# Operator algebra question

1. Sep 30, 2005

### cire

A is an operator, in the expression <m|A|n>|m><n|, can I insert the matrix element <m|A|n> between the |m> and <n| like:
|m><m|A|n><n|????????

2. Sep 30, 2005

### Tom Mattson

Staff Emeritus
Remember that <m|A|n> is just a complex number, so it commutes with everything you can imagine, including bras and kets. So placing it between |m> and <n| doesn't result in an illegal expression, but it is a strange way of writing it.

It's almost like writing the expression "x2y" instead of "2xy" in algebra. There's nothing wrong with either expression, but how often do you see coefficients sandwiched inbetween variables?

3. Sep 30, 2005

### DavidK

I would like to disagree with the previous poster about the strangeness of expressing <m|A|n>|m><n| as |m><m|A|n><n|. It can in fact be quite a usefull way of re-writing. If we for example sum over n and m, we get:

$\hat{A}=\hat{1}\hat{A}\hat{1}=\sum_{mn}|m><m|\hat{A}|n><n|=\sum_{mn}<m|\hat{A}|n>|m><n|$

4. Sep 30, 2005

### Tom Mattson

Staff Emeritus
Ah, but he didn't have the summation signs in there. I agree that your steps show the flow of logic of introducing the completeness relation most clearly. But even you moved the coefficient to the left of the outer product at the end.

5. Sep 30, 2005

### MalleusScientiarum

Do be careful if you have that constant inside another inner product, because the definition of the inner product requires that to pull it out you might have to take a complex conjugate.

6. Sep 30, 2005

### DavidK

My point, however, was only that the possibility of moving your complex number (<m|A|n>) wherever you please can be very useful when manipulating operator expressions. My example was maybe somewhat misleading

7. Oct 1, 2005

### George Jones

Staff Emeritus
The expression $|m><m|A|n><n|$ does have a somewhat straightforward interpretation, assuming $|m>$ and $|n>$ are normalized.

The expression is an operator that is the product of the 3 operators $|n><n|$, $A$, and $|m><m|$ and means: first project onto the state $|n>$; transform by A; project onto the state $|m>$.

Regards,
George

8. Oct 1, 2005

### robousy

n>

I thought <m|A|n> could be written as $$A_{mn}$$ which constitutes a matrix and thus will not commute with everything.

9. Oct 2, 2005

### DavidK

$A_{mn}$ do not constitute a matrix. It is just a complex number which is a component of a matrix, and as a complex number it commutes with operators and state vectors.