Dirac Notation: Understanding <m|x|n> and Its Relation to Eigenstates

[Marthius]

I have recently finished reading a section on this notation, and while i though i understood it, i now find myself lost

The question is to show that
<m|x|n>
Is zero unless m = n + or - 1


As I understand it so far <m| and |n> correspond to the eigenstates of an arbitrary system and x is just supposed to be the x operator

The only thing i could think to do with his was plug into
\intm*xn dx but that did not help me

I also susspect i need to use that
<m|n>=\delta_mn

If anyone could give me a nudge in the correct direction it would be much appreciated.

 

[borgwal]

This question is meant to be about the energy eigenstates of the harmonic oscillator, not an arbitrary system.

 

[Marthius]

I'm sry, but I still do not really understand where to go with this, I suppose I can use |n> = the nth eigenstate of the harmonic oscillator, but isn't the x operator just x?

 

[borgwal]

Haven't you seen the "ladder operators" a^+ and a^-?

 

[Marthius]

Haven't you seen the "ladder operators" a^+ and a^-?

I have, and if x were a ladder operator this would be trivial, but I thought it was supposed to be the x (position) operator...

 

[borgwal]

You can write the x operator in terms of the ladder operators. Your question is trivial, too.

 

[Marthius]

So then would i write
x=.5*\sqrt{2h/mw}(A+A^{+})?
and distribute out getting
(<m|A|n>+<m|A^+|n>) * some constant

 

[borgwal]

So then would i write
x=.5*\sqrt{2h/mw}(A+A^{+})?
and distribute out getting
(<m|A|n>+<m|A^+|n>) * some constant

Indeed! And all you need is to show for what m,n combinations <m|A|n>+<m|A^+|n> is nonzero, which as you said is trivial.

 

[Marthius]

It makes sense now, thank you for your help