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# 429: Variation Method for a Particle in a Box with an Internal Barrier

$$\psi _{1} (x) = \sqrt{2} \sin{ \pi x}$$

$$\psi _{2} (x) = \sqrt{2} \sin{3 \pi x}$$

$$V(x) = if [(x \geq .45) (x \leq .55), 3,0]$$

Plot trial wavefunctions and potential energy. Evaluate matrix elements for $$100 E_h$$ internal barrier:

$$S_{11} = \int_{0}^{1} \psi _{1} (x)^{2} dx$$ $$S_{11} = 1$$

$$S_{12} = \int_{0}^{1} \psi _{1} (x) \psi _{2} (x) dx$$ $$S_{12} = 0$$

$$S_{22} = \int_{0}^{1} \psi _{2} (x)^{2} dx$$ $$S_{22} = 1$$

\begin{align} H_{11} &= \int_{0}^{1} \psi _{1} (x) (- \dfrac{1}{2}) \dfrac{d^{2}}{dx^{2}} \psi _{1} (x) dx + \int_{.45}^{.55} \psi _{1} (x)~100~ \psi _{1} (x) dx\\[4pt] & \approx 24.7711\end{align}

\begin{align} H_{12} &= \int_{0}^{1} \psi _{1} (x) (- \dfrac{1}{2}) \dfrac{d^{2}}{dx^{2}} \psi _{2} (x) dx + \int_{.45}^{.55} \psi _{1} (x)~100~ \psi _{2} (x) dx \\[4pt] & \approx -19.1912\end{align}

\begin{align} H_{22} &= \int_{0}^{1} \psi _{2} (x) (- \dfrac{1}{2}) \dfrac{d^{2}}{dx^{2}} \psi _{2} (x) dx + \int_{.45}^{.55} \psi _{2} (x)~100~ \psi _{2} (x) dx \\[4pt] &\approx 62.9972 \end{align}

Solve the secular equations and normalization constraint for the energy and coefficients.

Seed values for energy and coefficient: E = 5 c1 = .5 c2 = .5

Given

$$(H_{11} - E S_{11})c_{1} + (H_{12} - E S_{12})c_{2} = 0$$

$$(H_{12} - E S_{12})c_{1} + (H_{22} - E S_{22})c_{2} = 0$$

$$c_{1}^{2} S_{11} + 2 c_{1} c_{2} S_{12} + c_{2}^{2} S_{22} = 1$$

$${\begin{pmatrix} E \\ c_{1} \\ c_{2} \end{pmatrix}} = Find(E, c_{1}, c_{2})$$

$${\begin{pmatrix} E \\ c_{1} \\ c_{2} \end{pmatrix}} = {\begin{pmatrix} 16.7989 \\ 0.9235\\ 0.3836 \end{pmatrix}}$$

Plot variational results:

$$\Phi (x) = c_{1} \Phi _{1} (x) + c_{2} \psi_{2} (x)$$ Calculate the probability the particle is in the barrier:

$$\int_{0.45}^{0.55} \Phi (x)^{2} dx = 0.0605$$

Calculate potential and kinetic energy:

$$V = 100 \int_{0.45}^{0.55} \Phi (x)^{2} dx$$ $$V = 6.0541$$

$$T = E - V$$

$$T = 10.7448$$

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