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'>?
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.
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>.
Homework Statement
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:
I would appreciate your help.
Homework Equations
int { int { m(x)* M(x,x') g(x') } dx}...
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'
I would appreciate your help.