convention the fact that in many textbooks state $|+\rangle$ has eigenvalue +1 and $|-\rangle$ -1? I thought that it was related to the fact that they are eigenstates of the $\sigma_x$ matrix with eigenvalues respectively $\pm 1$ and that there was a way to generalize that for states with $e^{i\phi}$ phase factor. – Hub One Jul 22 at 9:56
• It is exactly related to that, but it only makes sense if you specify the operator rather than the states. Similarly, you can take $\cos\phi\sigma_x+\sin\phi\sigma_y$. But it is just arbitrary. There's no reason I couldn't use $-\sigma_x$ instead of $\sigma_x$. I'd just end up with different labels, but they are just labels for the underlying measurements. – DaftWullie Jul 22 at 9:59