4.10: Measurements
( \newcommand{\kernel}{\mathrm{null}\,}\)
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, ψ, is such that a measurement of
is bound to yield a certain result,
? Well, expressing ψ as a linear combination of the eigenstates of
, we have
ψ=∑iciψi
where ψi is an eigenstate of corresponding to the eigenvalue ai. . If a measurement of
is bound to yield the result
then
⟨A⟩=a
and
σ2A=⟨A2⟩−⟨A⟩=0
Now it is easily seen that
⟨A⟩=∑i|ci|2ai⟨A2⟩=∑i|ci|2a2i
Thus, Eq. (268) gives
∑ia2i|ci|2−(∑iai|ci|2)2=0
Furthermore, the normalization condition yields
∑i|ci|2=1
For instance, suppose that there are only two eigenstates. The above two equations then reduce to |c1|2=x, and |c2|2=1−x where 0≤x≤1, and
(a1−a2)2x(1−x)=0
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 is one in which one of the |ci|2 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 ai is simply |ci|2, where ci 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 resulting in any possible answer is unity. Finally, if a measurement of
results in the eigenvalue ai then immediately after the measurement the system will be left in the eigenstate corresponding to ai.
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 AB≠BA) 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 ψi and ai are the normalized eigenstates and eigenvalues of
, respectively. It follows that
(AB−BA)ψi=(AB−Bai)ψi=(A−ai)Bψi=0
or
A(Bψi)=ai(Bψi)
Thus, Bψi is an eigenstate of corresponding to the eigenvalue ai (though not necessarily a normalized one). In other words, Bψi∝ψi, or
Bψi=biψi
where bi is a constant of proportionality. Hence, ψi 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)