4.10: Measurements

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)

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