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A tricritical point is the meeting of a line of continuous phase transitions and a line of first-order phase transitions. In the symmetric scalar Landau free energy, it occurs at $r=u=0$ with $v>0$. Two independent controls are needed to tune both coefficients. The continuous transition lies at $r=0,u>0$, and the tricritical three-phase line found above has $r=3u^2/(16v),u<0$; both approach the tricritical point. At zero conjugate field the three coexisting states on this first-order line are the disordered state and the two opposite ordered states.

Setting $u=H=0$ makes the leading stabilizing interaction sextic. The nonzero minimum obeys $M^4=-r/v$, so the order-parameter critical exponent along this tuned direction is ${\beta }_{\mathrm{tri}}=1/4$. Substitution givesand two temperature derivatives give the heat-capacity critical exponent ${\alpha }_{\mathrm{tri}}=1/2$. At $r=u=0$, the equation of state is $H=v{M}^5$, giving the critical-isotherm exponent ${\delta }_{\mathrm{tri}}=5$.

For magnetic susceptibility, differentiating $H=rM+v{M}^5$ gives ${\chi }^{-1}=r+5v{M}^4$. This is $r$ above the transition and $-4r$ below, hence ${\gamma }_{\mathrm{tri}}=1$. The quadratic fluctuation kernel still has a mass proportional to $|r|$, giving ${\nu }_{\mathrm{tri}}=1/2$ and ${\eta }_{\mathrm{tri}}=0$. Thus the mean-field tricritical exponent calculation yieldsDetuning $u$ gives tricritical crossover scaling; one must specify the approach to the tricritical point when quoting these critical exponents. The sextic interaction has upper critical dimension three, so sufficiently strong long-distance fluctuations change these mean-field predictions below three dimensions.

A tricritical point occurs when a first-order boundary meets a continuous-transition boundary. In the spin-inversion-symmetric sextic Landau free energy, two independent parameters must tune $r=u=0$, while $v>0$ keeps the potential stable. In the $h=0$ plane, the continuous boundary is $r=0,u>0$, and the first-order boundary is $r=3u^2/(16v),u<0$. They meet at the tricritical point; the order-parameter jump on the first-order side tends to zero as $u\uparrow 0$. The first panel shows this phase diagram, including the metastability limits derived above.

Along the tuned line $u=0,h=0$, the equation of state is $M(r+v{M}^4)=0$. Thus $M=(-r/v{)}^{1/4}$ on a selected ordered branch, rather than the ordinary square-root behavior. Differentiation gives inverse magnetic susceptibility $r$ above and $4|r|$ below, while at criticality $h=v{M}^5$. Hence the mean-field tricritical exponent calculation yields $\beta =1/4$, $\gamma =1$, $\delta =5$; the minimized potential $-|r{|}^{3/2}/(3\sqrt{v})$ also gives $\alpha =1/2$. These tricritical powers require tuning the quartic coefficient, not merely varying a generic temperature path. The ordinary scalar exponents requested separately are calculated next.

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.

At the tricritical point, tune $u=0$, hold $v>0$, and let $r=at$ cross zero. The equation of state becomes $h=rM+v{M}^5$. At $h=0,r<0$, $M^4=-r/v$, while the inverse susceptibility on the ordered branch is $r+5v{M}^4=4|r|$. On the disordered side it is $r$. At $r=0$, $M=\mathrm{sgn}(h)(|h|/v{)}^{1/5}$. Therefore $\beta =1/4$, $\gamma =1$ and $\delta =5$.

The minimized singular free-energy density isTwo temperature derivatives give a singular heat capacity proportional to $|t{|}^{-1/2}$ on the ordered side. Consequently the mean-field tricritical exponent calculation givesThe displayed scaling relations for critical exponents hold numerically:These are Landau theory values with both relevant coefficients tuned: along a generic path with fixed $u>0$, the asymptotic transition is ordinary rather than tricritical. The sextic upper critical dimension is three, so fluctuations can change the powers below three dimensions, and logarithmic corrections are possible at three. The vanishing disordered mean-field singular amplitude does not invalidate the ordered-side exponent.