Measurement

Suppose that $A$ 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 $A$ 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

$$\psi = \sum_i c_i \psi_i,$$ (266)

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

$$\langle A\rangle= a,$$ (267)

and

$$\sigma_A^{ 2} = \langle A^2\rangle - \langle A\rangle = 0.$$ (268)

Now it is easily seen that

$$\langle A\rangle = \sum_i \vert c_i\vert^2 a_i,$$ (269)

$$\langle A^2\rangle = \sum_i \vert c_i\vert^2 a_i^{ 2}.$$ (270)

Thus, Eq. (268) gives

$$\sum_i a_i^{ 2} \vert c_i\vert^2 - \left(\sum_i a_i \vert c_i\vert^2\right)^2=0.$$ (271)

Furthermore, the normalization condition yields

$$\sum_i \vert c_i\vert^2 =1.$$ (272)

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

$$(a_1-a_2)^2 x (1-x) = 0.$$ (273)

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 $\vert c_i\vert^2$ 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 Eq. (266), then it is clear from Eq. (269), and the general definition of a mean, that the probability of a measurement of $A$ yielding the eigenvalue $a_i$ is simply $\vert c_i\vert^2$, where $c_i$ is the coefficient in front of the $i$th eigenstate in the expansion. Note, from Eq. (272), that these probabilities are properly normalized: i.e., 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 Sect. 4.8, 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

$$(A B-B A) \psi_i = (A B-B a_i) \psi_i = (A-a_i) B \psi_i = 0,$$ (274)

or

$$A (B \psi_i) = a_i (B \psi_i).$$ (275)

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

$$B \psi_i = b_i \psi_i,$$ (276)

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).