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I am confused about how to calculate the probabilities of getting a certain result when measuring a Pauli observable on a Bell state. When you measure an observable the state is projected onto an eigenstate of the observable.

For example, given $|\psi\rangle = (|{+-}\rangle + |{-+}\rangle)/\sqrt{2}$ as the state and $\sigma_x = \begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$ as the observable, if the probability is $1/2$, how to calculate actually it? What is the state after measurement?

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General procedure

First, write down the eigendecomposition of your observable $A$

$$ A = \sum_{m} \lambda_m P_m $$

where $\lambda_m$ is the $m$th eigenvalue of $A$ and $P_m$ is the associated projector. Next, compute the projections of the input state $|\psi\rangle$ on the eigenspaces of $A$

$$ |\psi_m\rangle = P_m|\psi\rangle. $$

Note that $|\psi_m\rangle$ is not normalized. The square of its norm is the probability of the $m$th measurement outcome and normalized $|\psi_m\rangle$ is the post-measurement state.

Explicitly, the probability of obtaining the $m$th outcome is

$$ p(m) = \|\,|\psi_m\rangle\,\|^2 = \langle\psi_m|\psi_m\rangle = \langle\psi|P_m^2|\psi\rangle = \langle\psi|P_m|\psi\rangle \tag1 $$

which is known as the Born rule and the post-measurement state is

$$ |\psi_m'\rangle = \frac{|\psi_m\rangle}{\|\,|\psi_m\rangle\,\|} = \frac{P_m|\psi\rangle}{\sqrt{p(m)}}.\tag2 $$

Remark on measuring composite states

An important technicality is that when $|\psi\rangle$ is a composite state and we measure a subsystem, then $P_m$ acts only on the subsystem being measured. Thus, we can extend $P_m$ to a tensor product $P_m\otimes I$ to obtain an equivalent operator acting on the composite system.

Alternatively, we can use the fact that Dirac notation makes it easy to keep track of which bras and kets are acted on by the projector. This can be done implicitly by relying on the order of bra and ket labels or explicitly by adding subscripts to states and operators. Below, we do this implicitly, but add color for readability.

Example

Let us apply the above to the measurement of $A = \sigma_x$ on the first qubit of $|\psi\rangle = (|{+-}\rangle+|{-+}\rangle)/\sqrt{2}$. Eigendecomposition of $\sigma_x$ is

$$ A = \sigma_x = |+\rangle\langle+| - |-\rangle\langle-|. $$

Substituting into $(1)$, we get

$$ \begin{align} p(+1) &= \langle\psi\color{red}{|+\rangle\langle+|}\psi\rangle \\ &= \frac{1}{2}(\langle{+-}|+\langle{-+}|)\color{red}{|+\rangle\langle+|}(|{+-}\rangle+|{-+}\rangle) \\ &= \frac{1}{2}(\langle+\color{red}{|+\rangle\langle+|}+\rangle \langle-|-\rangle \\ &+ \langle+\color{red}{|+\rangle\langle+|}-\rangle \langle-|+\rangle \\ &+ \langle-\color{red}{|+\rangle\langle+|}+\rangle \langle+|-\rangle \\ &+ \langle-\color{red}{|+\rangle\langle+|}-\rangle \langle+|+\rangle) \\ &= \frac{1}{2}(1 + 0 + 0 + 0) \\ &= \frac{1}{2} \end{align} $$

where we the projector $|+\rangle\langle+|$ acts on the first qubit. From $(2)$, the post-measurement state is

$$ \begin{align} |\psi_+'\rangle &= \frac{|+\rangle\langle+|\psi\rangle}{\sqrt{p(m)}} \\ &= \frac{|+\rangle\langle+|\frac{|{+-}\rangle+|{-+}\rangle}{\sqrt{2}}}{\sqrt{\frac{1}{2}}} \\ &= |{+}\rangle\langle+|{+-}\rangle + |+\rangle\langle+|{-+}\rangle \\ &= |{+-}\rangle + 0 \\ &= |{+-}\rangle \end{align} $$

where as before the projector $|+\rangle\langle+|$ acts on the first qubit in the two-qubit kets.

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