I When can one clear the operator

[Ernesto Paas]

Hi all!
I'm having problems understanding the operator algebra. Particularly in this case:
Suppose I have this projection $\langle \Phi_{1} | \hat{A} | \Phi_{2} \rangle$ where the $\phi's$ have an orthonormal countable basis.
If I do a state expansion on both sides then I suppose I'd get this: \sum_{n,m} \langle n | b_{n}^* \, \hat{A} \, c_{m} | m \rangle
And what's to stop me from moving the operator to the left and getting a kronecker-delta?

 

[strangerep]

I presume your expansion states are eigenstates of $\hat A$? If so, then I'd say "yes..."

 

[Ernesto Paas]

I presume your expansion states are eigenstates of $\hat A$? If so, then I'd say "yes..."

Ah no... Sorry for not pointing that out. They are just of the same same arbitrary orthonormal basis.

 

[strangerep]

Oh, well, if your expansion states are not eigenstates of $\hat A$, then... what are they? E.g., what is $$\hat A |m\rangle ~=~ ?$$

 

[Ernesto Paas]

I don't know, that's the problem. I've seen bras and kets being moved around without being eigenstates of an operator. Like in this case $$ \langle a | b \rangle \langle c1 | A | c2 \rangle \langle d | e \rangle = \langle a | b \rangle \langle d | e \rangle \langle c1 | A | c2 \rangle $$

 

[Ernesto Paas]

Is the operator like glued to the left ket? Can I move the bra around?

 

[strangerep]

Think of the bra $\langle a|$ as a row vector "$a^T$'', the ket $|b\rangle$ as a column vector "$b$", and the operator $\hat A$ as a matrix $A$. I'll write it the overall expression as $a^T A b$. Does that much make sense?
(I'm not sure how much ordinary linear algebra you already know.)

 

[Ernesto Paas]

Ok it hadn't occurred to me to think of them in that way. It wouldn't make sense to do a commutation there if the rows didn't equal the columns. Nevertheless, moving scalars around wouldn't be a problem.

Thanks, it was actually really easy...

 

[strangerep]

More generally, see the second line of my signature below...

Ballentine, ch1, can help a lot with these sort of questions.