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1.13: 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 (see Sakurai, Section 1.1). 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$ which 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$ which 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$ which 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

 

$\displaystyle \xi\,\vert\xi' \eta'\rangle$ $\displaystyle = \xi' \,\vert\xi' \eta'\rangle,$ (63)
$\displaystyle \eta\,\vert\xi' \eta'\rangle$ $\displaystyle = \eta'\,\vert\xi' \eta'\rangle.$ (64)

 

 

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

 

$\displaystyle (\xi\,\eta -\eta\,\xi)\, \vert\xi' \eta'\rangle = \vert\rangle$ (65)

 

 

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

 

$\displaystyle (\xi\,\eta -\eta\,\xi)\, \vert A\rangle = \vert\rangle,$ (66)

 

 

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

 

$\displaystyle \xi\,\eta =\eta\,\xi.$ (67)

 

 

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

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