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The question is: Does the Non-linear $\sigma$-model have a Goldstone mode?

Consider a $O(N)$ mode for which the Hamiltonian is $$H=J\sum_{i,j}\vec{n}_i \cdot \vec{n}_j,$$ where $\vec{n}=(\vec{\pi},\sigma)$ is a $N$ component unit rotor with fixed length, say $\vec{n}^2=1$. If we consider the low energy excitation above symmetry breaking ground state $\vec{n}=(0,0,0,\sigma)$, writting $\sigma=\sqrt{1-\vec{\pi}^2}$, one can have the non-linear $\sigma$ model for small $\vec{\pi}$,

$$H=\int d^dx[\frac{J}{2}(\nabla\vec{\pi})^2+\frac{J}{2}(\vec{\pi}\cdot\nabla\vec{\pi})^2-\frac{\rho}{2}\vec{\pi}^2],$$

where the continuous limit is assumed and $\rho=N/V$. The first term in the bracket looks like a spin wave excitation in a $XY$ model, say a Goldstone mode. However, the second term seems reflect the interaction of those excitation which may open a gap, and the the third term seems a mass term. Does this mean there is no Goldstone mode in a $O(N)$ rotor mode whose ordered states do break continuous symmetry?

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You should be careful that the "mass term" you wrote is a perturbation of the main action.

To be more precise : we can write the Lagrangian of the non linear $\sigma$ model as ($K=J/T$ where $T$ is the temperature of the system)

$\cal{ L}$ $= \frac{K}{2}[(\nabla \vec\pi)^2+\frac{(\vec\pi\nabla \vec\pi)^2}{1-\vec \pi^2}]-\frac{\rho}{2}\log(1-\vec\pi^2)$.

Now, recall that this model makes sense in the limit where $K\gg1$, which correspond, in the classical spin language, to the limit where the system is well into the ordered phase. Therefore, only configurations with $\pi\lesssim 1/\sqrt{K}$ give an important contribution in the path integral. Let's rescale the field by $g=1/\sqrt{K}$, which gives

$\cal{ L}$ $= \frac{1}{2}[(\nabla \vec\pi)^2+g\frac{(\vec\pi\nabla \vec\pi)^2}{1-g\vec \pi^2}]-\frac{\rho}{2}\log(1-g\vec\pi^2)$.

We can expand the Lagrangian in power of $g$, which gives

$\cal{ L}$ $= \frac{1}{2}(\nabla \vec\pi)^2+\frac{g}{2}[(\vec\pi\nabla \vec\pi)^2+\rho\vec\pi^2]+\cdots$.

You see that the bare propagator (order $g^0$) is now gapless, and the term proportional to $\rho$ is now a perturbation to the bare action (i.e., the ``mass'' term is not included in the bare propagator). In fact, one can show that this term is needed to insure that the $\vec\pi$ stay gapless; this is precisely the role of the logarithm, which brings new interactions to insure the validity of Goldstone theorem order by order in $g$.

You should not get confused by the fact that this perturbation is quadratic in $\vec\pi$, which one would naively include in the bare propagator, because this a very peculiar kind of perturbation theory, which expansion parameter is $g$.

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  • $\begingroup$ but to derive the non-linear $\sigma$ model, the mass term just origin from expanding $log(1-\vec{\pi}^2)$. $\endgroup$ – R.Wigner Nov 15 '13 at 19:37
  • $\begingroup$ Hum, yes sorry. I guess what I meant is that you need this term to avoid the creation of a mass at each order in perturbation in $1/J$ (because in fact the log terms are perturbations). I'll rewrite the answer. $\endgroup$ – Adam Nov 15 '13 at 21:09
  • $\begingroup$ @hongchaniyi: Should be better now. Let me know if it's still not clear. $\endgroup$ – Adam Nov 15 '13 at 21:39
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The log term is coming from the delta function in the measure that enforces $|{\bf n}|=1$, At each point $x$ in the path integral $$ Z=\int d[{\bf n}(x)]\delta(|{\bf n}(x)|^2-1) e^{-S[{\bf n}]} $$ we write $$ d^3 n(x) \delta(|{\bf n}(x)|^2-1) = d^2 \pi(x) \sqrt{g}=\frac{d^2\pi(x)} {\sqrt{1-|{\bf \pi}|^2}} $$ where $g$ is the determinant of the metric. Putting this factor in the exponential with the action gives the term $$ \int d^d x[-\frac 12 \delta^d(0)\ln(1-|{\bf \pi}^2)] $$ The fact that the coefficient is $\delta^d(0)$ (interpreted as $N/V$ the density of spins) shows that it is not an ordinary mass term. It is there to preserve the rotational invariance (and hence the Goldstone mode) which would otherwise be broken by the co-ordinate choice in writing the measure as $d[\pi]$.

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