# Representation of second quantization

1. Oct 3, 2009

### Petar Mali

In coordinate representation in QM probality density is:

$$\rho(\vec{r})=\psi^*(\vec{r})\psi(\vec{r})$$

in RSQ representation operator of density of particles is

$$\hat{n}(\vec{r})=\hat{\psi}^{\dagger}(\vec{r})\hat{\psi}(\vec{r})$$

Is this some relation between this operator and density matrix?

Operator of number of particles is

$$\hat{N}=\int d^3\vec{r}\hat{\psi}^{\dagger}(\vec{r})\hat{\psi}(\vec{r})$$

Why I can now use

$$\hat{\psi}^{\dagger}(\vec{r})=\sum_k\hat{a}_k^{\dagger}\varphi^*_k(\vec{r})\qquad \hat{\psi}(\vec{r})=\sum_k\hat{a}_k\varphi_k(\vec{r})$$ ?

where $$\{\varphi_k\}$$ is complete ortonormal set.

Thanks

2. Oct 4, 2009

### alxm

Which operator? What's $$\hat{\psi}^{\dagger}$$? Do you mean $$a^\dagger|\psi>$$?

Anyway, the matrix elements for creation/annhilation operators are:
$$<m|a^\dagger|n> = \sqrt{n+1}\delta_{m,n+1}$$
$$<m|a|n> = \sqrt{n}\delta_{m,n-1}$$

They don't form a complete set; the creation/annihilation operators aren't self-adjoint.

3. Oct 4, 2009

### Petar Mali

I defined

$$\hat{\psi}^{\dagger}(\vec{r})=\sum_k\hat{a}_k^{\dagger}\varphi^*_k(\vec{r})\qquad \hat{\psi}(\vec{r})=\sum_k\hat{a}_k\varphi_k(\vec{ r})$$

$$\hat{a}_k^{\dagger},\hat{a}_k$$ are operators you talking about and $$\{\varphi_k\}$$ form complete set. You did not read my post.

4. Oct 4, 2009

### kanato

I usually think of the density matrix as something like
$$\hat\rho(r,r') = \hat\psi^\dagger(r) \hat\psi(r')$$
so your density operator would be the diagonal element.

I think your question about the transform relies on the assumption that you can write
$$\delta(r) = \sum_k \phi_k(r)$$
which certainly works for $$\phi_k$$ as plane waves but I am not sure about other basis sets.

It's late and I'm tired so I apologize if this doesn't make much sense.