===== Nov 04 (Wed) Laplacian in Spherical Coordinate, Legendre Polynomals =====
** Responsible party: Captain America, Cthulhu Food **
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Please try to include the following
* main points understood, and expand them - what is your understanding of what the points were.
* expand these points by including many of the details the class discussed.
* main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
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** Mathematical method to get Laplacian**
The Laplacian in Cartesian coordinates is:
We start the derivation of the Laplacian in spherical coordinates with the kinetic energy operator:
For the moment we can disregard because it is a multiplicative constant.
In 1-D:
In 3-D:
is proportional to
Where:
(yk says) Rather,
and
The above integral can also be written if we introduce a few tensors:
The first one, is a tensor, when it is bracketed by row and column vectors and it is form proper 3D distance, .
The 2nd tensor is similar to the first, but it will allow and to form a proper 3D dot product of two //gradients//, even though some of the differentiation do not have proper units (like ). This 2nd tensor takes the form of .
These two tensors are both expressed with , but one has indeces as subscripts, and the other, superscripts, in the convention we are using. In fact, these two matrices are inverse of each other.
In addition, a scaler quantity, is introduced, which is the determinant of the tensor . Using these notations, the expectation value for the kinetic energy in the spherical coordinates can be written as,
where
.
This expression can be equated to that for the normal Cartesian version of the same calculation, .
//i.e.//
.
Both sides of this equation can be modified (using integral by parts) to move the differentiation operator on the left to be operating on the "right" function. //i.e.//
.
At this point, if we replace with , we get:
.
Since this equation between two integrals hold for any wave functions, the integrand must be equal, too. Further, we can regard the equation for the operator in front of . From this, we can claim that
or
**Legendre Polynomials**
where is separation of the r-dependent part and is the separation of the dependent part.
Then the Differential equation form of replacing
is:
Then we can take
We can then take where and therefore
Then is also a differential equation where
From this we get
for all
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