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?

  • 1
    $\begingroup$ 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?) $\endgroup$ Commented Aug 17, 2021 at 15:52
  • $\begingroup$ However, there are highly recondite exceptions that "prove" (~test) the rule... the quantization transition is very subtle... $\endgroup$ Commented Aug 17, 2021 at 16:10
  • $\begingroup$ Wow! Now, how do you quantize (order) the quantum hamiltonian? $\endgroup$ Commented Aug 17, 2021 at 16:26
  • $\begingroup$ ok I see what you mean. Because of the non commutation canonical transformations are not so trivial $\endgroup$ Commented Aug 17, 2021 at 16:35
  • 1
    $\begingroup$ 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
    Commented Mar 9, 2022 at 0:17

2 Answers 2


I do not understand well the question, even in the specific case you consider.

Are you asking whether or not there is a unitary transformation $U: L^2(\mathbb{R}, dx) \to L^2(\mathbb{R}, dx)$ such that $$U\hat{P}U^{-1}= |\hat{X}|^{-1} \hat{P}\:, \quad U\hat{X}U^{-1}= (...)?$$

Obviously NO: $|\hat{X}|^{-1} \hat{P}$ is not even selfadjoint as it would be if $U$ were unitary.

There is first of all an interpretative issue concerning the "quantization" of functions of $x$ and $p$ as $|x|^{-1}p$. Without a precise definition, the question is not well-posed. On the other hand we know that there is no quantization map that satisfies all natural properties one expects, as established in the various no-go results popularly known as the "Groenewold-van Hove theroem".

In summary, it is not possible to answer without a precise interpretation of the quantization of functions of $x$ and $p$. On the other hand no quantization procedure exists which satisfies all expected requirments.

It is possible to reformulate the issue in the framework of some specific quantization procedures, but the answer strictly depends on the made choices. I think that at the end of the day many spurious arguments would enter the final mathematically correct issue making it quite far from the original spirt, based on naive viewpoints as the Dirac quantization procedure.

  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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.