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In a certain text, it was given that a function is given by $$\varphi_{n}(t)= \sqrt{\frac{(\eta)_n}{n !}} \tanh^{n} (\alpha t) \operatorname{sech}^{\eta}(\alpha t)$$ where $\eta$ and $\alpha$ are some constant number, $t$ is time. Further, it was written that at any fixed time $t$ and large $n$ the wavefunction has the form $$\left|\varphi_n(t)\right|^2 \sim e^{-n / \xi(t)}$$ where $\xi(t)$ is "delocalization length" that grows exponentially in time: $\xi(t) \sim e^{2 \alpha t}$ for $\alpha t \gg 1$.

How to get this behaviour for the it at large $n$? I tried to take the function and look at the asymptotic behaviour for large $t$ but then got stuck with the $\tanh$ part how to take the large limit of $n$ for it? Instead, I tried to get the delocalization length as a function of time with the limit $\alpha t \gg 1$ but couldn't get so as with this limit, the $\tanh$ part will be simply one. As a follow-up can someone also suggest a simple strategy to do such asymptotic analysis for other cases without messing up?

Erosannin
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  • What's the $(\eta)_n$ notation, is that a falling factorial? Also is the exponent on the sech actually $\eta$? – Ian Mar 21 '23 at 14:36
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    Also is the interest in $\alpha t$ large but $n$ even larger or $\alpha t$ arbitrary and $n$ large? The former is easier, because then you can look at $\tanh(x)$ as approximately $1-4e^{-2x}$. – Ian Mar 21 '23 at 14:45
  • $(\eta)_{n} = \frac{\Gamma(\eta+n)}{\Gamma(\eta)}$ is the Pochhamer symbol. Regarding the second part, it should be as follows: we first find the behaviour for finite "t" and large n and then find $\xi(t)$ from this. After that we take the $\alpha t \gg 1$ limit on the $\xi(t)$ to get the said exponential behaviour. – Erosannin Mar 21 '23 at 14:58
  • Yes the exponent of sech is $\eta$ – Erosannin Mar 21 '23 at 15:05
  • Just to be clear, you mean the rising factorial then, not the falling factorial, is that correct? – Ian Mar 21 '23 at 16:51
  • Yes rising factorial. – Erosannin Mar 21 '23 at 16:54
  • How does $\eta$ compare to $1$? I ask because doing some asymptotics on the quantity in the square root seems to yield $\eta-1$ which seems a bit weird. – Ian Mar 21 '23 at 16:56
  • Let us continue this discussion in chat. – Erosannin Mar 21 '23 at 17:54
  • Note for posterity that as Gary said I made a mistake in my two-term asymptotic for $\tanh$. – Ian Mar 22 '23 at 19:30

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First, for fixed $\eta$, $$ \frac{{(\eta )_n }}{{n!}} = \frac{{\Gamma (n + \eta )}}{{\Gamma (\eta)\Gamma (n + 1)}} \sim \frac{{n^{\eta - 1} }}{{\Gamma (\eta)}} $$ as $n\to+\infty$. Second, for $x>0$, $$ \tanh (x) = 1 - 2{\rm e}^{ - 2x} + 2{\rm e}^{ - 4x} + \ldots , $$ whence $$ \log \tanh (x) \sim \log (1 - 2{\rm e}^{ - 2x} ) \sim - 2{\rm e}^{ - 2x} $$ for large positive $x$. This gives $$ \tanh ^{2n} (\alpha t) = \exp (2n\log \tanh (\alpha t)) \sim \exp ( - 2n\exp ( - 2\alpha t)) $$ for large positive $\alpha t$. Third, for $x>0$, $$ \operatorname{sech}(x) = 2{\rm e}^{ - x} - 2{\rm e}^{ - 3x} + \ldots , $$ whence $$ \operatorname{sech}^{2\eta} (\alpha t)\sim 4^{\eta} {\rm e}^{ - 2\eta \alpha t} $$ for large positive $\alpha t$. Accordingly, $$ \left| {\varphi _n (t)} \right|^2 \sim \frac{{4^\eta }}{{\Gamma (\eta )}}{\rm e}^{ - 2\eta \alpha t} n^{\eta - 1} \exp ( - 2n\exp ( - 2\alpha t)) $$ for large positive $n$ and $\alpha t$ with fixed $\eta$.

Gary
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