# Why can you make $V$ stationary with respect to a parameter of the field in Derrick's theorem?

I'm going over Coleman's derivation of Derrick's theorem for real scalar fields in the chapter Classical lumps and their quantum descendants from Aspects of Symmetry (page 194).

Theorem: Let $\phi$ be a set of scalar fields (assembled into a big vector) in one time dimension and $D$ space dimensions. Let the dynamics of these fields be defined by,

\begin{align} \mathcal{L} &= \frac{1}{2}\partial_\mu\phi\partial^\mu\phi-U(\phi) \end{align}

and let $U$ be non-negative and equal to zero for the ground state(s) of the theory. Then for $D\geq2$ the only non-singular time-independent solutions of finite energy are the ground states.

Proof: Define $V_1 =\frac{1}{2}\int d^Dx(\nabla\phi)^2$ and $V_2 = \int d^DxU(\phi)$. $V_1$ and $V_2$ are both non-negative and are simultaneously equal to zero only for the ground states. Define
\begin{align} \phi(x,\lambda)\equiv\phi(\lambda x) \end{align} where $\lambda\in\mathbb{R}^+$. For these functions the energy is given by \begin{align} V(\lambda,\phi) = \lambda^{2-D}V_1 + \lambda^{-D}V_2 \end{align}

I know this should be stationary at $\lambda=1$. But I claim it should be stationary wrt the fields, not our parameter $\lambda$. So is the formal statement \begin{align} \frac{\partial V}{\partial\lambda}\biggr\rvert_{\lambda=1} = {\frac{\delta V}{\delta \phi}}\frac{\partial\phi}{\partial\lambda}\biggr\rvert_{\lambda=1} \end{align} and we know \begin{align} \frac{\delta V}{\delta\phi}\rvert_{\lambda=1}=0 \end{align} such that we can make $V$ stationary w.r.t $\lambda$ \begin{align} \frac{\partial V}{\partial\lambda}\biggr\rvert_{\lambda=1} = (D-2)V_1 + DV_2 = 0 \end{align}

Is this the correct way to argue it? I don't really understand how to mix functional and partial derivatives. But I imagine my chain rule between spaces is not okay.

I found https://math.stackexchange.com/q/476497/, and also Derrick’s theorem. They are related but do not answer the question as formally.

• You can make the mix of functionals and partials go away if you use the Euler-Lagrange equations instead of $\delta V/\delta \phi = 0$, I think. – ACuriousMind Mar 16 '16 at 10:28
• Interesting, I think the statement $\frac{\delta V}{\delta\phi}\biggr\rvert_{\lambda=1}=0$ is the Euler-Lagrange equation for this because the Lagrangian has no $\partial_\mu\phi$ dependence. I don't think I can think of $L$ as $L[\phi,\partial_\mu\phi; \lambda,\partial_\mu\lambda]$ or something since $\lambda$ it not in function space, no? – jeau_von_shrau Mar 16 '16 at 17:52

OK, perhaps the notation in Ref. 1 is a bit confusing. Let us elaborate on Derrick's No-Go theorem:

Derrick's No-Go theorem: For the number of spatial dimensions $D>2$, the only time-independent finite-energy solutions are ground states.

In a nutshell, the idea of the proof is to derive a necessary condition by a simple 1-parameter scaling argument. (Contrary to what one might naively expect, it is not particularly useful to set up Euler-Lagrange (EL) equations via functional differentiation because we do not know much about the potential $U$.)

Since we are considering time-independent field configurations, we should find stationary$^1$ field configurations for the total potential energy

$$V[\phi]~=~V_1[\phi]+ V_2[\phi], \tag{1}$$

where

$$V_1[\phi] ~:=~\frac{1}{2} \int \!\mathrm{d}^Dx~ (\nabla \phi)^2 ~\geq~ 0, \qquad V_2[\phi]~:=~ \int \!\mathrm{d}^Dx~ U(\phi)~\geq~ 0, \qquad U(\phi) ~\geq~ 0. \tag{2}$$

Now we want to check if some field configuration ${\bf x} \mapsto \phi_1({\bf x})$ is stationary. Define a 1-parameter family of field configurations

$$\phi_{\lambda}({\bf x})~:=~\phi_1(\lambda{\bf x}), \qquad \lambda ~\in~ [0,\infty[.\tag{3}$$

Note that $\phi_{\lambda=1}=\phi_1$, so the notation (3) is consistent. We then calculate (by change of variables in the integrals) that

$$V[\phi_{\lambda}]~=~\lambda^{2-D}V_1[\phi_1]+\lambda^{-D}V_2[\phi_1].\tag{4}$$

A necessary (but far from sufficient!) condition for $\phi_1$ to be stationary is to differentiate the function $\lambda \mapsto V[\phi_{\lambda}]$ wrt. the parameter $\lambda$ at the point $\lambda=1$,

$$0~\stackrel{?}{=}~ \left. \frac{d V[\phi_{\lambda}]}{d\lambda}\right|_{\lambda=1}~=~ (2-D) V_1[\phi_1] - D V_2[\phi_1]. \tag{5}$$

Varying along the 1-parameter family constitutes only one out of infinitely many possibilities to vary the field configuration $\phi_1$, but it is the only one we'll need!

