# Operation on a state vector

1. Sep 12, 2009

### snoopies622

If A is a matrix that represents an observation and v is a vector that represents a system state before being observed - and v is not an eigenvector of A - is there any physical significance to the product Av? At first I thought it was what v would become after the observation (I was reading a pretty bad book). Now I know that's completely wrong but I was just wondering if it meant anything at all.

Thanks.

2. Sep 12, 2009

### Fredrik

Staff Emeritus
"If A is an operator representing an observable"...is how your question should have started. You could also have said "measurable quantity" or something like that instead of "observable".

There's no useful interpretation of Av that holds for arbitrary states v.

If you want the state after the interaction, its $e^{-iHt}v\otimes w$ where w is the state of the measuring apparatus, H is the Hamiltonian describing the interaction between the two, and t is the time from "before the measurement" to "after the measurement".

3. Sep 13, 2009

### snoopies622

Thanks Fredrik. I didn't know that there is a vector associated with the measuring apparatus.

4. Sep 13, 2009

### atyy

(v*).Av or something like that is the average if you prepare multiple copies of v and measure A on it multiple times.

Often written <v|A|v>.

You can also see Ehrenfest's theorem.

5. Sep 15, 2009

### snoopies622

Hey thanks, atyy. The Ehrenfest theorem looks very interesting indeed.