Question

Calculate the expectation value of the position of the particle.

Short answer

The expectation value of the position of the particle is .

Step-by-step solution

Step 1:Understandingthe concept of the expectation value of the position of a particle.

This integral can be thought of as the averagevalue that would be obtained from a large number of measurements. It could also be thought of as the average position value for a large number of particles described by the same wavefunction.

The expectation value of the position is the integral of the wave function squared ${\mathit{\psi }}^{\mathbf{*}}\mathit{\psi }$multiplied by the position, X.

Given information.

The wave function of the particle is$\psi \left(x\right)$.

$\mathbf{\psi }\left(x\right)\mathbf{=}\{\begin{array}{l}2\sqrt{a^3{\mathrm{xe}}^{-\mathrm{ax}}}X>0\\ 0X<0\end{array}$

To find the expectation value of the position of the particle.

Using formula to calculate the expectation value.

The expectation value of the position is the integral of the wave function squared, multiplied by the position, X.

$\begin{array}{rcl}\mathbf{⟨}\mathbf{x}\mathbf{⟩}& \mathbf{=}& {\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\mathbf{\Psi }\mathbf{·}\mathbf{x}\mathbf{·}\mathbf{\Psi dx}\\ & \mathbf{=}& {\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{0}}\mathbf{\Psi }\mathbf{·}\mathbf{x}\mathbf{·}\mathbf{\Psi dx}\mathbf{+}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{\Psi }\mathbf{·}\mathbf{x}\mathbf{·}\mathbf{\Psi dx}\\ & \mathbf{=}& \mathbf{0}\mathbf{+}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{\Psi }\mathbf{·}\mathbf{x}\mathbf{·}\mathbf{\Psi dx}\\ & \mathbf{=}& {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\left(2\sqrt{a^3}{\mathrm{xe}}^{-\mathrm{ax}}\right)}^{\mathbf{2}}\mathbf{xdx}\\ & \mathbf{=}& {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{x}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}\\ & \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}{\left[-\frac{x^3}{2a}-\frac{3x^2}{4a^2}-\frac{6x}{8a^3}-\frac{6}{16a^4}\right]}_{\mathbf{0}}^{\mathbf{\infty }}\\ & \mathbf{=}& \mathbf{0}\mathbf{-}\left[4a^3\left(-\frac{3}{8a^4}\right)\right]\\ & \mathbf{=}& \frac{\mathbf{3}}{\mathbf{2}\mathbf{a}}\end{array}$

Hence, the expectation value of the position of the particle is .