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In a paper by Kitagawa & Ueda Squeezed spin states they give an argument that the minimum variance in one-axis twisting Hamiltonian scales like $V_{min} \propto S^{-2/3}$. I will shortly describe how it goes.

The exact expression for the variance is the following

$$\begin{align} V_{-} & = \frac{S}{2}\left[1+\frac{1}{2}\left(S-\frac{1}{2}\right)\left(A -\sqrt{A^2 + B^2}\right)\right] \\[3mm] A & = 1 - \cos^{2S-2}\mu \\[3mm] B & = 4\sin\frac{\mu}{2}\cos^{2S-2}\frac{\mu}{2} \end{align}$$

They approximate this equation in the limit $S \gg 1$ and $|\mu| \ll 1$ by

$$V_{-} \approx \frac{S}{2}\left(\frac{1}{4\alpha^2} + \frac{2}{3}\beta^2 \right)$$ with $\alpha = 1/2S\mu$, $\beta = 1/4 S\mu^2$. Additional assumptions: $|\alpha| > 1$, $\beta \ll 1$. Then You can minimize this approximate solution over $\mu$ and find the scalling.

I am struggling to derive this approximate expression from the exact solution, but without effort. This is not a simple Taylor expansion around $0$.
You can compare them in Mathematica and find that approximation nicely catches the minimum. Can Anyone help me get this approximation?

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    $\begingroup$ send $S\rightarrow\infty$, $\mu\rightarrow 0$ at constant $S\mu$. $\endgroup$
    – Carlo Beenakker
    Commented May 20, 2015 at 13:15
  • $\begingroup$ @Carlo: He says limit $S \ll 1$ ... so you claim that is a mistake and he intends $S \to \infty$ instead? $\endgroup$
    – Gerald Edgar
    Commented May 20, 2015 at 13:16
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    $\begingroup$ indeed, this is a mistake, S goes to infinity, not to zero, otherwise you cannot have $S\mu$ of order unity and $S\mu^2$ small (in other words, $\mu$ is of order $1/S$). $\endgroup$
    – Carlo Beenakker
    Commented May 20, 2015 at 13:31
  • $\begingroup$ Yes, my bad. I have corrected the question. $\endgroup$
    – WoofDoggy
    Commented May 20, 2015 at 15:21
  • $\begingroup$ @CarloBeenakker You mean expanding everything around $\mu=0$ and keep all terms of the order $S\mu$?. In this way I get a polynomial. I am missing something. My idea was to approximate the square root by $A + 2B^2 A/(4A^2+B^2)$. $\endgroup$
    – WoofDoggy
    Commented May 20, 2015 at 16:04

2 Answers 2

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So there are two parameters $\alpha$ and $\beta$ and a function $V(\alpha,\beta)$, obtained from your first equation by substituting $S=\alpha^2/\beta$ and $\mu=2\beta/\alpha$. With some effort we can make a Taylor series expansion of this function around $\beta=0$, to second order. The result is not pretty:

$$\frac{2}{S}V(\alpha,\beta)=\left(1 + 2 \alpha (\alpha - \sqrt{1 + \alpha^2})\right) + \left(-3 - 4 \alpha^2 + \alpha^{-1}\sqrt{ 1 + \alpha^2}+ 4 \alpha \sqrt{1 + \alpha^2}\right) \beta + \tfrac{1}{3} \left(34 + 3\alpha^{-2} + 16 \alpha^2 - 14\alpha^{-1} \sqrt{1 + \alpha^2}- 2 \alpha (13 + 8 \alpha^2)(1 + \alpha^2)^{-1/2}\right) \beta^2+{\rm order}(\beta^3)$$

I can imagine that the authors of the paper you mention were not happy with this formula, and made one further approximation, even though they did not explicitly say so. They write $|\alpha|>1$, but in fact they assume $|\alpha|\gg 1$, retaining terms of order $\beta^2$ and terms of order $1/\alpha^2$, but neglecting terms of order $(\beta/\alpha)^2$. That gives the desired result,

$$\frac{2}{S}V(\alpha,\beta)=\tfrac{1}{4}\alpha^{-2}+\tfrac{2}{3}\beta^2+\text{higher order terms}.$$

(So this is a Taylor series in $\epsilon$ if you identify $\beta=\beta'\epsilon$, $\alpha=\alpha'/\epsilon$; the "higher order terms" are all of order $\epsilon^3$ or smaller.)

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  • $\begingroup$ I still can't see it. Where did You get this $\alpha^{-2}$ ? You mean $\sqrt{1+\alpha^2} = \alpha\sqrt{1 + 1/\alpha^2} \approx \alpha (1 + 1/(2\alpha^2) - 1/(8\alpha^4))$... Why should I expand to second order? $\endgroup$
    – WoofDoggy
    Commented May 21, 2015 at 12:53
  • $\begingroup$ it's really simple: give the whole expression to Mathematica, substitute $\beta=\beta'\epsilon$, $\alpha=\alpha'/\epsilon$, and ask Mathematica to expand in powers of $\epsilon$ to second order; and you'll get the desired result. You expand to second order because there are no terms of lower order. $\endgroup$
    – Carlo Beenakker
    Commented May 21, 2015 at 13:59
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Only the assumption $S\to\infty$ is needed to verify that there is a local minimum near the place where $V_-$ has its minimum, namely at $\mu=\mu_0 = 12^{1/6} S^{-2/3}$, but it doesn't give the approximation $V_-$. I'm sure it gives a better approximation near the minimum.

Take the Taylor expansion of $V$ about $\mu=\mu_0$. Then take the asymptotic expansion of the first few coefficients as $S\to\infty$. It gives $$ V = (12^{2/3}S^{1/3}/16 + O(S^{-1/3})) + O(S^{1/3}) (\mu-\mu_0) + (12^{1/3}S^{5/3}/4 + O(S)) (\mu-\mu_0)^2 + \cdots\,. $$ The linear term is negligible near $\mu=\mu_0$ so there is a local minimum near there. By taking a few more terms, it is easy to find the minimum more accurately. The next approximation is $$ 12^{1/6} S^{-2/3} - \frac{7\cdot2^{2/3}\cdot3^{5/6}}{40} S^{-4/3}. $$

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