
#1
Nov205, 06:05 AM

P: 219

Lets imagine the wavefunction psy(x,y) and with this the probability distribution psi(x,y)^2 of the position of an electron is given. psy(x,y) is a two dimensional wave function.
So how can I find the electric field strength produced by this electon at position x=3, y=2? 



#2
Nov205, 06:07 AM

Math
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Thanks
PF Gold
P: 38,881

I'm confused. Assuming the wave function for the electron, how can you then say that the electron is at (3,2)?




#3
Nov205, 06:18 AM

P: 219

I don't say it is there. I want to know, if the wave function of an electron is given by psi(x,y) how can we then find the distribution of the electric field E(x,y) produce by this electron (described with psi(x,y)) ? This is my question.




#4
Nov205, 09:18 AM

P: 85

Electric field produced by electron 



#5
Nov205, 09:38 AM

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PF Gold
P: 6,238

The simplest one is the "effective potential", which is used in quantum chemistry: just calculate the psi^2, and put that equal to a charge density. Then calculate the Efield from this charge density. This method is not entirely correct however. A "better" way would be, if we can use an electrostatic approximation, to consider the "E field" to be in a superposition of different values, each corresponding to a different position of the electron, and with an amplitude given by the amplitude of the electron to be in that position (namely, psi(x,y)). In QFT, what you ask is a terribly tricky question ! I'm not sure that there is an entirely correct way of handling it. 



#6
Nov205, 09:49 AM

P: 219

Ah, thanks. I only need the easy one.
What you mean is this differential equation? dE(x,y)/dx+dE(x,y)/dy=psi(x,y)^2/e0 where e0 is the electric permittivity Seems to be a good approximation for me, better an easy one. Can I change this formula in that way, that I can get E(x,y) immediately? 



#7
Nov205, 02:47 PM

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PF Gold
P: 6,238

You better calculate the potential from the density: [tex] V(p) = 1/\epsilon_0 \int d^3r \psi^2/rp [/tex] (check this, typing from memory...) cheers, Patrick. 



#8
Nov205, 04:03 PM

P: 3

nov.02.2005
One must also consider whether the electron is a solid sphericle particle or hollow sphericle body to describe the field strength? dr_syed_ameen 



#9
Nov305, 04:09 AM

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P: 11,863

Nope, in this case, when the electron's described at quantum level, it's assumed a pointlike particle.
Poisson's equation for the electrostatic potential reads [tex] \nabla^{2}V\left(\vec{r}\right) =\frac{\rho\left(\vec{r}\right)}{\epsilon_{0}} [/tex] with the charge density given by Vanesch. Daniel. 



#10
Nov305, 09:42 AM

P: 219

Ok. lets assume I have the following wave function (not normalised):
psi(x,y)=exp[kx+ky] Sorry, I can't calculate this. What means "p" in Vanesch's equation? And why do I have to integrate this. I mean p(r) is the charge densitity and the charge density of an electron's wave function is just psi^2. And how can I get the electric field strength produced by the electron out of the electric potential, which I guess isn't static??? Could I also do this: Calculate psi^2 which is equal to the charge density. And then if the charge density at the point x=2 and y=3 (example) would be 0.33 I can immedialtely calculate the electic field strength on this point. 



#11
Nov405, 03:13 AM

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P: 11,863

That's actually "rho". Not that "immediate". You've been already told what to do. Daniel. 



#12
Nov405, 04:44 AM

P: 555

The solution I liked the most was that in which you consider the wave function as an instrument for you to know the state of quantum coherent superpositions of different field values at a given point.
Nothing more natural than this procedure. If the charge is in a quantum superposition of different spatial locations, then, its field must also be in a quantum superposition of different field values. state of field at (x0,y0) = int int PSI(x,y) . {field generated by the electron at (x,y) in the point (x0,y0)} dx dy where {field generated by the electron at (x,y) in the point (x0,y0)} is the well known Coulomb's expression: 1/(4 PI e0) Q_electron/ (distance between (x,y) and (x0,y0)) ^2. Best Regards DaTario 



#13
Nov405, 01:19 PM

P: 219

Ok dexter and Vanesch and all together. I will try to calculate it by myself.
The last important question: How can I find the direction of the A(vector) (picture) if I have found 'Ar'. (this is of course the nabla operator in spherical coordinates)? 



#14
Nov505, 12:49 AM

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PF Gold
P: 6,238

The thing I wrote down was in fact very simple: you have a (static) charge density in space, rho(X), with rho(X) here given by the psi^2, but that doesn't matter. When you have, in electrostatics, a given charge density rho(X) in space, you can calculate the potential from it using the integral I gave, which is nothing else but a solution to the differential equation Dexter wrote down. Of course, for most distributions, you will not be able to write down the solution in a closed analytic form. cheers, Patrick. 



#15
Nov505, 11:17 AM

P: 85





#16
Nov605, 06:57 AM

P: 219

I have got an absurd result. I calculated the Efield of a particle with charge e (electron) with a wave function proportional to this exp[r]. I solved the differential equation dexter gave me and then calculated out of this the electric field.
The result is: the electric field strength is everywhere E(r)>0. This cannot be because a particle of negative charge should eleiminate the field of a particle with positive charge. I tested if my solution is correct and of course it is (put it in dexters formula and got the right result). Can this be true? 



#17
Nov605, 07:37 AM

P: 1,667

Cheers, careful 



#18
Nov605, 09:49 AM

P: 219

Thanks you all, I could solve all what I wanted.
Thanks dexter for the Poisson equation and thanks Vanesch for this integral thing. hehe Careful. I don't understand anything what you say (you must be on an higher level in physics) 


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