Nail-Tinted Glasses

Feynman Hibbs 3-1: The Free Particle

Having finished grinding my way through another typo in Feynman Hibbs, it's time to celebrate by putting it all in one place.

Section 3-1 describes the process of computing the kernel of a free particle for a finite interval from the kernel of an infinitesmal.

The tool that we will be relying heavily upon is the Gaussian Integral.

Without going too deeply into detail: ∫ e^[-x^2] is solved by squaring the integral and transforming into polar coordinates. Having established that, ∫ e^[-Ax^2] is calculated with a simple change of variables, and ∫ e^[-Ax^2+bx] by completing the square.

∫ e^[-Ax^2+bx] = (π/A)^1/2 e^[b^2/4A]

e^[(i/ħ)S]
= e^[(i/ħ) ∫ L ]
= e^[(i/ħ) ε Σ L ( (x[i+1]-x[i])/ε , (x[i+1]-x[i]) / 2 ) ]

For a free particle, L = mx'^2/2

= 1/A e^[(i/ħ) ε Σ m/2 ( (x[i+1]-x[i])/ε ) ^2 ]
= 1/A e^[(mi/2ħε) Σ (x[i+1]-x[i]) ^2 ]
= 1/A e^-[(m/2iħε) Σ (x[i+1]-x[i]) ^2 ]

where A is just the factor required to get the correct answer to zero order. For this problem, it is (2πiħε/m)^1/2. This is, for those familiar with the story, the "analogous" bit.

So what value does this have when we go from x0 to x2? K(2,1)K(1,0)dx1. Which is to say, the product of two of those, integrated across all possible values of the middle term. The product of the terms is just taking the sum of the exponents, and the constants can be tossed out in front. The key part is going to be

(x2-x1)^2 + (x1-x0)^2

= x2^2 - 2x2x1 + x1^2 + x1^2 - 2x1x0 + x0^2

= 2x1^2 - 2[x2+x0]x1 + [x2^2+x0^2]

The last of these terms has no x1 dependence, and so can be brought outside the integral. The first two terms are the form of a Gaussian. When the integral is evaluated, we get an exponent in [x2+x0]^2, which happily cancels partially with the term out in front, setting up another Gaussian.

∫ (2πiħε/m)^2/2 e^-[(m/2iħε) [(x2-x1)^2 + (x1-x0)^2] ] dx1

= (m/2πiħε) ∫ e^-[(m/2iħε) [x2^2-2x2x1+x1^2 + x1^2-2x1x0 + x0^2 ] dx1

= (m/2πiħε) ∫ e^-[(m/2iħε) [x2^2 + x0^2 + 2x1^2 -2 [x2+x0]x1 ] dx1

= (m/2πiħε) ∫ e^-[(m/2iħε) [x2^2 + x0^2] e^-[(m/2iħε) 2x1^2 -2 [x2+x0]x1 ] dx1

= (m/2πiħε) e^-[(m/2iħε) [x2^2 + x0^2] ∫ e^-[(m/2iħε) 2x1^2 -2 [x2+x0]x1 ] dx1

= (m/2πiħε) e^-[(m/2iħε) [x2^2 + x0^2] (π/(2m/2iħε))^1/2 e^[((2m/2iħε)[x2+x0])^2/4(2m/2iħε)]

= (m/2πiħε) e^-[(m/2iħε) [x2^2 + x0^2] (π/(2m/2iħε))^1/2 e^[(2m/2iħε)[x2+x0]^2/4]

= (m/2πiħε) e^-[(m/2iħε) [x2^2 + x0^2] (π/(2m/2iħε))^1/2 e^[(m/2iħ2ε)[x2+x0]^2]

= (m/2πiħε) e^-[(m/2iħ2ε) [2x2^2 + 2x0^2] (π/(2m/2iħε))^1/2 e^[(m/2iħ2ε)[x2+x0]^2]

= (m/2πiħε)(π/(2m/2iħε))^1/2 e^-[(m/2iħ2ε) [2x2^2 + 2x0^2 - [x2+x0]^2]]

= (m/2πiħε)(π/(2m/2iħε))^1/2 e^-[(m/2iħ2ε) [x2^2 + x0^2 - 2x2x0] ]

= (m/2πiħε)(π/(2m/2iħε))^1/2 e^-[(m/2iħ2ε) [x2 - x0]^2 ]

= (m/2πiħε)(2πiħε/2m)^1/2 e^-[(m/2iħ2ε) [x2 - x0]^2 ]

= (m/2πiħ(2ε))^1/2 e^-[(m/2iħ(2ε)) [x2 - x0]^2 ]

The next step is to invoke a proof by induction: replace (2ε) by (nε) above, multiply by 1/A e^-[(m/2iħε) (x[n+1]-x[n]) ^2 ], integrate with respect to x[n]. When the integral is evaluated, the form is the same, but all of the n are replaced by (n+1).

The point is that when you grind enough of these out, you have an expression in terms of x(a), x(b), and Nε, where Nε = b - a. The ε drop out, replaced by coordinates strictly determined by the endpoints of the calculation, and now you can take the limit ε -> 0 without changing the result.

What is the typo? Hibbs substituted m/i, rather than -m/i, when writing out 3-4. Or he is using a notation which is inconsistent with the previous page.

January 25, 2004 12:02 PM | TrackBack