Phy5645/Free particle SE problem

Submitted by team 1

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Answer:

(a) Plane wave $\psi =e^{ikz}\!$ does not depend on $x\!$ or $y\!$. Therefore the Schrodinger equation becomes $\left(\partial _{z}^{2}+k^{2}\right)\psi =0\!$. Obviously this is a solution to the equation of

${\frac {\partial ^{2}}{\partial z^{2}}}\left(e^{ikz}\right)+k^{2}e^{ikz}=0.$

(b) In polar coordinates, the Laplacian can be rewritten as

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

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

$\left(\partial _{r}^{2}+{\frac {2}{r}}\partial _{r}+k^{2}\right)\psi =0$

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

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