Phy5645/Free particle SE problem: Difference between revisions

Latest revision as of 13:22, 18 January 2014

(a) The plane wave, $\psi =e^{ikz},\!$ does not depend on $x\!$ or $y\!$ . Therefore, the Schrödinger equation becomes $\left({\frac {d^{2}}{dz^{2}}}+k^{2}\right)\psi =0\!$ . We may easily see that this is a solution to the equation:

{\frac {d^{2}}{dz^{2}}}\left(e^{ikz}\right)+k^{2}e^{ikz}=0.

(b) In spherical coordinates, the Laplacian is given by

\nabla ^{2}=\partial _{r}^{2}+{\frac {2}{r}}\partial _{r}+{\frac {1}{r^{2}}}\partial _{\theta }^{2}+{\frac {\cot \theta }{r^{2}}}\partial _{\theta }+{\frac {1}{r^{2}\sin ^{2}\theta }}\partial _{\phi }^{2}.

The spherical wave $\psi ={\frac {e^{ikr}}{r}}\!$ does not depend on $\theta \!$ or $\phi \!$ . Therefore, the Schrödinger equation becomes

\left({\frac {d^{2}}{dr^{2}}}+{\frac {2}{r}}{\frac {d}{dr}}+k^{2}\right)\psi =0

\Rightarrow {\frac {d^{2}}{dr^{2}}}\left({\frac {e^{ikr}}{r}}\right)+{\frac {2}{r}}{\frac {d}{dr}}\left({\frac {e^{ikr}}{r}}\right)+k^{2}{\frac {e^{ikr}}{r}}={\frac {2e^{ikr}}{r^{3}}}-{\frac {2ike^{ikr}}{r^{2}}}+{\frac {2}{r}}\left(-{\frac {e^{ikr}}{r^{2}}}+{\frac {ike^{ikr}}{r}}\right)=0.

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-Submitted by team 1
+'''(a)''' The plane wave, <math> \psi = e^{ikz},\!</math> does not depend on <math> x\!</math> or <math> y\!</math>.  Therefore, the Schrödinger equation becomes <math> \left( \frac{d^2}{dz^2} + k^2 \right) \psi = 0 \!</math>.  We may easily see that this is a solution to the equation:
-------
-'''Question: A free particle Schrodinger Equation'''
-Time-independent Schrodinger equation for a free particle is given by
-:<math>
-\frac{1}{2m} \left( \frac{\hbar}{i} \frac{\partial}{\partial \mathbf{r}} \right)^2 \psi \left(\mathbf{r} \right) = E \psi\left(\mathbf{r} \right)
-</math>
-It is customary to write <math> E = \frac{\hbar^2 k^2}{2m} \!</math> to simplify the equation
-:<math>
- \left( \nabla^2 + k^2 \right) \psi \left( \mathbf{r} \right) = 0.
-</math>
-Show that (a) a plane wave <math> \psi\left(\mathbf{r} \right) = e^{ikz} \!</math>, and (b) a spherical wave <math> \psi\left(\mathbf{r} \right) = \frac{e^{ikr}}{r} \! </math> where <math> r = \sqrt{x^2 + y^2 + z^2} \! </math>, satisfy the equation. (In either case, the wave length of the solution is given by <math> \lambda = \frac{2\pi}{k} \!</math> and the momentum by de Broghie's relation <math> p = \hbar k \! </math>. )
---------
-'''Answer:'''
-(a) Plane wave <math> \psi = e^{ikz} \! </math> does not depend on <math> x \!</math> or <math> y \!</math>. Therefore the Schrodinger equation becomes <math> \left( \partial_z^2 + k^2 \right) \psi = 0 \!</math>. Obviously this is a solution to the equation of
 :<math>
-\frac{\partial^2}{\partial z^2} \left( e^{ikz} \right) + k^2 e^{ikz} = 0.
+\frac{d^2}{dz^2} \left( e^{ikz} \right) + k^2 e^{ikz} = 0.
 </math>
-(b) In polar coordinates, the Laplacian can be rewritten as
+'''(b)''' In spherical coordinates, the Laplacian is given by
 :<math>
-\nabla^2 = \partial_{r}^2 + \frac{2}{r} \partial_r + \frac{1}{r^2} \partial_{\theta}^2 + \frac{\cos\theta}{r^2 \sin\theta} \partial_{\theta} + \frac{1}{r^2 \sin^2\theta} \partial_{\phi}^2 .
+\nabla^2 = \partial_{r}^2 + \frac{2}{r} \partial_r + \frac{1}{r^2} \partial_{\theta}^2 + \frac{\cot\theta}{r^2} \partial_{\theta} + \frac{1}{r^2 \sin^2\theta} \partial_{\phi}^2 .
 </math>
-The spherical wave <math> \psi = \frac{ikr}{r} \! </math> does not depend on <math> \theta \!</math> or <math> \phi \!</math>. Therefore, the Schrodinger equation becomes
+The spherical wave <math> \psi = \frac{e^{ikr}}{r} \! </math> does not depend on <math> \theta \!</math> or <math> \phi \!</math>. Therefore, the Schrödinger equation becomes
 :<math>
-\left( \partial_r^2 + \frac{2}{r} \partial_r + k^2 \right) \psi = 0
+\left( \frac{d^2}{dr^2} + \frac{2}{r} \frac{d}{dr} + k^2 \right) \psi = 0
 </math>
 :<math>
-  \Rightarrow \frac{\partial^2}{\partial r^2} \left( \frac{e^{ikr}}{r} \right) + \frac{2}{r} \frac{\partial}{\partial r} \left( \frac{e^{ikr}}{r} \right) + k^2 \frac{e^{ikr}}{r} = \frac{2e^{ikr}}{r^3} - \frac{2ike^{ikr}}{r^2} + \frac{2}{r} \left( -\frac{e^{ikr}}{r^2} + \frac{ike^{ikr}}{r} \right) = 0.
+  \Rightarrow \frac{d^2}{dr^2} \left( \frac{e^{ikr}}{r} \right) + \frac{2}{r} \frac{d}{dr} \left( \frac{e^{ikr}}{r} \right) + k^2 \frac{e^{ikr}}{r} = \frac{2e^{ikr}}{r^3} - \frac{2ike^{ikr}}{r^2} + \frac{2}{r} \left( -\frac{e^{ikr}}{r^2} + \frac{ike^{ikr}}{r} \right) = 0.
 </math>
-Back to [[Stationary States]]
+Back to [[Stationary States#Problem|Stationary States]]

Phy5645/Free particle SE problem: Difference between revisions

Latest revision as of 13:22, 18 January 2014

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