I've read in David Tong's lecture notes on gauge theory that the Hamiltonian of Yang-Mills theory does not depend on the angular parameter $\theta$, because it can be absorbed in the electric field:

$$ \mathcal{H}=\frac{1}{g^2}\text{tr}(\mathbf{E}^2+\mathbf{B}^2)=g^2\text{tr}(\mathbf{\pi}-\frac{\theta}{8\pi^2}\mathbf{B})^2+\frac{1}{g^2}\text{tr}(\mathbf{B}^2). $$

Here, $g$ is the gauge coupling, $E_i=\dot{A}_i$ is the non-Abelian electric field, $B_i=-\frac{1}{2}\epsilon_{ijk}F^{jk}$ the non-Abelian magnetic field, ${F}_{\mu\nu}$ is the gluon field strength, and $$ \mathbf{\pi}=\frac{\partial \mathcal{L}}{\partial \mathbf{\dot{A}}}= \frac{1}{g^2}\mathbf{E}+\frac{\theta}{8\pi^2}\mathbf{B} $$ is the momentun conjugate to $\mathbf{A}$ (see pp. 39 and 40 of the lecture notes).

In contrast, it's well known that the Yang-Mills Lagrangian contains a topological $\theta$-term: $$ \mathcal{L}= -\frac{1}{2g^2}\text{tr}(F^{\mu\nu}F_{\mu\nu})+\frac{\theta}{16\pi^2}\text{tr}(F^{\mu\nu}\tilde{F}_{\mu\nu})=\frac{1}{g^2}\text{tr}(\mathbf{\dot{A}}^2-\mathbf{B}^2)-\frac{\theta}{4\pi^2}\text{tr}(\mathbf{\dot{A}} \mathbf{B}), $$ where $\tilde{F}_{\mu\nu}$ is the Hodge dual of $F_{\mu\nu}$, and the last equality holds for $A_0=0$ and $D_iE_i=0$.

How is this possible? Tong mentions that the $\theta$-dependence in the Hamiltonian formalism is somehow hidden in the structure of the Poisson bracket, but he gives no detailed explanation. Why doesn't it appear in the Hamiltonian itself? The $\theta$-term gives rise to the infamous strong CP problem, so how can we explicitly compute this $\theta$-dependence of Yang-Mills in this formalism?


1 Answer 1


It's not particularly strange or unusual for a term to seemingly not appear in the Hamiltonian, but still have a physical effect. For example, the Hamiltonian for a free particle is $$H = \frac{p^2}{2m} = \frac12 m v^2, \quad p = mv.$$ On the other hand, the Hamiltonian for a particle in a magnetic field is $$H = \frac{(p-eA)^2}{2m} = \frac12 m v^2, \quad p = m v + e A.$$ They are of course different functions of $(x, p)$, reflecting the fact that the dynamics are different, but if you naively write the Hamiltonian in terms of the "physical" variables $(x, v)$ then you get the same result in both cases, since magnetic fields do no work. There is of course still an effect, because $v$ is related to $p$ in a different way. Your example with the $\theta$-term is just a more complicated version of this, so there's nothing really weird to resolve.

  • $\begingroup$ Thanks for great answer! So the "physical" effect we see in the Hamiltonian formalism is the altered dispersion relation of the particle moving in the external field? $\endgroup$
    – Thomas
    Commented Apr 26, 2019 at 22:11
  • $\begingroup$ @LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$. $\endgroup$
    – knzhou
    Commented Apr 26, 2019 at 22:31

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