What is the link between path integral source terms and the propagator?
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Consider a diffusion process defined by
∂ϕ∂t=D∂2ϕ∂x2
where ϕ is the probability density of a particle's position and D is the diffusion coefficient (or whatever it's called). Denote the transition probability to go from (0,0) to (x,t) as W(x,t;0,0). It's easy enough to show that
W(xf,t;0,0)=14πDt−−−−√exp(−x2f4Dt).
We can imagine getting this result by summing the probabilities of all possible paths starting at (0,0) and ending at (x,t) [a]. We can approximate this sum over all paths by dividing the time axis into n+1 slices of length ϵ. Note that the 0th slice is the starting point where the position is constrianed to be x=0 and the (N+1)th slice is the end point where the position is constrained to be x=xf.
Using the known solution to the diffusion over each time slice, we can write [b]
W(x,t;0,0)==∫dx1⋯dxn(14πDϵ)n/2exp[−14Dϵ∑i=0n(xi+1−xi)2]∫dx1⋯dxn(14πDϵ)n/2exp[−14Dϵ(∑ijxiAijxj+∑ixiJi)−14Dϵx2f]
where
Aij=⎡⎣⎢⎢⎢⎢⎢⎢⎢2−1−12−1−1⋱−1−12−1−12⎤⎦⎥⎥⎥⎥⎥⎥⎥
is the discrete version of the second derivative, and J, which depends on the boundary conditions, is in this particular case Ji=2δi,nxf.
There's a known formula
∫dx1⋯dxnexp[−12∑ijAijxixj+∑iJixi]=(2π)ndetA−−−−−√exp(12∑ijJi(A−1)ijJj).
I don't know A−1, so I can't evaluate the expression. However, I am given to believe that there is a link between A−1 and the propagator W(xf,tf;xi,ti). If that's true, then I should be able to work out the term involving the J's because I know the propagator.
What is the link between A−1 and W, and how can I use that link to evaluate ∑ijJi(A−1)ijJj?
[a]: i.e. we can invent the path integral.
[b]: Yes, I'm using the known solution to rewrite the problem as an integral. This looks like circular reasoning but the point is to get things into a form suitable for more complex problems.
∂ϕ∂t=D∂2ϕ∂x2
where ϕ is the probability density of a particle's position and D is the diffusion coefficient (or whatever it's called). Denote the transition probability to go from (0,0) to (x,t) as W(x,t;0,0). It's easy enough to show that
W(xf,t;0,0)=14πDt−−−−√exp(−x2f4Dt).
We can imagine getting this result by summing the probabilities of all possible paths starting at (0,0) and ending at (x,t) [a]. We can approximate this sum over all paths by dividing the time axis into n+1 slices of length ϵ. Note that the 0th slice is the starting point where the position is constrianed to be x=0 and the (N+1)th slice is the end point where the position is constrained to be x=xf.
Using the known solution to the diffusion over each time slice, we can write [b]
W(x,t;0,0)==∫dx1⋯dxn(14πDϵ)n/2exp[−14Dϵ∑i=0n(xi+1−xi)2]∫dx1⋯dxn(14πDϵ)n/2exp[−14Dϵ(∑ijxiAijxj+∑ixiJi)−14Dϵx2f]
where
Aij=⎡⎣⎢⎢⎢⎢⎢⎢⎢2−1−12−1−1⋱−1−12−1−12⎤⎦⎥⎥⎥⎥⎥⎥⎥
is the discrete version of the second derivative, and J, which depends on the boundary conditions, is in this particular case Ji=2δi,nxf.
There's a known formula
∫dx1⋯dxnexp[−12∑ijAijxixj+∑iJixi]=(2π)ndetA−−−−−√exp(12∑ijJi(A−1)ijJj).
I don't know A−1, so I can't evaluate the expression. However, I am given to believe that there is a link between A−1 and the propagator W(xf,tf;xi,ti). If that's true, then I should be able to work out the term involving the J's because I know the propagator.
What is the link between A−1 and W, and how can I use that link to evaluate ∑ijJi(A−1)ijJj?
[a]: i.e. we can invent the path integral.
[b]: Yes, I'm using the known solution to rewrite the problem as an integral. This looks like circular reasoning but the point is to get things into a form suitable for more complex problems.
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the path integral formulation in the field of statistical mechanics. 2 Path Integral Method Define the propagator of a quantum system between two spacetime points (x ′ ,t ′ )
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