Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In Zworski's Semiclassical Analysis, he defines the following method of quantization: for a symbol $a = a(x,\xi) \in \mathscr{S}(\mathbb{R}^{2n})$ and $u \in \mathscr{S}(\mathbb{R}^n)$,

$$ Op_t(a)u(x) : = \frac{1}{(2\pi h)^n} \int_{\mathbb{R}^n \times \mathbb{R}^n} e^{\tfrac{i}{h}\langle x - y, \xi \rangle} a(tx + (1 - t)y, \xi) u(y) dy d\xi$$

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:

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?

Furthermore, perhaps I'm really more interested in

How can we interpret the role of the parameter $t$ in forming the pseudodifferential operators?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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).

With $t=0$, you start with the multiplications and then you take derivatives.

$t=1/2$ 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.

share|improve this answer
    
Any ideas where interesting problems/applications might arise where we use the quantization with various $t$ besides the ones mentioned? –  Christopher A. Wong Oct 20 '12 at 2:54
    
This business has to do with the $J^t=\exp{itD_x\cdot D_\xi}$: the adjoint of $op_t(a)=op(J^t a)$ is $op_{1-t}(\overline a)$, so somehow the only way to have full stability with respect to taking adjoints is indeed the Weyl choice. Of course, you have always asymptotic stability for good classes of symbols, say up to $h^\infty$ in the semi-classical case, but it is not very satisfactory at the algebraic level. –  Bazin Oct 21 '12 at 21:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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