Consider the following peculiar Lagrangian with two degrees of freedom $q_1$ and $q_2$

$$ L = \dot q_1 q_2 + q_1\dot q_2 -\frac12(q_1^2 + q_2^2) $$

and the goal is to properly quantize it, following Dirac's constrained quantization procedure. (This is a toy example related to Luttinger liquids and the fractional quantum Hall effect. The degrees of freedom $q_1$ and $q_2$ correspond to two bosonic modes $a_k$ and $b_k$.)

First, note that the equations of motion are

$$ \dot q_1 = -q_2 ,\quad \dot q_2 = -q_1 ,$$

which shows that this model is not complete nonsense as it carries interesting dynamics. (EDIT: Many thanks to the answerers for pointing out my silly mistake: these equations are wrong, the correct ones would be $q_1=q_2=0$.)

However, the question is

How to quantize the above Lagrangian in a systematic fashion?

(I'm actually trying to quantize a different model, but with similar difficulties, hence the emphasis on "systematic"). The usual procedure of imposing canonical commutation relations does not work because the velocities cannot be expressed in terms of the conjugate momenta. According to Dirac, we have to interpret the equations for the canonical momenta as constraints

$$ \phi_1 = p_1 - \frac{\partial L}{\partial \dot q_1} = p_1 - q_2 \approx 0 $$ $$ \phi_2 = p_2 - \frac{\partial L}{\partial \dot q_2} = p_2 - q_1 \approx 0 $$

The hamiltonian is

$$ H = \frac12 (q_1^2 + q_2^2) $$

Unfortunately, the constraints have poisson brackets $\lbrace\phi_1,\phi_2\rbrace = 0$ and the secondary constraints read

$$ \lbrace \phi_1 , H \rbrace = q_1 \approx 0$$ $$ \lbrace \phi_2 , H \rbrace = q_2 \approx 0$$

Clearly, these weirdo constraints no longer have any dynamics and no useful quantization will come out of them.

Is there a systematic method to quantize this theory, for example BRST quantization? Or did I simply make a mistake while trying to apply Dirac's constrained quantization procedure?

  • 2
    $\begingroup$ Somehow I could not get (classical) EOM from your Lagrangian. For instance dL/dq1 = q2' - q1. dL/dq1' = q2. Using dL/dq1 = d[dL/dq1']/dt I get q2' - q1 = q2', which does not look like a valid EOM. $\endgroup$
    – valdo
    Sep 18, 2011 at 15:26
  • $\begingroup$ @Greg,if you change the relative sign between the derivative terms, you would have got an interesting problem, where one can exercise the treatment of Dirac's first class constraints, or equivalently symplectic reduction or the projection onto the lowest landau level. $\endgroup$ Sep 19, 2011 at 9:28

2 Answers 2


The classical equations of motion are not affected by changing the Lagrangian

$$L \qquad \longrightarrow \qquad L' = L+ \frac{dF}{dt}$$

by a total time derivative. Put $F= -q_1 q_2$. Then

$$L' = -\frac{1}{2}(q_1^2 + q_2^2).$$

This Lagrangian $L'$ does not contain time derivatives, and thus there are no dynamics. The classical equations of motion are

$$ q_1=0 \qquad \mathrm{and}\qquad q_2=0, $$

in conflict with what is said in the original question formulation (v1).


The problem is that this Lagrangian is directly integrable with respect to time, and so the time-dependence is determined by a boundary condition, and there are no locally defined conjugage momenta.

Consider just the portion of this action that has explicit time derivatives

$$S=\int dt \int d^{3}x \,\,q_1 {\dot q_{2}} +q_{2} {\dot q_{1}}$$

Now, perform the transformation:

$$\begin{align*} q_{1}&=a Q_{1} + b Q_{2}\\ q_{2}&=c Q_{1} + d Q_{2} \end{align*}$$

If $bc+ad=0$, then this part of the action becomes:

$$\begin{align*} S&=\int d^{3}x\int dt\, A Q_{1}{\dot Q_{1}} + B Q_{2} {\dot Q_{2}}\\ &=\int d^{3}x\,\left[\frac{A}{2} Q_{1}^{2} + \frac{B}{2} Q_{2}^{2}\right]_{t_{0}}^{t^{f}} \end{align*}$$

Where $A$ and $B$ are constants that depend on the choice of $a,b,c,d$ under the constraint that $bc+ad=0$. Now, there are no explicit time derivatives in the action at all (and we would be free to transform back to the $q_{1},q_{2}$ coordinates if we wished, and formally, the conjugate momenta are zero. This is why you must be able to solve for the velocities when doing the Legendre transform. Otherwise, the conjugate momenta will be ill-defined and the Lagrange transformation will fail. This action is topological and doesn't have a well-defined local Hamiltonian dynamics.

  • $\begingroup$ Thanks! While a topological action would be interesting, I finally found the mistake I made in my original problem. $\endgroup$ Sep 19, 2011 at 11:26
  • $\begingroup$ @Greg: and yeah, Qmechanic's solution to this is much cleaner than mine. :) $\endgroup$ Sep 19, 2011 at 11:55

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