# Dirac's K

1. Jul 15, 2006

### wavemaster

In http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html" [Broken], he mentions this:

I'm trying to get the same result, but I'm stuck. Has anyone done this before?

Last edited by a moderator: May 2, 2017
2. Jul 16, 2006

### samalkhaiat

Write $$L= \frac{1}{2} (\frac{x-x'}{\lambda})^{2} - V(x)$$ , change integration variable to $$x-x'=y$$ , then expand $$V(x),\phi(x)$$ about $$y=0$$ and keep first order terms in $$\lambda$$ ;
$$\phi(x',t+\lambda)= A \int dy exp(\frac{iy^{2}}{2\lambda}) exp[-i\lambda (V(x') + yV'(x') + ..)][\phi(x',t) + y \frac{\partial {\phi(x')}}{\partial{x'}} + \frac{1}{2} y^{2}\frac{\partial^{2} \phi(x',t)}{\partial {x'}^{2}} +...]$$
or
$$\phi(x',t+\lambda) = A[1 - i \lambda V(x')] \int dy exp[\frac{iy^{2}}{2\lambda}][\phi(x',t) + (1/2) y^{2} \frac{\partial^{2}\phi}{\partial {x'}^{2}}]$$
we dropped the factor
$$exp[-iV'(x')y\lambda]$$
because when the integration over y is done this will be of order
$$\lambda^{3/2}$$
we also used
$$\int dy y exp(-a y^{2}) = 0$$

Now use the integrals:
$$\int dy exp(-a y^{2}) = (\pi/a)^{1/2}$$
$$\int dy y^{2} exp(-a y^{2}) = (\pi/2)^{1/2}(1/2a)$$

Schrodinger eq follows when you choose A so that the coefficient of
$\phi(x',t)$ is one, and take the limit $\lambda\rightarrow 0$;

$$\frac{\phi(x',t+\lambda)-\phi(x',t)}{\lambda} = \frac{\partial}{\partial{t}} \phi(x',t)$$

regards

sam

Last edited by a moderator: May 2, 2017
3. Jul 17, 2006

### wavemaster

I didn't know the bounds are infinities... Also, I couldn't get $$\lambda$$ to the 3/2th power, for $$-iV'(x') y \lambda$$, but as you did, I used only the first term for potential.

Anyway, I couldn't see how Schrödinger equation comes out from what you've written, so
I've tried to work out the details for myself, but I ended up a weird equation that is far from Schrödinger's. Here it goes:

(You seem to've cranked up $$\hbar$$ and m to 1, but for keeping a better track of things, I'm leaving them there).

$$L = \frac{m}{2} {(\frac{x-x'}{\epsilon})}^2 - V(x)$$
$$y = x-x'$$, keeping $$x'$$ fixed, $$dy = dx$$ (I guess this was what you meant).

Then,

$$\varphi(x',t+\epsilon) = A \int_{-\infty}^{\infty} e^{\frac{im}{\hbar \epsilon}y^2} e^{-\frac{i}{\hbar} \epsilon V(x')} (\varphi(x',t) + y\frac{\partial \varphi(x',t)}{\partial x} + \frac{y^2}{2}\frac{\partial^2 \varphi(x',t)}{\partial x^2}) dy$$

$$\varphi(x',t+\epsilon) = A e^{-\frac{i}{\hbar} \epsilon V(x')} (\varphi(x',t)\sqrt{\frac{\pi}{-im/\hbar \epsilon}} + \frac{1}{2}\frac{\partial^2 \varphi(x',t)}{\partial x^2} \sqrt{\frac{\pi}{2}} \frac{1}{2(-im/\hbar \epsilon)})$$

To make coefficient of $$\varphi(x',t)$$ 1, A should obey:

$$A e^{-\frac{i}{\hbar} \epsilon V(x')} \sqrt{\frac{\pi}{-im/\hbar \epsilon}} = 1$$

By making A so, I got:

$$i\hbar \frac{\partial \varphi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \varphi}{\partial x^2} \frac{i}{2\sqrt2} \sqrt{\frac{\hbar \epsilon}{im}}$$

which has some weird coefficients for spatial derivative, and missing the potential term. Am I (or maybe you) doing something wrong??

Last edited: Jul 17, 2006
4. Jul 17, 2006

### samalkhaiat

5. Jul 17, 2006

### wavemaster

Ups. I missed that bit. So my A became

$$A \sqrt{\frac{\pi}{-im/\hbar \epsilon}} = 1$$. $$V(x') \frac{\partial^2 \varphi}{\partial x^2}...$$ term is dropped since it has an second order $$\epsilon$$. OK, by doing so, I had the potential term nicely, but I still have a weird extra coefficent in the same place. This time, it's this:

$$\sqrt{\frac{im}{8 \hbar \epsilon}}$$.

Can you spot the error this time?..

6. Jul 17, 2006

### eljose

Umm.it seems too easy if all physics were so easy we shouldn't have problem finding Quantum Gravity:

As far as i know the Integral equation satisfied by the wave is:

$$\psi(x',t')=\int_{-\infty}^{\infty}dxdtG(x,x',t,t')\psi(x,t)$$

with $$G(x,x',t,t')=A\int D[x]e^{iS(x)]/\hbar$$

$$S(x)=\int_{a}^{b}dt [(1/2)m(\dot x)^{2}-V(x)]$$

However the Functional integral above can't be done except for "Semiclassical Approach" or when V(x)=0 or V(x)=x^{2}

7. Jul 17, 2006

### samalkhaiat

the error was in
$$\int dy e^{-a y^{2}} = (\pi/2)^{1/2}(1/2a)$$
it should be
$$\int dy e^{-a y^{2}} = (\pi/a)^{1/2}(1/2a)$$
I think my miss-keying is to blame

regards

sam

8. Jul 18, 2006

### wavemaster

Horray! Huge thanks, finally Schrödinger equation did come out!
One last question, how could you (or Feynman) see which term you had to expand? I rather tried expanding $$e^{iLdt/\hbar}$$ directly, without expanding the potential term around a fixed point, and without leaving a $$e^{imv^2/2\hbar}$$. It was all a mess. I really wonder how you did see?

eljose, is that equation for QG?

9. Jul 18, 2006