Re: Weird Path Integral question
From: Igor Khavkine (k_igor_k_at_lycos.com)
Date: 06/17/04
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Date: 17 Jun 2004 05:19:12 -0400
Eduard Antonyan <eddi@uchicago.edu> wrote in message news:<2bc1d0hppr43q34e522cfm7pchj9j7f6jl@4ax.com>...
> Does anyone know of a way to calculate the path integral of the
> following Lagrangian:
>
> L = (x')^4/4 [1/4 factor inserted for simplicity]
>
> ?
First go to the Hamiltonian formalism:
p = dL/d(x') = (x')^3 <=> x' = p^(1/3),
H = px' - L = (3/4) p^(4/3).
If the Hamiltonian is not quadratic in the momenta, the path integral
must be expressed as
Z = Int Dx[t] Dp[t] exp[i/hbar Int (px' - (3/4) p^(4/3)) dt].
To evaluate the path integral we discretize the integral in the exponent
Sum_{i=0 to N-1} p(s_i) (x(t_{i+1}) - x(t_i)) - (3/4) p(s_i)^(4/3) dt
= p(s_{N-1}) y - p(t_0) x
- Sum_{i=1 to N-1} x(t_i) (p(s_{i+1}) - p(s_i))
- Sum_{i=0 to N-1} (3/4) p(s_i)^(4/3) dt,
where x = x(t_0) and y = x(t_N) are fixed, t_i < s_i < t_{i+1},
and dt = (t_N-t_0)/N = T/N. The path integral breaks up into an ordinary
multiple integral. For each x(t_i) integral we get a delta function
2pi delta(p(s_{i+1}) - p(s_i)) for i=0 to N-2.
Because of the delta functions, the p(s_i) integrals for i=1,...,N-1
are trivial and the only integral that remains is (p=p(t_0))
Int exp[i/hbar p (y-x) - T (3/4) p^(4/3)] dp.
Besides a prefactor of (2pi)^N, which should be cancelled
anyway, this is basically the path integral Z, since at this
point we can freely take N to infinity. Not surprisingly,
it is just the Fourier transform the time evolution operator
corresponding to the Hamiltonian H = (3/4) p^(4/3), which is
diagonal in the momentum basis.
As for the actual integral, I'm afraid I can only give a series
solution:
Z = (3/2) exp(-i 3pi/8) (4hbar/3T)^(3/4)
Sum_{n=0 to oo} Gamma(3(n+1)/4)/n!
[(4(y-x)/3T)^(3n/4)+(4(x-y)/3T)^(3n/4)].
Hope this helps.
Igor
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