# "Hidden" theta-term in Hamiltonian formulation of Yang-Mills theory

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?

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.

• 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? Commented Apr 26, 2019 at 22:11
• @LCF Sure, I suppose you can put it that way. In any case, the Hamiltonian is just a different function of $x$ and $p$. Commented Apr 26, 2019 at 22:31