Recent content by 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...

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

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 $$ \langle a | b \rangle \langle c1 | A | c2 \rangle \langle d | e \rangle = \langle a | b \rangle \langle d | e \rangle \langle c1 | A | c2 \rangle $$

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

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...