In Hamiltonian mechanics, a canonical transformation is a change of canonical coordinates $(q, p, t) \rightarrow (Q, P, t)$ that preserves the form of Hamilton's equations.

Now in quantum mechanics canonical transformation should be replaced by unitary transformation. My question is are all canonical transformations unitary transformations?

To clarify what I mean suppose we we have quantum Hamiltonian $$ {H}=H({q}, {p})=\frac{1}{2}\left({p}^{2}+\omega^{2} {q}^{2}\right) $$ Performing a transformation of coordinates $ T\left(x^{\prime}, p^{\prime}\right)=$ $(x, p)$ where $$ x=\left\{\begin{array}{rl} \sqrt{\left|2 x^{\prime}\right|}, & x^{\prime}>0 \\ -\sqrt{\left|2 x^{\prime}\right|}, & x^{\prime}<0 \end{array}, \quad p=p^{\prime} \sqrt{\left|2 x^{\prime}\right|}\right. \text {, } $$ with the inverse $T^{-1}$ in the form $$ x^{\prime}=\left\{\begin{array}{rl} \frac{1}{2} x^{2}, & x>0 \\ -\frac{1}{2} x^{2}, & x<0 \end{array}, \quad p^{\prime}=|x|^{-1} p\right. $$ the Hamiltonian $H$ transforms to the following function $$ H^{\prime}\left(x^{\prime}, p^{\prime}\right)=H\left(T\left(x^{\prime}, p^{\prime}\right)\right)=\left|x^{\prime}\right| p^{\prime 2}+\omega^{2}\left|x^{\prime}\right| $$ Is $T$ a unitary transformation?

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    $\begingroup$ Classical canonical transformations preserve Poisson Brackets. Quantum canonical transformations preserve the Born commutators, and are unitary. The two are connected in a subtle way. Is that the question? Motion is a canonical transformation, both classical and quantum. Is that the question? (Surely you are not asking about classical CTs in Hilbert space?) $\endgroup$ Aug 17, 2021 at 15:52
  • $\begingroup$ However, there are highly recondite exceptions that "prove" (~test) the rule... the quantization transition is very subtle... $\endgroup$ Aug 17, 2021 at 16:10
  • $\begingroup$ Wow! Now, how do you quantize (order) the quantum hamiltonian? $\endgroup$ Aug 17, 2021 at 16:26
  • $\begingroup$ ok I see what you mean. Because of the non commutation canonical transformations are not so trivial $\endgroup$ Aug 17, 2021 at 16:35
  • $\begingroup$ All unitary transformations are canonical, not all canonical transformations are unitary. A trivial example is a change in scale: multiply all the x's by two and divide all the p's by 2, that is canonical, but not unitary. $\endgroup$
    – AndresB
    Mar 9 at 0:17

1 Answer 1

  1. Recall that the Poisson bracket is related to the Groenewold-Moyal star product via $$ A\star B~:=~A \exp\left(\stackrel{\leftarrow}{\partial_I}\frac{i\hbar}{2} \omega^{IJ} \stackrel{\rightarrow}{\partial_J}\right) B~=~AB+\frac{i\hbar}{2}\{A,B\}_{PB} +{\cal O}(\hbar^2), \tag{1} $$ cf. e.g. this related Phys.SE post.

  2. Moreover, the algebra of functions/symbols $A,B,\ldots $ on phase space (equipped with the star product $\star$) is isomorphic to the algebra of operators $\hat{A},\hat{B},\ldots$ (using composition $\circ$), cf. e.g. this related Phys.SE post.

  3. Returning to OP's title question, if a function/symbol $H$ is a generator of a canonical transformation/symplectomorphism/Hamiltonian vector field/Hamiltonian flow, then the corresponding self-adjoint operator $\hat{H}$ is a generator of a unitary transformation $e^{\frac{i}{\hbar}\hat{H}}$ in the quantum theory, cf. above comment by Cosmas Zachos.


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