# 10.10: Variation Method for a Particle in a Gravitational Field

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The particle of unit mass in a gravitational field for which g = 1 has the energy operator shown below.

$$- \frac{1}{2} \frac{d^2}{dz^2} \blacksquare + z \blacksquare$$

$$\int_{0}^{ \infty} \blacksquare dz$$

The following trial wave function for this problem is:

$$\Phi ( \alpha, z) := 2 ( \frac{2 \alpha}{ \pi ^{ \frac{1}{3}}})^{ \frac{3}{4}} z~exp ( - \alpha z^{2})$$

Determine whether or not the wave function is normalized.

$$\int_{0}^{ \infty} \Psi ( \alpha , z)^{2} dz |_{simplify}^{assume,~ \alpha > 0} \rightarrow 1$$

Evaluate the variational energy integral.

$$E ( \alpha ) := \int_{0}^{ \infty} \Psi ( \alpha , z) - \frac{1}{2} \frac{d^2}{dz^2} \Phi ( \alpha , z) dz ... |_{simplify}^{assume,~ \alpha >0} \rightarrow \frac{1}{2 \pi ^{ \frac{1}{2}}} \frac{3 \pi ^{ \frac{1}{2}} \alpha ^{2} + 2 (2)^{ \frac{1}{2}} \alpha ^{ \frac{1}{2}}}{ \alpha} + \int_{0}^{ \infty} z \Psi ( \alpha , z)^{2} dz$$

Minimize the energy with respect to the variational parameter $$\alpha$$ and report its optimum value and the ground‐state energy.

$$\alpha := 1$$ $$\alpha := Minimize(E, \alpha )$$ $$\alpha = 0.4136$$ $$E( \alpha ) = 1.8611$$ $$E_{exact} := 1.8558$$

Plot the wave function with the distance of the particle from the surface on the vertical axis.

Find that distance below which there is a 90% probability of finding the particle.

$$\alpha := 1$$

Given $$\int_{0}^{a} \Psi ( \alpha , z)^{2} dz = .90$$

Find (a) = 1.9440

Find the most probable value of the position of the particle from the surface.

$$\frac{d}{dz} \Psi (0.4136, z) = 0 |_{float,~3}^{solve,~z} \rightarrow {\begin{pmatrix} -1.10 \\ 1.10 \end{pmatrix}}$$

Calculate the probability that the particle will be found below the most probable distance from the surface.

$$\int_{0}^{1.10} \Psi ( \alpha , z)^{2} dz = 0.4279$$

Calculate the probability that tunneling is occurring: $$\int_{1.861}^{ \infty} \Psi ( \alpha , z)^{2} dz = 0.1256$$

Kinetic energy: $$\int_{0}^{ \infty} \Psi ( \alpha , z) - \frac{1}{2} \frac{d^2}{dz^2} \Psi ( \alpha , z) dz = 0.6204$$

Potential energy: $$\int_{0}^{ \infty} z \Psi ( \alpha , z)^{2} dz = 1.2407$$

What is the apparent virial theorem for this system: $$E = 3T = \frac{3}{2} V$$

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