I´m a bit stuck with an exercise I have to do for a class of mine. We have been given a Hamiltonian

$$\hat{H}=\hbar\omega\hat{a}^{\dagger}\hat{a}+\hbar\theta\left(\hat{a}^2+\hat{a}^{\dagger 2}\right)$$

and were asked to calculate the time evolution of $\hat{x}(t)$ and $\hat{p}(t)$. I tried to first calculate the time evolution of the ladder operators of which they consist using Heisenberg's equations of motion $$\frac{i}{\hbar}[\hat{H},\hat{a}]=\frac{d}{dt}\hat{a}(t).$$ I then receive two coupled differential equations:


$$\frac{d}{dt}\hat{a}^{\dagger}(t)=i(\omega\hat{a}^{\dagger}+2\theta\hat{a}). $$

However, I don't really know how to solve these kind of equations. I would be very grateful for any of your help,

  • 2
    $\begingroup$ This is a standard first order ODE with constant coefficients ($\dot{y} = C y$). I'm sure you have seen it before. $\endgroup$ – lcv Jan 17 '19 at 16:30
  • $\begingroup$ Are you familiar with the concept of normal modes? You can write your system of differential equations in matrix form and then diagonalize it... $\endgroup$ – DanielSank Jan 17 '19 at 17:49
  • $\begingroup$ Thank you for your answers. I am not too familiar with this way of solving coupled differential equations but had already found a page about normal modes at the time I posted this. The problem is that the matrix that connects the time derivatives of the ladder operators and the operators themselves isn't a normal matrix so it cannot be diagonalized. Or am I missing something? $\endgroup$ – Franz Jan 17 '19 at 18:49
  • 1
    $\begingroup$ The matrix you have written in your coupled differential equations is indeed normal and can be diagonalized. Hint, the eigenvalues are $\pm\sqrt{\omega^2 - 4\theta^2}$ $\endgroup$ – user1936752 Jan 17 '19 at 19:37
  • $\begingroup$ Thank you for the help! Before starting to diagonalize the matrix I checked if it commuted with its transposed conjugate to make sure I could solve the equation trough finding the normal modes. I probably just messed up there. Thanks again I will try to flag this as solved $\endgroup$ – Franz Jan 17 '19 at 19:51

Although this might be a bit hifalutin, there is an algebraic structure to this problem since the operators $$ K_0=\frac{1}{2}(\hat a^\dagger a+\hat a\hat a^\dagger)\, ,\qquad K_+=\hat a^\dagger\hat a^\dagger\, ,\qquad K_-=\hat a\hat a $$ close on the Lie algebra $\mathfrak{su}(1,1)$. One can verify that $\{K_0,K_\pm\}$ actually close under commutation, and the commutation relations differ from those of angular momentum by an single sign. You can find details on $\mathfrak{su}(1,1)$ in

  1. Ban, Masashi. "Decomposition formulas for su (1, 1) and su (2) Lie algebras and their applications in quantum optics." JOSA B 10.8 (1993): 1347-1359.
  2. Wodkiewicz, K., and J. H. Eberly. "Coherent states, squeezed fluctuations, and the SU (2) am SU (1, 1) groups in quantum-optics applications." JOSA B 2.3 (1985): 458-466.

    both of which emphasize the similarities and differences with angular momentum theory. There is also useful material in

  3. Ui, Haruo. "su (1, 1) quasi-spin formalism of the many-boson system in a spherical field." Annals of Physics 49.1 (1968): 69-92.

Thus your Hamiltonian can be rewritten as $$ H= \alpha K_0+\beta K_x $$ where $K_x=\frac{1}{2}(K_++K_-)$. A transformation $T(\tau)$ generated by $K_y=\frac{1}{2i}(K_+-K_-)$: $$ T(\tau)=e^{-i \tau K_y} $$ with suitable $\tau$ will bring your Hamiltonian to diagonal form.

In the literature, $T(\tau)$ is often referred to as a squeezing transformation, or Bogoliubov transformation. Its effect is to transform $\hat a$ and $\hat a^\dagger$ into linear combinations $\hat b$ and $\hat b^\dagger$ such that $$ \left(\begin{array}{c} \hat b \\ \hat b^\dagger\end{array} \right)= \left(\begin{array}{cc} \cosh\frac{1}{2}\tau&\sinh\frac{1}{2}\tau\\ \sinh\frac{1}{2}\tau&\cosh\frac{1}{2}\tau\\ \end{array}\right) \left(\begin{array}{c} \hat a \\ \hat a^\dagger\end{array} \right)\, . $$ In terms of the position and momentum, the transformation scales $x\to X= \zeta x$ and $p\to p/\zeta$, which is equivalent to changing the frequency of the oscillator. The actual calculation would be in the spirit of diagonalizing $H=\alpha L_z+\beta L_x$ using a rotation generated by $L_y$.


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