Physics Library
 An open source physics library
Encyclopedia | Forums | Docs | Random | Template Test |  
Login
create new user
Username:
Password:
forget your password?
Main Menu
Sections

Talkback

Downloads

Information
time independent Schrödinger equation in spherical coordinates (Definition)

When writing the time independent Schrödinger equation in spherical coordinates, we need to plug the Laplacian in Spherical Coordinates into the time independent Schrödinger equation. The Laplacian was found to be

$\displaystyle \nabla _{sph}^{2} = \frac{1}{r^2} \frac{\partial}{\partial r}\lef... ...\theta}\right) + \frac{1}{r^2 sin^2 \theta} \frac{\partial^2}{\partial \phi^2} $

Using the three dimensional Schrödinger equation we then have

$\displaystyle \hat{H} \psi(r,\theta, \phi) = -\frac{\hbar^2}{2m} \left [ \frac{... ...^2} \right ] + V(r,\theta, \phi) \psi(r,\theta, \phi) = E\psi(r, \theta, \phi) $
We can gain insight into this somewhat ugly equation by rewriting it using the square of the angular momentum operator in spherical polar coordinates:

$\displaystyle \hat{L}^2 = {1 \over \sin\theta} {\partial\over\partial\theta}\le... ...artial\theta}\right) +\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} $
This leads to

$\displaystyle \left(-\frac{\hbar^2}{2m}\left({1 \over r^2} {\partial\over\parti... ...}{r^2}+V(r, \theta, \phi)\right)\psi(r, \theta, \phi) = E\psi(r, \theta, \phi) $

Spherically symmetric separable solution

This equation is only exactly solvable if $V=V(r)$, a function without angular dependence. We then write $\psi(r, \theta, \phi) = R(r)Y(\theta, \phi)$ leading to the following equation:
$\displaystyle \left(-\frac{\hbar^2}{2m}\left({1 \over r^2} {\partial\over\parti... ...t{L}^2}{r^2}+V(r, \theta, \phi)\right)\psi(r, \theta, \phi)R(r) Y(\theta, \phi)$ $\displaystyle = E R(r) Y(\theta, \phi)$    
$\displaystyle -\frac{\hbar^2}{2m} \left( Y(\theta, \phi) \left( \frac{1}{r^2}{\... ...frac{R(r)}{2m}\frac{\hat{L}^2 \,Y(\theta, \phi)}{r^2} + V(r)R(r)Y(\theta, \phi)$ $\displaystyle = E R(r)(Y(\theta, \phi)$    
To solve this equation we need to remove the angular dependence. This is simply done by substituting the eigenfunctions of $\hat{L}^2$ into the equation. These are known to be the spherical harmonics, $Y^m_l(\theta, \phi)$. We also know that these have eigenvalues $\hbar^2 l(l+1)$, i.e.

$\displaystyle \hat{L}^2 \,Y_l^m(\theta, \phi) = \hbar^2 l(l+1) Y^m_l(\theta, \phi) $
We now substitute this result into the Schrödinger equation and divide through by a common factor of $Y_l^m(\theta, \phi)$

$\displaystyle \left( -\frac{\hbar^2}{2m} \left(\frac{1}{r^2}{\partial \over \pa... ...rtial r}\right) +\frac{\hbar^2l(l+1)}{r^2}\right) + V(r) \right) R(r) = E R(r) $
This is the radial equation.
References

[1] Griffiths, D. "Introduction to Quantum Mechanics" Prentice Hall, New Jersey, 1995.



"time independent Schrödinger equation in spherical coordinates" is owned by bloftin. [ full author list (2) ]

View style:

See Also: time independent Schrödinger equation, radial equation


Cross-references: radial equation, function, operator, angular momentum, square, Laplacian, time independent Schrödinger equation, Laplacian in Spherical Coordinates
There is 1 reference to this object.

This is version 6 of time independent Schrödinger equation in spherical coordinates, born on 2006-01-15, modified 2008-09-13.
Object id is 108, canonical name is TimeIndependentSchrodingerEquationInSphericalCoordinates.
Accessed 3544 times total.

Classification:
Physics Classification: 03.65.-w (Quantum mechanics )
Pending Errata and Addenda
None.
[ View all 3 ]
Discussion
Style: Expand: Order:

No messages.

Testing some escape charachters for html category with a generator has an injective cogenerator" now escape ” with "