4.10: Measurements
- Page ID
- 1167
Suppose that is an Hermitian operator corresponding to some dynamical variable. By analogy with the discussion in Sect. 3.16, we expect that if a measurement of yields the result then the act of measurement will cause the wavefunction to collapse to a state in which a measurement of is bound to give the result . What sort of wavefunction, \(\begin{equation}\psi\end{equation}\), is such that a measurement of is bound to yield a certain result, ? Well, expressing \(\begin{equation}\psi\end{equation}\) as a linear combination of the eigenstates of , we have
\begin{equation}\psi=\sum_{i} c_{i} \psi_{i}\end{equation}
where \(\begin{equation}\psi_{i}\end{equation}\) is an eigenstate of corresponding to the eigenvalue \(\begin{equation}a_{i}\end{equation}\). . If a measurement of is bound to yield the result then
\begin{equation}\langle A\rangle=a\end{equation}
and
\begin{equation}\sigma_{A}^{2}=\left\langle A^{2}\right\rangle-\langle A\rangle=0\end{equation}
Now it is easily seen that
\begin{equation}\begin{aligned}
\langle A\rangle &=\sum_{i}\left|c_{i}\right|^{2} a_{i} \\
\left\langle A^{2}\right\rangle &=\sum_{i}\left|c_{i}\right|^{2} a_{i}^{2}
\end{aligned}\end{equation}
Thus, Eq. (268) gives
\begin{equation}\sum_{i} a_{i}^{2}\left|c_{i}\right|^{2}-\left(\sum_{i} a_{i}\left|c_{i}\right|^{2}\right)^{2}=0\end{equation}
Furthermore, the normalization condition yields
\begin{equation}\sum_{i}\left|c_{i}\right|^{2}=1\end{equation}
For instance, suppose that there are only two eigenstates. The above two equations then reduce to \(\begin{equation}\left|c_{1}\right|^{2}=x, \text { and }\left|c_{2}\right|^{2}=1-x\end{equation}\) where \(\begin{equation}0 \leq x \leq 1\end{equation}\), and
\begin{equation}\left(a_{1}-a_{2}\right)^{2} x(1-x)=0\end{equation}
The only solutions are \(\begin{equation}x=0 \text { and } x=1\end{equation}\). This result can easily be generalized to the case where there are more than two eigenstates. It follows that a state associated with a definite value of is one in which one of the \(\begin{equation}\left|c_{i}\right|^{2}\end{equation}\) is unity, and all of the others are zero. In other
words, the only states associated with definite values of are the eigenstates of . It immediately follows that the result of a measurement of must be one of the eigenvalues of . Moreover, if a general wavefunction is expanded as a linear combination of the eigenstates of , like in Eq. (266), then it is clear from Eq. (269), and the general definition of a mean, that the probability of a measurement of yielding the eigenvalue \(\begin{equation}a_{i} \text { is simply }\left|c_{i}\right|^{2}, \text { where } c_{i}\end{equation}\) is the coefficient in front of the \(\begin{equation}i \text { th }\end{equation}\) eigenstate in the
expansion. Note, from Eq. (272), that these probabilities are properly normalized: i.e., the probability of a measurement of resulting in any possible answer is unity. Finally, if a measurement of results in the eigenvalue \(\begin{equation}a_{i}\end{equation}\) then immediately after the measurement the system will be left in the eigenstate corresponding to \(\begin{equation}a_{i}\end{equation}\).
Consider two physical dynamical variables represented by the two Hermitian operators and . Under what circumstances is it possible to simultaneously measure these two variables (exactly)? Well, the possible results of measurements of and are the eigenvalues of and , respectively. Thus, to simultaneously measure and (exactly) there must exist states which are simultaneous eigenstates of and . In fact, in order for and to be simultaneously measurable under all circumstances, we need all of the eigenstates of to also be eigenstates of , and vice versa, so that all states associated with unique values of are also associated with unique values of , and vice versa.
Now, we have already seen, in Sect. 4.8, that if and do not commute (i.e., if \(\begin{equation}A B \neq B A\end{equation}\)) then they cannot be simultaneously measured. This suggests that the condition for simultaneous measurement is that and should commute. Suppose that this is the case, and that the \(\begin{equation}\psi_{i}\end{equation}\) and \(\begin{equation}a_{i}\end{equation}\) are the normalized eigenstates and eigenvalues of , respectively. It follows that
\begin{equation}(A B-B A) \psi_{i}=\left(A B-B a_{i}\right) \psi_{i}=\left(A-a_{i}\right) B \psi_{i}=0\end{equation}
or
\begin{equation}A\left(B \psi_{i}\right)=a_{i}\left(B \psi_{i}\right)\end{equation}
Thus, \(\begin{equation}B \psi_{i}\end{equation}\) is an eigenstate of corresponding to the eigenvalue \(\begin{equation}a_{i}\end{equation}\) (though not necessarily a normalized one). In other words, \(\begin{equation}B \psi_{i} \propto \psi_{i}\end{equation}\), or
\begin{equation}B \psi_{i}=b_{i} \psi_{i}\end{equation}
where \(\begin{equation}b_{i}\end{equation}\) is a constant of proportionality. Hence, \(\begin{equation}\psi_{i}\end{equation}\) is an eigenstate of , and, thus, a simultaneous eigenstate of and . We conclude that if and commute then they possess simultaneous eigenstates, and are thus simultaneously measurable (exactly).
Contributors
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)
\( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)