3.9: Measurement

Suppose that \(A\) is an Hermitian operator corresponding to some dynamical variable. By analogy with the discussion in Section [scoll], we expect that if a measurement of \(A\) yields the result \(a\) then the act of measurement will cause the wavefunction to collapse to a state in which a measurement of \(A\) is bound to give the result \(a\). What sort of wavefunction, \(\psi\), is such that a measurement of \(A\) is bound to yield a certain result, \(a\)? Well, expressing \(\psi\) as a linear combination of the eigenstates of \(A\), we have

\[ \psi = \sum_i c_i\,\psi_i, \label{e4.128} \]

where \(\psi_i\) is an eigenstate of \(A\) corresponding to the eigenvalue \(a_i\). If a measurement of \(A\) is bound to yield the result \(a\) then

\[\langle A\rangle= a, \nonumber \]

and

\[ \sigma_A^2 = \langle A^2\rangle - \langle A\rangle = 0. \label{e4.130} \]

Now, it is easily seen that

\[\begin{align} \label{e4.131} \langle A\rangle &= \sum_i |c_i|^2\,a_i,\\[4pt] \langle A^2\rangle &= \sum_i |c_i|^2\,a_i^2.\end{align} \]

Thus, Equation \ref{e4.130} gives

\[\sum_i a_i^2\,|c_i|^2 - \left(\sum_i a_i\,|c_i|^2\right)^2=0. \nonumber \]

Furthermore, the normalization condition yields

\[ \sum_i |c_i|^2 = 1. \label{e4.134} \]

For instance, suppose that there are only two eigenstates. The previous two equations then reduce to \(|c_1|^2=x\), and \(|c_2|^2=1-x\), where \(0\leq x\leq 1\), and

\[ (a_1-a_2)^2\,x\,(1-x) = 0. \label{e4.126} \]

The only solutions are \(x=0\) and \(x=1\). 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 \(A\) is one in which one of the \(|c_i|^2\) is unity, and all of the others are zero. In other words, the only states associated with definite values of \(A\) are the eigenstates of \(A\). It immediately follows that the result of a measurement of \(A\) must be one of the eigenvalues of \(A\). Moreover, if a general wavefunction is expanded as a linear combination of the eigenstates of \(A\), like in Equation \ref{e4.128}, then it is clear from Equation \ref{e4.131}, and the general definition of a mean, that the probability of a measurement of \(A\) yielding the eigenvalue \(a_i\) is simply \(|c_i|^2\), where \(c_i\) is the coefficient in front of the \(i\)th eigenstate in the expansion. Note, from Equation \ref{e4.134}, that these probabilities are properly normalized: that is, the probability of a measurement of \(A\) resulting in any possible answer is unity. Finally, if a measurement of \(A\) results in the eigenvalue \(a_i\) then immediately after the measurement the system will be left in the eigenstate corresponding to \(a_i\).

Consider two physical dynamical variables represented by the two Hermitian operators \(A\) and \(B\). Under what circumstances is it possible to simultaneously measure these two variables (exactly)? Well, the possible results of measurements of \(A\) and \(B\) are the eigenvalues of \(A\) and \(B\), respectively. Thus, to simultaneously measure \(A\) and \(B\) (exactly) there must exist states which are simultaneous eigenstates of \(A\) and \(B\). In fact, in order for \(A\) and \(B\) to be simultaneously measurable under all circumstances, we need all of the eigenstates of \(A\) to also be eigenstates of \(B\), and vice versa, so that all states associated with unique values of \(A\) are also associated with unique values of \(B\), and vice versa.

Now, we have already seen, in Section 1.8, that if \(A\) and \(B\) do not commute (i.e., if \(A\,B\neq B\,A\)) then they cannot be simultaneously measured. This suggests that the condition for simultaneous measurement is that \(A\) and \(B\) should commute. Suppose that this is the case, and that the \(\psi_i\) and \(a_i\) are the normalized eigenstates and eigenvalues of \(A\), respectively. It follows that

\[(A\,B-B\,A)\,\psi_i = (A\,B-B\,a_i)\,\psi_i = (A-a_i)\,B\,\psi_i = 0, \nonumber \]

or

\[A\,(B\,\psi_i) = a_i\,(B\,\psi_i). \nonumber \]

Thus, \(B\,\psi_i\) is an eigenstate of \(A\) corresponding to the eigenvalue \(a_i\) (though not necessarily a normalized one). In other words, \(B\,\psi_i\propto \psi_i\), or

\[B\,\psi_i = b_i\,\psi_i, \nonumber \]

where \(b_i\) is a constant of proportionality. Hence, \(\psi_i\) is an eigenstate of \(B\), and, thus, a simultaneous eigenstate of \(A\) and \(B\). We conclude that if \(A\) and \(B\) commute then they possess simultaneous eigenstates, and are thus simultaneously measurable (exactly).