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.


1 Answer 1


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 $\rho_B$.

The physical significance of what you have found is that no matter what Alice does on her side (in this case measurement), this cannot change Bob's reduced state. In other words, shared entanglement is not enough to allow Alice to communicate with Bob.


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