# Canonical Transformation in Quantum Phase Space

I am looking for a unitary representation $$\hat T$$ of the following canonical transformation

\begin{align} q_1&\rightarrow q_2 &p_1&\rightarrow p_2\\ q_2 &\rightarrow -q_1&p_2&\rightarrow -p_1 \end{align}

which is a 90°-rotation in the $$(q_1,q_2)$$-subspace of a 4-dim phase-space. It is therefore a point-transformation, since it does not mix positions and momenta. $$\hat T$$ acts as

$$\hat T \hat q_1 \hat T^\dagger =\hat q_2 \quad \hat T \hat p_1 \hat T^\dagger =\hat p_2\\ \hat T \hat q_2 \hat T^\dagger =-\hat q_1 \quad \hat T \hat p_2 \hat T^\dagger =-\hat p_2$$

One guess of mine is

$$\hat T = e^{-i( p_1(q_1-q_2)-p_2(q_2+q_1))}$$

but I do not know a way of proofing it, apart from expanding the exponentials and then computing everything brute-force, e.g. in

$$\bigg( \sum_n^\infty \frac{i^j( p_1(q_1-q_2)-p_2(q_2+q_1))^n}{n!}\bigg) \hat q_1 \bigg( \sum_m^\infty \frac{i^m( p_1(q_1-q_2)-p_2(q_2+q_1))^m}{m!}\bigg) = \hat q_2.$$

One would have to commute $$\hat q_1$$ to the left, which seems ridiculously laborious to me. Is there an easy way to find $$\hat T$$ for such a point-transformation? And if one must resort to guessing, is there an easy way to proof that what one has found acts in the right way?

I am deeply grateful for any help!

• You will have ordering problems since not all operators in your exponential commute. Nov 3, 2020 at 22:52
• To calculate that expression, start from calculating simple and general commutator $[e^A, B]$. Since the only thing in '$A$' that does not commute with $q_1$ is $p_1$, calculation is not that cumbersome. Nov 4, 2020 at 0:30

Observe the obvious invariants $$I=q_1^2+ q_2^2, ~~~ J= p_1^2+p_2^2.$$ Observe the hermitian operator $$r=q_1p_2-q_2p_1$$ commutes with both of them, so it's worth considering its effect on your four variables, $$[r, q_1]=iq_2 \\ [r, q_2]=-iq_1 \\ [r, p_1]=ip_2 \\ [r, p_2]=-ip_1.$$
But this is the precise rotations you are after a π/2 rotation for, so $$T= e^{-i\pi r/2}$$ will do the trick, by the Hadamard identity, $$T q_1 T^\dagger = q_1 + (-i\pi/2) [r,q_1] + \frac{1}{2!} (-i\pi/2)^2 [r,[r,q_1]]+... \\ = q_1 \cos\pi/2 +q_2 \sin \pi/2= q_2,\\ T q_2 T^\dagger =- q_1, \\ T p_1 T^\dagger = p_2, \\ T p_2 T^\dagger =- p_1.$$