Canonical transformation

Chen

Hello,

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

[tex]H(p, q) = p^2 + 3pq + q^2[/tex]

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

[tex]p = P \cos{\theta} + Q \sin{\theta}[/tex]
[tex]q = -P \sin{\theta} + Q \cos{\theta}[/tex]

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

Applying this transformation, I get:

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

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

[tex]H(P,Q) = \frac{P^{2}}{2} + kQ^{2}[/tex]

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

krab

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