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Chapter 7. The Eigenvalue Problem

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<strong>7.</strong> <strong>The</strong> <strong>Eigenvalue</strong> <strong>Problem</strong>, December 17, 2009 3<br />

This gives<br />

α =<br />

e iφ<br />

<br />

ψ | ψ<br />

(9)<br />

where φ is a real number, which cannot be determined by solving Eq. 8. <strong>The</strong><br />

ket<br />

e iφ<br />

|η = |ψ (10)<br />

ψ | ψ<br />

is normalized and therefore is a pure state of  corresponding to the eigenvalue<br />

λ. If the system is in the state |η and we measure A, we are certain<br />

that the result is λ.<br />

In practice, most computer programs that calculate eigenstates do not<br />

normalize them. If the computer produces |ψ as the eigenket corresponding<br />

to eigenvalue λ, youmustuseEq.10tofind the eigenstate |η that is a pure<br />

state. Since e iφ in Eq. 10 is an irrelevant phase factor, we can drop it (i.e. take<br />

φ = 0); the pure state corresponding to λ is<br />

|η =<br />

|ψ <br />

ψ | ψ<br />

(11)<br />

§ 4 Orthogonality. <strong>The</strong> pure states must also be orthogonal. Because an<br />

operator Â, representing an observable A, is always Hermitian, one can prove<br />

that eigenstates corresponding to different eigenvalues must be orthogonal.<br />

That is, if<br />

Â|η i = λ i |η i and Â|η j = λ j |η j (12)<br />

and<br />

then<br />

<strong>The</strong>proofiseasy.Wehave<br />

λ i = λ j (13)<br />

η i | η j =0 (14)<br />

Âη i | η j −η i | Âη j =0 (15)<br />

because  is Hermitian. Using Eqs. 12 in Eq. 15 gives<br />

(λ i − λ j )η i | η j =0

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