Phy5645/Problem 1D sample: Difference between revisions

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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.
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.


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The Schroedinger's equation takes the form:
The Schroedinger's equation takes the form:
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\begin{align}
\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] \\
-\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)  
+ \left[X(x)+Y(y)+Z(z)\right]\Phi(x) \Delta(y) \Omega (z) &= E\Phi(x) \Delta(y) \Omega (z)  
\end{align}
\end{align}
</math>
</math>
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:<math>
:<math>
\begin{align}
-\frac{\hbar^2}{2m} \frac{1}{\Phi(x)} \frac{d^2\Phi(x)}{dx^2} + X(x)
-\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}{\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)  
-\frac{\hbar^2}{2m} \frac{1}{\Omega(z)} \frac{d^2\Omega(z)}{dz^2} + Z(z)  
= E   
= E   
\end{align}
</math>
</math>



Revision as of 00:46, 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:

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:


Assuming that can be write like:


So,


Dividing by

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

where , and are constant and


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