I am having a hard time in understanding a well known statement always made in the context of field theory.


Consider a classical real scalar field theory with Lagrangian density given by $$\mathcal{L}[\phi]=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi+\frac{1}{2}\mu^2\phi^2-\frac{\lambda}{4!}\phi^4$$ The Hamiltonian density for this theory is given by a Legendre transform to be $$\mathcal{H}[\phi,\pi]=\frac{1}{2}\pi^2+\frac{1}{2}(\vec{\nabla}\phi)^2-\frac{1}{2}\mu^2\phi^2+\frac{\lambda}{4!}\phi^4$$ and we are asked to find the field configuration of minimum energy.

Usually one says that since $\mathcal{H}[\phi,\pi]$ is composed of a quadratic (hence always positive) part which is $\frac{1}{2}\pi^2+\frac{1}{2}(\vec{\nabla}\phi)^2$ then $\mathcal{H}[\phi,\pi]$ is smaller when this part is zero. Then one, in order to make this part vanish, assumes the field to be constant $\phi(x)=a \ \in \mathbb{R}$ and then minimizes the potential $$V[\phi]=-\frac{1}{2}\mu^2\phi^2+\frac{\lambda}{4!}\phi^4$$ now seen to be not anymore a functional of fields, but merely a $V: \mathbb{R} \to \mathbb{R}$ funcion which sends $a \mapsto V(a)$.

The question

My question is the following: How do we say that this is true? Why can not there be a field configuration $\phi(x)=\chi(x)$ which is not constant and therefore does not make the quadratic part vanish, but still gives a much smaller potential and therefore an overall a smaller energy?

Some more reasoning I did:

Thinking of the analogy in calculus, it seems to me that this statement is similar to the following:

"In order to find the minimum of the function $F(x)=f^2(x)+g(x)$ look first at which values of $\tilde{x}_i \in \mathbb{R}$ you have $f(\tilde{x}_i)=0$. Then among these choose the one for which you have the minimum of $g(x)$. Let say with no loss of generality that is is $x_1$. Then $x_1$ is the global minimum of the function $F(x)$."

This statement is of course false and it is not hard to find counterexamples, such as $$F(x)=[(x-1)(x-2)]^2+e^x$$ In fact applying the statement above I have that $\tilde{x_1}=1$ and $\tilde{x}_2=2$ are the zeroes of the quadratic part. Therefore I look among them and find $F(1)=g(1)=e$ and $F(2)=g(2)=e^2$ as candidate minima. Therefore I would conclude saying $x=1$ is the minimum but this is clearly false since since taking $x=\frac{1}{2}$ gives $F(1/2)<F(1)$

What is wrong with my reasoning?

Note (for all of you that posses Peskin-Schroeder) the question arises from reading the statements in page 348.

  • $\begingroup$ It seems there is an error in the lagrangian: it should be $-\frac{1}{2}\mu^2 \phi^2$, so the hamiltonian turns out to be a sum of nonnegative terms. $\endgroup$
    – user91126
    Commented Nov 3, 2013 at 20:17
  • $\begingroup$ It is not an error, I wrote a negative mass term like this on purpose. I am basically copying the Lagrangian from Peskin, page 348. $\endgroup$ Commented Nov 3, 2013 at 20:30
  • $\begingroup$ Oh sorry, I misread. Anyway, you have to consider that taking $\phi(x) = a$ is not an imposition, but a result: if you require that $V[\phi]$ has a minimum, then you get that $\phi$ must be a uniform field, given by (11.2) of Peskin. As you know, if the hamiltonian is not bounded from below, the theory is inconsistent (there are many way to see it). Moreover, the minimum state energy must be (from Wightman's postulates) invariant under the action of the Poincaré group, so it can't depend on the point. $\endgroup$
    – user91126
    Commented Nov 3, 2013 at 20:57
  • $\begingroup$ In order to find the minimum energy configuration, I have to require that $\mathcal{H}[\phi,\pi]$ has a minimun, not tat $V[\phi]$ has one. Therefore I think I can not understand your reasoning. Can you prove that the minimum for the functional $\mathcal{H}[\phi,\pi]$ is obtained only for the constant field configurations? $\endgroup$ Commented Nov 3, 2013 at 21:05
  • $\begingroup$ Sorry, probably I can't understand the real meaning of your question. It seems to me quite trivial that any constant field minimizes the kinetic energy, since it is semipositive-definite and it contains only the derivates of a field ($\pi$ is not really independent.). Moreover, it turns out that $V[\phi]$ is minimized by a constant field too, so the same field minimizes both the contributions. This is consistent with Poincaré invariance of vacuum at a profound level. I repeat, probably I haven't understand very well the point. Anyway, I hope my comments were useful to you as yours are to me. $\endgroup$
    – user91126
    Commented Nov 3, 2013 at 22:26

