(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Suppose that |i> and |j> are eigenkets of some Hermitian operator. Under what condition can we conclude that |i> + |j> is also an eigenket of A? Justify your answer.

2. Relevant equations

It seems that all that is needed is for "A" to be a linear operator and for |i> and |j> to have the same eigenvalue. Justification:

A(u + v) = Au + Av (that's A's linearity at work)

Au = u[0]*u

Av = v[0]*v

If we have u[0] = v[0] (matching eigenvalues), then:

A(u + v) = u[0]*(u + v) = v[0]*(u + v)

...and thus:

A(u + v) = [some common scalar]*(u + v)

...meaning (u + v) is eigenstate of linear operator "A".

My question: qed?

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# Homework Help: Sakurai, p. 59, Pr 1.6 - critique the proof

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