David Tong SFT Sheet 1 Solutions: Landau Theory

Source: David Tong, Statistical Field Theory, Example Sheet 1; 8 worked problems.

These prompts paraphrase David Tong's Statistical Field Theory, Example Sheet 1 (October 2018). Problem numbers, starred status, equations, and guidance follow the official sheet.

1 Sheet 1, Problem 1: Transfer Matrix for the One-Dimensional Ising Model

Take $N$ Ising spins with $s_{N+1}=s_1$ and partition function \[Z=\sum_{s_1=\pm1}\cdots\sum_{s_N=\pm1} \prod_{i=1}^{N} \exp\!\left[\beta J s_i s_{i+1} +\frac{\beta B}{2}(s_i+s_{i+1})\right].\tag{1.1}\] Show that $Z=\operatorname{tr}T^N$ for the transfer matrix \[T=\begin{pmatrix} e^{\beta J-\beta B}&e^{-\beta J}\\ e^{-\beta J}&e^{\beta J+\beta B} \end{pmatrix}.\tag{1.2}\] Hint: Express the $i$th factor as $T_{s_i,s_{i+1}}=\exp[\beta J s_i s_{i+1} +\beta B(s_i+s_{i+1})/2]$, with $s_i$ labelling a column and $s_{i+1}$ a row. Find the two eigenvalues, \[\lambda_\pm=e^{\beta J}\cosh(\beta B) \pm\sqrt{e^{2\beta J}\cosh^2(\beta B)-2\sinh(2\beta J)},\tag{1.3}\] and explain why $Z\simeq\lambda_+^N$ as $N\to\infty$. At $B=0$, deduce that the magnetisation is always zero and that temperature causes no phase transition in this one-dimensional system.

2 Sheet 1, Problem 2: Weiss Mean Field Theory

Expand an Ising nearest-neighbour product about the equilibrium magnetisation $m$ as \[s_i s_j=(s_i-m)(s_j-m)+m(s_i-m)+m(s_j-m)+m^2.\tag{2.1}\] Neglect the product of fluctuations. For the source's nearest-neighbour parameter $q$, derive \[Z=e^{-\beta JNqm^2/2} 2^N\cosh^N(\beta Jqm+\beta B),\tag{2.2}\] and obtain the self-consistency equation \[m=\tanh(\beta B+\beta Jqm).\tag{2.3}\] At $B=0$, compare the graphical solutions above and below $T_c=Jq$. For nonzero $B$, find the high-temperature asymptotic solution.

Here, we shall treat $m$ merely as the equilibrium distribution. Keeping terms linear in fluctuation, we have \[s_i s_j \approx m(s_i-m)+m(s_j-m)+m^2 =m(s_i +s_j) - m^2.\tag{2.4}\] Then, \[\sum_{\langle ij\rangle}s_i s_j \approx -\frac{Nqm^2}{2}+qm\sum_i s_i,\tag{2.5}\] which gives \[E[s_i] \approx -(B+Jmq)\sum_i s_i + \frac{NJqm^2}{2} = -(B+Jmq)\bar s + \frac{NJqm^2}{2},\tag{2.6}\] where $\bar s = \sum_i s_i$. Summing over all spin configurations (including the binomial multiplicity of each value of $\bar s$), the partition function factorises: \[Z \approx e^{-\beta JNqm^2/2}\prod_{i=1}^{N}\sum_{s_i=\pm1}e^{\beta(B+Jqm)s_i} =e^{-\beta JNqm^2/2}2^N\cosh^N(\beta Jqm+\beta B).\tag{2.7}\] The factorisation is exact within the mean-field approximation for either parity of $N$.

The self-consistency equation immediately follows from \[m = \frac{1}{N\beta}\frac{\partial \log Z}{\partial B}.\tag{2.8}\]

Solving this equation would give us the allowed value for $m$. If we differentiate the RHS, we find \[\frac{\beta Jq}{\cosh^2(\beta J qm)} \leq \beta J q.\tag{2.9}\] For $T > T_c$, $\beta Jq <1$, and hence \[\frac{\beta Jq}{\cosh^2(\beta J qm)} \leq 1,\tag{2.10}\] which means that $m = 0$ is the only solution to the self-consistency equation. However, for $T < T_c$, there can be two extra solutions to the equation. This is a continuous (second-order) phase transition: the stable nonzero solutions approach zero as $T\to T_c^-$.

