# 10.19: Trigonometric Trial Wave Function for the 3D Harmonic Potential Well

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Trial wave function: $$\Psi (r, \beta) := \sqrt{ \frac{3 \beta^3}{ \pi^3}} sech( \beta r)$$

Integral: $$\int_{0}^{ \infty} \blacksquare 4 \pi r^2 dr$$

Kinetic energy operator: $$T = \frac{-1}{2r} \frac{d^2}{dr^2} (r \blacksquare )$$

Potential energy operatory: $$V = \frac{1}{2} k r^2$$

a. Demonstrate the wave function is normalized.

$$\int_{0}^{ \infty} \Psi (r, \beta )^2 4 \pi r^2 dr |_{simplify}^{assume,~ \beta > 0} \rightarrow 1$$

b. Evaluate the variational integral.

$$E( \beta ) := \int_{0}^{ \infty} \Psi (r, \beta ) [ \frac{-1}{2r} \frac{d^2}{dr^2} (r \Psi (r, \beta ))] 4 \pi r^2 dr + \int_{0}^{ \infty} \Psi (r, \beta ) \frac{1}{2} r^2 \Psi (r, \beta ) 4 \pi r^2 dr$$

c. Minimize the energy with respect to the variational parameter $$\beta$$.

$$\beta$$ := 1 $$\beta$$ := Minimize (E, $$\beta$$) $$\beta$$ = 1.471 E( $$\beta$$) = 1.597

d. The exact ground state energy for the 3D harmonic oscillator is 1.5 Eh. Calculate the percent error.

$$\frac{E( \beta ) - 1.5}{1.5} = 6.488$$%

e. Compare the optimized trial wave function with the exact solution by plotting the radial distribution functions.

$$\Phi (r) := ( \frac{1}{ \pi})^{ \frac{3}{4}} exp( \frac{r^2}{2})$$

h. Calculate the overlap integral between the trial wave function and the exact wave function.

$$\int_{0}^{ \infty} \Psi (r, \beta ) \Phi (r) 4 \pi r^2 dr = 0.989$$

i. Calculate the probability that tunneling is occurring.

Classical turning point:

$$1.597 = \frac{1}{2} r^2 |_{float,~3}^{solve,~r} \rightarrow {\begin{pmatrix} -1.79 \\ 1.79 \end{pmatrix}}$$

Tunneling probability:

$$\int_{1.79}^{ \infty} \Psi (r, \beta )^2 4 \pi r^2 dr = 12.598$$%

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