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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.

I start with the Lagrangian density:

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

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!

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    $\begingroup$ 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? $\endgroup$ – gj255 Apr 11 '17 at 16:59
  • $\begingroup$ @gj255 What do you mean exactly? $\endgroup$ – Thomas Russell Apr 11 '17 at 17:41
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    $\begingroup$ 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? $\endgroup$ – gj255 Apr 11 '17 at 17:53
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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, $$ {\cal H} = \tfrac{1}{2}\left ( (\partial_x \phi)^2 + \frac{4a}{b} \sin ^2 \frac{b \phi}{2} \right )= \tfrac{1}{2}\left (\partial_x \phi-2\sqrt{\frac{a}{b}}\sin \frac{b\phi}{2}\right )^2 - \frac{4a^{1/2}}{b^{3/2}}\partial_x \left (\cos \frac{b\phi}{2}\right). $$

The solution you have nullifies the first term, the perfect square--that's how it is easiest found, in the first place.

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.

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