1

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!

Viktor Zelezny
  • 115
  • You will have ordering problems since not all operators in your exponential commute. – ZeroTheHero Nov 03 '20 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. – hwang Nov 04 '20 at 00:30

1 Answers1

1

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

Cosmas Zachos
  • 62,595