Commuting set of operators (misunderstanding)

[Somali_Physicist]

I don’t see how the definition of |a_n> transmorphs into the statement involving the kroneck delta functions.

 

[PeterDonis]

I don’t see how the definition of |a_n> transmorphs into the statement involving the kroneck delta functions.

What definition and what statement? Please give specific references.

 

[Somali_Physicist]

What definition and what statement? Please give specific references.

Apologies

 

[stevendaryl]

Apologies

So we have two different complete bases:

$|a_n\rangle$

$|b_m\rangle$

If we let $C_{nm} = \langle b_m|a_n\rangle$, then you can write:

$|a_n\rangle = \sum_m C_{nm} |b_m\rangle$

At this point, they are just defining $|(a_n) b\rangle$ to be $\sum_m C_{nm} \ \delta_{b, b_m}|b_m\rangle$. The point of the $\delta_{b, b_m}$ is to include only those terms such that $b_m = b$. It's just a fact that:

$\sum_m C_{nm} |b_m\rangle = \sum_b \sum_m C_{nm} \ \delta_{b, b_m}|b_m\rangle = \sum_b |(a_n) b\rangle$

So:

$|a_n\rangle = \sum_b |(a_n) b\rangle$