Suppose that we wish to simultaneously measure two observables, and , of a microscopic system? Let us assume that we possess an apparatus that is capable of measuring , and another that can measure . For instance, the two observables in question might be the projection in the - and -directions of the spin angular momentum of a spin one-half particle. These could be measured using appropriate Stern-Gerlach apparatuses (see Sakurai, Section 1.1). Suppose that we make a measurement of , and the system is consequently thrown into one of the eigenstates of , , with eigenvalue . What happens if we now make a measurement of ? Well, suppose that the eigenstate is also an eigenstate of , with eigenvalue . In this case, a measurement of will definitely give the result . A second measurement of will definitely give the result , and so on. In this sense, we can say that the observables and simultaneously have the values and , respectively. Clearly, if all eigenstates of are also eigenstates of then it is always possible to make a simultaneous measurement of and . Such observables are termed compatible.
Suppose, however, that the eigenstates of are not eigenstates of . Is it still possible to measure both observables simultaneously? Let us again make an observation of which throws the system into an eigenstate , with eigenvalue . We can now make a second observation to determine . This will throw the system into one of the (many) eigenstates of which depend on . In principle, each of these eigenstates is associated with a different result of the measurement. Suppose that the system is thrown into an eigenstate , with the eigenvalue . Another measurement of will throw the system into one of the (many) eigenstates of which depend on . Each eigenstate is again associated with a different possible result of the measurement. It is clear that if the observables and do not possess simultaneous eigenstates then if the value of is known (i.e., the system is in an eigenstate of ) then the value of is uncertain (i.e., the system is not in an eigenstate of ), and vice versa. We say that the two observables are incompatible.
We have seen that the condition for two observables and to be simultaneously measurable is that they should possess simultaneous eigenstates (i.e., every eigenstate of should also be an eigenstate of ). Suppose that this is the case. Let a general eigenstate of , with eigenvalue , also be an eigenstate of , with eigenvalue . It is convenient to denote this simultaneous eigenstate . We have
We can left-multiply the first equation by , and the second equation by , and then take the difference. The result is
for each simultaneous eigenstate. Recall that the eigenstates of an observable must form a complete set. It follows that the simultaneous eigenstates of two observables must also form a complete set. Thus, the above equation implies that
where is a general ket. The only way that this can be true is if
Thus, the condition for two observables and to be simultaneously measurable is that they should commute.
- Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)