1 Answer 1


The key insight is that the field configuration that is constant and whose constant value minimizes the potential energy density simultaneously minimizes both the potential energy functional and the kinetic energy functional individually. Let' see how this works:

Let $T$ and $V$ be the kinetic and potential functionals whose sum is the Hamiltonian; \begin{align} T[\pi,\phi] &= \int d^3x \left[\frac{1}{2}\pi(x)^2 + \frac{1}{2}(\nabla\phi(x))^2\right] \\ V[\phi] &= \int d^3x\left[-\frac{1}{2}\mu^2\phi(x)^2 + \frac{\lambda}{4!}\phi(x)^4\right] = \int d^3x\, \mathscr V(\phi(x)) \end{align} Let $\phi_0$ be any real number that minimizes the potential density \begin{align} \mathscr V(\phi_0)\leq \mathscr V(\phi), \qquad \text{for all $\phi\in\mathbb R$} \end{align} Let $\mathcal C$ denote the set of all admissible field configurations $\phi$, and let $\mathcal P$ be the set of all admissible momentum configurations $\pi$. Then I claim that if $\phi_*$ is the field configuration whose value is $\phi_0$ everywhere, then \begin{align} H[\pi, \phi_*]\leq H[\pi, \phi], \qquad \text{for all $\phi\in\mathcal C$ and $\pi\in \mathcal P$}. \end{align} The proof goes as follows. Notice that given any field configuration $\phi$, we have \begin{align} \mathscr V(\phi_0) \leq \mathscr V(\phi(x)), \qquad \text{for all $x\in\mathbb R^3$}. \end{align} By integrating both sides of this inequality, we immediately find that $\phi_*$ is a minimum of the potential energy functional; $V[\phi_*] \leq V[\phi]$ for all $\phi \in \mathcal C$. Next, notice that since $(\nabla\phi)^2$ is manifestly non-negative, given any $\pi\in P$ we have $T[\pi, \phi]\geq T[\pi, \bar \phi]$ for any $\bar\phi$ constant, and for all $\phi\in\mathcal C$. In particular, $T[\pi, \phi_*]\leq T[\pi, \phi]$ for all $\pi\in \mathcal P$ and $\phi\in\mathcal C$. In other words, for any $\pi$, the configuration $\phi_*$ is also a minimum of the kinetic energy functional. Putting these facts together, we find that \begin{align} V[\phi_*] + T[\pi, \phi_*]\leq V[\phi] + T[\pi, \phi], \qquad \text{for all $\phi\in\mathcal C$ and $\pi\in \mathcal P$}. \end{align} as desired.

  • $\begingroup$ So basically one has to reason the other way around. First you prove $\phi(x)=\phi_0$ is a minimum for $V[\phi]$ and then you show it is also a minimum for $T[\phi,\pi]$. Right? $\endgroup$ Commented Nov 3, 2013 at 21:14
  • 1
    $\begingroup$ @FedericoCarta Well we could just as easily have proven that $\phi(x) = \phi_0$ minimizes $T$ and then proven that it minimizes $V$. The point is that it minimizes both independently and therefore minimizes their sum as well. But yeah, the way it's presented is kind of confusing; we don't need to first prove that the field needs to be constant, and then find the smallest admissible constant value. Instead, we simply note that the field that is constant and which has that value is a global minimizer of the Hamiltonian. $\endgroup$ Commented Nov 3, 2013 at 21:19
  • $\begingroup$ So the basic point is that a constant field configuration minimizes both and independently $T$ and $V$ and therefore is a global minimizer of $H$. I believe I understand now, thank you. $\endgroup$ Commented Nov 3, 2013 at 21:21
  • $\begingroup$ @FedericoCarta Yes. When I first read your question, I tried to first just show that the field configuration needed to be constant to be a minimizer of $H$, but failed and reached the same confusion. Then I thought, "well we don't so much care about that fact in isolation, we care that a particular constant field configuration minimizes $H$." Then I realized that clearly such a field configuration minimizes the kinetic energy, and then I realized that it minimizes the potential energy too since its precisely the configuration where the potential density has its minimal value everywhere. $\endgroup$ Commented Nov 3, 2013 at 21:28
  • 1
    $\begingroup$ Yes, it is quite simple in the end. I guess it is the way in which usually this fact is presented that makes it hard to understand the first time. $\endgroup$ Commented Nov 3, 2013 at 21:32

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