# Dirac notation

1. Nov 3, 2007

### raj2004

Hey guys,
I am having difficulty interpreting M(x,x') into dirac notation. How do i write M(x,x') in dirac notation? The actual problem is to write the following in dirac notation:

int { int { m(x)* M(x,x') g(x') } dx} dx'

2. Nov 3, 2007

### raj2004

1. The problem statement, all variables and given/known data
Hey guys,
I am having difficulty interpreting M(x,x') into dirac notation. How do i write M(x,x') in dirac notation? The actual problem is to write the following in dirac notation:

2. Relevant equations
int { int { m(x)* M(x,x') g(x') } dx} dx'

3. The attempt at a solution
i tried to use M(x,x') = m(x) m(x'). would that be appropriate?

3. Nov 3, 2007

### raj2004

could any one help me on this ?

4. Nov 3, 2007

### mjsd

what is what in
?

5. Nov 3, 2007

### raj2004

int means integration

6. Nov 3, 2007

### mjsd

and what about m , M, g? are the states, operators .. what is what?

7. Nov 4, 2007

### CompuChip

Consider the inner product $\langle m | M | g \rangle$ and write it out in the position basis (hint, insert the completeness relation twice, once at each vertical bar).

8. Nov 4, 2007

### CompuChip

For your attempt at solution, I don't know what M and m are supposed to be, but if they are what I think they are (M is some sort of propagator and m is a state) then I don't think you can write this.

Last edited: Nov 4, 2007
9. Nov 4, 2007

### raj2004

I think m(x) = <x|M> and m(x)* = <M|x>. But i don't know what to write for M(x,x') in dirac notation. Here x and x' are two different bases. Also, g(x') = <x'|g>.

10. Nov 4, 2007

### CompuChip

11. Nov 4, 2007

### raj2004

Ok, now i tried it. Following what you said, I found above integral equals to <m|M|g>. But i don't understand what you mean by M(x,x') is propagator? In dirac notation does M(x,x') equal to <x|M|x'> . It works out fine if i make that assumption.

12. Nov 4, 2007

### malawi_glenn

What does your original integral equal? And what do you mean by "I think m(x) = <x|M> .. " ? You dont know?

13. Nov 4, 2007

### raj2004

No. I was not sure about that statement. But now it's clear. m(x) = <x|m> , not <x|M>. Thanks for your help guys. I get the answer <m|M|g>. But i still don't understand why M(x,x') = <x|M|x'>?

14. Nov 4, 2007

### CompuChip

OK, let me put it this way: do you know what $\int \mathrm{d}x |x\rangle\langle x|$ is?