aaaa202 said:
I was wondering about this: The identity operator writes a vector in the basis that is used to express the identity operator:
1 = Ʃlei><eil
But if you are to apply it to a vector in a given basis A should the lei> then be expressed in terms of their own basis or in terms of A?
It doesn't really matter as long as you can calculate the inner products.
Maybe an example might help you. (Since I'm not a physicist, I don't know what the correct notation is for bra-ket, but I'll try).
Take the following basis
e_1=(1,0,0),~~ e_2=(1,1,0),~~ e_3= (1,1,1)
Take the inner product
<(a,b,c)|(x,y,z)>=ax+by+cz
This formulation of the inner product is going to be dependent on the basis of course.
Anyway, let's calculate
(|e_2><e_2|)|a>
where a is given by (1,2,3). Then we get
<e2|a>=1*1+1*2+0*3=3
So |e_2><e_2|a>= 3|e_2>=(3,3,0).
Now what if a is given in terms of another basis? So what if
a=e_1+2e_2+3e_3
One possibility is to write a in the normal basis, then we get
a=(1,0,0)+(2,2,0)+(3,3,3)=(6,5,3)
and then calculate as usual:
|e_2><e_2|a>=(11,11,0)
Another possibility is to write the inner product in terms of the basis \{e_1,e_2,e_3\}. That is, find a formula for
<ae_1+be_2+ce_3|xe_1+ye_2+ze_3>
and then you can calculate |e_2><e_2|a> without converting to another basis.
Does this answer your question?