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Chemistry LibreTexts

Particle in a 3-dimensional box

The Time-Independent Schrodinger Equation

\[-\dfrac{\hbar^{2}}{2m}\nabla^{2}\psi(r) + V(r)\psi(r) = E\psi(r) \label{3.9.1}\]

Since we are dealing with a 3-dimensional figure, we need to add the 3 different axes into the Schrondinger equation:

\[-\dfrac{\hbar^{2}}{2m}\left(\dfrac{d^{2}\psi(r)}{dx^{2}} + \dfrac{d^{2}\psi(r)}{dy^{2}} + \dfrac{d^{2}\psi(r)}{dz^{2}}\right) = E\psi(r) \label{3.9.2}\]

The easiest way in solving this partial differential equation is by having the wavefunction equal to a product of individual function for each independent variable (e.., the Separation of Variables technique):

\[\Psi{(x,y,z)} = X(x)Y(y)Z(z) \label{3.9.3}\]

Now each function has its own variable:

  • \(X(x)\) is a function for variable \(x\) only
  • \(Y(y)\) function of variable \(y\) only
  • \(Z(z)\) function of variable \(z\) only

Now substitute Equation \(\ref{3.9.3}\) into Equation \(\ref{3.9.2}\) and divide it by the product, x, y, z:

  • \(\dfrac{d^{2}\psi}{dx^{2}} = YZ\dfrac{d^{2}X}{dx^{2}} \Rightarrow \dfrac{1}{X}\dfrac{d^{2}X}{dx^{2}}\)
  • \(\dfrac{d^{2}\psi}{dy^{2}} = XZ\dfrac{d^{2}Y}{dy^{2}} \Rightarrow \dfrac{1}{Y}\dfrac{d^{2}Y}{dy^{2}}\)
  • \(\dfrac{d^{2}\psi}{dz^{2}} = XY\dfrac{d^{2}Z}{dz^{2}} \Rightarrow \dfrac{1}{Z}\dfrac{d^{2}Z}{dz^{2}}\)

\[-\dfrac{\hbar^{2}}{2mX} \dfrac{d^{2}X}{dx^{2}} + \left(-\dfrac{\hbar^{2}}{2mY} \dfrac{d^{2}Y}{dy^{2}}\right) + \left(-\dfrac{\hbar^{2}}{2mZ} \dfrac{d^{2}Z}{dz^{2}}\right) = E \label{3.9.4}\]

\(E\) is an energy constant, and is the sum of x, y, and z. For this to work, each term must equal its own constant. For example, 

\[\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} E_{x}X = 0\]

Now separate each term in Equation 3.9.4 to equal zero:

\[\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} E_{x}X = 0 \label{3.9.5a}\]

\[\dfrac{d^{2}Y}{dy^{2}} + \dfrac{2m}{\hbar^{2}} E_{y}Y = 0 \label{3.9.5b}\]

\[\dfrac{d^{2}Z}{dz^{2}} + \dfrac{2m}{\hbar^{2}} E_{z}Z = 0 \label{3.9.5c}\]

Now we can add all the energies together to get the total energy:

\[E_{x}+ E_{y} + E_{z} = E \label{3.9.6}\]

Do these equations look familiar? They should because we have now reduced the 3D box into a particle in a box problem!

\[\dfrac{d^{2}X}{dx^{2}} + \dfrac{2m}{\hbar^{2}} E_{x}X = 0 \approx \dfrac{d^{2}\psi}{dx^{2}} = -\dfrac{4\pi^{2}}{\lambda^{2}}\psi \label{3.9.7}\]

Now the equations are very similar to a 1-D box and the boundary conditions are identical.

\[n = 1, 2,..\infty\]

Use the normalization wavefunction equation for each variable:

Normalization wavefunction Equation

\[\psi(x) = \sqrt{2/L}\sin{{n \pi x}/L}\]

  • \(Limit: 0 \leq x \leq L\)

\[\psi(x) = 0\]

  • \(Limit: L < x < 0\)

Normalization wavefunction Equation for each variable

\[X(x) = \sqrt{\dfrac{2}{a}} \sin \left( \dfrac{n_{x}\pi x}{a} \right) \label{3.9.8a}\]

\[Y(y) = \sqrt{\dfrac{2}{b}} \sin \left(\dfrac{n_{y}\pi y}{b} \right) \label{3.9.8b}\]

\[Z(z) = \sqrt{\dfrac{2}{c}} \sin \left( \dfrac{n_{z}\pi z}{c} \right) \label{3.9.8c}\]

The limits of the three quantum numbers

  • \(n_{x} = 1, 2, 3 ...\infty\)
  • \(n_{y} = 1, 2, 3 ...\infty\)
  • \(n_{z} = 1, 2, 3 ...\infty\)

For each constant use the de Broglie Energy equation:

\[E_{x} = \dfrac{n_{x}^{2}h^{2}}{8ma^{2}} \label{3.9.9}\]

\[n_{x} = 1...\infty\]

Do the same for variables \(n_y\) and \(n_z\).

Combine Equation \(\ref{3.9.3}\) with Equations \(\ref{3.9.8a}\) -  \(\ref{3.9.8c}\) to find the wavefunctions inside a 3D box.

\(\psi(r) = \sqrt{\dfrac{8}{V}}\sin \left( \dfrac{n_{x} \pi x}{a} \right) \sin \left(\dfrac{n_{y} \pi y}{b}\right) \sin  \left(\dfrac{ n_{z} \pi z}{c} \right)\)

\[V = a \times b \times c = volume \; of \; box\]

To find the Total Energy level, add Equation \(\ref{3.9.9}\) and Equation \(\ref{3.9.6}\).

\[E = \dfrac{h^{2}}{8m}\left(\dfrac{n_{x}^{2}}{a^{2}} + \dfrac{n_{y}^{2}}{b^{2}} + \dfrac{n_{z}^{2}}{c^{2}}\right) \label{3.9.10}\]