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Tip of a spreading wave-packet: asymptotics beyond all orders of a saddle point expansion

+ 3 like - 0 dislike

This is a technical question coming from mapping of an unrelated problem onto dynamics of a non-relativistic massive particle in 1+1 dimensions. This issue is with asymptotics dominated by a term beyond all orders of a saddle point expansion (singular terms of an asymptotic series), like in the problem of the lifetime of a bound state in 1+0 negative coupling $\phi^4$ toy model.

Consider a particle with an initial (normalized) wave-function $$\psi_0(x) = e^{-(x+e^{-x})/2}$$ This specific shape defines a natural unit for $x$, note the double-exponential asymptotics of $\psi_0(x)$ as $ x \to -\infty$.

Time evolution under the Hamiltonian $\mathcal{H}=-\frac{1}{2}\partial_x^2 $ transforms the wave-function to (using the textbook propagator)

$$\psi(x,t) = (2 \pi i t)^{-1/2} \int e^{i (x-x')^2/(2t)} \psi_0(x') d x'$$

My question is about the asymptotics of this integral, especially the leading front propagating to the left. Here is where I've hit the wall:

The saddle point expansion in $t^{-1}$ gives $$t |\psi(x,t)|^2 \sim e^{-e^{-x}} \left [1 + (e^{3x} -2 e^{2x}) \, t^{-1} /8 + O(t^{-2}) \right ] $$ which converges nicely (checked numerically) for $x \gtrsim 1$, but fails to capture the terms of order $e^{x/t}$ that dominate over the double exponential at negtavie $x$.

For $t \to +\infty$ the solution becomes symmetric, $$|\psi(x,t \to \infty)|^2=\frac{1}{t \cosh (\pi x/t)}$$

Any ideas/hints will be appreciated.

This post has been migrated from (A51.SE)

asked Mar 6, 2012 in Theoretical Physics by Slaviks (610 points) [ revision history ]
retagged Mar 25, 2014 by dimension10
Minor typo in the question(v3): The square root $(2\pi it)^{1/2}$ in the second formula should be in the denominator.

This post has been migrated from (A51.SE)
@Qmechanic Thanks, fixed this one! I'm prone to typos :(

This post has been migrated from (A51.SE)

1 Answer

+ 3 like - 0 dislike

After some struggle and a useful hint form a colleague, the problem finally cracked:

  1. Going to momentum space gives $$\psi_{0k} = \int \psi_0(x) e^{i k x} dx= 2^{-ik+1/2} \Gamma(-ik+1/2)$$

  2. Applying time evolution with $\mathcal{H}=k^2/2$ gives $$\psi(x,t) = \frac{1}{2\pi} \int e^{-i k^2 t/2-i k x} \psi_{0k} dk$$

  3. Using large-$z$ asymptotic expansion for $\Gamma(z+1)$ and identifying a stationary phase point near $k \approx -x/t$ give the leading order which coincides with the asymptotics of $t \gg 1$ solution: $$\psi(x,t) \sim (2/t) e^{\pi x/t} $$

  4. Finally, the prefactor is recovered by keeping all leading logs in the expansion of $\Gamma(z+1)$ and solving for the stationary point using Lambert W function (for which Mathematica nicely handles the asymptotics), gives the final answer $$\psi(x,t) = (2/t) e^{\pi x/t} (-2 x/t)^{\pi/t} + O(t^{1+\epsilon})$$ with $\epsilon>0$, valid for $x \ll -t$ both for $t$ large and small.

Thanks to all who paid attention and helped with advice.

This post has been migrated from (A51.SE)
answered Mar 8, 2012 by Slaviks (610 points) [ no revision ]

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