Particle Number Operator (Hermitian?)

1. Nov 17, 2013

Mimb8

Particle Number Operator (Hermitian??)

Hey guys,

I'm studying the quantic harmonic oscillator and I'm using "Cohen-Tannoudji Quantum Mechanics Volume 1".

At some point he introduced the particle number operator, N, such that:

N=a+.a , where a+ is the conjugate operator of a.

The line of proof that they used to show that N is hermitian is the following:

N+=a+.(a+)+=N , where N+ is the conjugate operator of N.

Somehow, in ways I cannot understand, this proves it.

But everytime I try and prove it myself I get to the conclusion that N is not hermitian.

Can someone be kind enough and help me with this :)

Thank you

2. Nov 17, 2013

George Jones

Staff Emeritus
Did you mean to write this? It should be N^+ = (a^+ a)+ = a+ (a+)+ = a+ a = N.

3. Nov 17, 2013

Mimb8

Hi George,

I'm not sure I get the transition:

(a+. a)+=a+.(a+)+

Back me up here, if I have the operator (AB) and I apply the conjugate it gives me:

(AB)+=A+.B+

Right?

(I seem to be missing something really obvious here :s)

4. Nov 17, 2013

George Jones

Staff Emeritus
No, (AB)+ = B+ A+

5. Nov 17, 2013

Mimb8

Oh...I seem to have misread that property (shame on me).

Thank you :)