3.2: Partial Measurements

Let us recall how measurements work in single-particle quantum theory. Each observable $Q$ is described by some Hermitian operator $\hat{Q}$, which has an eigenbasis $\{|q_i\rangle\}$ such that

$$\hat{Q}|q_i\rangle = q_i |q_i\rangle.$$

For simplicity, let the eigenvalues $\{q_i\}$ be non-degenerate. Suppose a particle initially has quantum state $|\psi\rangle$. This can always be expanded in terms of the eigenbasis of $\hat{Q}$:

$$|\psi\rangle = \sum_i \psi_i\, |q_i\rangle, \;\;\mathrm{where}\;\;\,\textrm{and}\;\, \psi_i = \langle q_i|\psi\rangle.$$

The measurement postulate of quantum mechanics states that if we measure $Q$, then (i) the probability of obtaining the measurement outcome $q_i$ is $P_i = |\psi_i|^2$, the absolute square of the coefficient of $|q_i\rangle$ in the basis expansion; and (ii) upon obtaining this outcome, the system instantly “collapses” into state $|q_i\rangle$.

Mathematically, these two rules can be summarized using the projection operator

$$\hat{\Pi}(q_i) = |q_i\rangle\langle q_i|.$$

Applying this operator to $|\psi\rangle$ gives the non-normalized state vector

$$|\psi'\rangle = |q_i\rangle \langle q_i|\psi\rangle.$$

From this, we glean two pieces of information:

1. The probability of obtaining this outcome is $\langle\psi'|\psi'\rangle = |\langle q_i|\psi\rangle|^2$.

2. The post-collapse state is obtained by the re-normalization $|\psi'\rangle \rightarrow |q_i\rangle$.

For multi-particle systems, there is a new complication: what if a measurement is performed on just one particle?

Consider a system of two particles A and B, with two-particle Hilbert space $\mathscr{H}_A \otimes \mathscr{H}_B$. We perform a measurement on particle $A$, corresponding to a Hermitian operator $\hat{Q}_A$ that acts upon $\mathscr{H}_A$ and has eigenvectors $\{|\mu\rangle\; |\; \mu = 1, 2, \dots\}$ (i.e., the eigenvectors are enumerated by some index $\mu$). We can write any state $|\psi\rangle$ using the eigenbasis of $\hat{Q}_A$ for the $\mathscr{H}_A$ part, and an arbitrary basis $\{|\nu\rangle\}$ for the $\mathscr{H}_B$ part:

$$\begin{align} \begin{aligned} |\psi\rangle &= \sum_{\mu\nu} \psi_{\mu\nu}\, |\mu\rangle |\nu\rangle \\ &= \sum_\mu |\mu\rangle |\varphi_\mu \rangle, \;\;\;\mathrm{where}\;\;\; |\varphi_\mu\rangle\equiv \sum_\nu \psi_{\mu\nu}\,|\nu\rangle \;\in\; \mathscr{H}_B. \end{aligned} \label{psi_twop}\end{align}$$

Unlike the single-particle case, the “coefficient” of $|\mu_i\rangle$ in this basis expansion is not a complex number, but a vector in $\mathscr{H}_B$.

Proceeding by analogy, the probability of obtaining the outcome labelled by $\mu$ should be the “absolute square” of this “coefficient”, $\langle\varphi_\mu|\varphi_\mu\rangle$. Let us define the partial projector

$$\hat{\Pi}(\mu) \,=\, |\mu\rangle\langle \mu| \otimes \hat{I}.$$

The $A$ slot of this operator contains a projector, $|\mu\rangle\langle \mu|$, while the $B$ slot leaves the $\mathscr{H}_B$ part of the two-particle space unchanged. Applying the partial projector to the state given in Equation $\eqref{psi_twop}$ gives

$$|\psi'\rangle \,=\, \hat{\Pi}(\mu)\, |\psi\rangle \,=\, |\mu\rangle |\varphi_\mu\rangle.$$

Now we follow the same measurement rules as before. The outcome probability is

$$P_\mu = \langle\psi'|\psi'\rangle = \langle \mu|\mu\rangle\, \langle \varphi_\mu|\varphi_\mu\rangle = \sum_\nu |\psi_{\mu\nu}|^2.$$

The post-measurement collapsed state is obtained by the re-normalization

$$|\psi'\rangle \;\;\rightarrow\;\; \frac{1}{\sqrt{\sum_{\nu'} |\psi_{\mu\nu'}|^2}}\; \sum_{\nu} \psi_{\mu\nu} |\mu\rangle |\nu\rangle.$$

Example $\PageIndex{1}$

A system of two spin-$1/2$ particles is in the “singlet state”

$$|\psi\rangle = \frac{1}{\sqrt{2}} \Big(|\!+\!z\rangle|\!-\!z\rangle \,-\, |\!-\!z\rangle|\!+\!z\rangle\Big).$$

For each particle, $|\!+\!z\,\rangle$ and $|\!-\!z\,\rangle$ denote eigenstates of the operator $\hat{S}_z$, with eigenvalues $+\hbar/2$ and $-\hbar/2$ respectively. Suppose we measure $S_z$ on particle A. What are the probabilities of the possible outcomes, and the associated post-collapse states?

• First outcome: $+\hbar/2$.
  • The partial projector is $|\!+\!z\rangle\langle+z| \otimes \hat{I}$.

  • Applying the projection to $|\psi\rangle$ yields $|\psi'\rangle = (1/\sqrt{2})\,|\!+\!z\rangle|\!-\!z\rangle$.

  • The outcome probability is $\displaystyle P_+ = \langle \psi'|\psi'\rangle = \frac{1}{2}$.

  • The post-collapse state is $\displaystyle \frac{1}{\sqrt{P_+}} |\psi'\rangle = |\!+\!z\rangle |\!-\!z\rangle$

• Second outcome: $-\hbar/2$.
  • The partial projector is $|\!-\!z\rangle\langle-z| \otimes \hat{I}$.

  • Applying the projection to $|\psi\rangle$ yields $|\psi'\rangle = (1/\sqrt{2})\,|\!-\!z\rangle|\!+\!z\rangle$.

  • The outcome probability is $\displaystyle P_- = \langle \psi'|\psi'\rangle = \frac{1}{2}$.

  • The post-collapse state is $\displaystyle = \frac{1}{\sqrt{P_-}} |\psi'\rangle = |\!-\!z\rangle |\!+\!z\rangle$.

The two possible outcomes, $+\hbar/2$ and $-\hbar/2$, occur with equal probability. In either case, the two-particle state collapses so that $A$ is in the observed spin eigenstate, and $B$ has the opposite spin. After the collapse, the two-particle state is no longer entangled.