Why are first class constraints harder to quantize than second class constraints?

Question

I understand that the well known system with the second class constraints: \begin{align} &q_1 = 0 \\ &p_1 = 0 \end{align}

has the apparent problem when performing quantization using the Poisson brackets when demanding $\hat{q}_1 \psi = 0$ and $\hat{p}_1\psi = 0$: $$ [\hat{q}_1, \hat{p}_1]\psi = 0 \neq i \hbar \{q_1,p_1\}_{PB}\psi = i\hbar \psi $$

The solution is to remove $q_1$ and $p_1$ from the Poisson brackets to obtain the Dirac bracket, in which we then have: $$ [\hat{q}_1, \hat{p}_1]\psi = 0 = i \hbar \{q_1,p_1\}_{DB}\psi $$

Now suppose we consider a different system with the constraints: \begin{align} &q_1 = 0 \\ &q_2 = 0 \end{align}

These are clearly first class constraints since: $$ \{q_1, q_2\}_{PB} = \frac{\partial q_1}{\partial q_a} \frac{\partial q_2}{\partial p_a} - \frac{\partial q_2}{\partial q_a} \frac{\partial q_1}{\partial p_a} = 0 $$

So there is no inconsistency in quantization here, since if we demand that $\hat{q}_1 \psi = 0$ and $\hat{q}_2 \psi = 0$, then we have: $$ [\hat{q}_1, \hat{q}_2]\psi = 0 = i \hbar \{q_1,q_2\}_{PB}\psi $$

So shouldn't quantizing first class systems be easier since we have $\{\phi_m, \phi_{m'}\} \approx 0$ for all first class systems? I have read about operator ordering ambiguity but I don't understand it very well, for example, I can't really see any operator ordering ambiguity in the example above. Also, if operator ordering ambiguity is a problem for first class constraints, wouldn't it be a problem for second class constraints as well?

Answer 1

First-class constraints generate gauge transformations (assuming the Dirac conjecture), i.e. map physically equivalent states onto each other. Even if you do not assume the Dirac conjecture, then first-class constraints still generate transformations that map allowed states, i.e. those on the constraint surface, to other allowed states.

Second-class constraints don't do that - their transformations have no physical significance because the transformations generated by them map states on the constraint surface to states outside of it. This is precisely because they are second-class, so their transformations do not preserve the fulfillment of all constraints.

The construction of the Dirac bracket shows that we can effectively impose these constraints as "strong" relations that not only hold on-shell but in general. Note that the Dirac bracket crucially relies on the invertibility of the matrix $\{\chi_i,\chi_j\}$ of the Poisson brackets of second-class constraints. This is identically zero by definition for first-class constraints, so you cannot construct a Dirac bracket for these.

You can often view second-class constraints as "gauge-fixations" of some underlying first-class constraint. Your system with $q_1 = 0, p_1 = 0$ is equivalent to a system with the single first-class constraint $p_1 = 0$ generating gauge transformations $q_1 \mapsto q_1 + a$. Choosing the gauge fixation $q_1 = 0$ yields the original system.

So the fact that it's easy to eliminate second-class constraints tells us that the "easy" way to treat first-class constraints is to fix their gauges completely, and then apply the Dirac bracket procedure. This complete gauge fixing is not always possible or desirable (loss of manifest covariance under a symmetry we "like", Gribov ambiguities, etc.), and it is in these cases that the first-class constraints become "difficult". Note that in the $q_1 = 0, q_2 = 0$ example, you can choose the gauge fixations $p_1 = 0, p_2 = 0$ and arrive at a system with purely second-class constraints to which you can apply the "easy" Dirac bracket procedure.

A good reference for these claims is, as for many issues surrounding gauge theories and constrained quantization, "Quantization of Gauge Systems" by Henneaux and Teitelboim, in this case chapter 1.

Answer 2

ACuriousMind has already given as good answer. Let me just add that it is typically the other way around, i.e. that first-class systems are simpler than second-class systems, at least at the formal level.

• This is e.g. because it is often difficult to invert the matrix of constraints to construct the Dirac bracket, especially in field theory, where it typically leads to non-local operators.

• Another argument is that one could in principle always demote a gauge theory with $n$ first-class constraints $(\phi^1,\ldots, \phi^n)$ and $n$ gauge-fixing constraints $(\chi_1,\ldots, \chi_n)$ to a non-gauge theory with $2n$ second-class constraints. The opposite process (called conversion in the literature) is more challenging (and introduces new variables).