I have two coupled quantum harmonic oscillators given by the following Hamiltonian:

$$H=\frac{p_{x}^{2}}{2}+\frac{\omega^{2} x^{2}}{2}+\frac{p_{y}^{2}}{2}+\frac{\Omega^{2} y^{2}}{2}+\frac{C p_{x} y}{2}.$$

As you can see, the coupling is done over the position variable of one oscillator and the momentum of the other. I need to find the wavefunction of states where either of the two oscillators (or both) is excited.

What I tried to do: In general when I have couple harmonic oscillators where the coupling term is of the form $C (x_1-x_2)^2$ I start by diagonalising the Hamiltonian, then define normal coordinates and write down the Hamiltonian using those. I.e., I write the Hamiltonian under the form

$$H=\frac{1}{2} \sum_{i=1}^{2} p_{i}^{2}+\frac{1}{2} \sum_{i, j=1}^{2} x_{i} K_{i j} x_{j}.$$

Then I diagonalise $K$,

$$K_D = UKU^T.$$

And define normal coordinates

$$\left(\begin{array}{l} x_{+} \\ x_{-} \end{array}\right) \equiv U\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right),$$

leading to

$$H=\frac{1}{2}\left[p_{+}^{2}+p_{-}^{2}+\omega_{+}^{2} x_{+}^{2}+\omega_{-}^{2} x_{-}^{2}\right].$$

I am unable to use this here since I do not know how to diagonalise this Hamiltonian with the term $C p_x y$ present.

  • $\begingroup$ Since you have posted a bounty asking for a more detailed answer, could you clarify what about Mike Stone's answer is insufficient? It looks rather comprehensive to me. $\endgroup$ Jun 27, 2021 at 4:43
  • $\begingroup$ @RichardMyers MS’s answer is great but there are a few details in it that I do not understand how to implement exactly. For instance how to I get $R$? How exactly do I compute $H^{-1/2}$? $\endgroup$
    – Y2H
    Jun 27, 2021 at 7:22
  • $\begingroup$ @RichardMyers another issue is that, when I define $D$ as Mike Stones notes and assume no constraints on $S$ at all, Mathematica simply shows that the equation $S^T H S$ cannot be satisfied. $\endgroup$
    – Y2H
    Jun 27, 2021 at 12:10

3 Answers 3


Your problem is basically that of a charged particle in a magnetic field in the Landau gauge, but I'll explain the general theory for such systems as it's both interesting and suprisingly complicated --- and seldom found in textbooks. It's essentially the theory of Bose Bogoluibov transformations.

Suppose we are given classical (or quantum) quadratic Hamiltonian $$ {\mathcal H}[{\bf p},{\bf x}]= \frac 12 M_{ij} p_ip_j+ \frac 12 V_{ij} x_ix_j + K_{ij}\,p_i x_j \nonumber\\ =\frac 12 \left[\matrix{{\bf p}^T& {\bf x}^T}\right] H \left[\matrix{ {\bf p}\cr {\bf x}}\right], \nonumber $$ where and $H$ is the real, positive definite, symmetric $2N$-by-$2N$ matrix $$ H= \left[\matrix{M& K\cr K^T &V}\right]. $$ (If $H$ is not positive definite the diagonalization is not possible as the energy of the system is not bounded below and the eigenfrequencies are pure imaginary.)

We seek a transformation $$ \left[\matrix{ {\bf p}\cr {\bf x}}\right]= S \left[\matrix{ {\bf P}\cr {\bf X}}\right] $$ that diagonalizes $H$ via the congruence transformation ${H}\to S^T HS $ and preserves the classical Poisson brackets and/or the quantum commutation relations. This preservation requires $$ S^T JS=J, \quad J= \left[\matrix{0&- {\mathbb I}_n \cr {\mathbb I}_n&\phantom{-} 0}\right], $$ so $S\in {\rm Sp}[2N, {\mathbb R}]$, the non-compact symplectic group of linear canonical transformations. That positive definiteness of $H$ is sufficient to ensure that we can find such an $S$ is the statement of Williamson's theorem. The resulting frequencies are called the symplectic spectrum of $H$. They are unrelated to the eigenvalues of $H$. Further matrix $S$ is neither orthogonal nor unitary, and so does not obey $S^{-1}=S^\dagger$

