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# 428: Variation Method for a Particle in a Semi-Infinite Potential Well

This problem deals with the variational approach to the particle in the semi-infinite potential well.

Kinetic energy operator: $$- \frac{1}{2} \frac{d^2}{dx^2} \blacksquare$$

Integral: $$\int_{0}^{ \infty} \blacksquare dx$$

Potential energy: $$V(x) := if[( x \leq 2), 0 , 2]$$ Trial wave function: $$\Phi (x, \beta ) := 2 \beta ^{ \frac{3}{2}}~x~exp(- \beta x)$$

If the trial wave function is not normalized, normalize it.

$$\int_{0}^{ \infty} \Phi (x, \beta )^{2} dx~ assume,~ \beta > 0 \rightarrow 1$$

Evaluate the variational energy integral.

$$E( \beta ) := \int_{0}^{ \infty} \Phi (x, \beta ) (- \frac{1}{2}) \frac{d^2}{dx^2} \Phi (x, \beta ) dx ... |_{simplify}^{assume,~ \beta >0} \rightarrow \frac{1}{2} \beta^{2} + 16 \beta^{2} e^{-4 \beta} + 8 \beta e^{-4 \beta} + 2 e^{- 4 \beta} + \int_{2}^{ \infty} 2 \Phi (x, \beta )^{2} dx$$

Minimize the energy with respect to $$\beta$$:

$$\beta$$ := .3 $$\beta$$ := Minimize $$(E, \beta )$$ $$\beta = 1.053$$ $$E( \beta ) = 0.972$$

Display optimized trial wave function and potential energy: Calculate average position and most probable position of the particle:

$$\int_{0}^{ \infty} x \Phi (x, \beta )^{2} dx = 1.425$$

$$\frac{d}{dx} \Phi (x, \beta ) = 0 |_{solve,~x}^{float,~3} \rightarrow \frac{1}{ \beta} = 0.95$$

Calculate the probability of the particle in the barrier.

$$\int_{2}^{ \infty} \Phi (x, \beta )^{2} dx = 20.891%$$

Calculate the potential energy, and the kinetic energy.

$$V := \int_{2}^{ \infty} 2 \Phi (x, \beta )^{2} dx$$ $$V = 0.418$$

$$T := E ( \beta ) - V$$ $$T = 0.554$$

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