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| classes:2009:fall:phys4101.001:lec_notes_1014 [2009/10/14 23:14] – czhang | classes:2009:fall:phys4101.001:lec_notes_1014 [2009/10/15 23:37] (current) – fixed k and kappa per Yuichi's request prestegard |
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| If we divide Equation 2 by Equation 1, we get: | If we divide Equation 2 by Equation 1, we get: |
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| <math>k\cot{ka} = -\kappa \Rightarrow \cot{ka} = -\frac{\kappa}{k}</math> | <math>k\cot{ka} = -\kappa \Rightarrow -\cot{ka} = \frac{\kappa}{k}</math> |
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| To avoid the confusion of k and <math>\kappa</math>, | To avoid the confusion of k and <math>\kappa</math>, |
| we will introduce new different variables Z and <math>Z_0</math>. | we will introduce 2 new variables Z and <math>Z_0</math>. |
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| Since <math>k^2=\frac{-2mE}{h^2}</math> and <math>\kappa^2=\frac{-2m(E+V_0)}{h^2}</math>, | |
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| we define<math>Z_0^2=\frac{-2ma^2V_0}{h^2}</math> and <math>Z^2=k^2a^2=\frac{-2ma^2(E+V_0)}{h^2}</math> | Since <math>\kappa^2=\frac{-2mE}{\hbar^2}</math> and <math>k^2=\frac{2m(E+V_0)}{\hbar^2}</math>, |
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| | we define<math>Z_0^2=\frac{-2ma^2V_0}{\hbar^2}</math> and <math>Z^2=k^2a^2=\frac{-2ma^2(E+V_0)}{\hbar^2}</math> |
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| | thus <math>\kappa^2a^2=Z^2-Z_0^2</math> |
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| | thus <math>\frac{\kappa}{k}=\sqrt{\frac{\kappa^2a^2}{k^2a^2}}=\sqrt{\frac{Z^2-Z_0^2}{Z^2}}=\sqrt{(\frac{Z_0}{Z})^2-1}\</math> |
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| | <math>\Rightarrow -\cot{Z}=\frac{\kappa}{k}= \sqrt{(\frac{Z_0}{Z})^2-1}</math> |
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| | thus we are looking for the solution of Z for the 2 transcendental equation <math>\Rightarrow -\cot{Z}=\frac{\kappa}{k}= \sqrt{(\frac{Z_0}{Z})^2-1}</math> |
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| | Since it is really hard to solve it analytically, we can solve it graphically. |
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| | {{:classes:2009:fall:phys4101.001:qm_lec_pic.jpg|}}\\ **the cotangent part of the graph does not look quite right! //Yuichi//** |
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| | as we can see from the graph, the periodic functions are CotZ, and the other one is <math>\sqrt{(\frac{Z_0}{Z})^2-1}</math> |
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| | The physical meaning of the solution, which is the intersection point, represents one bound state. In other words, the total number of intersection points tell us the number of bound state of the system. |
| | Say when <math>Z_0=1</math>, it yields one intersection, which means we only have one bound state. |
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| | The other line is when <math>Z_0=10</math>, which has more intersections denoting more bound states. |
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| | Remember that <math>Z_0^2=\frac{2ma^2V_0}{\hbar^2}</math>, the magnitude of Z0 is determined by the production of a^2 and V0, a is the width of potential well, and V0 is the depth of the potential well. If we keep a constant , raise the potential to infinity, Z0 goes to infinity, then we have infinite square well, which corresponds with infinite intersection on graph, and we would have infinite bound states. |
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| | If we keep potential constant, and increase the width of the well, it would also increase the number of bound states. |
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