Free particle in spherical polar coords

In summary, we are considering the time-independent Schrodinger equation in spherical polar coordinates for a free particle with azimuthal quantum number l=0. We solve the radial equation to find the unnormalized radial wavefunction u(r)=rR(r). Normalizing R(r) using the definition of the dirac delta function, we find the solution to be R(r)=(A/r)sin(kr) with A as a constant.
  • #1
cscott
782
1

Homework Statement



Consider the time-independent Schrodinger equation in spherical polar coordinates for a free particle, in the case where we have an azimuthal quantum number [itex]l=0[/itex].

(a) Solve the radial equation to find the (unnormalized) radial wavefunction [itex]R(r)[/itex].
(b) Normalize [itex]R(r)[/itex], using the definition of the dirac delta function [itex]\delta(k'-k)[/itex].

Homework Equations



Radial Equation:

[tex]u(r) = rR(r)[/tex]

[tex]-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right]u = Eu[/tex]

The Attempt at a Solution



For a free particle, [itex]V=0[/itex], and with [itex]l = 0[/itex] the radial equation reduces to,

[tex]\frac{d^2u}{dr^2}=-k^2u,~~k=\frac{\sqrt{2mE}}{\hbar}[/tex],

with solution,

[tex]u = A\sin(kr) + B\cos(kr)[/tex],

but [itex]u(r)=rR(r)[/itex], so [itex]B=0[/itex] for a normalizable wavefunction (considering r->0). Therefore,

[tex]R(r) = \frac{A}{r}\sin(kr)[/tex]

and to normalize,

[tex]\int_{0}^{\inf} r^2|R(r)|^2~dr = 1[/tex]

[tex]|A|^2 \int_{0}^{\inf} \sin^2(kr)~dr=1[/tex]

How do I use the dirac delta function?
 
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  • #2
Should I keep [itex]u[/itex] as,

[tex]u(r)=Ae^{ikr}[/tex]

[tex]R(r)=\frac{A}{r}e^{ikr}[/tex]
 
  • #3
Hi

Sorry, but I don't see why do you need to use the delta of k'-k. I would integrate the sine square but the problem is that you forgot the integration for the angular variables. Thus there is a "4 times pi" factor missed which should be at RHS of the last equation as denominator of 1/(4*pi) since the 4*pi is the result of the integration of

[tex]\int sin\theta d\theta \int d\varphi[/tex]

This comes from the fact that to normalize you have to integrate in a volume whose differential element is

[tex]dv= r^2 sin\theta dr d\theta d\varphi[/tex]

But I don't see anything else to be added.
 

1. What is a free particle in spherical polar coordinates?

A free particle in spherical polar coordinates is a physical system in which a particle is allowed to move without any external forces acting upon it, and its position is described using the spherical polar coordinate system.

2. What are the three main coordinates used in spherical polar coordinates?

The three main coordinates used in spherical polar coordinates are the radial distance (r), the azimuth angle (θ), and the polar angle (φ).

3. How is the position of a free particle in spherical polar coordinates determined?

The position of a free particle in spherical polar coordinates is determined by the values of its three coordinates (r, θ, φ) which represent the distance from the origin, the angle from the positive x-axis in the xy-plane, and the angle from the positive z-axis in the three-dimensional space, respectively.

4. How do you convert coordinates from Cartesian to spherical polar?

To convert coordinates from Cartesian to spherical polar, you can use the following equations:

r = √(x² + y² + z²)

θ = arctan(y/x)

φ = arccos(z/r)

5. What are the advantages of using spherical polar coordinates to describe the position of a free particle?

One advantage of using spherical polar coordinates is that it simplifies the mathematical description of certain physical systems, such as those with spherical symmetry. It also allows for a more intuitive understanding of the position of a particle in three-dimensional space compared to Cartesian coordinates. Additionally, it can be useful in solving certain types of differential equations in physics and engineering.

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