# Dirac's K

## Main Question or Discussion Point

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

...So I simply put them equal, taking the simplest example where the Lagrangian is $$\frac{m}{2}\dot{x}^2 - V(x)$$ but soon found I had to put a constant of proportionality A in, suitably adjusted. When I substituted $$Ae^{i \epsilon L / \hbar}$$ for K to get

$$\varphi(x',t+\epsilon) = \int A e^{\frac{i}{\hbar} L(\frac{x'-x}{\epsilon},x) \epsilon} \varphi(x,t) dx$$

and just calculated things out by Taylor series expansion, out came the Schrödinger equation.
I'm trying to get the same result, but I'm stuck. Has anyone done this before?

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samalkhaiat
wavemaster said:
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?
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

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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??

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samalkhaiat
wavemaster said:

$$\varphi(x',t+\epsilon) = A e^{-\frac{i}{\hbar} \epsilon V(x')} (\varphi(x',t)\sqrt{\frac{\pi}{-im/\hbar \epsilon}} + \frac{1}{2}\frac{\partial^2 \varphi(x',t)}{\partial x^2} \sqrt{\frac{\pi}{2}} \frac{1}{2(-im/\hbar \epsilon)})$$
Write
$$e^{-\frac{i}{\hbar} \epsilon V(x')}= 1 - \frac{i}{\hbar} \epsilon V(x')$$
so that you dot miss the potential term ,
then make equal to one the coefficient of $$\varphi(x',t)$$ NOT $$\varphi(x',t) V(x')$$

now try it regards

sam

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?..

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}

samalkhaiat
wavemaster said:
Can you spot the error this time?..
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

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?

samalkhaiat
wavemaster said:
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?
Let us write

$$L(\frac{x-x'}{\epsilon} , x) = L(\frac{y}{\epsilon} , y+x') = T(\frac{y}{\epsilon}) - V(y+x')$$

since we know the functional form of T;

$$T= \frac{m}{2}(\frac{y}{\epsilon})^{2}$$

and this can be integrated out, then we only need to expand $$V(y+x')$$ .

We also do not expand the e first;

$$e^{iL\epsilon} = 1+ iL \epsilon + ...$$

because this would lead to divergent integrals.
I hope this helps.

sam