# Showing raising operator raising!

1. Jan 16, 2013

### Sekonda

Hey,

I have a question on showing how the raising operator in QM raises a particular eigenstate by 1 unit, the question is showed below:

I think I know how to do this but not sure if what I'm doing is sufficient:

$$\hat{N}a^{\dagger}|n>=([\hat{N},a^{\dagger}]+a^{\dagger}\hat{N})|n>$$

So I considered the N operator acting on the raising operator, we know the commutation relation stated on the RHS, so this simplifies to:

$$\hat{N}a^{\dagger}|n>=(a^{\dagger}+a^{\dagger}\hat{N})|n>$$

Letting N act on the state N we attain:

$$\hat{N}a^{\dagger}|n>=(1+n)a^{\dagger}|n>$$

I'm not sure if it's enough to say now that by definition of the raising operator 'a-dagger', it raises the state n to n+1 and we conclude that 'a-dagger' acting on a state n is equal to some constant multiplied by (1+n) and the state 1+n.

Cheers,
SK

2. Jan 16, 2013

### Simon Bridge

I'd be more explicit than that:

from (given) $[\hat{N},a^\dagger]|n\rangle = a^\dagger |n\rangle$ ... expand the commutator

$\Leftrightarrow \hat{N}[a^\dagger|n\rangle ] - a^\dagger [\hat{N}|n\rangle ] = a^\dagger |n\rangle$ ... since: $\hat{N}|n\rangle = n|n\rangle$ (given)

$\Leftrightarrow \hat{N}[a^\dagger|n\rangle] - n[a^\dagger|n\rangle] = a^\dagger |n\rangle$

$\Rightarrow \hat{N}[a^\dagger|n\rangle] = a^\dagger |n\rangle+na^\dagger|n\rangle = (n+1)[a^\dagger |n\rangle]$

i.e. $a^\dagger |n\rangle$ is an eigenstate of $\hat{N}$ with eigenvalue n+1 ...

$\Rightarrow a^\dagger |n\rangle=|n+1\rangle$

... is |n+1> normalized already?

Last edited: Jan 16, 2013
3. Jan 16, 2013

### Simon Bridge


\renewcommand{\bra}[1]{\langle {#1} |}
\renewcommand{\ket}[1]{|{#1}\rangle}
\renewcommand{\braket}[1]{ \langle #1 \rangle }

so that \bra{n}\; \ket{n,l,m,s}\; \braket{\psi}\; \bra{\psi}H\ket{\psi} gives you: $$\bra{n}\; \ket{n,l,m,s}\; \braket{\psi}\; \bra{\psi}H\ket{\psi}$$

4. Jan 17, 2013

### Sekonda

Ahh yes, that's better. I think the ket 'n+1' is not normalised already but that's only due to the normalisation factor 'c' they've put in in the question.

I think what you have written is sufficient, they do not ask you to determine 'c' - so I presume we don't have to in this case.

Also your second post coding hasn't all come up properly on my screen!