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I'd like to define (and calculate the properties of) an object like $e^{\boldsymbol{A\cdot\nabla}}f\left(\boldsymbol{x}\right)$, where $A$ is a collection of matrices written in a vector form, say $\boldsymbol{A\cdot\nabla}=A_{x}\partial_{x}+A_{y}\partial_{y}$ for instance (I think that's the simplest non-trivial example), in which case $f\left(\boldsymbol{x}\right)\equiv f\left(x,y\right)$. The difficulty is that $\left[A_{x},A_{y}\right]\neq0$. $f$ has all the properties one needs (smoothness and co ; it comes from a problem of physics), and it commutes with the $A$'s. The $A$'s do not depend on the coordinates (in the example, this property would read $\partial_{x}A_{x,y}=\partial_{y}A_{x,y}=0$)

Does this object make any sense ? Does it have a name ?

For instance, I'd like to show something like $$e^{\boldsymbol{A\cdot\nabla}}f\left(\boldsymbol{x}\right)e^{-\boldsymbol{A\cdot\nabla}}g\left(\boldsymbol{x}\right)=f\left(\boldsymbol{x}+\boldsymbol{A}\right)g\left(\boldsymbol{x}\right)$$ but this is obviously wrong by direct expansion (it is true when e.g. $A_{y}=0$ in the example above). At second order there are some commutators $\left[A_{i},A_{j}\right]$ appearing.

Any help is warm welcome. (I'm not even sure the tags below are well chosen :-(

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  • $\begingroup$ related $\endgroup$
    – Ben Grossmann
    Aug 8, 2016 at 21:43
  • $\begingroup$ Also related $\endgroup$
    – Ben Grossmann
    Aug 8, 2016 at 21:47
  • $\begingroup$ @Omnomnomnom Thanks you very much for adding tags. I'm not sure I understand your previous comment. In the case when there is only one matrix $A$, the formula I gave in the question applies, i.e. $ e^{A\partial_{x}}f\left(x\right)e^{-A\partial_{x}}g\left(x\right) = f \left( x+A \right) g \left( x \right) $. This is just because all the $A$'s commute and the direct expansion of the exponential into a series works straightforwardly. $\endgroup$
    – FraSchelle
    Aug 9, 2016 at 3:12
  • $\begingroup$ Oh, I misunderstood the question. Ignore my comment. $\endgroup$
    – Ben Grossmann
    Aug 9, 2016 at 3:16
  • $\begingroup$ The difficulty is to deal with, say $e^{A_{x}\partial_{x}+A_{y}\partial_{y}}f\left(x,y\right)$ with $A_{x,y}=\sigma_{x,y}$, the $\sigma$ being Pauli matrices for instance. $\endgroup$
    – FraSchelle
    Aug 9, 2016 at 3:18

1 Answer 1

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Let $l(x,\xi) = \langle x, x^*\rangle + \langle \xi, \xi^*\rangle$ for some fixed $x^*, \xi^* \in \mathbb{R}^d$. Then we get with Weyl quantization that $$(e^{il})^w(x,D) = e^{il(x,D)},$$ where $$e^{il(x,D)}u(x) := e^{i\langle x^*, x\rangle + i/2\langle x^*, \xi^*\rangle}u(x + \xi^*)$$ is the (unique) solution to $i\partial_t v + l(x,D)v = 0$ with initial data $v(0) = u$.

The main point is that you have to use Weyl quantization to get rid of the commutators. The usual pseudodifferential (Kohn-Nirenberg) quantization is not good in that respect.

You can find the above theorem in Zworski - Semiclassical Analysis. Pseudodifferential operators can be found in various books on PDE.

I hope this helps.

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  • $\begingroup$ Surely it helps somehow. The object I'm trying to define indeed come from semi-classical quantization with a gauge-field (for instance $A_{i}=A_{i}^{a}\sigma^{a}/2$ with $\sigma$ the Pauli matrices), and certainly to identify the correct flow is a crucial step in using the object. But I'm not really familiar with all this. I'll try to check in the book you mention. Thank you very much for the help. $\endgroup$
    – FraSchelle
    Aug 9, 2016 at 11:17
  • $\begingroup$ If you use semiclassical operators you can even do better: To any reasonable symbol $p$ and its Hamiltonian flow $\kappa_t$ you can associate a semiclassical evolution equation $hD_t u + p^w(x,hD) u = 0$ and the solution operator $S(t) : u_0 \mapsto u(t)$ satisfies the semiclassical Egorov theorem: $S(-t) a^w(x,hD) S(t) = b^w(x,hD)$ with $b = \kappa_t^* a + O(h)$. This is Thm 11.1 in the book. The intuition is always that the dynamics are governed by the Hamiltionian flow. $\endgroup$
    – mcd
    Aug 9, 2016 at 14:57

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