We next assume$^2$ that $D>2$. Then eq. (5) is only possible if

$$V_1[\phi_1]~=~0~=~V_2[\phi_1].\tag{6}$$

Eq. (6a) implies that $$\nabla \phi_1 \equiv 0,\tag{7}$$

or equivalently that

$$\phi_1 \text{ is {\bf x}-independent}.\tag{8}$$

Eq. (6b) and (8) then imply that

$$\phi_1 \text{ is a ground state}.\tag{9}$$

References:

1. S. Coleman, Aspects of Symmetry, 1985; p. 194.

2. R. Rajaraman, Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, 1987; Section 3.2 & 3.3.

--

$^1$ Stationary in the sense that they satisfy EL equations for $V$, i.e. the functional derivative vanishes,

$$0~\approx~\frac{\delta V[\phi]}{\delta \phi}~=~ -\nabla^2\phi + \frac{\partial U(\phi)}{\partial \phi}.\tag{10}$$

[Here the $\approx$ symbol means equality modulo EL equations.]

$^2$ The case $D=2$. Contrary to what Ref. 1 claims, the Derrick's No-Go theorem does not hold for $D=2$. Let us here consider just the $D=2$ case. We claim that one can no longer conclude that $V_1[\phi_1]$ should be zero. Ref. 1 gives the following wrong proof (using our notation):

For $D = 2$, however, Eq. (5) only implies the vanishing of $V_2[\phi_1]$, and a small amount of further argument is required. If $V_2[\phi_1]$ vanishes it is stationary, since zero is its minimum value. Thus we may apply Hamilton's principle to $V_1[\phi_1]$ alone, from which it trivially follows that $V_1[\phi]$ also vanishes. Q.E.D.

The potential energy term $V_2[\phi_1]=0$ must still be zero, cf. eq. (5). In other words, the $\phi_1$-image

$${\rm Im}(\phi_1) ~\subseteq ~U^{-1}(\{0\}) \tag{11}$$

must lie in the zero locus $U^{-1}(\{0\})$ (or preimage of $\{0\}$) of the potential $U$, i.e. the set of minimum points for the potential $U$ in target space.

Lagrange multiplier. As an aside, if a constraint $\chi(\phi)\approx 0$ emulates the zero locus $U^{-1}(\{0\})=\chi^{-1}(\{0\})$, one may effectively replace the functional (1) with $$\tilde{V}[\phi,\Lambda]~=~V_1[\phi]+ \int \!\mathrm{d}^Dx~\Lambda~ \chi(\phi) , \tag{12}$$

where $\Lambda=\Lambda({\bf x})$ is a Lagrange multiplier. The EL equations read

$$0~\approx~\frac{\delta\tilde{V}[\phi,\Lambda]}{\delta \phi}~=~ -\nabla^2\phi + \Lambda ~\frac{\partial \chi(\phi)}{\partial \phi}.\tag{13}$$

It is impossible to know for sure how Ref. 1 reached the wrong conclusion $V_1[\phi_1]=0$, but it might be partly spurred by forgetting to properly take into account the constraint force, i.e. the last term in eq. (13).

One-point compactification of space. In polar coordinates, the term $V_1[\phi_1]$ reads

$$0~\leq~V_1[\phi_1]~=~\int_0^{2\pi} \! \mathrm{d}\theta \int_0^{\infty} \! \mathrm{d}r~\left(r \left(\frac{\partial \phi_1}{\partial r}\right)^2+ \frac{1}{r} \left(\frac{\partial \phi_1}{\partial \theta}\right)^2 \right)~<~\infty.\tag{14}$$

For each $\theta\in[0,2\pi[$, under mild regularity conditions, we can assume that the limit

$$\lim_{r\to \infty}\frac{\partial \phi_1}{\partial \theta}\tag{15}$$

exists. To keep the energy (14) finite, the limit (15) must be zero. In other words,

$$\phi_1(r\!=\!\infty,\theta) \text{ does not depend on } \theta.\tag{16}$$

So we can effectively one-point compactify the space $\mathbb{R}^2\cup \{\infty\}\cong S^2$ into a 2-sphere $S^2$.

Counterexample. One counterexample with finite $V_1[\phi_1]>0$ is a so-called baby skyrmion in a $2D$ $O(3)$ model with a mexican-hat-like potential. The target space is here $\mathbb{R}^3$, and the zero locus

$$U^{-1}(\{0\})~=~\{\phi \in \mathbb{R}^3 |~ |\phi| =1\}~\cong~S^2\tag{17}$$

for the potential $U$ forms a 2-sphere. Because $\pi_2(S^2)\cong \mathbb{Z}$, the field configuration $\phi_1:S^2\to S^2$ is protected by a topological charge

$$V_1[\phi_1]~\geq~ 4\pi |Q|.\tag{18}$$

If $Q\neq 0$, we conclude that $\phi_1$ is not a ground state. See e.g. Ref. 2 for further details.

• You speak of eq. (6a) and (6b), but you have only a single eq. (6). – ACuriousMind Mar 17 '16 at 17:08
• @ACuriousMind: Yeah, I was referring to the two equality signs in eq. (6), respectively. – Qmechanic Mar 17 '16 at 17:10
• @Qmechanic: This is a fantastic answer. But I do not understand this "necessary (but not sufficient!)" line in equation 5. Why is it reasonable to take a regular derivative of a functional $V$. Can you explain explicitly? – jeau_von_shrau Mar 19 '16 at 0:59
• The map $\phi\mapsto V[\phi]$ is indeed a functional, while $\lambda \mapsto V[\phi_{\lambda}]$ is just a function. Therefore an ordinary derivative $\frac{d }{d\lambda}$ suffice in eq. (5) (as opposed to a functional derivative $\frac{\delta }{\delta\phi}$). – Qmechanic Mar 19 '16 at 1:08