In order to find the eigenstates of this Hamilatonian

$$ H = \sum_{j=1}^3 \left( - \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x_j^2}\right) + \frac{1}{2} m \omega^2 \left( (x_1 - x_2)^2 + (x_1 - x_3)^2 + (x_2 - x_3)^2\right) $$

I want to do a coordinate transformation $y_1 = x_1 - x_2$, $y_2 = x_1 - x_3$, $y_3 = x_2 - x_3$, but I am not sure how to transform the differential operators. Are they just $\frac{\partial^2}{\partial y_j^2}$? Or do they transform to something else? How do I calculate this?

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    $\begingroup$ Beware of the Jacobian! $\endgroup$ – Torsten Hĕrculĕ Cärlemän Jan 6 '14 at 8:10
  • $\begingroup$ It occurs to me that this might be more of a math question than a physics question... we'll see what people think. $\endgroup$ – David Z Jan 6 '14 at 8:17
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    $\begingroup$ Comment to the question (v2): Note that OP's proposed change of variables in his example is not bijective since $y_1=y_2-y_3$. $\endgroup$ – Qmechanic Jan 6 '14 at 13:05

Since you're changing from $x_j$ to $y_j$, you have to consistently make this change to the $x_j$'s everywhere they occur, including in differential operators. To do this, you need to use the Jacobian matrix. The rule can be written like this:

$$\begin{pmatrix} \frac{\partial}{\partial x_1} \\ \frac{\partial}{\partial x_2} \\ \frac{\partial}{\partial x_3} \end{pmatrix} = \underbrace{ \begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_2}{\partial x_1} & \frac{\partial y_3}{\partial x_1} \\ \frac{\partial y_1}{\partial x_2} & \frac{\partial y_2}{\partial x_2} & \frac{\partial y_3}{\partial x_2} \\ \frac{\partial y_1}{\partial x_3} & \frac{\partial y_2}{\partial x_3} & \frac{\partial y_3}{\partial x_3} \end{pmatrix} }_\text{Jacobian matrix} \begin{pmatrix} \frac{\partial}{\partial y_1} \\ \frac{\partial}{\partial y_2} \\ \frac{\partial}{\partial y_3} \end{pmatrix}$$

but if you go through the matrix multiplication, you'll see that this is just three instances of the chain rule. So it's equivalent to solving for the $y_j$'s in terms of the $x_j$'s and then using the chain rule on those formulas.

Note that in some cases, you'll see a change of variables where one or more of the new variables has the same label as one of the original variables. For example, changing Cartesian coordinates $x,y,z$ to cylindrical coordinates $\rho,\theta,z$. It's easy to think you can get away with leaving the common variable(s), $z$ in this case, out of the Jacobian, but you can't. In this situation, I find it best to invent a completely new set of names for the new variables - for example, think of it as changing from $x,y,z$ to $\rho,\theta,q$ - and then when you're done, just rename $q$ to $z$ in the formulas you find.

  • $\begingroup$ But when I do this, I get mixed partial derivates in the first sum of the Hamiltonian which do not cancel. For example $\frac{\partial^2}{\partial x_1^2} = \frac{\partial^2}{\partial y_1^2} + 2\frac{\partial^2}{\partial y_1 y_2} + \frac{\partial^2}{\partial y_2^2}$. Is there something wrong with my coordinate transformation? I do not want to have any mixed terms. $\endgroup$ – iblue Jan 6 '14 at 8:51
  • $\begingroup$ Just because you don't want there to be mixed derivatives doesn't mean there won't be. In many cases, there are. You could check your results against this for example. $\endgroup$ – David Z Jan 6 '14 at 8:55
  • $\begingroup$ On the excercise sheet, it says that the Hamiltonian can be transformed to a form with no mixed terms, neither in the derivatives nor in the coordinates themselves. $$H = \sum_{j=1}^3 \left(\alpha_j \frac{\partial^2}{\partial y_j^2} + \beta_j y_j^2\right)$$. Is this possible? Or is there a mistake on the excercise sheet? $\endgroup$ – iblue Jan 6 '14 at 9:18

There is no mistake on your exercise sheet.

To find a solution, you have to postulate a linear dependence of the $y^i$ from the $x^j$, $y^i= a^i_j x^j$,where the matrix $a$ of coefficients $a^i_j$ is constant. You have $x^j = (a^{-1})^j_i y^i$.

Now, simply express the initial terms as a function of $y^i$ and express diagonal constraints, you have :

$\sum\limits_{j=1}^3a_j^i a_j^k= \delta ^{ik} \tag{1}$ for the differential term constraint, and :

$\sum\limits_{j=1}^3(a^{-1})^j_i (a^{-1})^{[j+1]}_k= 0 \tag{2}$ for the quadratic $y^2$ term constraint, where $[j+1]=(j+1) \mod 3$

By noting that : $a_j^k = (a^t) _k^j$, and $(a^{-1})^{[j+1]}_k = (a^{(-1) t})_{[j+1]}^k$, $(1)$ and $2$ can be rewritten :

$\sum\limits_{j=1}^3a_j^i (a^t) _k^j= \delta ^i_k\tag{3}$ and :

$\sum\limits_{j=1}^3(a^{-1})^j_i (a^{(-1)t})_{[j+1]}^k= 0 \tag{4}$

The equation $3$ means simply that $aa^t= \mathbb Id$, so $a$ is an orthogonal matrix. For simplicity, we may search for a matrix so that $a=a^t=a^{-1}$, so we have only to check the equation $(4)$, rewritten :

$\sum\limits_{j=1}^3a^j_i a_{[j+1]}^k= 0 \tag{5}$

A solution is :

$a = \dfrac{1}{3}\begin {pmatrix} 1&-2&-2\\-2&1&-2\\-2&-2&1 \end {pmatrix}$


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