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.