An Easier Method for Solving Schroedinger Equation

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In general solving the eigenvalue problem of three-dimensional Schroedinger equation is difficult unless the shape of potential is particularly simple. I tried to develop a generic method to solve that problem:

 

 

 

In general solving Schroedinger equation is much harder than modeling quantum phenomenon with that. For example, the movement of electron around proton is quantised with the following Schroedinger equation.

 

(ih/(2p))¶f/¶t = -{h2/((2p)22m)}Df-e2f/r

 

To solve it as eigenvalue problem it is necessary to express the solution with the spherical function and Laguerre function. Apart from Laguerre function, calculating practically spherical function is laboured task. Are there any easier methods for solving the Schroedinger equation especially on the eigenvalue problem?

 

One simple way is to specify the coordinates of electron on y- and z-axis. Then the Schroedinger equation becomes one-dimensional as,

 

(ih/(2p))¶f(x,y0,z0)/¶t = -{h2/((2p)22m)}¶2f(x,y0,z0)/¶x2-e2f/r(x,y0,z0)

 

As the eigenvalue problem, it is rewritten as,

 

-{h2/((2p)22m)}¶2f(x,y0,z0)/¶x2-e2f/r(x,y0,z0) = W(y0,z0)f

 

This is just ordinary differential equation. So we can solve it relatively easily (approximately in some cases, though.) However since the potential is the function of not only x but also y and z generally, the "eigenvalue" W depends on the coordinates of electron on y- and z-axis, i.e,

 

W(y0,z0) ¹ W(y0',z0') where (y0,z0) ¹ (y0',z0')

 

Is it impossible to obtain the three dimensional solution which has specific eigenvalue from one dimensional solution mentioned above?

 

Let's go back to classical dynamics. The movement of point mass x = (x1(t),x2(t),x3(t)) under the potential U is described as,

 

(1/2)m|dx/dt|2+U(x) = E

 

In this case the total energy E doesn't change through time as you know. Now we describe the components x1, x2 and x3 of x with virtually independent three parameters as,

 

x = (x1(t1),x2(t2),x3(t3))

 

Then the velocity vector v of x is,

 

v = (dx1/dt1,dx2/dt2,dx3/dt3)

 

It should be t1 = t2 = t3 = t in order to coincide v with ordinary velocity. The partial derivative of kinetic energy for tj is,

 

(¶/¶tj)(1/2)m|dx/dt|2 = (¶/¶tj)(1/2)m(dxj/dtj)2 = m(d2xj/dtj2)(dxj/dtj) (j = 1,2,3)

 

while the partial derivative of potential U for tj is,

 

¶U/¶tj = Sk(¶U/¶xk)(dxk/dtj) = (¶U/¶xj)(dxj/dtj)

 

On the other hand, Newton's equation is expressed with the parameter tj as,

 

m(d2xj/dtj2) = -¶U/¶xj

 

Hence,

 

(¶/¶tj){(1/2)m(dxj/dtj)2+U}

= m(d2xj/dtj2)(dxj/dtj)+(¶U/¶xj)(dxj/dtj)

= {m(d2xj/dtj2)+¶U/¶xj}(dxj/dtj)

= 0

 

Namely when let Ej the j-component of kinetic energy of mass point, i.e.

 

Ej = (1/2)m(dxj/dtj)2 (caution: not take the sum regarding the index j)

 

then the above equation indicates that ((Ej + potential energy)) doesn't change through the time tj. Although the time tj is imaginary parameter, the preservation of that sum is not meaningless physically. For example, say j = 1, it can be rewritten with ordinary time t as,

 

(d/dt){(1/2)m(dx/dt)2+U(x(t),y(t0),z(t0))} = 0

 

where the coordinate y and z of mass point are fixed as y(t0) and z(t0).

 

The above preservation of the j-component kinetic energy plus potential energy is quantised as,

 

(ih/(2p))¶fj/¶tj = -{h2/((2p)22m)}¶2fj/¶xj2+Ufj

 

This is nothing other than ordinary one-dimensional Schroedinger equation. Therefore the eigenvalue problem on three-dimensional Schroedinger equation is decomposed into one-dimensional eigenvalue problem as,

 

-{h2/((2p)22m)}¶2fj/¶xj2+Ufj = wjfj

 

where wj corresponds to the j-component kinetic energy plus potential energy in the case of classical dynamics mentioned above. The solution fj is regarded as the function of coordinate xj and eigenvalue wj as,

 

fj = fj(wj,xj)

 

However as I mentioned previously, "eigenvalue" wj of one-dimensional Schroedinger equation depends on the other coordinates. Therefore we must take the convolution of wj for getting eigenfunction f(W,x,y,z) of Schroedinger equation on the three-dimensional space as,

 

f(W,x,y,z) = òdw2dw3f1(W-w2-w3,x)f2(w2,y)f3(w3,z)

 

where W is the eigenvalue of three-dimensional Schroedinger equation. Actually W is obtained from solving one-dimensional eigenvalue problem on Schroedinger equation for each component.

[iNow]

What you copied from another site didn't paste well. Can you correct that?

 

 

Also (thanks to another thread post elsewhere by Lockheed), you might enjoy the below:

 

http://users.wpi.edu/~ck/feynman_Nobel_Lecture.pdf