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| classes:2009:fall:phys4101.001:lec_notes_1102 [2009/11/04 00:39] – jbarthel | classes:2009:fall:phys4101.001:lec_notes_1102 [2009/11/07 21:45] (current) – yk |
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| ===== Nov 02 (Mon) ===== | ===== Nov 02 (Mon) Main Topics in Chap 4, Separation variables for Spherical coordinate ===== |
| ** Responsible party: Captain America, David Hilbert's hat ** | ** Responsible party: Captain America, David Hilbert's hat ** |
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| The kinetic energy term, <math>-\frac{\hbar^2}{2m}\frac{\partial^2}{ \partial x^2}</math>, must model the 3-Dimensional kinetic energy of the system, and therefore turns into: | The kinetic energy term, <math>-\frac{\hbar^2}{2m}\frac{\partial^2}{ \partial x^2}</math>, must model the 3-Dimensional kinetic energy of the system, and therefore turns into: |
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| <math> [-\frac{\hbar^2}{2m}\nabla^2+ V(x)]\psi=E \psi</math> | <math> [-\frac{\hbar^2}{2m}\nabla^2+ V(x,y,z)]\psi=E \psi</math> |
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| Where <math>\nabla^2</math> is equal to <math> \frac{ \partial^2}{ \partial x^2} + \frac{ \partial^2}{ \partial y^2} + \frac{ \partial^2}{ \partial z^2}</math> | Where <math>\nabla^2</math> is equal to <math> \frac{ \partial^2}{ \partial x^2} + \frac{ \partial^2}{ \partial y^2} + \frac{ \partial^2}{ \partial z^2}</math>. You can see that if you restrict this 3 dimensional case to one dimension, our original one dimensional Schrodinger equation comes out. |
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| **Separation of Variables** | **Separation of Variables** |
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| Using spherical coordinates will be useful for future problems that we will be solving, so it is necessary to transform the Schrodinger equation into spherical coordinates. The Laplacian will take the form of: | Using spherical coordinates will be useful for future problems that we will be solving, so it is necessary to transform the Schrodinger equation into spherical coordinates. Using the relations <math> x = r sin \theta cos \phi </math>, <math> y = r sin\theta sin\phi</math>, and <math>z = r cos\theta </math>, you can derive the Laplacian in spherical coordinates. |
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| | The Laplacian will take the form of: |
| <math>\nabla^2=\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r}) + \frac{1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} (sin(\theta) \frac{\partial}{\partial \theta}) + \frac{1}{r^2 sin^2(\theta)} (\frac{\partial^2}{\partial \phi^2})</math> | <math>\nabla^2=\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial}{\partial r}) + \frac{1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} (sin(\theta) \frac{\partial}{\partial \theta}) + \frac{1}{r^2 sin^2(\theta)} (\frac{\partial^2}{\partial \phi^2})</math> |
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| <math>-\frac{\hbar^2}{2 m} [\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial \psi}{\partial r}) + \frac{1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} (sin(\theta) \frac{\partial \psi}{\partial \theta}) + \frac{1}{r^2 sin^2(\theta)} (\frac{\partial^2 \psi}{\partial \phi^2})] + V \psi = E \psi</math> | <math>-\frac{\hbar^2}{2 m} [\frac{1}{r^2} \frac{\partial}{\partial r}(r^2 \frac{\partial \psi}{\partial r}) + \frac{1}{r^2 sin(\theta)} \frac{\partial}{\partial \theta} (sin(\theta) \frac{\partial \psi}{\partial \theta}) + \frac{1}{r^2 sin^2(\theta)} (\frac{\partial^2 \psi}{\partial \phi^2})] + V \psi = E \psi</math> |
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| **To Be Finished:** | |
| *Discussion on The Angular Equation | |
| *Legendre Polynomials | |
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| | Suppose that the wavefunction is a separable solution. It has the form: |
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| | <math>\psi(r,\theta,\phi) = R(r)Y(\theta,\phi). </math> |
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| | What we want to do is plug this form into the Schrodinger equation, and use the fact that |
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| | <math> \frac{\partial \psi}{\partial r} = \frac{\partial (RY)}{\partial r} = Y \frac{\partial R}{\partial r} </math>, <math> \frac{\partial \psi}{\partial \theta} = \frac{\partial (RY)}{\partial \theta} = R \frac{\partial Y}{\partial \theta} </math>, <math> \frac{\partial \psi}{\partial \phi} = \frac{\partial (RY)}{\partial \phi} = R \frac{\partial Y}{\partial \phi} </math> |
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| | To reduce the Schrodinger equation into one side dependent on r,<math> \frac{\partial R}{\partial r} </math> and the other side dependent on <math> \phi </math>, <math> \theta </math>, <math> \frac{\partial Y}{\partial \theta} </math>, and <math> \frac{\partial Y}{\partial \phi} </math>. Because each side of the equation is dependent on R or Y alone, you can set both sides equal to a constant and turn the Schrodinger equation into a set of solvable differential equations. |
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