Reading canonical commutation relations from the action (QHE)

[binbagsss]

Hi

I'm sure i understood this a week or so ago, and I've forgot the idea now. I'm just really confused, again, how you read the commutator relationships of from the action ?

many thanks

(source http://www.damtp.cam.ac.uk/user/tong/qhe/five.pdf)

 

[vanhees71]

This is called "canonical quantization" and is a recipe to make a guess for a quantum theory analogue of a classical dynamical model formulated with the action principle. This is of course very dangerous, because it works sometimes but sometimes it doesn't. E.g., if you "canonically quantize" the spinning top (in non-relativistic QT) you run in trouble when doing it too naively. Safer ground is the analysis in terms of group theory a la Noether.

With that caveat, the logic of canonical quantization (for field theories) is that you have a Lagrangian involving some fields $\phi_k$ of the form
$$\mathcal{L}=\mathcal{L}(\phi_k, \dot{\phi}_k,\vec{\nabla} \phi_k).$$
Then you define the canonical conjugated field momenta,
$$\Pi_k=\frac{\partial \mathcal{L}}{\partial \dot{\phi}_k},$$
and then do the usual "translation" from (functional) Poisson brackets to commutators (bosons) or anticommutators (fermions) of field operators. In the Heisenberg picture, it's the equal-time commutator,
$$[\hat{\phi}_k(t,\vec{x}),\hat{\Pi}_l(t,\vec{y})]=\mathrm{i} \hbar \delta^{(3)}(\vec{x}-\vec{y}).$$
This defines the observable algebra, and if you are lucky that algebra makes physical sense ;-)).

 

[Demystifier]

E.g., if you "canonically quantize" the spinning top (in non-relativistic QT) you run in trouble when doing it too naively.

I didn't know that, can you give some reference?

 

[samalkhaiat]

Hi

I'm sure i understood this a week or so ago, and I've forgot the idea now. I'm just really confused, again, how you read the commutator relationships of from the action ?
View attachment 252531
View attachment 252532

many thanks

(source http://www.damtp.cam.ac.uk/user/tong/qhe/five.pdf)

From the action, you read off the Lagrangian \mathcal{L} = \frac{me^{2}}{4\pi \hbar} \ \epsilon^{\mu\nu\rho} \ a_{\mu} \partial_{\nu} a_{\rho} . Now, calculate \frac{\partial \mathcal{L}}{\partial (\partial_{\sigma}a_{\tau})} = \frac{me^{2}}{2\pi \hbar} \ \epsilon^{\mu \sigma \tau} \ a_{\mu} . Now, setting \sigma = 0 and \tau = 1, you obtain the conjugate momentum \pi^{1}(x) = \frac{me^{2}}{2\pi \hbar} \epsilon^{012} \ a_{2} (x), or \pi^{1} (x^{\prime}) = \frac{me^{2}}{2\pi \hbar} \ a_{2}( x^{\prime} ) .The result follows If substitute this in the postulated canonical commutation relation \big[ a_{1} (x) , \pi^{1} (x^{\prime}) \big] = i\hbar \delta^{2} (x - x^{\prime}).

 

[vanhees71]

I didn't know that, can you give some reference?

I think it's an example in Hagen Kleinert's book on path integrals.