I How to find the wavefunction in this case?

[Kashmir]

We've a two interacting particle system, with Hamiltonian as:
$H_{s y s}=\frac{\mathbf{p}_{1}^{2}}{2 m_{1}}+\frac{\mathbf{p}_{2}^{2}}{2 m_{2}}+V\left(\mathbf{r}_{1}, \mathbf{r}_{2}\right)$

we reduce it to two non interacting fictitious particles,one moving freely other in a central field, thus the system has Hamiltonian: $H_{s y s}=\frac{\mathbf{P}^{2}}{2 M}+\frac{\mathbf{p}_{r e l}^{2}}{2 \mu}+V(\mathbf r)$ with the operator relations as:

$\mathbf{R}=\frac{m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}}{m_{1}+m_{2}}$$\mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}$

$\mathbf{p}_{r e l}=\frac{m_{1} \mathbf{p}_{2}-m_{2} \mathbf{p}_{1}}{m_{1}+m_{2}}$

$\mathbf{P}=\mathbf{p}_{1}+\mathbf{p}_{2}$


We solve $H|E\rangle=E|E\rangle$ for the two non interacting particles and find the wavefunction $\psi_{s y s}(\mathbf{R}, \mathbf{r})=\psi_{C M}(\mathbf{R}) \psi_{r e l}(\mathbf{r})$

How do we find the wavefunction in terms of $\mathbf r_1,\mathbf r_2$? We can't just invert $\mathbf{R}=\frac{m_{1} \mathbf{r}_{1}+m_{2} \mathbf{r}_{2}}{m_{1}+m_{2}}$$\mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}$ and use them because they are operator relations.

 

[vanhees71]

Why do you think that? The wave function in terms of $\vec{r}_1$ and $\vec{r}_2$ is just the wave function you've written down with $\vec{R}$ and $\vec{r}$ expressed in terms of $\vec{r}_1$ and $\vec{r}_2$.

 

[Kashmir]

Why do you think that? The wave function in terms of $\vec{r}_1$ and $\vec{r}_2$ is just the wave function you've written down with $\vec{R}$ and $\vec{r}$ expressed in terms of $\vec{r}_1$ and $\vec{r}_2$.

The eigenstate is
$\int|R\rangle \otimes|r\rangle \psi(R, r) d R d r=|E\rangle$

In $r_1,r_2$ representation I need to find $\left\langle r_{1}| \otimes\left\langle r_{2}|\left|\int\right| R\right\rangle \otimes \mid r\right\rangle \psi(R, r) d R d r$

 

[andresB]

Not sure what you mean,

(1) Not sure what you mean, in the position representation the position operators act just as numbers.
(2) Even in the operator sense, r₁ and r₂ most definitely can be given in terms of R_CM and r_rel. Try to find the inverse transformation.

I'll take the opportunity to point out to this channel

 

[Kashmir]

Not sure what you mean,

(1) Not sure what you mean, in the position representation the position operators act just as numbers.
(2) Even in the operator sense, r₁ and r₂ most definitely can be given in terms of R_CM and r_rel. Try to find the inverse transformation.

I'll take the opportunity to point out to this channel

I do use this channel. Professor M is nice :). Thank you

 

[Kashmir]

Why do you think that? The wave function in terms of $\vec{r}_1$ and $\vec{r}_2$ is just the wave function you've written down with $\vec{R}$ and $\vec{r}$ expressed in terms of $\vec{r}_1$ and $\vec{r}_2$.

So I simply write $R, r$ in terms of $r_1,r_2$ in $\psi_{s y s}(\mathbf{R}, \mathbf{r})=\psi_{C M}(\mathbf{R}) \psi_{r e l}(\mathbf{r})$ and get my wavefunction in terms of $r_1,r_2$. Right?

 

[vanhees71]

Right!