We observe that $H$ being positive definite ensures that the matrices $H^{\pm 1/2}$ are well defined. We then see that
the matrix $$ \tilde J= H^{-1/2}JH^{-1/2} $$ is skew symmetric and hence an element of $\mathfrak{so}[2N]$. Therefore there exists an $R\in {\rm O}(2N)$ that conjugates $\tilde J$ into the Cartan algebra of $\mathfrak{so}(2n)$ --- i.e. $$ R^TH^{-1/2} J H^{-1/2} R= \left[\matrix{0&- \Omega^{-1} \cr \Omega^{-1} &0}\right], $$ with $\Omega^{-1}$ diagonal. As we are allowing $R\in {\rm O}[2N]$ rather than demanding $ R\in {\rm SO}[2N]$, we can ensure that $\Omega^{-1}$ possesses strictly positive entries. This being so, we define
$$ D= \left[\matrix{\Omega^{1/2}&0 \cr 0& \Omega^{1/2}}\right], $$ so that $$ DR^T H^{-1/2} J H^{-1/2} RD=J,\nonumber\\ DR^T H^{-1/2}H H^{-1/2} R D= D^2.\nonumber $$ Thus $$ S= H^{-1/2} RD \in {\rm Sp}[2N, {\mathbb R}], $$ and $$ S^T HS= D^2 = \left[\matrix{\Omega&0 \cr 0& \Omega}\right] $$ where $\Omega= {\rm diag}(\omega_1,\ldots \omega_N)$. We have $$ {\mathcal H}[{\bf P}, {\bf Q}]=\frac 12 \left[\matrix{ {\bf P}^T& {\bf X}^T}\right]\left[\matrix{ \Omega &0\cr 0&\Omega}\right]\left[\matrix{ {\bf P}\cr {\bf X}}\right]= \sum_i \frac {\omega_i}{2} (P_i^2+X_i^2), $$ which is a set of decoupled harmonic oscillators with frequency $\omega_i$.

We can of course scale $X_i\to \sqrt{\omega} X_i$ and $P_i\to P_i/\sqrt{\omega}$ without changing the commutation relations, and so
$$ \sum_i \frac {\omega_i}{2} (P_i^2+X_i^2)\to \sum_i \frac {1}{2} (P_i^2+\omega_i^2X_i^2), $$ which probably looks more familiar.

The eigenvalues $\omega_i$ are most easily found from those of $JH$ which are $\pm i\omega_i$.

Note added: I see that this same strategy is discussed in a math stack exchange answer: https://math.stackexchange.com/questions/1171842/finding-the-symplectic-matrix-in-williamsons-theorem

  • $\begingroup$ Never seen such computations. Very cool! +1 $\endgroup$ Jun 13, 2021 at 13:43
  • $\begingroup$ I am unsure If one could find a symmetric K matrix such that the proposed form of the Hamiltonian matches the question's Hamiltonian since there is only a $p_x y$ term. Also, is it 100% sure that the $p_k$ in the first equation should not be a $p_j$ ? Aside from that, thank you very much for posting this since it looks like it works beautifully when there are xipj and xjpi terms. $\endgroup$
    – elscan
    Jun 13, 2021 at 17:31
  • $\begingroup$ Thanks for catching typo (now fixed)! I think that you have $H=(1/2)(p_x+\lambda y)^2- (1/2)\lambda^2 y^2 +\ldots $ which is why I was suggesting that you regard $\lambda y=A_x$ as a gauge field. YOu can also add a $xp_y-yp_x$ angular momentum term to make symmetric but may nt help much. $\endgroup$
    – mike stone
    Jun 13, 2021 at 18:19
  • 1
    $\begingroup$ I don't know of one. I chased this stuff down in papers. In particular Simon, Chaturvedi, Srinivasan "Congruences and Canonical Forms for a Positive Matrix: Application to the Schweinler-Wigner Extremum Principle" arXiv:math-ph/9811003 $\endgroup$
    – mike stone
    Jun 14, 2021 at 11:49
  • 1
    $\begingroup$ Juts look at my notes oin this: people.physics.illinois.edu/stone/bose_bogoliubov.pdf $\endgroup$
    – mike stone
    Jun 26, 2021 at 13:10

