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I have a particle of mass $m$ that moves in 2-d in the potential $V(x,y)=\frac{1}{2}m\omega^2(6x^2-2xy+6y^2)$. I have to use the coordinates $u=\frac{x+y}{\sqrt 2}$ and $w=\frac{x-y}{\sqrt2}$ to show that the schrodinger equation in $(u,w)$ is given by $$-\frac{\hbar^2}{2m}\left(\frac{d^2}{du^2}+\frac{d^2}{dw^2}\right)\psi(u,w)+\bar{V}(u,w)\psi(u,w) = E\psi(u,w)$$

and to find $\bar{V}(u,w)$ and $\frac{d^2}{du^2},\frac{d^2}{dw^2}$.

I've done questions regarding particles moving in 2-d in the potential $V(x,y)$ before, but I'm not sure how to make the variable change as required above. I know the schrodinger equation in $(x,y)$ is given by: $$-\frac{\hbar^2}{2m}\left(\frac{d^2}{dx^2}+\frac{d^2}{dy^2}\right)\psi(x,y)+V(x,y)\psi(x,y) = E\psi(x,y)$$ But how would i find $\bar{V}(u,w)$ and $\frac{d^2}{du^2},\frac{d^2}{dw^2}$?

I tried rearranging $u=\frac{x+y}{\sqrt 2}$ in terms of $x$ and $y$ and substituting it into $v(x,y)$ but I still end up with $x$ and $y$ terms.

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I think you should first see that $x = \frac{u + v}{\sqrt{2}}$ and $y = \frac{u - v}{\sqrt{2}}$. If you do the substitutions you should no $x$ or $y$. To replace the derivatives, just use the chain rule.

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thanks, now i see it! I think you just made a typo though, its $x = \frac{u + w}{\sqrt{2}}$. – Ray Jan 16 '12 at 18:44

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