QuantumPhysics.dvi

(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.

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