- #1
TimeRip496
- 254
- 5
- a|n>=C|n-1>
- a+|n>=D|n+1>
And because |n-1> is normalized, <n-1|n-1>=1: (<n|a+)(a|n>)=C2
Thus, <n|a+a|n>=C2
Where a is the annihilation operator and a+ is the creation operator
I don't understand this as isn't <n|a+=<n+1|D , thus <n|a+a|n> =<n+1|DC|n-1> instead?
This is from the quantum physics for dummies.
- a+|n>=D|n+1>
And because |n-1> is normalized, <n-1|n-1>=1: (<n|a+)(a|n>)=C2
Thus, <n|a+a|n>=C2
Where a is the annihilation operator and a+ is the creation operator
I don't understand this as isn't <n|a+=<n+1|D , thus <n|a+a|n> =<n+1|DC|n-1> instead?
This is from the quantum physics for dummies.