Weird Path Integral question [Archive]

Eduard Antonyan

Jun16-04, 05:42 PM

Igor Khavkine

Jun17-04, 04:19 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nEduard Antonyan <eddi@uchicago.edu> wrote in message news:<2bc1d0hppr43q34e522cfm7pchj9j7f6jl@4ax.com>. ..\n> Does anyone know of a way to calculate the path integral of the\n> following Lagrangian:\n>\n> L = (x\')^4/4 [1/4 factor inserted for simplicity]\n>\n> ?\n\nFirst go to the Hamiltonian formalism:\n\np = dL/d(x\') = (x\')^3 <=> x\' = p^(1/3),\n\nH = px\' - L = (3/4) p^(4/3).\n\nIf the Hamiltonian is not quadratic in the momenta, the path integral\nmust be expressed as\n\nZ = Int Dx[t] Dp[t] exp[i/hbar Int (px\' - (3/4) p^(4/3)) dt].\n\nTo evaluate the path integral we discretize the integral in the exponent\n\nSum_{i=0 to N-1} p(s_i) (x(t_{i+1}) - x(t_i)) - (3/4) p(s_i)^(4/3) dt\n= p(s_{N-1}) y - p(t_0) x\n- Sum_{i=1 to N-1} x(t_i) (p(s_{i+1}) - p(s_i))\n- Sum_{i=0 to N-1} (3/4) p(s_i)^(4/3) dt,\n\nwhere x = x(t_0) and y = x(t_N) are fixed, t_i < s_i < t_{i+1},\nand dt = (t_N-t_0)/N = T/N. The path integral breaks up into an ordinary\nmultiple integral. For each x(t_i) integral we get a delta function\n\n2pi delta(p(s_{i+1}) - p(s_i)) for i=0 to N-2.\n\nBecause of the delta functions, the p(s_i) integrals for i=1,...,N-1\nare trivial and the only integral that remains is (p=p(t_0))\n\nInt exp[i/hbar p (y-x) - T (3/4) p^(4/3)] dp.\n\nBesides a prefactor of (2pi)^N, which should be cancelled\nanyway, this is basically the path integral Z, since at this\npoint we can freely take N to infinity. Not surprisingly,\nit is just the Fourier transform the time evolution operator\ncorresponding to the Hamiltonian H = (3/4) p^(4/3), which is\ndiagonal in the momentum basis.\n\nAs for the actual integral, I\'m afraid I can only give a series\nsolution:\n\nZ = (3/2) exp(-i 3pi/8) (4hbar/3T)^(3/4)\nSum_{n=0 to oo} Gamma(3(n+1)/4)/n!\n[(4(y-x)/3T)^(3n/4)+(4(x-y)/3T)^(3n/4)].\n\nHope this helps.\n\nIgor\n]\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>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
]