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| classes:2009:fall:phys4101.001:lec_notes_1104 [2009/11/05 14:37] – jbarthel | classes:2009:fall:phys4101.001:lec_notes_1104 [2009/11/07 21:44] (current) – yk |
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| ===== Nov 04 (Wed) ===== | ===== Nov 04 (Wed) Laplacian in Spherical Coordinate, Legendre Polynomals ===== |
| ** Responsible party: Captain America, Cthulhu Food ** | ** Responsible party: Captain America, Cthulhu Food ** |
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| ** Mathematical method to get Laplacian** | ** Mathematical method to get Laplacian** |
| *needs to be finished | |
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| | The Laplacian in Cartesian coordinates is: |
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| | <math>\bigtriangledown^2=\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2}</math> |
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| | We start the derivation of the Laplacian in spherical coordinates with the kinetic energy operator: |
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| | <math>\ <KE> = <\Psi|\frac{p^2}{2m}|\Psi> = \frac{1}{2m}*<\Psi|{p^2}|\Psi> </math> |
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| | For the moment we can disregard <math>\ \frac{1}{2m} </math> because it is a multiplicative constant. |
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| | In 1-D: |
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| | <math>\ <\Psi|p^2|\Psi> = <p\Psi|p\Psi> = \int (\frac{\hbar}{i} \frac{\partial\Psi}{\partial\Psi})^*(\frac{\hbar}{i} \frac{\partial\Psi}{\partial\Psi})dx </math> |
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| | In 3-D: |
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| | <math> <KE> </math> is proportional to <math> \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV </math> |
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| | Where: |
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| | <math> \bigtriangledown=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}+\frac{\partial\Psi}{\partial z}</math> (yk says) Rather, <math> \bigtriangledown\Psi=\begin{bmatrix} |
| | \frac{\partial\Psi}{\partial x}\\ |
| | \frac{\partial\Psi}{\partial y}\\ |
| | \frac{\partial\Psi}{\partial z}\end{bmatrix}</math> |
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| | and |
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| | <math> dV = r^2 sin\theta dr d\theta d\phi </math> |
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| | The above integral can also be written if we introduce a few tensors: |
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| | The first one, <math>g_{ij}=\begin{bmatrix} |
| | 1 & 0 & 0\\ |
| | 0 & r^2 & 0\\ |
| | 0 & 0 & r^2\sin\theta\end{bmatrix} </math> is a tensor, when it is bracketed by row and column vectors <math>\begin{bmatrix}dr & d\theta & d\phi\end{bmatrix}</math> and <math>\begin{bmatrix}dr \\ d\theta \\ d\phi\end{bmatrix}</math> it is form proper 3D distance, <math> ds^2 = dr^2 + r^2 d\theta^2 + (r\sin\theta)^2 d\phi^2</math>. |
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| | The 2nd tensor is similar to the first, but it will allow <math>\begin{bmatrix}\frac{d}{dr} & \frac{d}{d\theta} & \frac{d}{d\phi}\end{bmatrix}</math> and <math>\begin{bmatrix}\frac{d}{dr} \\ \frac{d}{d\theta} \\ \frac{d}{d\phi}\end{bmatrix}</math> to form a proper 3D dot product of two //gradients//, even though some of the differentiation do not have proper units (like <math>d\theta</math>). This 2nd tensor takes the form of <math>g^{ij}=\begin{bmatrix} |
| | 1 & 0 & 0\\ |
| | 0 & \frac{1}{r^2} & 0\\ |
| | 0 & 0 & \frac{1}{r^2\sin\theta}\end{bmatrix} </math>. |
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| | These two tensors are both expressed with <math>g</math>, 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. |
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| | In addition, a scaler quantity, <math>g</math> is introduced, which is the determinant of the tensor <math>g_{ij}</math>. Using these notations, the expectation value for the kinetic energy in the spherical coordinates can be written as, |
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| | <math> \int (\bigtriangledown_i \Psi)^* g^{ij} (\bigtriangledown_j \Psi) g^{(\frac{1}{2})} dr d\theta d\phi </math> |
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| | where |
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| | <math> g^{(\frac{1}{2})} = r^2 sin(\theta) </math> |
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| | <math> i,j = r,\theta,\phi </math>. |
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| | This expression can be equated to that for the normal Cartesian version of the same calculation, <math> \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV = \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) g^{\frac{1}{2}} dr d\theta d\phi</math>. |
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| | //i.e.// |
| | <math> \int (\bigtriangledown_i \Psi)^* g^{ij} (\bigtriangledown_j \Psi) g^{(\frac{1}{2})} dr d\theta d\phi = \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV</math>. |
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| | 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.// |
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| | <math> \int (\Psi)^* \bigtriangledown_i [g^{ij} g^{(\frac{1}{2})} (\bigtriangledown_j \Psi) dr d\theta d\phi = \int (\Psi)^*\bigtriangledown(\bigtriangledown \Psi) dV = \int (\Psi)^*(\bigtriangledown^2 \Psi) dV</math>. |
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| | At this point, if we replace <math>dV</math> with <math>g^{\frac{1}{2}} dr d\theta d\phi</math>, we get: |
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| | <math> \int (\Psi)^* \bigtriangledown_i [g^{ij} g^{(\frac{1}{2})} (\bigtriangledown_j \Psi) dr d\theta d\phi = \int (\Psi)^*(\bigtriangledown^2 \Psi) g^{\frac{1}{2}} dr d\theta d\phi</math>. |
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| | 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 <math>\Psi</math>. From this, we can claim that |
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| | <math> \bigtriangledown_i g^{ij} g^{\frac{1}{2}} \bigtriangledown_j = \bigtriangledown^2 g^{\frac{1}{2}} </math> or |
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| | <math> \bigtriangledown^2 = \frac{1}{g^{\frac{1}{2}}}\bigtriangledown_i g^{ij} g^{\frac{1}{2}} \bigtriangledown_j </math> |
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| **Legendre Polynomials** | **Legendre Polynomials** |
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| We can then take <math>\xi=\frac{1-z}{2}</math> where <math>-1 <= z <= </math> and therefore <math>0 < z < 1</math> | We can then take <math>\xi=\frac{1-z}{2}</math> where <math>-1 < z < 1 </math> and therefore <math>0 < z < 1</math> |
| Then <math>U(\xi)</math> is also a differential equation where | Then <math>U(\xi)</math> is also a differential equation where |
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