Phy5645/Problem 1D sample

(Submitted by team 1. Based on problem 3.19 in Schaum's Theory and problems of Quantum Mechanics)

Consider a particle of mass m in a three dimensional potential: $V(x,y,z)=X(x)+Y(y)+Z(z)$

Using the Schroedinger's equation show that we can treat the problem like three independent one-dimensional problems. Relate the energy of the three-dimensional state to the effective energies of one-dimensional problem.

______________________________________________________________________________________________________________________________________

The Schroedinger's equation takes the form: $-{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}\Psi (x,y,z)}{dx^{2}}}+(X(x)+Y(y)+Z(z))\Psi (x,y,z)=E\Psi (x,y,z)$

Assuming that $\Psi$ can be write like: $\Psi (x,y,z)=\Phi (x)\Delta (y)\Omega (z)$

So,

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

Dividing by $\Psi (x,y,z)$

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

Perfectly we can separate the right hand side in three parts, where only one depends of x, only one of y and only one of z. Then each of these parts must be equal to a constant. So:

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

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

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

Ex, Ey and Ez are constant where: $E=Ex+Ey+Ez$

Hence the three-dimensional problem has been divided in three one-dimensional problems where the total energy E is the sum of the energies Ex, Ey and Ez in each dimension.