# Sakurai 1.27 - Transformation Operators

1. Dec 17, 2008

### Domnu

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

The attempt at a solution
Here's what I have... I'm not sure if the last step is correct, but apart from that, I'm sure everything else is right. So,

$$\langle b''|A|b'\rangle = \sum_{a'} \sum_{a''} \langle b'' | a'' \rangle \langle a'' | A | a' \rangle \langle a' | b' \rangle$$
$$= \sum_{a'} \sum_{a''} \delta(a'-a'') a' \langle b'' | a'' \rangle \langle a' | b' \rangle$$
$$= \sum_{a'} a' \langle b'' | a' \rangle \langle a' | b' \rangle$$

And then I said that the above was

$$= a' \langle b'' | b' \rangle = a' \delta(b' - b'') \Leftrightarrow \langle b'' | f(A) | b' \rangle = f(a') \delta(b' - b'')$$

where $$a'$$ was the eigenvalue of corresponding eigenket to $$b'$$.

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

2. Dec 19, 2008

### turin

I would say that, while your first step is correct, you can do better than that. I suggest that you think of an even better first step which will obviate your suspect final step.

BTW, your problem statement did not specify that the inner product is the Dirac delta. Is this stated somewhere else?

3. Dec 19, 2008

### Domnu

It's the Dirac delta because they're eigenkets of some operator (okay well to make the argument more rigorous, the eigenkets a and b are eigenkets of some Hermitian operator where there is no degeneracy).

4. Dec 19, 2008

### turin

This remains insufficient to determine that $$\langle{}a'|a''\rangle=\delta\left(a'-a''\right)$$, etc.. But, at any rate, you can avoid this issue entirely (see my previous post).