For fixed nonzero $B$ and $T\to\infty$, linearising the self-consistency equation gives $m\simeq B/(T-Jq)\sim B/T$ (with $k_B=1$).

3 Sheet 1, Problem 3: Exact Mean Field Limit of a Long-Range Ising Model

Let every spin interact with every other spin, including itself, through \[E=-B\sum_i s_i-\frac{J}{2N} \left(\sum_i s_i\right)\left(\sum_j s_j\right).\tag{3.1}\] Explain the factor $1/N$. Establish the Gaussian identity \[e^{\beta J\alpha^2/(2N)} =\sqrt{\frac{N\beta J}{2\pi}} \int_{-\infty}^{\infty}dx\, e^{-N\beta Jx^2/2+\alpha\beta Jx},\tag{3.2}\] and use it to express the partition function as \[Z=\sqrt{\frac{N\beta J}{2\pi}} \int_{-\infty}^{\infty}dx\,e^{-NS(x)}, \qquad S(x)=\frac{\beta Jx^2}{2} -\log\!\left[2\cosh\bigl(\beta(B+Jx)\bigr)\right].\tag{3.3}\] Find the saddle $x_\star$ that dominates at large $N$, up to its prefactor. Derive its stationarity equation and show that $m=x_\star$.

Since we expect the energy to scale as $N$, while $\sum_i s_i$ also scales as $N$. Therefore, we must have a factor of $1/N$.

The Gaussian identity is straightforward. Let us define \[\sum_i s_i \equiv \alpha.\tag{3.4}\] Then, \[\begin{aligned} Z &= \sum_{\alpha} e^{\beta J \alpha^2/(2N)+\beta B \alpha}\\ &=\sqrt{\frac{N\beta J}{2\pi}}\sum_\alpha e^{\beta B \alpha}\int_{-\infty}^{\infty}dx\,e^{-N\beta Jx^2/2+\alpha\beta Jx}\\ &=\sqrt{\frac{N\beta J}{2\pi}}\int_{-\infty}^{\infty}dx\,e^{-N\beta Jx^2/2}\sum_\alpha \,e^{\beta(B + Jx)\alpha}\\ &=\sqrt{\frac{N\beta J}{2\pi}}\int_{-\infty}^{\infty}dx\,e^{-N\beta Jx^2/2} \cdot 2^N\cosh^N\beta(B+ Jx) \\ &=\sqrt{\frac{N\beta J}{2\pi}} \int_{-\infty}^{\infty}dx\,e^{-NS(x)}. \end{aligned}\tag{3.5}\] The saddle is given by \[S'(x_\star) = 0 \implies x_\star = \tanh (\beta B + \beta J x_\star),\tag{3.6}\] which satisfies the same equation as $m$ as we have seen. Therefore, we have $m = x_\star$, which means that the long-range Ising model realises the mean field theory exactly.

4 Sheet 1, Problem 4*: First-Order and Tricritical Transitions

Study \[f(m)=\alpha_2(T)m^2+\alpha_4m^4+\alpha_6m^6, \qquad \alpha_4<0,\quad\alpha_6>0,\tag{4.1}\] as $\alpha_2(T)$ decreases through zero. Sketch the changing free-energy landscape, prove that the first-order transition occurs at $\alpha_2=\alpha_4^2/(4\alpha_6)$, and find the magnetisation jump. Draw the phase diagram in the $(\alpha_2,\alpha_4)$ plane. At $\alpha_4=0$ the transition at $\alpha_2(T_c)=0$ is tricritical. Calculate its mean field exponents $\alpha,\beta,\gamma,\delta$; include a term proportional to $Bm$ when finding the last two.

$f(m)$ crosses the $m$-axis at \[m = 0, \quad m^2 = \frac{-\alpha_4 \pm \sqrt{ \alpha_4^2 - 4\alpha_2 \alpha_6}}{2\alpha_6},\tag{4.2}\] where the latter exist when $\alpha_2 < \alpha^2_4 /(4\alpha_6)$. While the turning points of $f(m)$ is at \[m= 0, \quad m^2 = \frac{-\alpha_4 \pm \sqrt{\alpha_4^2 - 3\alpha_6 \alpha_2}}{3\alpha_6},\tag{4.3}\] where the latter exists when $\alpha_2 < \alpha^2_4 /(3\alpha_6)$ before $f(m)$ crosses the axis. The two minima are degenerate at $\alpha_2=\alpha_4^2/(4\alpha_6)$; below this value the nonzero minima are favoured. At coexistence, the magnetisation jumps from $m=0$ to $m=\pm\sqrt{-\alpha_4/(2\alpha_6)}$.

