Phy5645/Problem 1D sample: Difference between revisions

Revision as of 00:34, 5 December 2009

(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}}}+\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,

-{\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)

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)=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 constant and $E=E_{x}+E_{y}+E_{z}\!$

Hence the three-dimensional problem has been divided in 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.

@@ Line 2: / Line 2: @@
 Consider a particle of mass m in a three dimensional potential:
-<math>V(x,y,z)=X(x) + Y(y) + Z(z)</math>
+<math>V(x,y,z) = X(x)+Y(y)+Z(z)\!</math>
 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.
@@ Line 9: / Line 9: @@
 The Schroedinger's equation takes the form:
-<math>-\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)</math>
+:<math>-\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)</math>
-Assuming that <math>\psi</math> can be write like:
+Assuming that <math>\psi\!</math> can be write like:
-<math>\psi(x,y,z)=\Phi(x) \Delta(y) \Omega (z)  </math>
+:<math>\psi(x,y,z)=\Phi(x) \Delta(y) \Omega (z) \!</math>
 So,
-<math>-\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)  </math>
+:<math>
+-\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)  </math>
-Dividing by <math>\psi(x,y,z) </math>
+Dividing by <math>\psi(x,y,z) \!</math>
-<math>-\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  </math>
+:<math>
+-\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
+</math>
 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:
+:<math>
+-\frac{\hbar^2}{2m} \frac{1}{\Phi(x)} \frac{d^2\Phi(x)}{dx^2} + X(x) = E_x  </math>
-<math>-\frac{\hbar^2}{2m}\frac{1}{\Phi(x)}   \frac{d^2\Phi(x)}{dx^2}     + X(x)     = Ex  </math>
+:<math>
+-\frac{\hbar^2}{2m} \frac{1}{\Delta(y)} \frac{d^2\Delta(y)}{dy^2} + Y(y) = E_y  </math>
-<math>-\frac{\hbar^2}{2m}\frac{1}{\Delta(y)} \frac{d^2\Delta(y)}{dy^2}   + Y(y)     = Ey  </math>
-<math>-\frac{\hbar^2}{2m}\frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2}   + Z(z)     = Ez  </math>
+:<math>
+-\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z) = E_z  </math>
-Ex, Ey and Ez are constant where: <math>  E = Ex+Ey+Ez  </math>
+where E_x, E_y and E_z are constant and <math> E = E_x+E_y+E_z \!</math>
-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.
+Hence the three-dimensional problem has been divided in three one-dimensional problems where the total energy E is the sum of the energies <math> E_x \!</math>, <math> E_y \!</math> and <math> E_z \!</math> in each dimension.

Phy5645/Problem 1D sample: Difference between revisions

Revision as of 00:34, 5 December 2009

Navigation menu

Search