- #1
AxiomOfChoice
- 533
- 1
Suppose I have a Schrodinger equation for two interacting particles located at x and y; so, something like
[tex]
\left( i \frac{\partial}{\partial t} + \frac{1}{2m_x} \frac{\partial^2}{\partial x^2} + \frac{1}{2m_y} \frac{\partial^2}{\partial y^2} + V(x-y) \right) \psi(x,y,t) = 0.
[/tex]
Now, I want to shift to center of mass coordinates R and r and write the Hamiltonian in terms of them. The resulting Schrodinger equation looks something like
[tex]
\left( i \frac{\partial}{\partial t} + H(R,r) \right) \varphi(R,r,t) = 0.
[/tex]
Question: is it true that
[tex]
\psi(x,y,t) = \varphi \left(\frac{m_x x + m_y y}{m_x + m_y},x-y,t\right)?
[/tex]
If so, why? I have been trying to prove this but have so far just been going around in circles it seems...
[tex]
\left( i \frac{\partial}{\partial t} + \frac{1}{2m_x} \frac{\partial^2}{\partial x^2} + \frac{1}{2m_y} \frac{\partial^2}{\partial y^2} + V(x-y) \right) \psi(x,y,t) = 0.
[/tex]
Now, I want to shift to center of mass coordinates R and r and write the Hamiltonian in terms of them. The resulting Schrodinger equation looks something like
[tex]
\left( i \frac{\partial}{\partial t} + H(R,r) \right) \varphi(R,r,t) = 0.
[/tex]
Question: is it true that
[tex]
\psi(x,y,t) = \varphi \left(\frac{m_x x + m_y y}{m_x + m_y},x-y,t\right)?
[/tex]
If so, why? I have been trying to prove this but have so far just been going around in circles it seems...