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| classes:2009:fall:phys4101.001:lec_notes_1102 [2009/11/03 02:41] – 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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| **Class input on main points of the beginning of Chapter 4:** | **Class input on main points of the beginning of Chapter 4:** |
| * <math>\nabla\^2</math> spherical coordinates | * <math>\nabla^2</math> spherical coordinates |
| * <math>L^2, L_z</math>, and why they have discrete values | * <math>L^2, L_z</math>, and why they have discrete values |
| * The Hydrogen atom model | * The Hydrogen atom model |
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| **Using 3-D Coordinates:** | **Using 3-D Coordinates:** |
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| From the one-dimensional Schrodinger Equation: | From the one-dimensional Schrodinger Equation: |
| <math> [-\frac{\hbar^2}{2m}\frac{ \partial^2}{ \partial x^2} + V(x)]\psi=E \psi</math> | <math> [-\frac{\hbar^2}{2m}\frac{ \partial^2}{ \partial x^2} + V(x)]\psi=E \psi</math> |
| 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>. 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** |
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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. 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> |
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| | Plugging this into the Schrodinger equation we get: |
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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> |
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| | Suppose that the wavefunction is a separable solution. It has the form: |
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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> | <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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| **To Be Finished:** | <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> |
| *Discussion on The Angular Equation | |
| *Legendre Polynomials | |
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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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