# Measuring multi-qubit state with respect to an arbitrary basis

Suppose Alice and Bob each hold qubits; they have a joint state

$$|\psi\rangle = \frac{1}{\sqrt 3}\big(|00\rangle + |01\rangle + |11\rangle\big).$$

Alice measures the first qubit in some basis (say $$|+\rangle, |-\rangle$$); I want to see what Bob's qubit looks like post-measurement. The result should be a a probabilistic ensemble (mixed state). How do we do this in practice?

I thought about it this way: measurement in a basis is equivalent to applying an appropriate unitary matrix before measurement. So for $$|+\rangle, |-\rangle$$, we'd apply

\begin{align*} (H \otimes I)|\psi\rangle &= \frac{1}{\sqrt 3}\bigg(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)|0\rangle + \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)|1\rangle + \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)|1\rangle\bigg) \\ &= \frac{1}{\sqrt 6}\bigg(|0\rangle|0\rangle + |1\rangle|0\rangle + 2|0\rangle|1\rangle\bigg) \end{align*}

But when we measure the first qubit, we get "$$|0\rangle$$" (really $$|+\rangle$$) with probability $$\frac 16 + \frac{2^2}{6} = \frac 56$$, leaving the state as $$\frac{\frac{1}{\sqrt 6}\big(|0\rangle|0\rangle + 2|0\rangle|1\rangle\big)}{\sqrt{\frac 16 +\frac 46}}$$ (is this correct to say?). Likewise we could get "$$|1\rangle$$" (really $$|-\rangle$$) with probability $$\frac 16$$, leaving the state as $$\frac{\frac{1}{\sqrt 6}|1\rangle|0\rangle}{\sqrt{\frac 16}}$$.

This would mean that post-measurement, Bob's qubit has a state described by the density matrix

$$\rho = \frac 56\frac{|0\rangle + 2|1\rangle}{\sqrt 5}\frac{\langle0| + 2 \langle 1|}{\sqrt 5} + \frac 16|0\rangle \langle 0|$$

Does this make sense? In particular, does the $$H \otimes I$$ operation make sense?

I ask because when I take $$\operatorname{tr}_A |\psi\rangle \langle \psi|$$ I get $$\rho_B = \left( \begin{matrix} \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} \end{matrix} \right)$$ which seems to be a different density matrix altogether.

The measurement operation you have is correct. The state that Bob holds after the measurement that you denote as $$\rho$$ is
$$\rho=\frac{5}{6} \frac{|0\rangle+2|1\rangle}{\sqrt{5}} \frac{\langle 0|+2\langle 1|}{\sqrt{5}}+\frac{1}{6}|0\rangle\langle 0| = \frac{1}{3} \vert 0 \rangle\langle 0\vert + \frac{1}{3} \vert 0 \rangle\langle 1\vert+ \frac{1}{3} \vert 1 \rangle\langle 0\vert+ \frac{2}{3} \vert 1 \rangle\langle 1\vert.$$
In the $$\{\vert 0\rangle, \vert 1\rangle\}$$ basis, the matrix representation is exactly what you denote as $$\rho_B$$.