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WoofDoggy
WoofDoggy
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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 \ll 1$$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?

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 $S \ll 1$ 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?

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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WoofDoggy
WoofDoggy
  • 237
  • 1
  • 9

Find the expansion of the exact solution (beyond Taylor)

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 $S \ll 1$ 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?