Solution to Exercise 16

We want to diagonalize the bosonic Hamiltonian

$${\mathcal H}= \frac{1}{2} \left( \begin{array}{cc} a^{\da... ... \left( \begin{array}{c} a \\ a^{\dagger}\end{array} \right)$$

by appropriate canonical transformation. Here $c$, $b$ are real and $c > b$. The operators obey bosonic commutation relation $[a, a^{\dagger}] = 1$. We consider a transformation of the form

$$\left( \begin{array}{c} a \\ a^{\dagger}\end{array} \right)... ... \left( \begin{array}{c} c \\ c^{\dagger}\end{array} \right).$$ (1)

Inverting the above transformation we get $[ c , c^{\dagger}] = 1/ (u^2 - v^2)$. Since, the transformation has to be canonical, we get $u^2 - v^2 = 1$. This is satisfied by $u = \cosh \theta$ and $v = \sinh \theta$. Thus, the two unknown parameters have been reduced to one. We will choose this unknown $\theta$ such that the Hamiltonian is diagonal in terms of the $c, c^{\dagger}$ operators. From equation (1) we get

$${\mathcal H} = \frac{1}{2} \left( \begin{array}{cc} c^{\dagger}& c \end{array} \... ...end{array} \right) \left( \begin{array}{c} c \\ c^{\dagger}\end{array} \right)$$

$$= \frac{1}{2} \left( \begin{array}{cc} c^{\dagger}& c \end{array} \... ...nd{array} \right) \left( \begin{array}{c} c \\ c^{\dagger}\end{array} \right).$$ (2)

The value of $\theta$ that diagonalizes the Hamiltonian is given by $\tanh 2\theta = - b/c$. Then, $\cosh 2\theta = c/\sqrt{c^2 - b^2}$ and $\sinh 2\theta = -b/ \sqrt{c^2 - b^2}$. We get,

$${\mathcal H} = \frac{1}{2} \left( \begin{array}{cc} c^{\dagger}& c \end{array} \... ...end{array} \right) \left( \begin{array}{c} c \\ c^{\dagger}\end{array} \right)$$

$$= \hbar \omega \left( c^{\dagger}c + \frac{1}{2} \right).$$ (3)

Thus, the spectrum is the usual harmonic oscillator spectrum with frequency $\hbar \omega = \sqrt{c^2 - b^2}$.