Consider a particle on the real line with:

$L=\frac{1}{2}(\partial_0q)^2 + f(q)\partial_0q$

the equation of motion is that of a free particle $\partial_0^2q=0$. In fact $\delta[f(q)\partial_0q]=0$. Is this right?


1) I assume that OP means that $q=q(t)$ is a single dynamical variable for a classical Lagrangian function

$$ L~=~ \frac{1}{2}m\dot{q}^2 + f(q)\dot{q}, $$

where $f=f(q)$ is a given function. I also assume that the problem is well-posed, i.e., that the problem has consistent boundary conditions. Say, the $q$ variable is fixed at initial and final time,

$$q(t_i)=q_i \qquad\text{and} \qquad q(t_f)=q_f.$$

2) Here we will only address the classical aspect of OP's question(v1). Yes, under mild assumptions, we may assume that $f=f(q)$ is the derivative of some function $F=F(q)$, i.e.,

$$ f(q)~=~ F^{\prime}(q).$$

Then the last term in the Lagrangian becomes just a total time derivative

$$ L~=~ \frac{1}{2}m\dot{q}^2+ \frac{d F(q)}{dt}. $$

Classically, the equation of motion do not depend on total derivative terms in the Lagrangian, so an equivalent Lagrangian is the Lagrangian for a free particle

$$ \tilde{L}~=~ \frac{1}{2}m\dot{q}^2. $$

Hence the equation of motion is just


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  • $\begingroup$ There's an ordering subtlety in these things, which is important to clarify. Only the Stratonovich centered difference obeys the chain rule, and allows the time derivative term to be ignored. $\endgroup$ – Ron Maimon May 3 '12 at 6:10
  • $\begingroup$ Yes, good point, which becomes important at the quantum mechanical level. $\endgroup$ – Qmechanic May 3 '12 at 11:07

This question is surprisingly ambiguous! You have a product of noncommuting operators in the path integral, which means that the answer is sensitive to the discretization.

First, you should note that

$$ {d\over dt} F(q(t)) = F'(q) \dot{q}$$

So by choosing F to be the antiderivative of f, you learn that the term is a perfect derivative, and is therefore irrelevant, because the integral of a perfect derivative only depends on the endpoints.

But when you have the product of non-commuting operators in a Lagrangian (even though they are not operators in the Lagrangian!), you need to be careful about the discretization order. If you use a forward discretization, the product $f\dot{q}$ contributes a potential energy $f'(q)/2m$. If it is backwards, there is $-f'(q)/2m$ potential. Only in centered difference is the equation of motion that of the free particle.

The equation of motion is therefore:

$$ m\ddot{x} \pm f''(q)/2m = 0$$

or else

$$ {m\ddot{x} = 0}$$

depending on the operator order.

Generally, the convention for Lagrangians in physics is that when a derivative multiplies a function, the derivative is interpreted as a centered difference, so that your term is actually perfect derivative. But the centered difference convention is not optimal for everything, since it leads to the "Fermion doubling problem". You should just know the ambiguity, and how it works. It is described in detail for the stochastic path integral case (which is exactly analogous and slightly more intuitive) in the answer to this question: Cross-field diffusion from Smoluchowski approximation .

Here is the difference between the orderings in a nutshell:

$$f(q(t+\epsilon)) {q(t+\epsilon) - q(t)\over \epsilon} - f(q(t+\epsilon){q(t+\epsilon)-q(t)\over \epsilon } $$ $$ = { (f(q(t+\epsilon)-f(q(t))(q(t+\epsilon)-q(t))\over\epsilon}$$

And this is equal to (up to higher order terms that vanish as $\epsilon$ goes to zero)

$$ f'(q(t)) { (q(t+\epsilon)-q(t))^2\over \epsilon}$$

A Brownian motion (or a typical contributing path to a quantum path integral, by analytic continuation in time), has the property that the squared displacement is proportional to the time for infintiesimal increments of time. So the quotient in parentheses averages to a constant.

The value of the constant is clarified by understanding canonical commutation in the path integral.

$$ q(t+\epsilon) m\dot{q}(t) - q(t)m\dot{q} = 1 $$

For a Brownian motion (this is the square displacement law, you can work it out by expanding the derivative). It is saying that the position is actually nontrivially correlated with the (infinite) value of the velocity, so that the future position is actually finitely correlated with the average velocity given the past position. This is intuitive if you think about it, and it is what the canonical commutation relations look like in the path integral.

From this, you learn that the commutator term is ${1\over m} f'(q)$, and half this is added to make a forward difference, subtracted to make a backward difference, or do nothing for a centered difference.

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  • $\begingroup$ +1; Operator ordering ambiguities become important at the quantum mechanical level. See also this answer. $\endgroup$ – Qmechanic May 3 '12 at 12:00

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