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 :-(

  • $\begingroup$ related $\endgroup$ Aug 8 '16 at 21:43
  • $\begingroup$ Also related $\endgroup$ Aug 8 '16 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 '16 at 3:12
  • $\begingroup$ Oh, I misunderstood the question. Ignore my comment. $\endgroup$ Aug 9 '16 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 '16 at 3:18

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.

  • $\begingroup$ 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 '16 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 '16 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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