Phy5645/Problem 1D sample: Difference between revisions

Latest revision as of 13:28, 18 January 2014

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.

Back to Motion in One Dimension

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-(Submitted by team 1. Based on problem 3.19 in Schaum's Theory and problems of Quantum Mechanics)
+The Schrödinger equation takes the form,
-Consider a particle of mass m in a three dimensional potential:
+:<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>
-<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.
+Let us assume that <math>\psi</math> has the form, <math>\psi(x,y,z)=\Phi(x) \Delta(y) \Omega (z).\!</math>  Then
-______________________________________________________________________________________________________________________________________
+:<math>
+\begin{align}
+-\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{align}
+</math>
-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>
+Dividing by <math>\psi(x,y,z),\!</math> we obtain
+:<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>
-Assuming that <math>\Psi</math> can be write like:
+We may now separate the left-hand side into three parts, each depending on only one of the three coordinates <math>x,\,y,</math> and <math>z.\!</math>  Each of these parts must be equal to a constant.  Therefore,
-<math>\Psi(x,y,z)=\Phi(x) \Delta(y) \Omega (z)  </math>
+:<math>
+-\frac{\hbar^2}{2m}\frac{d^2\Phi(x)}{dx^2} + X(x)\Phi(x) = E_x\Phi(x)  </math>
+:<math>
+-\frac{\hbar^2}{2m}\frac{d^2\Delta(y)}{dy^2} + Y(y)\Delta(y) = E_y\Delta(y)  </math>
-So,
+:<math>
+-\frac{\hbar^2}{2m}\frac{d^2\Omega(z)}{dz^2} + Z(z)\Omega(z) = E_z\Omega(z),  </math>
-<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>
+where <math> E_x,\!</math>, <math> E_y,\!</math> and <math> E_z\!</math> are constants and <math> E = E_x+E_y+E_z.\!</math>
+Hence, the three-dimensional problem has been divided into three one-dimensional problems where the total energy <math>E</math> is the sum of the energies <math> E_x,\!</math> <math>E_y,\!</math> and <math>E_z\!</math> in each dimension.
-Dividing by <math>\Psi(x,y,z) </math>
+Back to [[Motion in One Dimension#Problem|Motion in One Dimension]]
-<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)     = Ex  </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>
-Ex, Ey and Ez are constant where: <math>  E = Ex+Ey+Ez  </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.

Phy5645/Problem 1D sample: Difference between revisions

Latest revision as of 13:28, 18 January 2014

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