Phy5645/Free particle SE problem

Submitted by team 1

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Question: A free particle Schrodinger Equation

Time-independent Schrodinger equation for a free particle is given by

${\frac {1}{2m}}\left({\frac {\hbar }{i}}{\frac {\partial }{\partial \mathbf {r} }}\right)^{2}\psi \left(\mathbf {r} \right)=E\psi \left(\mathbf {r} \right)$

It is customary to write $E={\frac {\hbar ^{2}k^{2}}{2m}}\!$ to simplify the equation

$\left(\nabla ^{2}+k^{2}\right)\psi \left(\mathbf {r} \right)=0.$

Show that (a) a plane wave $\psi \left(\mathbf {r} \right)=e^{ikz}\!$, and (b) a spherical wave $\psi \left(\mathbf {r} \right)={\frac {e^{ikr}}{r}}\!$ where $r={\sqrt {x^{2}+y^{2}+z^{2}}}\!$, satisfy the equation. (In either case, the wave length of the solution is given by $\lambda ={\frac {2\pi }{k}}\!$ and the momentum by de Broghie's relation $p=\hbar k\!$. )

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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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