Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

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

+ 2 like - 0 dislike
76 views

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 in Theoretical Physics by Void (1,505 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 by Arnold Neumaier (12,385 points) [ revision history ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysi$\varnothing$sOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...