# 10.18: Trigonometric Trial Wave Function for the Harmonic Potential Well

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Definte potential energy: V(x) := $$\frac{s^2}{2}$$

Display potential energy:

Choose trial wave function: $$\Psi (x, \beta ) := \sqrt{ \frac{ \beta}{2}} sech( \beta x)$$

Set up variational energy integral:

$$E( \beta ) := \frac{ \int_{- \infty}^{ \infty} \Psi (x, \beta ) \frac{-1}{2} \frac{d^2}{dx^2} \Psi (x, \beta ) dx + \int_{- \infty}^{ \infty} \Psi (x, \beta ) \frac{x^2}{2} \Psi (x, \beta ) dx}{ \int_{0}^{ \infty} \Psi (x, \beta )^2 dx} |^{assume,~ \beta > 0}_{simplify} \rightarrow \frac{1}{24} \frac{4 \beta ^2 + \pi^2}{ \beta^2}$$

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

$$\beta$$ := 0.2 $$\beta$$ := Minimize (E, $$\beta$$) $$\beta$$ = 1.253 E( $$\beta$$) = 0.524

Display wave function in the potential well.

Calculate the probability that the particle is in the potential barrier.

$$2 \int_{0}^{ \infty} \Psi (x, \beta )^2 dx = 1$$

Define quantum mechanical tunneling.

Tunneling occurs when a quon (a quantum mechanical particle) has probability of being in a nonclassical region. In other words, a region in which the total energy is less than the potential energy.

Calculate the probability that tunneling is occurring.

Calculate the classical turning point.

$$\frac{x^2}{2} = 0.524 |_{float,~4}^{solve,~x} \rightarrow {\begin{pmatrix} -1.024 \\ 1.024 \end{pmatrix}}$$

$$2 \int_{1.024}^{ \infty} \Psi (x, \beta )^2 dx = 0.143$$

Calculate the kinetic and potential energy contributions to the total energy.

Kinetic energy:

$$\int_{- infty}^{ \infty} \Psi (x, \beta ) \frac{-1}{2} \frac{d^2}{dx^2} \Psi (x, \beta ) dx = 0.262$$

Potential energy:

$$\int_{- \infty}^{ \infty} V(x) \Psi (x, \beta )^2 dx = 0.262$$

Is the virial theorem satisfied?

Yes, for the harmonic potential the virial theorem is T = V = E/2.

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