Finding the energy of a solution to the Sine-Gordon equation

I am delving into Quantum-Field Theory, and am stuck trying to work out how to compute the energy of a soliton solution to the Sine-Gordon equation in 1-1 spacetime.

$$\mathcal{L}=\frac{1}{2}(\partial_{t}\phi)^{2}-\frac{1}{2}(\partial_{x}\phi)^{2}-\frac{a}{b}[1-\cos(b\phi)]$$

Where $a,b$ are arbitrary constants. Using this, and the Euler-Lagrange equations, we can see that the equation of motion is:

$$\partial_{tt}\phi-\partial_{xx}\phi+a\sin(b\phi)=0$$

To which an appropriate stationary solution is:

$$\phi(x) = \frac{4}{b}\arctan\left(\exp\left((ab)^{1/2}x\right)\right)$$

However, I want to work out the energy of this solution.

I figured that I can compute the Hamiltonian and then use the relationship $\hat{H}\left|\phi\right\rangle = E\left|\phi\right\rangle$ to compute the energy of the solution (which I am informed is of the form $E = ca^{1/2}$).

The Hamiltonian I can get from the Hamiltonian density: $$\hat{\mathcal{H}}(\phi)=\Pi^{0}\partial_{0}\phi-\mathcal{L} = \frac{1}{2}(\partial_{t}\phi)^{2} + \frac{1}{2}(\partial_{x}\phi)^{2} + \frac{a}{b}[1-\cos(b\phi)]$$

Therefore:

$$\hat{H}\phi(x) = \int \hat{\mathcal{H}} \:\mathrm{d}x$$

However, from here I am stuck!

• It seems to me that $\phi(x)$ is not a quantum state here, so you shouldn't try to act on it with a linear operator. Have you tried to compute the classical Hamiltonian for this solution? – gj255 Apr 11 '17 at 16:59
• @gj255 What do you mean exactly? – Thomas Russell Apr 11 '17 at 17:41
• Perhaps you could explain what exactly is causing you problems? Are you struggling to compute the integral at the bottom of your question or do you have some conceptual difficulty? – gj255 Apr 11 '17 at 17:53

I suspect you took an unfortunate left turn. Your hamiltonian density is fine, and for a stationary solution it is just the Bogomol'nyi trick, $${\cal H} = \tfrac{1}{2}\left ( (\partial_x \phi)^2 + \frac{4a}{b} \sin ^2 \frac{b \phi}{2} \right )= \tfrac{1}{2}\left (\partial_x \phi-2\sqrt{\frac{a}{b}}\sin \frac{b\phi}{2}\right )^2 - \frac{4a^{1/2}}{b^{3/2}}\partial_x \left (\cos \frac{b\phi}{2}\right).$$
So, integrating the energy density, the surface term, integrated from minus to plus infinity, by your solution yields equal and opposite answers at the upper and lower bounds, and so $E=8\pi a^{1/2}/b^{3/2}$. It is a classical problem, so far.