2
$\begingroup$

I am reading the book "Quantum Field Theory" by Jean-Bernard Zuber and Claude Itzykson.

I encounter great difficulties from page 457 section 9-3-1, which introduces Dirac's constrained system in classical mechanics. To begin with, consider a classical systems with $2n$ degrees of freedom submitted to a constraint $$f(p,q)=0. \tag{1}$$ So far, as far as I understand in field theory, examples of such system are gauge theories. Call $C$ the $2n-1$-dimensional manifold in phase space characterized by equation (1). Let $\mathcal{R}$ be the ring of differentiable functions which vanishes on $C$. We use the notation $F\sim 0$ to mean $F\in\mathcal{R}$. For any two functions $f$ and $F$ that vanish on $C$, there must exists some function $\alpha(p,q)$ such that $F(p,q)=\alpha(p,q)f(p,q)$.

We then introduce a Lagrange multiplier $\lambda(t)$ to incorporate the constraint (1) into the action $$S=\int dt\left\{p\dot{q}-H(p,q)-\lambda(t)f(p,q)\right\}. \tag{2}$$

The equations of motion are $$\dot{q}^{i}=\frac{\partial H}{\partial p_{i}}+\lambda\frac{\partial f}{\partial p_{i}}, \quad \dot{q}_{i}=-\frac{\partial H}{\partial q^{i}}-\lambda\frac{\partial f}{\partial q^{i}}, \quad 1\leqslant i\leqslant n. \tag{3}$$

Next, we pick up an arbitrary member $F\in\mathcal{R}$, and define an equivalence relation $\mathcal{E}$ on $C$. The author says the following:

Consider the flow generated by $F$ in phase space. In infinitesimal form, it is described by the equations $$\frac{dq^{i}}{du}=\frac{\partial F}{\partial p_{i}},\quad \frac{dp_{i}}{du}=-\frac{\partial F}{\partial q^{i}},\quad F\sim 0. \tag{4}$$

Two points on $C$ are equivalent iff they belong to the same trajectory of the flow (4).

My first question is: What is a flow generated by a function on the phase space? What's the mathematical definition? Why does it satisfy equations (4)?

The author then proves that the above definition of the equivalence relation is independent of time evolution and the choice of $F$ in $\mathcal{R}$. Thus, the surface $C$ is split into time-independent eqquivalence classes $\mathcal{E}$, and the quotient space $C/\mathcal{E}$ is the real physical phase space, which is an $2n-2$-dimensional manifold. From my understanding, such flows corresponds to gauge orbits in QFT, and all physical observables must be invariant along these flows. It is therefore sufficient to fix the gauge by choosing an auxilliary constraint $$g(p,q)=0 \tag{5}$$

which intersects with $C$ transversely. Then the authors say that to ensure $g(p,q)$ varies monotonically along each flow line, $g$ must satisfy the condition $$\left\{f,g\right\}_{P}\neq 0, \tag{6}$$

where $\left\{,\right\}_{P}$ represents the Poisson bracket.

My second question is: why does (6) ensure that $g$ varies monotonically along each flow line?

$\endgroup$
2
  • $\begingroup$ I've corrected what must be a typo in eq. (6) - if it wasn't yours, it's one in the book. $\endgroup$
    – ACuriousMind
    Jan 14, 2022 at 19:34
  • $\begingroup$ Background reading. $\endgroup$ Jan 14, 2022 at 22:00

1 Answer 1

4
$\begingroup$

The flow of a classical Hamiltonian observable is the flow of its associated Hamiltonian vector field. The differential equation that an integral curve $(q(u),p(u))$ of this flow/vector field obeys is precisely your eq. (4). You should think of these integral curves as the motions you would get if the associated observable $f$ was your Hamiltonian - for $f=H$ eq. (4) is just Hamilton's equations of motion!

Note that your eq. (4) can be equivalently written as $\dot{q} = \{f,q\}$ and $\dot{p} = \{f,p\}$ and more generally as $\dot{g} = \{f,g\}$ for the change $\dot{g}$ of any function $g(q(u),p(u))$ under the flow. The explains your eq. (6): When $\{f,g\} = 0$, then the derivative of $g$ along the flow is zero and may change sign, hence $\neq 0$ guarantees monotonicity of $g(q(u),p(u))$ along every integral curve.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.