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Suppose that [tex]f(A)[/tex] is a function of a Hermitian operator [tex]A[/tex] with the property [tex]A|a'\rangle = a'|a'\rangle[/tex]. Evaluate [tex]\langle b''|f(A)|b'\rangle[/tex] when the transformation matrix from the [tex]a'[/tex] basis to the [tex]b'[/tex] basis is known.

The attempt at a solution

Here's what I have... I'm not sure if thelast stepis correct, but apart from that, I'm sure everything else is right. So,

[tex]\langle b''|A|b'\rangle = \sum_{a'} \sum_{a''} \langle b'' | a'' \rangle \langle a'' | A | a' \rangle \langle a' | b' \rangle[/tex]

[tex]= \sum_{a'} \sum_{a''} \delta(a'-a'') a' \langle b'' | a'' \rangle \langle a' | b' \rangle[/tex]

[tex]= \sum_{a'} a' \langle b'' | a' \rangle \langle a' | b' \rangle[/tex]

And then I said that the above was

[tex]= a' \langle b'' | b' \rangle = a' \delta(b' - b'') \Leftrightarrow \langle b'' | f(A) | b' \rangle = f(a') \delta(b' - b'')[/tex]

where [tex]a'[/tex] was the eigenvalue of corresponding eigenket to [tex]b'[/tex].

Is the last step correct? That was the only part I was a bit shaky on... there is an [tex]a'[/tex] index inside the summation...

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# Homework Help: Sakurai 1.27 - Transformation Operators

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