   Next: Continuous Eigenvalues Up: Fundamentals of Quantum Mechanics Previous: Eigenstates and Eigenvalues

# 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   (269)   (270)

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 2010-07-20