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I'm reading A. Zee's "Quantum Field Theory in a Nutshell" section I.4 in which he used path integrals to calculate the energy of a real scalar field and 2 sources depending only on the position. I.e., the lagrangian for the real scalar field is:

$$\mathcal{L} = \frac{1}{2}((\partial \phi)^2 - m^2 \phi^2) + J(\vec{x}) \phi$$

Where $$J(\vec{x}) \equiv J_1(\vec{x}) + J_2(\vec{x}) \equiv \delta^{(3)}(\vec{x}-\vec{x}_1)+\delta^{(3)}(\vec{x} - \vec{x}_2)$$ is a sum of 2 sources (delta functions).

Then he proceeds to do intensive calculations using path integrals to show that the energy is:

$$E = -\frac{e^{-mr}}{4\pi r}.\tag{I.4.7}$$

My first question is: what does this energy represent? The energy of the field itself (Analogous to $$\frac{1}{2}(E^2 + B^2)$$ for the electromagnetic field) or of the sources in the field (analogous to $$\frac{1}{4 \pi \epsilon_0} \frac{e^2}{|r_2-r_1|}$$ for the electromagnetic field).

My other question is what insight does the path integral give? Would we reach the same conclusion if I calculated the energy density of a field using a classical method?

Qmechanic
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Habouz
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2 Answers2

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Both are equivalent. In electrostatics, say you have two charges generating respectively the EM fields $E_1,B_1$ and $E_2,B_2$ giving the total EM field $E_1+E_2,B_1+B_2$. The total field energy is: $$ \begin{align} U &= \frac{1}{2}\int (E^2+B^2)d^3x \\ &= \frac{1}{2}\int (E_1^2+B_1^2)d^3x+\frac{1}{2}\int (E_2^2+B_2^2)d^3x+\int (E_1\cdot E_2+B_1\cdot B_2)d^3x \end{align} $$ Note that this clean identification of contribution works because the energy is quadratic, or equivalently Maxwell's equations are linear. Field theories are not always so simple.

You interpret the first two terms as the the respective self energies of the charges due to the EM field. The last term is the interaction term which for stationary charges gives the Coulomb energy (stationary so no magnetic field): $$ \int E_1\cdot E_2 d^3x = \frac{q_1q_2}{4\pi\epsilon_0 r_{12}} $$

Thus both approaches are the same. Zee therefore calculates the LHS in the context of this new field theory to deduce the interaction energy between two particles.

Hope this helps.

LPZ
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  1. The energy $E$ in eq. (I.4.7) is the value$^1$ of the Hamiltonian $H$ of the real scalar field $$\begin{align} H~=~&\int d^3r ~{\cal H}, \qquad \pi~=~\partial_t\phi, \tag{A}\cr {\cal H}~=~&\pi\partial_t\phi-{\cal L}\tag{B}\cr ~=~&\frac{1}{2}\pi^2 +\frac{1}{2}(\vec{\nabla}\phi)^2+\frac{1}{2}m^2\phi^2 -J\phi \tag{C}\cr ~\stackrel{(E)}{\approx}~&\frac{3}{2}\pi^2 -\frac{1}{2}(\vec{\nabla}\phi)^2-\frac{1}{2}m^2\phi^2\tag{D}\end{align}$$ for a static (=time-independent) solution to the Euler-Lagrange (EL) eq. (=Klein-Gordon eq.) $$(\partial_t^2- \vec{\nabla}^2+m^2)\phi~\approx~J. \tag{E}$$

  2. We can in principle use the classical EOM (E) to eliminate the field $\phi$ in the Hamiltonian $H$ and rewrite it as a quadratic expressions in the sources $J$, cf. OP's question.

  3. By the way, notice from the minus sign in eq. (I.4.7) that for a scalar field theory like charges attract (which is the opposite of E&M where like charges repel).

  4. To see how to calculate the path integral, see e.g. my related Phys.SE answer here.

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$^1$ The infinite/singular self-interaction terms in the Hamiltonian $H$ have been removed in the energy $E$ in eq. (I.4.7).

Qmechanic
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