# Creation and annihilation operators

1. Apr 17, 2015

### TimeRip496

- 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.

2. Apr 17, 2015

### Orodruin

Staff Emeritus
No, the annihilation operator raises ket states, which it acts on from the left, but lowers bra states. The relation is simply the conjugation of a|n> = C|n-1>. Conjugation switches the order of the expressions, turn bras into kets, kets into bras, and conjugates operators.