QuantumPhysics.dvi
wang
(Wang)
#1
compatible observables, [px,py] = 0, while the spin componentsSxandSyare not, [Sx,Sy] =
i ̄hSz. The role played by compatibility of observables during measurementis expressed by
the following,
Theorem 3
(i) Two self-adjoint operatorsAandB, which satisfy[A,B] = 0, may be diagonalized in the
same basis, i.e. with common eigenspaces;
(ii) Two compatible observables may be observed simultaneously.
Proof (i) SinceAis self-adjoint, we may decompose it in a sum of projection operators,
A=
∑
i
aiPi
∑
i
Pi=IH (4.9)
where by assumption, the eigenvalues satisfyai 6 =aj wheni 6 =j. By multiplyingBto the
left and to the right by the identity operatorIH, we also have
B=
∑
i,j
PiBPj (4.10)
The commutator relation is easily computed in this basis,
0 = [A,B] =
∑
i,j
(ai−aj)PiBPj (4.11)
which implies thatPiBPj= 0 wheneveri 6 =j, and as a result
B=
∑
i
Bi Bi=PiBPi (4.12)
Inside the eigenspaceEi, the projection operatorPireduces to the identity operatorIi(which
manifestly commutes withBi). SinceBiis self-adjoint, it may be written as a direct sum of
projection operators weighted by its eigenvalues,
Bi=
∑
mi
bi,miPi,mi
∑
mi
Pi,mi=Ii (4.13)
wheremiis an index for matrixBiwhich labels all the distinct eigenvaluesbimiofBi. The
decomposition of both operators is then given by
A=
∑
i
∑
mi
aiPi,mi B=
∑
i
∑
mi
bi,miPi,mi (4.14)
(ii) It is manifest that an observation ofAwill produce an eigenstate ofA|φi〉with eigenvalue
ai, which with probability 1 will produce one of the eigenstates ofB in sectori, so that
simultaneous measurements ofAandBcan indeed be made.