Interpretation of a parameter in forming a pseudodifferential operator - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:52:12Z http://mathoverflow.net/feeds/question/109356 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109356/interpretation-of-a-parameter-in-forming-a-pseudodifferential-operator Interpretation of a parameter in forming a pseudodifferential operator Christopher A. Wong 2012-10-11T05:47:12Z 2012-10-14T13:36:29Z <p>In Zworski's <em>Semiclassical Analysis</em>, he defines the following method of quantization: for a symbol $a = a(x,\xi) \in \mathscr{S}(\mathbb{R}^{2n})$ and $u \in \mathscr{S}(\mathbb{R}^n)$,</p> <p>$$Op_t(a)u(x) : = \frac{1}{(2\pi h)^n} \int_{\mathbb{R}^n \times \mathbb{R}^n} e^{\tfrac{i}{h}\langle x - y, \xi \rangle} a(tx + (1 - t)y, \xi) u(y) dy d\xi$$</p> <p>where $t \in [0,1]$ is some parameter. The presence of $h$ is only because this form of quantization is motivated by quantum mechanics, and the Weyl Quantization is equal to $Op_{1/2}(a)$, with $t = 1/2$. It's also useful to notice that the above family of quantizations obeys the adjoint formula $$Op_t(a)^* = Op_{1 - t}(\bar{a}),$$ from which we can see that Weyl quantization on real symbols give self-adjoint operators, as desired in quantum mechanics. However, for $t \neq 1/2$, we lose this self-adjointness, and the only other value of $t$ treated in the text (as far as I know) is $t = 1$, because the quantization formula is very simple, and it is also one of the traditional ways to form pseudodifferential operators. My question is:</p> <blockquote> <p>What are contexts in which we might use a quantization of the above form with $t \neq \tfrac{1}{2}, 1$? Are there any natural situations in which they arise?</p> </blockquote> <p>Furthermore, perhaps I'm really more interested in</p> <blockquote> <p>How can we interpret the role of the parameter $t$ in forming the pseudodifferential operators?</p> </blockquote> http://mathoverflow.net/questions/109356/interpretation-of-a-parameter-in-forming-a-pseudodifferential-operator/109610#109610 Answer by Bazin for Interpretation of a parameter in forming a pseudodifferential operator Bazin 2012-10-14T13:36:29Z 2012-10-14T13:36:29Z <p>Let me answer to your second query and make $h=1$. You have $$Op_1(a(x) \xi)= a(x) D_x,\quad \text{with D_x=-i\partial_x},$$ $$Op_0(a(x) \xi)= D_x a(x),$$ $$Op_{1/2}(a(x) \xi)= \frac 12D_x a(x)+\frac 12a(x)D_x.$$ With $t=1$, you start with the derivations and then you multiply by the coefficients (in the case of a differential operator).</p> <p>With $t=0$, you start with the multiplications and then you take derivatives.</p> <p>$t=1/2$ is a symmetric compromise between the two bad solutions above. Note that the most important property of Weyl quantization is its symplectic invariance and not only the fact that real-valued symbols (Hamiltonians) get quantized by (formally) selfadjoint operators.</p>