370 Quantum mechanics
operators: A physical measurement must yield arealnumber, and Hermitian operators
have strictly real eigenvalues. In order to prove this, consider the eigenvalue problem
forAˆcast in Dirac notation:
Aˆ|ak〉=ak|ak〉, (9.2.13)
where|ak〉denotes an eigenvector ofAˆwith eigenvalueak. For a generalAˆ, the cor-
responding equation cast in Dirac bra form would be
〈ak|Aˆ†=〈ak|a∗k. (9.2.14)However, sinceAˆ†=Aˆ, this reduces to
〈ak|Aˆ=〈ak|a∗k. (9.2.15)Thus, if we multiply eqn. (9.2.13) by the bra vector〈ak|and eqn. (9.2.15) by the ket
vector|ak〉, we obtain the following two equations:
〈ak|Aˆ|ak〉=ak〈ak|ak〉〈ak|Aˆ|ak〉=a∗k〈ak|ak〉. (9.2.16)Consistency between these two relations requires thatak =a∗k, which proves that
the eigenvalues are real. Note that the operatorAˆcan be expressed in terms of its
eigenvalues and eigenvectors as
Aˆ=∑
kak|ak〉〈ak|. (9.2.17)The product|ak〉〈ak|is known as theouterortensorproduct between the ket and
bra vectors and could be used to project an arbitrary vector|φ〉along the direction of
|ak〉.
Another important property of Hermitian operators is that their eigenvectors form
a complete orthonormal set of vectors that span the Hilbert space. In order to prove
orthonormality of the eigenvectors, we multiply eqn. (9.2.13) by thebra vector〈aj|,
which gives
〈aj|Aˆ|ak〉=ak〈aj|ak〉. (9.2.18)
On the other hand, if we start with the bra equation (remembering thatAˆ=Aˆ†and
aj=a∗j)
〈aj|Aˆ=aj〈aj| (9.2.19)and multiply by the ket vector|ak〉, we obtain
〈aj|Aˆ|ak〉=aj〈aj|ak〉. (9.2.20)Subtracting eqn. (9.2.18) from (9.2.20) gives
0 = (ak−aj)〈aj|ak〉. (9.2.21)If the eigenvalues ofAˆare not degenerate, then fork 6 =j,ak 6 =aj, and it is clear
that〈aj|ak〉= 0. Ifk=j, then (aj−aj) = 0, and〈aj|aj〉can take on any value. This