Phy5645/Problem 1D sample

(Submitted by team 1)

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The Schrödinger equation takes the form,

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

Let us assume that $\psi$ has the form, $\psi (x,y,z)=\Phi (x)\Delta (y)\Omega (z).\!$ Then

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

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

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

We may now separate the left-hand side into three parts, each depending on only one of the three coordinates $x,\,y,$ and $z.\!$ Each of these parts must be equal to a constant. Therefore,

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

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

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

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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