Theta-dependence of massive Schwinger model

I've read in Coleman's paper on the massive Schwinger model (and in other papers on the same topic, like this one) that the model's Hamiltonian contains a topological $$\theta$$-term. However, if I write down the Lagrangian of the model (see p. 362 of Tong's lecture notes on gauge theory),

$$\mathcal{L}= \frac{1}{2e^2}E_x^2 +\frac{\theta}{2\pi}E_x+i\bar{\psi}_xD_x\psi_x - i m \bar{\psi}_x\psi_x,$$ and perform a Legendre transformation $$\mathcal{H}(E_x)=E_x\pi_x-\mathcal{L}$$ with

$$\mathbf{\pi}_x=\frac{\partial \mathcal{L}}{\partial E_x}= \frac{1}{e^2}E_x+\frac{\theta}{2\pi},$$

then the resulting Hamiltonian is independent of $$\theta$$:

$$\mathcal{H}= \frac{1}{2e^2}E_x^2 -i\bar{\psi}_xD_x\psi_x + i m \bar{\psi}_x\psi_x.$$

Now my question is: Is there any Hamiltonian formulation of the Schwinger model in terms of the electric field $$E_x$$, which is equivalent to the above Lagrangian formulation, while both formulations contain a $$\theta$$-term? This question is somewhat related to my previous question.

1 Answer

The last expression written in the question is not yet the Hamiltonian. It is the energy function. Although it is numerically equal to the Hamiltonian, but it is expressed in terms of the electric field which is the time derivative of the vector potential $$A_x(x)$$. $$\mathcal{E}= \frac{1}{2e^2}E_x^2 -i\bar{\psi}_xD_x\psi_x + i m \bar{\psi}_x\psi_x$$ In a Hamiltonian, all the time derivatives of the canonical variables should be replaced by their expressions in terms of the canonical momenta by means of the Legendre transform. Therefore, the expression for the Hamiltonian density should be: $$\mathcal{H}= \frac{ e^2}{2}(\pi_x - \frac{\theta}{2\pi})^2 -i\bar{\psi}_xD_x\psi_x + i m \bar{\psi}_x\psi_x$$ Now, the Hamiltonian is expressed in terms of the correct variables and the dependence on the theta parameter is explicit.

This substitution is not just formal, because in the quantum theory, the canonical momentum is represented by the derivative of the canonical coordinate. (In our case, it is the functional derivative, because it is field theory), thus:

$$\mathbf{\hat{\pi}}_x(x)=i\hbar\frac{\delta}{\delta A_x(x)}$$ Thus:

$$\mathcal{\hat{H}}= \frac{ e^2}{2}(i\hbar\frac{\delta}{\delta A_x(x)} - \frac{\theta}{2\pi})^2 -i\hat{\bar{\psi}}_xD_x\hat{\psi}_x + i m \hat{\bar{\psi}}_x\hat{\psi}_x$$

Geometrically, the energy function is a function over the tangent bundle $$TM$$ of the configuration space, spanned be the coordinates and velocities, while the Hamiltonian is a function on the cotangent bundle $$T^*M$$, spanned by the canonical coordinates and momenta. This fact is thoroughly explained in Foundation of mechanics by Abraham and Marsden (Chapter 3).

In $$1+1$$ dimensions, when the base manifold is the circle $$S^1$$, the effect of the theta term is manifested solely in the zero mode: $$\phi(t) = \int_0^{2\pi R} A_x(x, t) dx$$ The Lagrangian, restricted to the zero mode is given on page 321 of Tong's lectures. This is an Aharonov-Bohm Lagrangian on a ring. It has a quantum mechanical theta term. The above Hamiltonian analysis was performed for this quantum mechanical problem by Tong and the Schrödinger equation was explicitly solved.

• Thanks for your answer! I know that the Hamiltonian is explicitly theta-dependent when expressed in terms of the canonical variables. However, Coleman and others express the Hamiltonian in terms of the electric field and then add an electric background field proportional to theta. This is what I don't understand. – Thomas May 2 at 22:28
• I don't have Coleman's article, but as I see in the second reference by Zache et al. (quoting Coleman), they use an alternative Hamiltonian in which the theta term is absorbed in the fermion mass term (as a complex mass) (Equation 1). Do you need an explanation how to get this form of the Hamiltonian? because it was not mentioned in the question. – David Bar Moshe May 5 at 6:33
• It is possible to use the Hamiltonian written in the last equation in the question. But, in this case the boundary conditions of the wave functions (assuming space is a circle: $x\in S^1$) must be changed in order to obtain the same spectrum of the Hamiltonian written in terms of the canonical momenta. For example, for the zero-mode theory we must have: $\Psi(\phi = 2\pi) = e^{i\frac{\phi\theta}{2 \pi}}\Psi(\phi = 0)$… – David Bar Moshe May 26 at 12:36
• The answer to the question, why for the Hamiltonian written in terms of the canonical momenta we use the ordinary periodic wave functions, while in the Hamiltonian written with the electric field we use wave functions with the shifted boundary conditions has no mathematical answer…. – David Bar Moshe May 26 at 12:37
• Its justification lies solely in the fact that Aharonov-Bohm effect has been observed, please see for example Landsman and Linden math.ru.nl/~landsman/NPB2.pdf , page 156. Anyway, keeping the rules of the standard canonical analysis scheme would lead directly to the correct Hamiltonian . – David Bar Moshe May 26 at 12:38