What is the Correct Angle for Canonical Transformation?

[Chen]

Hello,

I need to solve the Hamiltonian of a one-dimensional system:

$$H(p, q) = p^2 + 3pq + q^2$$

And I've been instructed to do so using a canonical transformation of the form:

$$p = P \cos{\theta} + Q \sin{\theta}$$
$$q = -P \sin{\theta} + Q \cos{\theta}$$

And choosing the correct angle so as to the get the Hamiltonian of an harmonic oscillator.

Applying this transformation, I get:

$$H(P, Q) = P^2 + Q^2 - 3/2 (P^2 - Q^2) \sin{2 \theta} + 3 P Q \cos{2 \theta}$$

And as far as I can see, no choice of angle will get me to an Hamiltonian of an harmonic oscillator.

Am I correct? Can someone please check my calculation?

Thanks.

 

[AlphaNumeric]

I got the same expression for H(P,Q) as you, and I can't see what angle would give a nice form either. You'd want to get something like

$$H(P,Q) = \frac{P^{2}}{2} + kQ^{2}$$

from which you can then get the frequency from, but that would want $$\sin 2\theta = \frac{1}{3}$$ which isn't going to drop the 3PQ term you've got.

 

[krab]

All you need is to get $\cos 2\theta=0$. So what about $\theta=\pi/4$?