===== Nov 04 (Wed) Laplacian in Spherical Coordinate, Legendre Polynomals ===== ** Responsible party: Captain America, Cthulhu Food ** **To go back to the lecture note list, click [[lec_notes]]**\\ **previous lecture note: [[lec_notes_1102]]**\\ **next lecture note: [[lec_notes_1106]]**\\ **Main class wiki page: [[home]]** 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). * Other classmates can step in and clarify the points, and expand them. * How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture. * wonderful tricks which were used in the lecture.\\ \\ ** Mathematical method to get Laplacian** The Laplacian in Cartesian coordinates is: \bigtriangledown^2=\frac{\partial^2\Psi}{\partial x^2}+\frac{\partial^2\Psi}{\partial y^2}+\frac{\partial^2\Psi}{\partial z^2} We start the derivation of the Laplacian in spherical coordinates with the kinetic energy operator: \ = <\Psi|\frac{p^2}{2m}|\Psi> = \frac{1}{2m}*<\Psi|{p^2}|\Psi> For the moment we can disregard \ \frac{1}{2m} because it is a multiplicative constant. In 1-D: \ <\Psi|p^2|\Psi> = = \int (\frac{\hbar}{i} \frac{\partial\Psi}{\partial\Psi})^*(\frac{\hbar}{i} \frac{\partial\Psi}{\partial\Psi})dx In 3-D: is proportional to \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV Where: \bigtriangledown=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}+\frac{\partial\Psi}{\partial z} (yk says) Rather, \bigtriangledown\Psi=\begin{bmatrix} \frac{\partial\Psi}{\partial x}\\ \frac{\partial\Psi}{\partial y}\\ \frac{\partial\Psi}{\partial z}\end{bmatrix} and dV = r^2 sin\theta dr d\theta d\phi The above integral can also be written if we introduce a few tensors: The first one, g_{ij}=\begin{bmatrix} 1 & 0 & 0\\ 0 & r^2 & 0\\ 0 & 0 & r^2\sin\theta\end{bmatrix} is a tensor, when it is bracketed by row and column vectors \begin{bmatrix}dr & d\theta & d\phi\end{bmatrix} and \begin{bmatrix}dr \\ d\theta \\ d\phi\end{bmatrix} it is form proper 3D distance, ds^2 = dr^2 + r^2 d\theta^2 + (r\sin\theta)^2 d\phi^2. The 2nd tensor is similar to the first, but it will allow \begin{bmatrix}\frac{d}{dr} & \frac{d}{d\theta} & \frac{d}{d\phi}\end{bmatrix} and \begin{bmatrix}\frac{d}{dr} \\ \frac{d}{d\theta} \\ \frac{d}{d\phi}\end{bmatrix} to form a proper 3D dot product of two //gradients//, even though some of the differentiation do not have proper units (like d\theta). This 2nd tensor takes the form of g^{ij}=\begin{bmatrix} 1 & 0 & 0\\ 0 & \frac{1}{r^2} & 0\\ 0 & 0 & \frac{1}{r^2\sin\theta}\end{bmatrix} . These two tensors are both expressed with g, 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, g is introduced, which is the determinant of the tensor g_{ij}. Using these notations, the expectation value for the kinetic energy in the spherical coordinates can be written as, \int (\bigtriangledown_i \Psi)^* g^{ij} (\bigtriangledown_j \Psi) g^{(\frac{1}{2})} dr d\theta d\phi where g^{(\frac{1}{2})} = r^2 sin(\theta) i,j = r,\theta,\phi . This expression can be equated to that for the normal Cartesian version of the same calculation, \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV = \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) g^{\frac{1}{2}} dr d\theta d\phi. //i.e.// \int (\bigtriangledown_i \Psi)^* g^{ij} (\bigtriangledown_j \Psi) g^{(\frac{1}{2})} dr d\theta d\phi = \int (\bigtriangledown \Psi)^*(\bigtriangledown \Psi) dV. 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.// \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. At this point, if we replace dV with g^{\frac{1}{2}} dr d\theta d\phi, we get: \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. 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 \Psi. From this, we can claim that \bigtriangledown_i g^{ij} g^{\frac{1}{2}} \bigtriangledown_j = \bigtriangledown^2 g^{\frac{1}{2}} or \bigtriangledown^2 = \frac{1}{g^{\frac{1}{2}}}\bigtriangledown_i g^{ij} g^{\frac{1}{2}} \bigtriangledown_j **Legendre Polynomials** \Theta(\theta):\alpha,\beta = m^2 where \alpha is separation of the r-dependent part and \beta is the separation of the \phi dependent part. Then the Differential equation form of \Theta(\theta)= P_l^m (cos\theta) replacing z=cos\theta is: (1-z^2)\frac{d^2U}{dz^2} - 2z \frac{dU}{dz} + \alpha U = 0 Then we can take U(z)= \sum_{n=0}^\infty a_n z^n We can then take \xi=\frac{1-z}{2} where -1 < z < 1 and therefore 0 < z < 1 Then U(\xi) is also a differential equation where U(\xi)=\sum_{n=0}^\infty a_n \xi^2 From this we get const. U'' + const. U' + const. U= const for all \xi ------------------------------------------- **To go back to the lecture note list, click [[lec_notes]]**\\ **previous lecture note: [[lec_notes_1102]]**\\ **next lecture note: [[lec_notes_1106]]**\\