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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.

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  • $\begingroup$ 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. $\endgroup$ – ACuriousMind Mar 16 '16 at 10:28
  • $\begingroup$ 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? $\endgroup$ – jeau_von_shrau Mar 16 '16 at 17:52
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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.

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$^1$ 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.

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  • $\begingroup$ You speak of eq. (6a) and (6b), but you have only a single eq. (6). $\endgroup$ – ACuriousMind Mar 17 '16 at 17:08
  • $\begingroup$ @ACuriousMind: Yeah, I was referring to the two equality signs in eq. (6), respectively. $\endgroup$ – Qmechanic Mar 17 '16 at 17:10
  • $\begingroup$ @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? $\endgroup$ – jeau_von_shrau Mar 19 '16 at 0:59
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    $\begingroup$ 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 $\frac{\delta }{\delta\phi}$). $\endgroup$ – Qmechanic Mar 19 '16 at 1:08

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