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  How to find the extra constants in quantization in action-angle coordinates?

+ 2 like - 0 dislike

Consider a harmonic oscillator
$$H = \frac{1}{2}(p^2 + x^2)$$

We make a canonical transformation to $I = (p^2 + x^2)/2$ and $\varphi = \arctan(p/x)$. It is then easy to see that $\{I,\varphi \} = 1$ and the Hamiltonian reduces to

$$H = I$$

We now canonically quantize this Hamiltonian and we obtain that $\hat{I} = -i\hbar \partial_\varphi$. The stationary Schrödinger equation then simply reads

$$-i\hbar \partial_\varphi \psi = E\psi$$

with the obvious solution $\sim e^{i E_n \varphi/\hbar}, \,E_n = hn$. However, this is different from the result $E_n = h(n+1/2)$ we get by quantizing in the usual phase-space coordinates $p,x$.

This is a general pattern - it is very easy to find energy levels and other quantum numbers in action-angle coordinates but this will introduce a shift in the results as compared to the initial coordinate system.

This is probably a consequence of the nonlinear $\sim p^n x^k $ nature of the transformation and will thus introduce $\mathcal{O}(\hbar)$ differences due to the operator ordering ambiguity of canonical quantisation.This makes me believe that these shifts will always be only shifts by a constant.

It is not clear to me

  1.  Whether the observable consequences (such as $E_n - E_m = h (n-m)$) are truly always independent of the phase-space coordinate system in which we execute canonical quantisation.
  2. How to compute the value of the shifts to the quantum numbers induced by the coordinate transform.

Does anyone know the answer to this?

asked Apr 22, 2017 in Theoretical Physics by Void (1,645 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

In general, canonical quantization doesn't work in any phase-space coordinate system but only in local Cartesian coordinates. The situation of completely integrable systems ist very special. I am still waiting for recasting your question hinted at in https://www.physicsoverflow.org/38781.

In addition, the constant shift has no meaning as generally, only energy differences or gradients figure in expressions comparable with experiment. One can always renormalize the Hamiltonian so that the ground state energy is zero, and indeed this must be done in quantum field theory (where otherwise meaningless infinite constants appear). 

answered Apr 23, 2017 by Arnold Neumaier (15,787 points) [ revision history ]

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