- #1
jhosamelly
- 128
- 0
We know that
a|n> = √n | √(n-1)>
and
a' |n> = √(n+1) | n + 1 >
so, If we use this to find
<n|a'a|n>
it would be equal to n
<n|a'a|n> = n
Am I correct?
I'm not really sure about my calculations.
I operate with a first so.
<n|a'a|n>
<n|a' √n | √(n-1)>
= n
?
Can someone please help me with the complete solution?
a|n> = √n | √(n-1)>
and
a' |n> = √(n+1) | n + 1 >
so, If we use this to find
<n|a'a|n>
it would be equal to n
<n|a'a|n> = n
Am I correct?
I'm not really sure about my calculations.
I operate with a first so.
<n|a'a|n>
<n|a' √n | √(n-1)>
= n
?
Can someone please help me with the complete solution?