Compatible Observables

Suppose that we wish to simultaneously measure two observables, $\xi$ and $\eta$ , of a microscopic system. Let us assume that we possess an apparatus that is capable of measuring $\xi$ , and another that can measure $\eta$ . For instance, the two observables in question might be the projection in the $x$ - and $z$ -directions of the spin angular momentum of a spin one-half particle. These could be measured using appropriate Stern-Gerlach apparatuses [95]. Suppose that we make a measurement of $\xi$ , and the system is consequently thrown into one of the eigenstates of $\xi$ , $\vert\xi'\rangle$ , with eigenvalue $\xi'$ . What happens if we now make a measurement of $\eta$ ? Well, suppose that the eigenstate $\vert\xi'\rangle$ is also an eigenstate of $\eta$ , with eigenvalue $\eta'$ . In this case, a measurement of $\eta$ will definitely give the result $\eta'$ . A second measurement of $\xi$ will definitely give the result $\xi'$ , and so on. In this sense, we can say that the observables $\xi$ and $\eta$ simultaneously have the values $\xi'$ and $\eta'$ , respectively. Clearly, if all eigenstates of $\xi$ are also eigenstates of $\eta$ then it is always possible to make a simultaneous measurement of $\xi$ and $\eta$ . Such observables are termed compatible.

Suppose, however, that the eigenstates of $\xi$ are not eigenstates of $\eta$ . Is it still possible to measure both observables simultaneously? Let us again make an observation of $\xi$ that throws the system into an eigenstate $\vert\xi'\rangle$ , with eigenvalue $\xi'$ . We can now make a second observation to determine $\eta$ . This will throw the system into one of the (many) eigenstates of $\eta$ that depend on $\vert\xi'\rangle$ . In principle, each of these eigenstates is associated with a different result of the measurement. Suppose that the system is thrown into an eigenstate $\vert\eta'\rangle$ , with the eigenvalue $\eta'$ . Another measurement of $\xi$ will throw the system into one of the (many) eigenstates of $\xi$ that depend on $\vert\eta'\rangle$ . Each eigenstate is again associated with a different possible result of the measurement. It is clear that if the observables $\xi$ and $\eta$ do not possess simultaneous eigenstates then if the value of $\xi$ is known (i.e., the system is in an eigenstate of $\xi$ ) then the value of $\eta$ is uncertain (i.e., the system is not in an eigenstate of $\eta$ ), and vice versa. We say that the two observables are incompatible.

We have seen that the condition for two observables $\xi$ and $\eta$ to be simultaneously measurable is that they should possess simultaneous eigenstates (i.e., every eigenstate of $\xi$ should also be an eigenstate of $\eta$ ). Suppose that this is the case. Let a general eigenstate of $\xi$ , with eigenvalue $\xi'$ , also be an eigenstate of $\eta$ , with eigenvalue $\eta'$ . It is convenient to denote this simultaneous eigenstate $\vert\xi' \eta'\rangle$ . We have

$$\xi\,\vert\xi' \eta'\rangle = \xi' \,\vert\xi' \eta'\rangle,$$ (1.66)

$$\eta\,\vert\xi' \eta'\rangle = \eta'\,\vert\xi' \eta'\rangle.$$ (1.67)

We can left-multiply the first equation by $\eta$ , and the second equation by $\xi$ , and then take the difference. The result is

$$(\xi\,\eta -\eta\,\xi)\, \vert\xi' \eta'\rangle = \vert\rangle$$ (1.68)

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 previous equation implies that

$$(\xi\,\eta -\eta\,\xi)\, \vert A\rangle = \vert\rangle,$$ (1.69)

where $\vert A\rangle$ is a general ket. The only way that this can be true is if

$$\xi\,\eta =\eta\,\xi.$$ (1.70)

We conclude that the condition for two observables $\xi$ and $\eta$ to be simultaneously measurable is that they should commute.