Norm of a vector related to coherent states

[JorisL]

Hi,

For the past couple of days I've been attempting to derive the equality (for any normalised $\varphi$)
||(a+za^\dagger+ \lambda)\varphi ||^2 = ||(a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi ||^2 + (|z|^2-1)||\varphi ||^2
First of the summation seemed like a typo. So I first tried to prove this but got nowhere. As such I tried to change it to an equality. ( I entertained this idea because I found, and verified, some other typos in previous equations )Below is an outline of my attempt to derive the second equality (assuming a typo)

Spoiler

Here we consider a system of bosons with only one mode hence $[ a, a^\dagger ] = 1$.
I expanded the norm as $||(a+za^\dagger+ \lambda)\varphi ||^2 = \langle (a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi , (a+za^\dagger+ \lambda)\varphi \rangle$

I just cannot get a factor $|z|^2$ out of there. Because the inner product is defined as $\langle\alpha \psi,\phi\rangle = \bar{\alpha}\langle \psi, \phi \rangle$ while $\langle \psi,\alpha\phi\rangle = \alpha\langle \psi, \phi \rangle$.

So the straightforward way doesn't work. Next I tried to use the first equality.
Lets call the two norms in this equality A resp. B, could I maybe use A-B was my idea.
Because in the text the commutation relations were mentioned which (could) show up directly this way.
I'm not going to write everything I tried (a lot) down here but I tried various tricks etc.

Thanks

Joris

 

[fzero]

$||(a+za^\dagger+ \lambda)\varphi ||^2 = \langle (a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi , (a+za^\dagger+ \lambda)\varphi \rangle$

It's a little clearer if you write it in the notation

$$||(a+za^\dagger+ \lambda)\varphi ||^2 = \langle \varphi| (a^\dagger+\bar{z}a+ \bar{\lambda} )(a+za^\dagger+ \lambda)|\varphi \rangle.$$

Then you will see that

$$||(a+za^\dagger+ \lambda)\varphi ||^2 - || (a^\dagger+\bar{z}a+ \bar{\lambda} )\varphi ||^2 = \langle \varphi| [a^\dagger+\bar{z}a+ \bar{\lambda} ,a+za^\dagger+ \lambda]|\varphi \rangle$$

and a bit of algebra on the commutator should lead you to the result.

 

[JorisL]

Thanks I will check this tomorrow since it's late already.
Unbelievable I didn't notice that, guess its one of those case where prolonged staring makes it worse :)

Joris