In a classical field theory problem Lagrangian density is given as

$${\cal L}=\frac{1}{2}\dot{\phi }^{2}-\frac{1}{2}\left ( \bigtriangledown \phi \right )^{2}-\frac{1}{2}m^{2}\phi ^{2}\tag{2.6}$$

Then total Hamiltonian will be

$$H=\int \mathrm d^{3}x\left [ \frac{1}{2}\pi ^{2}+\frac{1}{2}\left ( \bigtriangledown \phi \right ) ^{2}+\frac{1}{2}m^{2}\phi ^{2} \right ]\tag{2.8}$$

Upto this everything is straight forward. But after that author says1

We can think of the above three terms, respectively, as the energy cost of "moving" in time, the energy cost of "shearing" in space, and the energy cost of having the field at all.

Please someone explain.

1M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory (Addison-Wesley, 1995), chapter 2.

  • $\begingroup$ I think second term will be (1/2)k(------)^2.But I can't understand the difference between second and third term. $\endgroup$ Mar 5, 2016 at 14:04

1 Answer 1


IMHO asking for the intuitive meaning of expressions is in general sterile. The Langrangian $(2.6)$ you wrote down is just the most general covariant/renormalisable scalar that can be written down for a scalar field. This is just mathematics, nothing deep behind it. From that Lagrangian we get the Hamiltonian $(2.8)$. Trying to ask for the physical meaning of $H$ is worthless. Anything that we could say about the physical meaning of $(2.8)$ cannot be generalised to other important Hamiltonians, and so we learn nothing from it. That being said, I'll answer your question.

First of all, note that the system described by $(2.6)$ has zero vev. This means: the equilibrium configuration is realised at $\phi=0$. If we wanted the minimum to be located at $\phi=\phi_0$ for a non-zero $\phi_0$ we should take $$ \frac{1}{2}m^2\phi^2\to \frac{1}{2}m^2(\phi-\phi_0)^2 $$ in the both the Lagrangian and Hamiltonian. We will do this here because it'll make the arguments more transpartent. In the end, you can take $\phi_0=0$. In any case, physics are independent of fields redefinitions, so the change $\phi\to\phi-\phi_0$ leaves the physics invariant. Again, nothing deep going on here: it's just because this will make some points more clear.

Classical field theory can be thought as the continuum limit of point mechanics, and so we will do here. In the discrete case, we have $$ \begin{aligned} H&=\sum_i \delta x\left[\frac{1}{2c^2}\dot x^2_i+\frac{1}{2}(x_{i-1}-2x_i+x_{i+1})^2+\frac{1}{2}m^2 (x_i-x_0)^2 \right]=\\ &=\sum_i \frac{1}{2}\mu^2\dot x^2_i+\frac{1}{2}k(x_{i-1}-2x_i+x_{i+1})^2+\frac{1}{2}\kappa (x_i-x_0)^2 \end{aligned} \tag{1} $$ where $k=\delta x$, $\mu=c^{-2}k$ and $\kappa=km^2$.

The system described by the Hamiltonian $(1)$ could be a string, for example. We can imagine an infinite number of point masses at positions $x_i(t)$, connected to each other by springs with constants $k$.

With this, the interpretation of the terms is straightforward: the first term is just the kinetic energy of each mass. Equivalently, it is the energy we must transfer to the system if we want each mass to move with velocity $\dot x_i$.

The second term is just what we would expect for masses connected to its nearest neighbours by springs with constant $k$. If you want to move a certain mass $x_i$ away from its neibourghs $x_{i-1},x_{i+1}$, you need some energy, because these masses pull one another.

The third term is similar to the second one, in the sense that it is also a kind of spring-like energy: it is the energy of an oscillator displaced from the equilibrium position $x_0$. This means: you can take the mass $i$ away from the position $x_0$ by pulling from it, but this requires a certain amount of energy. Therefore, you need a non-zero amount of eerngy if you want that $x_i\neq x_0$. This is the energy for just being. If you want, you can take $x_0=0$, i.e., $\phi_0=0$. This means that we are setting the origin at the equilibrium position.

The difference between the second and third terms is: the second term is the energy to move a particle away from its neighbours, and the third is the energy you need to move a particle away from the equilibrium position. You could make a rigid displacement of the particles, $x_i\to x_i-a\ \forall i$ for constant $a$, and this wouldn't change the second term: every particle is just as far as its neightbours as before. There is no shearing. But the third terms does change: a rigid displacement takes every particle away from $x_0$, so that you need a lot of energy to do this. The second and third tems are different: they account for different energies.

  • $\begingroup$ Can we say that the last term is a generalization of the rest mass energy? $\endgroup$ Jun 5, 2017 at 16:55
  • 1
    $\begingroup$ @TheQuantumMan in a sense, yes: without it the excitations would be massless, i.e., there would be no mass gap: $E(p)\to 0$ as $p\to 0$. $\endgroup$ Jun 5, 2017 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.