At the tricritical point, we have \[f(m) \approx -Bm + a(T-T_c)m^2 + \alpha_6m^6.\tag{4.4}\] The minima are at \[m^4 = - \frac{2a}{6\alpha_6}(T-T_c),\tag{4.5}\] and hence $\beta = 1/4$. The heat capacity is given by \[c = \beta ^2 \frac{\partial^2 \log Z}{\partial \beta^2}.\tag{4.6}\] Plugging the temperature dependent minima into the partition function, we find \[f(m_\mathrm{min}) \sim (T_c-T)^{3/2} \implies c \sim f'' \sim |T-T_c|^{-1/2}.\tag{4.7}\] Hence, the critical exponent $\alpha = 1/2$. With the presence of an external source, the minima satisfies \[B = 2a(T-T_c)m + 6\alpha_6 m^5 \implies m \sim B^{1/5}\tag{4.8}\] which gives $\delta = 5$. Then, \[\frac{1}{\chi} = \frac{\partial B}{\partial m}\bigg|_{B=0} =2a(T-T_c) + 30\alpha_6 m^4 .\tag{4.9}\] For $T > T_c$, $m =0$, and hence $\gamma = 1$. For $T< T_c$, $m^4\propto T_c-T$, so the two terms in $1/\chi$ are both linear in $|T-T_c|$ and $\gamma=1$ below $T_c$ as well.

5 Sheet 1, Problem 5: The XY Order Parameter

For a complex superfluid order parameter with \[f(\psi)=\alpha_2|\psi|^2+\alpha_4|\psi|^4,\tag{5.1}\] sketch $f$ when $\alpha_2<0<\alpha_4$. Identify the possible ground states, explain spontaneous symmetry breaking here, and compare the mean field exponents $\alpha$ and $\beta$ at $\alpha_2=0$ with those of the Ising model.

6 Sheet 1, Problem 6: Ising and XY Orders in an Anisotropic Spin Model

At each lattice site put a unit three-vector $\mathbf s_i=(s_i^x,s_i^y,s_i^z)$ and use \[E=-J\sum_{\langle ij\rangle}\mathbf s_i\cdot\mathbf s_j +g\sum_i\left[(s_i^z)^2 -\frac{(s_i^x)^2+(s_i^y)^2}{2}\right], \qquad J>0.\tag{6.1}\] Use mean field reasoning to account for the phase diagram below,

Hint: Compare the ground states as $g$ crosses zero at low $T$, then raise $T$ separately on either side of $g=0$. Detailed path-integral calculations are unnecessary.

7 Sheet 1, Problem 7: A Spatially Modulated Condensate

For a complex scalar field, consider \[F=\int dx\left[ \alpha_2(T)|\psi|^2+\alpha_4|\psi|^4 -\gamma\left|\frac{d\psi}{dx}\right|^2 +\kappa\left|\frac{d^2\psi}{dx^2}\right|^2\right], \qquad \alpha_4,\gamma,\kappa>0.\tag{7.1}\] Using a single nonzero Fourier mode $\psi_k=A_ke^{ikx}$ with $k=\pm k_0$, determine $k_0$ and show that the modulated phase appears at $\alpha_2=\gamma^2/(4\kappa)$. Identify the symmetries broken in the ordered phase.

8 Sheet 1, Problem 8: Multicritical Scaling and Upper Critical Dimension

Take \[f(m)=\alpha_2m^2+\alpha_{2n}m^{2n}, \qquad n\in\mathbb Z_+.\tag{8.1}\] Find $\beta$ from $m\sim(T_c-T)^\beta$ in the ordered phase. Insert the result into the free energy to show that the heat-capacity exponent obeys $\alpha=1-2\beta$. Compare this with the contribution to the free energy from fluctuations in the Gaussian path integral and establish the threshold $d_c=2n/(n-1)$ above which the mean field term dominates near criticality.