Undergrad Form of potential operator of two interacting particles

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In the context of two interacting particles in 3D, the Hilbert space is represented as the tensor product of the individual Hilbert spaces. The potential operator corresponding to the interaction is expressed as V(ℎ{r}_1 - ℎ{r}_2), which acts on the wave function in the position representation. This operator simplifies to V(ℎ{r}_1 - ℎ{r}_2) ψ(ℎ{r}_1, ℎ{r}_2) = V(ℎ{r}_1 - ℎ{r}_2) ψ(ℎ{r}_1, ℎ{r}_2). An abstract expression involving the tensor product is also provided, demonstrating the operator's action on the basis states. The discussion emphasizes the mathematical framework for describing two-particle systems and their interactions.
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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?
 
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It's most simple in the position representation
$$V(\hat{\vec{r}}_1-\hat{\vec{r}_2}) \psi(\vec{r}_1,\vec{r}_2)=V(\vec{r}_1-\vec{r}_2) \psi(\vec{r}_1,\vec{r}_2).$$
 
vanhees71 said:
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.
 
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,
$$\psi(\vec{r}_1,\vec{r}_2) = \langle \vec{r}_1| \otimes \langle \vec{r}_2|\psi \rangle.$$
 
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