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:
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The probability of obtaining this outcome is
\(\langle\psi'|\psi'\rangle = |\langle q_i|\psi\rangle|^2\)
.
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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?
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First outcome:
\(+\hbar/2\)
.
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The partial projector is
\(|\!+\!z\rangle\langle+z| \otimes \hat{I}\)
.
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Applying the projection to
\(|\psi\rangle\)
yields
\(|\psi'\rangle = (1/\sqrt{2})\,|\!+\!z\rangle|\!-\!z\rangle\)
.
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The outcome probability is
\(\displaystyle P_+ = \langle \psi'|\psi'\rangle = \frac{1}{2}\)
.
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The post-collapse state is
\(\displaystyle \frac{1}{\sqrt{P_+}} |\psi'\rangle = |\!+\!z\rangle |\!-\!z\rangle\)
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The partial projector is
\(|\!+\!z\rangle\langle+z| \otimes \hat{I}\)
.
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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 partial projector is
\(|\!-\!z\rangle\langle-z| \otimes \hat{I}\)
.
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.