Are there general methods, tips or tricks for choosing the correct change of coordinates so that the Hamiltonian of a quantum system becomes separable?

Referring to Shankar's Principles of Quantum Mechanics, we consider the Hamiltonian of a two coupled oscillators system $$ \mathscr{H} = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \frac{1}{2}m\omega^2(x_1^2 + x_2^2 + (x_1 - x_2)^2),$$

and by making the change of coordinates

$$ x_{\pm} = \frac{x_1 \pm x_2}{\sqrt{2}}, $$ $$ p_{\pm} = \frac{p_1 \pm p_2}{\sqrt{2}}, $$

it's easy to see that the transformed Hamiltonian is separable $ \mathscr{H} = \mathscr{H}_+ + \mathscr{H}_-$.

How would one have found in this case this change of coordinates?

How about a general Hamiltonian?

  • $\begingroup$ FWIW, the normal form of quadratic Hamiltonians are classified in Appendix 6 of Arnold's book. $\endgroup$
    – Qmechanic
    Nov 24, 2018 at 16:09

1 Answer 1


Here is a method that should work for any hamiltonian which is a polynomial of degree two or less in $x_1, ...x_N$, $p_1, ...p_N$, so it can have stuff like $x_1 p_2 $ or $p_1 p_3$, but not $x_1 p_2^2$

Then this hamiltonian can be written in matrix multiplication, for some matrix A whose coefficients are chosen so that the result equals the hamiltonian:

$$H=\begin{bmatrix} 1 & x_1 & ... & x_N & p_1 & ... & p_N \end{bmatrix}\begin{bmatrix} & & \\ & A & \\ & & \end{bmatrix} \begin{bmatrix} 1\\ x_1\\ ...\\ x_N \\ p_1 \\ ...\\ p_N \end{bmatrix}$$

There is not a unique choice for $A$. This can be seen if you write in index notation. Let's define the vector to the right of $A$ as $x$.Then in index notation, the sum is equal to

$$H=x^T A x=\sum_{i,j} x_i A_{ij} x_j$$

Since we sum over all $i$ and $j$, the sum can be split in to pairs like

$$x_1 A_{12} x_2 + x_2 A_{21} x_1=x_1 x_2 (A_{12}+A_{21})$$

This means the sum is not sensitive to each individual element of $A$, only the sum of pairs which are across the diagonal from each other. But we can choose an $A$ so that


And then there is a unique $A$ which gives you back your hamiltonian. This is the definition of a symmetric matrix, in components: $A^T = A$. Any symmetric matrix can be diagonalized, so by finding the eigenvectors and eigenvalues of $A$, $A$ can be written

$$A = UDU^T$$

where U is an orthogonal matrix, so $U U^T=1$, and D is a diagonal matrix. Plugging this into $H=x^T A x$, we get:

$$H = x^T U D U^T x$$

Now if you define new coordinates $y= U^T x$, the hamiltonian reads

$$H = y^T D y$$.

Since D is diagonal, this is equal to

$$H = \sum_i y_i D_{ii}y_i=\sum_i y_i^2 D_{ii}$$

This is of the form you asked about.

Application to your Example

I will leave it to you to fill in the details (that's the policy on this site), but for you the problem is simpler than the general case. You only need to write the $x_1,x_2$ part of the hamiltonian as a matrix, and the matrix which works is proportional to

$$A = \begin{bmatrix} 1 & -\frac{1}{2}\\ \frac{1}{2} & 1 \end{bmatrix}$$

Finding the eigenvectors/values of this matrix will give you the same coordinate transformation as you chose (I checked).

In simple cases like this, you can guess how to factor the hamiltonian and it'll be quicker than finding eigenvectors/values. But the eigenvectors/values approach will always work.

For polynomials of degree three or more, I don't believe this is always possible. And in fact Hamiltonians which are not separable in the way you described evolve non-entangled states into entangled states. This is why the "free" (non-interacting) part of a hamiltonian is the degree 2 or less part.


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