AxiomOfChoice
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Suppose I have a Schrödinger 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 Schrödinger 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 Schrödinger 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...