I am liking the answers which have been posted, but figure to offer an answer which uses more primitive means. This system has a classical analog with Lagrangian of the form:

$$ L=\frac{1}{2}\cdot \dot{x}\cdot M\cdot \dot{x}-\frac{1}{2}\cdot x\cdot K\cdot x\mp \frac{1}{2}\cdot \dot{x}\cdot C\cdot x $$ I say $\mp \frac{1}{2}\cdot \dot{x}\cdot C\cdot x$ since I am not 100 % certain that the $p_x y$ term is to be considered momentum - potential energy as is the case in electrodynamics.

Lagrange' s equation $\frac{d}{dt}\cdot \frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}$ produces:

$$ \left( M^T+M \right) \cdot \ddot{x} \mp C\cdot \dot{x}=\mp C^T\cdot \dot{x}-\left(K^T+K\right)\cdot x $$

Which produces an eigenvalue equation and I assume that all who are reading this know how to find the normal modes of this system such that the Lagrangian becomes:

$$ L=\frac{1}{2} \underset{\alpha }{\Sigma } \left(\dot{Q}_{\alpha }^2-\omega _{\alpha }^2 Q_{\alpha }^2\right) $$

where the $\left\{\overset{\to }{Q}_{\alpha }\right\}$ and $\left\{\omega _{\alpha }\right\}$ are the eigenmodes and eigenfrequencies. And so the classical Hamiltonian is:

$$ H=\frac{1}{2} \underset{\alpha }{\Sigma } \left(p_{\alpha }^2+\omega _{\alpha }^2 Q_{\alpha }^2\right) $$

And the only step left is to use canonical quantization :

$$ Q_{\alpha }\to \hat{Q}_{\alpha },p_{\alpha }\to \hat{p}_{\alpha }=-i\hbar \cdot \frac{\partial }{\partial Q_{\alpha }} $$

where I use the hats to indicate that $\hat{Q}$ and $\hat{p}$ are quantum operators.


I started off by assuming a general transformation of the form:

$$\left\{\begin{array}{l} p_{x}=v p_{-}+c p_{+}+d x_{+}, \\ p_{y}=k p_{-}+l p_{+}, \\ x=n x_{+}+q x_{-}, \\ y=u x_{+}+ax_{-}+b p_{+} , \end{array}\right.$$

with ${p_+, p_-,x_+, x_-}$ the new variables. To avoid confusion, I am also renaming the coupling coefficient $C$ to $\gamma$. Next I plugged the above transformations in the Hamiltonian and demanded that no off-diagonal terms survive. This constraints the transformation coefficients. I also demand that the coefficients next to $p_+^2 $ and $p_-^2$ in the Hamiltonian be $1$. This leads to the set of transformations

$$\left\{\begin{array}{l} p_{x}=-\sqrt{1-l^{2}} p_{+}+l p_{m}-\frac{a \gamma}{2} x_{-}, \\ p_{y}=k p_{+}+\sqrt{1-l^{2}} p_{-}, \\ x=n x_{+}, \\ y=a x_{-} . \end{array}\right.$$

We are now left with the set of coefficients ${a, l, n}$ to be determined. This can be done by enforcing the canonical commutation relations on the new and old sets of coordinates. We then find

$$[x, p_y]=0 \Rightarrow l=0, \\ [x, p_x]=i \hbar \Rightarrow n=-1, \\ [y, p_y] = i \hbar \Rightarrow a=1.$$

The resulting set of transformations is hence

$$\left\{\begin{array}{l} p_{x}=-p_{+}- \frac{\gamma}{ 2} x . \\ p_{y}=p_{-}, \\ x=-x_{+}, \\ y=x . \end{array}\right.$$

leading to the Hamiltonian

$$H = p_{+}^{2}+p_{-}^{2}+\omega^{2} x_{+}^{2}+\left(\Omega^{2}-\gamma^{2} / 4\right) x_{-}^{2}.$$

The normal modes are obviously

$$\omega_+ = \pm \omega, \,\,\,\,\,\,\,\, \omega_- = \sqrt{\left(\Omega^{2}-\gamma^{2} / 4\right)}.$$


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