Phy5645/Free particle SE problem: Difference between revisions

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'''(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:
 
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(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:
:<math>
:<math>
\frac{d^2}{dz^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>
</math>


(b) In spherical coordinates, the Laplacian is given by
'''(b)''' In spherical coordinates, the Laplacian is given by
:<math>
:<math>
\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 .  
\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 .  

Revision as of 16:49, 12 August 2013

(a) The plane wave, does not depend on or . Therefore, the Schrödinger equation becomes . We may easily see that this is a solution to the equation:

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

The spherical wave does not depend on or . Therefore, the Schrödinger equation becomes

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