Why don't we use Hamilton-Jacobi method in QM?

Question

In classical mechanics, we usually try to find a set of coordinates by Hamilton-Jacobi method to transform the Hamiltonian to zero such that the coordinates are conservations.

However, we never try by similar steps to transform the Hamiltonian to zero because what we can get is nothing from the zero-Hamiltonian.

Why don't we use Hamilton-Jacobi method in QM?

Answer 1

I) Well semiclassically, the Hamilton-Jacobi equation famously appears to lowest order in $\hbar$ in a WKB expansion of the Schrödinger equation. See also Ref. 1.

II) The quantum concept of a canonical transformation (CT)

$$ z^I~=~(\hat{q}^i; \hat{p}_i) \quad\longrightarrow\quad Z^J~=~(\hat{Q}^j; \hat{P}_j) \tag{1}$$

(where the old and new canonical variables both satisfy the CCR) typically becomes very difficult to implement to all quantum orders unless we are talking about an affine transformation

$$ z^I \quad\longrightarrow\quad Z^J ~=~ A^J{}_I z^I +b^J. \tag{2}$$

Thus the great flexibility of classical CT is (at least from a practice computational point of view, although not from a theoretical point of view) replaced by rigidity at the full quantum level.

III) In practice, when quantizing a theory, one would first seek out the simplest classical formulation of the problem (which yields the same classical equations of motions, and which is most susceptible to quantization), and then try to quantize that.

E.g. if the classical Lagrangian contains an overall square root, one would typically try to find an equivalent classical Lagrangian that is quadratic in the fundamental variables before attempting to quantize the system.

References:

1. B.S. DeWitt, The Global Approach to QFT, Vol 1, 2003; eq. (13.12).