A question regarding a Hamiltonian.

[MathematicalPhysicist]

In page (3) of the next link:http://www.phy.bris.ac.uk/people/berry_mv/the_papers/Berry401.pdf
he writes that one can transform the hamiltonian H=xp to p^2-x^2 with a simple rotation transformation.

Which one is that?

 

[dextercioby]

In the phase space (x,p), consider the transformations/rotations

$$2p'=p+x$$

$$2x'=p-x$$

What is H(x(x',p'),p(x',p')) equal to ?

 

[MathematicalPhysicist]

But aren't x and p not commutable? ($$[x,p]=i\hbar$$).

I mean $$p'^2-x'^2=1/4 (p^2+x^2+px+xp- p^2-x^2 +px+xp)=1/2 \{x,p\}$$

 

[Bill_K]

When he writes H = xp, he means a classical Hamiltonian. In QM it's not Hermitian until you symmetrize it.

"... its quantum counterpart (obtained by symmetrization)..."

 

[MathematicalPhysicist]

Ok, thanks.
So in QM we would take (xp+px)/2.

 

[dextercioby]

Ok, thanks.
So in QM we would take (xp+px)/2.

Actually $\frac{1}{2}\left(\bar{\displaystyle{\hat{x}\hat{p}+\hat{p}\hat{x}}}\right)$ (the bar should extend on both terms in the bracket), but for practical purposes, the operator without the bar is enough.