Struggling to Get the Same Result: Solving Schrödinger Equation

[wavemaster]

In http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html" , 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?

 

[samalkhaiat]

In http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html" , 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

 

[wavemaster]

Thanks for the reply!
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??

 

[samalkhaiat]

Thanks for the reply!

\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

 

[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?..

 

[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}

 

[samalkhaiat]

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

 

[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?

 

[samalkhaiat]

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