# Single-mode hamiltonian

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:

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

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

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

• This is a standard first order ODE with constant coefficients ($\dot{y} = C y$). I'm sure you have seen it before. – lcv Jan 17 '19 at 16:30
• Are you familiar with the concept of normal modes? You can write your system of differential equations in matrix form and then diagonalize it... – DanielSank Jan 17 '19 at 17:49
• 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? – Franz Jan 17 '19 at 18:49
• The matrix you have written in your coupled differential equations is indeed normal and can be diagonalized. Hint, the eigenvalues are $\pm\sqrt{\omega^2 - 4\theta^2}$ – user1936752 Jan 17 '19 at 19:37
• 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 – 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
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 $$L_y$$.