# Simple coupled quantum harmonic oscillator

I am having trouble finding the eigenvalues for the Hamiltonian

$$H = \frac{P_1^2}{2M} + \frac{P_2^2}{2m} + \frac{K}{2}x_1^2 + \frac{k}{2}(x_1 - x_2)^2$$

Even though I can find a basis where the $x_1$ and $x_2$ coordinates are decoupled I then get products of momenta in my new Hamiltonian. I think this problem should be transformable into two decoupled oscillators. Am I wrong about that?

Structurally, it looks like this:

• Start by rescaling one or both of the position/momenta pairs such that the kinetic-energy term has both masses equal (while also retaining $[x_i,p_i]=i\hbar$), and
• then find a rotation in the rescaled $x_1,x_2$ plane that will eliminate the coupling terms.
• That rigid rotation will be mirrored in the momentum plane, but because the masses are symmetric, it will no longer introduce momentum couplings.

Then you're done - you've got two decoupled oscillators whose position and momentum variables are canonically conjugate to each other and given as explicit linear combinations of the old ones.

• Even though i have already accepted your answer, i would like to know if there is some reason why we have to rescale one of the parameters, beyond the algebraic one that this simply makes the coupling term disappear. I can see that this works but is there some physical reason that makes the scaling necessary? Feb 8, 2018 at 20:04
• @Zarathustra Not particularly, that I can see. Ultimately it's all just one big symplectic transformation (as in Gec's answer, really) that gets things into shape, but the rescaling-then-rotation structure makes it easier to implement. Feb 8, 2018 at 20:14
• @Zarathustra - The reason is that only after the scaling to equal factors of the kinetic energy quadratic form, this form also stays a quadratic form (corresponding to a circle) with equal coefficients and no cross terms after applying the orthogonal transformation to diagonalize the quadratic form of the potential energy. See my expanded answer below. Feb 11, 2018 at 3:44

The Hamiltonian is a quadratic form both in the momenta and in the coordinates. You can decouple the problem by using the principal axis theorem on the coordinates to obtain generalized coordinates. For this, first a scaling of the coordinates to equalize the coefficients of the kinetic energy is necessary. The whole approach corresponds to "diagonalizing" both the matrix of the potential energy and of the kinetic energy by finding a suitable linear transformation consisting of a scaling and an subsequent orthogonal transformation. The latter involves finding the eigenvalues and eigenvectors of the potential energy matrix.

Note: Following the comment by @Emilio Pisanty, I edited the previous cursory answer and removed some ambiguities pointed out by him. This is directed at physicists who have absolved only a standard curriculum, and thus probably have not indulged in "symplectomorphisms".

(1) The problem corresponds to the typical small oscillation problem of coupled oscillators which is treated in most classical mechanics courses, usually in the framework of the Lagrange function, which can easily be translated to the corresponding Hamilton function. The decoupling of the given Hamilton function into a sum of uncoupled linear oscillator Hamilton functions is the problem of finding the "normal coordinates" for these oscillators.

This is described in detail, e.g. in H. Goldstein's book, Classical Mechanics, Chapter 6. Oscillations, especially, section 6.2 The Eigenvalue Equation and the Principal Axis Transformation. In this Harvard lecture, you'll find a shorter exposition of Goldstein's approach. It is completely sufficient to perform the transformations on the classical Hamiltonian (or Lagrange function). Once you have the decoupled Hamiltonian, you can quantize it. Goldstein designates the resulting necessary total transformation as "similarity transformation" to realize the "principal axis transformation".

(2) In order to decouple the oscillators, you have to simultaneously diagonalize the symmetric quadratic form (matrix) of the potential energy and the symmetric quadratic form (matrix) of the kinetic energy of the Lagrangian $L=T-V$ (expressed in velocities) by a suitable linear coordinate transformation. If the masses in the kinetic energy term were equal, you could do this with a single orthogonal coordinate transformation, usually called principal axis transformation.

(3) But here the masses are not equal. Therefore, you have first to scale the coordinates so that the coefficients of $v_1^2$ and $v_2^2$ become equal. This means, that the kinetic energy matrix is not only diagonal but has also the same coefficients. Then, after a subsequent orthogonal transformation, the quadratic form of the kinetic energy will stay a sum of the squares of the velocity with the same factor. This is described in an easily understandable way (directly applicable to the present 2-D case) in P. K. Aravind, Geometrical interpretation of the simultaneous diagonalization of two quadratic forms, American Journal of Physics 57, 309 (1989). I don't know whether there is any no-pay option to get this beautiful paper.

(4) Now you perform the orthogonal transformation ( this case, rotation in the $x_1$, $x_2$ plane) to diagonalize the symmetric matrix of the potential energy. The matrix of the kinetic energy stays diagonalized. Thus you end up with both matrices diagonalized so that the Lagrangian becomes a sum of separate oscillator Lagrangians.

(5) You can find the resulting Hamiltonian $H$ in the new generalized coordinates from the Lagrangian $L$ in the new generalized coordinates by the known relation $$H= \sum_i p_i \dot x_i-L$$ and $$p_i= \frac {\partial L}{\partial \dot x_i}$$ This ensures that your new generalized momenta $p_i$ are the canonical conjugates of the new generalized coordinates $x_i$.

(6) If you need the QM (Schrödinger) Hamiltonian, you quantize the classical one by substituting $p \to -i\hbar \nabla$.

• Thank you, i found the links you provided very helpful. Especially the principal axis theorem. Since i can only accept one answer all i can do is up vote you now. Feb 8, 2018 at 17:17
• This isn't quite accurate. The use of the principal axis theorem and its associated orthogonal transformation is either insufficient (if kept to the position plane, in which case a momentum coupling will appear) or incorrect (if extended to include the momenta, in which case the transformation will not preserve the canonical commutator / the Poisson bracket). If one insists on using one single transformation, the correct requirement is that it be symplectic, not orthogonal. Feb 9, 2018 at 10:49

Another method to solve the problems of this kind is to consider operators of the form $$\hat{a} = c x_1 + d x_2 + e P_1 + f P_2.$$ Here $c$, $d$, $e$ and $f$ are numerical coefficients. If you find solutions of the equation $$\left[ \hat{a}, H \right] = \varepsilon \hat{a},$$ you will get creation and annihilation operators and energies of quanta.

• This is correct, but it needs to be supplemented with the requirement that the new annihilation operators and their conjugates obey the bosonic commutation relation. Feb 9, 2018 at 10:53
• @Emilio Pisanty You are completely right. Thanks for your clarification. It is much easier for me to write short messages.
– Gec
Feb 9, 2018 at 13:49

More generally, given a semipositive definite quadratic real Hamiltonian $$H=\frac{1}{2}z^I H_{IJ} z^J~\geq~ 0 \tag{1}$$ in $2n$ canonical coordinates $$(z^1, \ldots z^{2n})~=~(q^1,\ldots, q^n,p_1, \ldots, p_n),\tag{2}$$ one may show that there always exists a real & linear symplectic transformation $$Z~=~ S z, \qquad S\in Sp(2n,\mathbb{R}),\tag{3}$$ that brings the Hamiltonian on diagonal form, cf. Ref. 1.

Note that the coordinate transformation must be symplectic in order to preserve the canonical commutation relations (CCR). There is a related Phys.SE post here.

References:

1. V.I. Arnold, Mathematical methods of Classical Mechanics, 2nd eds., 1989; Appendix 6.