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r1 - 23 Jul 2007 - 19:44:26 - ThanhTruongYou are here: TWiki >  Main Web > LectureNotes > UndergraduatePChem > Chapter5 > Particle3D

Particle in a Three-Dimensional Box

Let generalize the above solution to a more physical system, namely the particle in a 3D box.  In this case, the potential energy is given by

\[V(x,y,z) = \left \lbrace \begin{array}{l} 0 \; \; \; \; \; 0 \leq x \leq a; \; 0 \leq y \leq b;\; 0 \leq z \leq c \\ \infty \; \; \;otherwise \end{array} \right. \]

The Schrödinger equation in this case becomes

\[- \frac{ \hbar ^2}{ 2m} \left[ \frac{ \partial ^2}{ \partial x^2} + \frac{ \partial ^2}{ \partial y^2} + \frac{ \partial ^2}{ \partial z^2} \right ] \psi (x,y,z) = E \psi (x,y,z) \]

To solve this differential equation, we use the separation variable technique by assuming the $\psi (x,y,z)$ has the form:

\[ \psi \left ( x,y,z\right ) = X(x) Y(y) Z(z)\]

Such assumption is valid when the potential energy has the form $V(x,y,z) = V_x(x) + V_y(y) + V_z(z)$. Substitute the wavefunction into the differential Schrödinger equation and then divide both sides by $X(x) Y(y) Z(z)$ to obtain

\[- \frac{ \hbar ^2}{ 2m} \left[ \frac{ 1}{ X(x)} \frac{ d ^2 X(x)}{ d x^2} + \frac{ 1}{ Y(y)} \frac{ d^2 Y(y)}{ dy^2} + \frac{ 1}{ Z(z)} \frac{ d^2 Z(z)}{ d z^2}\right ] = E\]

Note that each of the three terms in the bracket on the left side of the above equation depends on only x, or y, or z independent variable and thus can vary independently. The only possibility for their sum equals to the constant E is when each of these terms also equals to a constant so the above equation can be valid for all x, y, and z. Thus, we can write

\[E=E_x + E_y + E_z\]

The original Schrödinger equation now can be separated into three separate ones as

\[- \frac{ \hbar ^2}{ 2m} \frac{ 1}{ X(x)} \frac{ d ^2 X(x)}{dx^2} = E_x\]

\[- \frac{ \hbar ^2}{ 2m} \frac{ 1}{ Y(y)} \frac{ d ^2 Y(y)}{dy^2} = E_y\]

\[- \frac{ \hbar ^2}{ 2m} \frac{ 1}{ Z(z)} \frac{ d ^2 Z(z)}{dz^2} = E_z\]

Each of these equations can be solved in the same way as for the particle in 1D box and it also has similar solution. The total wavefunction $\psi$ has the form

\[ \psi (x,y,z) = N \sin \left ( \frac{n_x \pi x }{a } \right ) \sin \left ( \frac{n_y \pi y }{b } \right )\sin \left ( \frac{n_z \pi z }{c } \right ) \]

where the normalization constant $N = \sqrt { \frac{ 8}{ abc} } $.

The total energy of the system is given by

\[E = \frac{ h^2}{ 8m} \left ( \frac{ n_x^2}{a^2 } + \frac{ n_y^2}{b^2 } + \frac{ n_z^2}{c^2 }\right ) \]

So each energy level is specified by three quantum numbers $(n_x, n_y, n_z)$ where $n_x, n_y, n_z = 1, 2, 3, ...$.

Note that if a=b=c then there exists a set of unique $(n_x, n_y, n_z)$ that have the same total energy. For an example, (2,1,1), (1,2,1), and (1,1,2) all gives the same energy level. In this case, we say that the energy level is degenerate with the degeneracy being the number of states that have the same energy, i.e. degeneracy = 3 in this example.

-- ThanhTruong - 23 Jul 2007

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