Form of potential operator of two interacting particles

[Kashmir]

Considering two interacting particles in 3d, the corresponding Hilbert space $H$ is the tensor product of the two individual Hilbert spaces of the two particles.
If the particle interaction is given by a potential $V(\mathbf r_1 -\mathbf r_2)$ ,what is the corresponding potential operator for it?

 

[vanhees71]

It's most simple in the position representation
$$V(\hat{\vec{r}}_1-\hat{\vec{r}_2}) \psi(\vec{r}_1,\vec{r}_2)=V(\vec{r}_1-\vec{r}_2) \psi(\vec{r}_1,\vec{r}_2).$$

 

[Kashmir]

It's most simple in the position representation
$$V(\hat{\vec{r}}_1-\hat{\vec{r}_2}) \psi(\vec{r}_1,\vec{r}_2)=V(\vec{r}_1-\vec{r}_2) \psi(\vec{r}_1,\vec{r}_2).$$

Thank you. I was looking for the abstract expression involving the tensor product of individual spaces.

 

[vanhees71]

If you write it in this way, it just reads
$$V(\hat{\vec{r}}_1 \otimes \hat{1} - \hat{1} \otimes \hat{\vec{r}}_2) |\vec{r}_1 \rangle \otimes |\vec{r}_2 \rangle = V(\vec{r}_1-\vec{r}_2) |\vec{r}_1 \rangle \otimes |\vec{r}_2 \rangle.$$
Now you use it in the definition of the two-particle wave function,
$$\psi(\vec{r}_1,\vec{r}_2) = \langle \vec{r}_1| \otimes \langle \vec{r}_2|\psi \rangle.$$