 Solution

A critical exponent is the leading power of a singular observable as a control variable approaches a critical value. Use $r\propto t=(T-T_c)/T_c$ and distinguish the exponents $\alpha ,\beta ,\gamma ,\delta ,\nu ,\eta$ from coefficient names in a Hamiltonian. Their standard definitions areThe Landau approximation predicts these powers by minimizing the Landau free energy and differentiating the equation of state. At an ordinary quartic point, $M^2=-r/u$, ${\chi }^{-1}=r$ above and $2|r|$ below, and $h=u{M}^3$ on the critical isotherm. The equilibrium singular free energy is proportional to $-r^2$ below, giving a heat-capacity jump. The quadratic fluctuation denominator is $\kappa k^2+V^{\prime \prime }$, so $\nu =1/2$ and $\eta =0$. Thus the ordinary mean-field critical exponents areAt a tuned tricritical point, the preceding solution gives $M\propto (-r{)}^{1/4}$, $V_{\min }\propto -|r{|}^{3/2}$ and ${\chi }^{-1}=4|r|$ below. At $r=0$ the equation of state is $h=v{M}^5$. The curvature still vanishes linearly, and the Gaussian gradient term still gives zero anomalous dimension. ThereforeThese mean-field tricritical exponent calculations require approach along the tuned quartic-zero direction. Fluctuations modify the powers below the appropriate upper critical dimension, and marginal effects can add logarithms at it.