Phy5645/Problem 1D sample

(Submitted by team 1)

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The Schroedinger's equation takes the form:

$-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\psi (x,y,z)}{dx^{2}}}+\left(X(x)+Y(y)+Z(z)\right)\psi (x,y,z)=E\psi (x,y,z)$

Assuming that $\psi \!$ can be write like:

$\psi (x,y,z)=\Phi (x)\Delta (y)\Omega (z)\!$

So,

${\begin{aligned}-{\frac {\hbar ^{2}}{2m}}\left[{\frac {d^{2}\Phi (x)}{dx^{2}}}\Delta (y)\Omega (z)+\Phi (x){\frac {d^{2}\Delta (y)}{dy^{2}}}\Omega (z)+\Phi (x)\Delta (y){\frac {d^{2}\Omega (z)}{dz^{2}}}\right]\\+\left[X(x)+Y(y)+Z(z)\right]\Phi (x)\Delta (y)\Omega (z)&=E\Phi (x)\Delta (y)\Omega (z)\end{aligned}}$

Dividing by $\psi (x,y,z)\!$

$-{\frac {\hbar ^{2}}{2m}}{\frac {1}{\Phi (x)}}{\frac {d^{2}\Phi (x)}{dx^{2}}}+X(x)-{\frac {\hbar ^{2}}{2m}}{\frac {1}{\Delta (y)}}{\frac {d^{2}\Delta (y)}{dy^{2}}}+Y(y)-{\frac {\hbar ^{2}}{2m}}{\frac {1}{\Omega (z)}}{\frac {d^{2}\Omega (z)}{dz^{2}}}+Z(z)=E$

We can perfectly separate the right hand side into three parts, where it will only depend on $x\!$, or on $y\!$ or only on $z\!$. Then each of these parts must be equal to a constant. So:

$-{\frac {\hbar ^{2}}{2m}}{\frac {1}{\Phi (x)}}{\frac {d^{2}\Phi (x)}{dx^{2}}}+X(x)=E_{x}$

$-{\frac {\hbar ^{2}}{2m}}{\frac {1}{\Delta (y)}}{\frac {d^{2}\Delta (y)}{dy^{2}}}+Y(y)=E_{y}$

$-{\frac {\hbar ^{2}}{2m}}{\frac {1}{\Omega (z)}}{\frac {d^{2}\Omega (z)}{dz^{2}}}+Z(z)=E_{z}$

where $E_{x}\!$, $E_{y}\!$ and $E_{z}\!$ are constants and $E=E_{x}+E_{y}+E_{z}\!$

Hence, the three-dimensional problem has been divided into three one-dimensional problems where the total energy $E\!$ is the sum of the energies $E_{x}\!$, $E_{y}\!$ and $E_{z}\!$ in each dimension.

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