satisfying thenormalization condition
〈p′|p〉=δ(p−p′)For afree particle Hamiltonian, both momentum and parity commute withH. So we can make
simultaneous eigenfunctions.
[H,p] = 0
[H,P] = 0Wecannot make eigenfunctions of all three operatorssince
[P,p] 6 = 0.So we have thechoiceof theeikxstates which are eigenfunctions ofHand ofp, but contain positive
and negative parity components. or we have the sin(kx) and cos(kx) states which contain two
momenta but are eigenstates ofHand Parity. These are just different linear combinations of the
same solutions.
8.7 Derivations and Computations
8.7.1 Eigenfunctions of Hermitian Operators are Orthogonal
We wish to prove that eigenfunctions of Hermitian operators are orthogonal. In fact we will first do
thisexcept in the case of equal eigenvalues.
Assume we have aHermitian operatorAand two of its eigenfunctionssuch that
Aψ 1 =a 1 ψ 1
Aψ 2 =a 2 ψ 2.Now we compute〈ψ 2 |A|ψ 1 〉two ways.
〈ψ 2 |Aψ 1 〉 = a 1 〈ψ 2 |ψ 1 〉
〈ψ 2 |Aψ 1 〉 = 〈Aψ 2 |ψ 1 〉=a 2 〈ψ 2 |ψ 1 〉Remember theeigenvalues are realso there’s no conjugation needed.
Now wesubtract the two equations. The left hand sides are the same so they give zero.
0 = (a 2 −a 1 )〈ψ 2 |ψ 1 〉Ifa 16 =a 2 then
〈ψ 2 |ψ 1 〉= 0.