Cubic approximation

05 Dec 2018 in Technical on Tap

The most general one-dimensional cubic approximation to the free energy per spin along the line connecting a minimum and the corresponding saddle looks like

$$\begin{aligned} f(x) &= c\int (x-x_m)(x-x_s)\,dx + d, \end{aligned}$$

where $x_m,x_s$ are the positions of the minimum and maximum, $c,d$ are constants and $x$ is the position along the line. The distance between the minimum and maximum is $r = |x_m-x_s|$, obviously, and the barrier height is $B= N\frac c6 r^3$. The factor of $N$ comes from the fact that at the barrier height is computed in the total free energy. The second derivative at the minimum is $λ := cr$ and $-λ=-cr$ at the maximum. The barrier height may hence also be expressed as $B = N\frac{λ^3}{6c^2}$.

The expectation that the barrier height is finite for a finite band edge means that $λ \to \text{const.}>0$ and $B\to \text{const.}>0$ as $N\to\infty$, which implies $c \sim N^{1/2}$ and consequently $r \sim N^{-1/2}$. Incidentally, since $r = |\sqrt{Nq_m} - \sqrt{Nq_s}| = \sqrt{N}\frac{|q_m-q_s|}{\sqrt{q_m}+\sqrt{q_s}}$, this means that $q_m-q_s\sim N^{-1}$. This is assuming that the minimum-saddle pair lies on the straight line from the origin in $m$-space to the minimum, and $q_m$ and $q_s$ are the values of $q$ at the minimum and maximum, respectively.

On the other hand, if a vanishing band gap is to have a large barrier height, e.g. $B\sim N^{1/3}$, then $λ \to 0$ and $B \to\infty$ as $N\to\infty$. This means that the behaviour of both $c$ and $r$ must be different from the finite band edge case. One possibility could be (just guessing here, of course) $c\sim N^{-1/6}$ and $r \sim N^{-1/6}$.

Somehow this doesn’t feel right to me. Perhaps a cubic equation is not flexible enough. For example, it restricts the minimum and saddle to have the same second derivative, apart from the sign. But a more flexible function will very likely be too flexible such that no new insight can be obtained.

Update: A more sensible combination for the vanishing band edge case would be $c\sim N^{-2/3}$ and $r=\mathcal O(1)$. That would agree with $λ$ going to 0 as $N^{-2/3}$, as we now seem to believe. It would also give rise to $q_m-q_s\sim N^{-1/2}$. Maybe the $N$-dependence of $q_m-q_s$ could be measured, and a difference between vanishing and nonvanishing band edge cases be established.

Update 4 Feb 2019: I now realized $c\sim N^{-2/3}$ and $r=\mathcal O(1)$ is in fact enforced by the relation $B = N\frac{λ^3}{6c^2}$ and the assumptions $λ \sim N^{-2/3}$ and $B\sim N^{1/3}$.

Update 5 Feb 2019: This whole cubic approximation is built on the assumption that the minimum $\mathbf m_m$ and the saddle $\mathbf m_s$ are connected by the softest eigenvector $\mathbf e$, i.e. $\begin{aligned} \mathbf m_m - \mathbf m_s &= r\mathbf e. \end{aligned}$ Multiply this by $\mathbf m_m$ to obtain

$$\begin{aligned} r(\mathbf e,\mathbf m_m) &= N q_m - (\mathbf m_s,\mathbf m_m) \\ &= N q_m - (\mathbf m_s,\mathbf m_s + \mathbf m_m - \mathbf m_s) \\ &= N q_m - N q_s - (\mathbf m_s, r\mathbf e) \end{aligned}$$

or

$$\begin{aligned} r(\mathbf e,\mathbf m_m+\mathbf m_s) &= N q_m - N q_s. \end{aligned}$$

Figure 41, which I sent some time ago, seemed to show that $(\mathbf e,\mathbf m_m) = \text{const.}\times ||\mathbf m_m|| N^{-1/6} = \text{const.}\times \sqrt{q_m} N^{1/3}$. It appears reasonable to expect the same for $(\mathbf e,\mathbf m_s)$. Collecting, we get

$$\begin{aligned} \text{const.}\times N^{-2/3}r(\sqrt{q_m}+\sqrt{q_s}) &= q_m - q_s. \end{aligned}$$