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Measurement
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, , is such that a measurement of is
bound to yield a certain result, ? Well, expressing as
a linear combination of the eigenstates of , we have

(266) 
where is an eigenstate of corresponding to the eigenvalue . If a measurement of is bound to yield the result then

(267) 
and

(268) 
Now it is easily seen that
Thus, Eq. (268) gives

(271) 
Furthermore, the normalization condition yields

(272) 
For instance, suppose that there are only two eigenstates. The above two
equations then reduce to , and , where ,
and

(273) 
The only solutions are and . 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 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
is simply , where is the coefficient in front of
the 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 then
immediately after the measurement the system will be left in the
eigenstate corresponding to .
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 ) 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 and are the normalized eigenstates and
eigenvalues of , respectively. It follows that

(274) 
or

(275) 
Thus, is an eigenstate of corresponding to
the eigenvalue (though not necessarily a normalized one). In other words,
, or

(276) 
where is a constant of proportionality.
Hence, 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).
Next: Continuous Eigenvalues
Up: Fundamentals of Quantum Mechanics
Previous: Eigenstates and Eigenvalues
Richard Fitzpatrick
20100720