Energy contributions of Hamiltonian density

Question

In Lancaster and Blundell, Quantum Field Theory for the Gifted Amateur, p.99, the Hamiltonian density is \begin{equation} \mathcal{H}=\frac{1}{2}[\partial_0\phi(x)]^2+\frac{1}{2}[\nabla\phi(x)]^2+\frac{1}{2}m^2[\phi(x)]^2,\tag{11.5} \end{equation} and it tells us that the energy has contributions from

1. a kinetic energy term reflecting changes in the configuration in time,

2. a 'shear term' giving an energy cost for spatial changes in the field, and

3. a 'mass' term reflecting the potential energy cost of there being a field in space at all.

In the equation above, i think the first term is the same as the classical mechanics. But i don't understand why second (shear) and third (mass) term are represent potential energy.

Answer 1

You can discretize the Klein-Gordon (KG) field $\phi$ as a displacement variable in a lattice. Let us for simplicity consider 1+1D, i.e. we have an 1D equidistant lattice of point-particles. Let $\phi(x)$ denote the transversal$^1$ displacement of the point-particle with equilibrium position $x$.

1. The kinetic energy maps to kinetic energy.

2. The potential gradient term can be reproduced by springs between nearest neighbors.

3. The KG mass term can be reproduced by a spring between each point-particle and its equilibrium position.

For details, see this related Phys.SE post and links therein.

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$^1$ Alternatively, instead of transversal displacement, one may consider longitudinal displacement.

Answer 2

Each of those terms represent the price to pay, in terms if energy, to have some specific configuration of the field:

• configurations that change with time (price is estimated by time derivative)
• configurations that change with space (price is estimated by the gradient)
• magnitude of the field (mass plays a role in enhancing this effect)

All those terms appear with a plus sign, suggesting that the most energetically favorable configurations are those where the field is more uniform, stabile an small

Answer 3

Actually the first two terms are kinetic part, and the last term is the potential energy.

To get a correspondence to the classical mechanics, just consider the vibration of a string. When the string is displaced from equilibrium, a segment associated with the interval dx has a length: $$dl = \sqrt{dx^2+d\psi^2}=dx\sqrt{1+(\frac{d\psi}{dx})^2}\approx dx\left[1+\frac{1}{2}(\frac{d\psi}{dx})^2\right]$$. ($\psi$ here is the displacement)

The potential is proportional to the string deformation: $$dU\propto dl - dx = dx(\frac{d\psi}{dx})^2$$ This potential is due to the deformation of the string. The second term in the Hamiltonian is corresponding to this, which is related to the spatial changes in the field.

I am not sure about the correspondence for the last term (potential term), but it's maybe corresponding to the potential of a oscillating spring (correct me if it's wrong): $$T=\frac{1}{2}m\omega^2 x^2$$