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Canonical Transformation in Quantum Phase Space

Physics Asked on May 30, 2021

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!

One Answer

Skip the silly hats--everything is an operator.

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 $$ T= e^{-ipi r/2} $$ will do the trick, by the Hadamard identity, $$ T q_1 T^dagger = q_1 + (-ipi/2) [r,q_1] + frac{1}{2!} (-ipi/2)^2 [r,[r,q_1]]+... = q_1 cospi/2 +q_2 sin pi/2= q_2, T q_2 T^dagger =- q_1, T p_1 T^dagger = p_2, T p_2 T^dagger =- p_1. $$

Correct answer by Cosmas Zachos on May 30, 2021

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