What is the probability that the particle would be found between x = 0and x = 1/a?

To find the probability amplitude for the particle to be found in the up state, we take the inner product for the up state and the down state. Square the amplitude. The probability is the modulus squared. Remember that the modulus squared means to multiply the amplitude with its complex conjugate.

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 probability that the particle would be in between x=0 and x = 1/a.

Formula Used:

Probability between x = 0 and x = x/a

$\mathbf{P}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}\mathbf{/}\mathbf{a}}{\mathbf{\psi }}^{\mathbf{*}}\mathbf{\psi dx}$

Substitute $2\sqrt{a^3}x{e}^{-ax}$for ${\psi }^{*}$and $\psi$in $\mathrm{P}={\int }_0^{1/\mathrm{a}}{\mathrm{\psi }}^{*}\mathrm{\psi dx}$

$\begin{array}{rcl}\mathbf{P}& \mathbf{=}& {\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}\mathbf{/}\mathbf{a}}{\left(2\sqrt{a^3}{\mathrm{xe}}^{-\mathrm{ax}}\right)}^{\mathbf{2}}\mathbf{dx}\\ & \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{1}\mathbf{/}\mathbf{a}}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}\end{array}$

Using integral by parts for${\int }_0^{\frac{1}{\mathrm{a}}}{\mathrm{x}}^2{\mathrm{e}}^{-2\mathrm{ax}}\mathrm{dx}$, where$\mathrm{u}={\mathrm{e}}^{-2\mathrm{ax}}$and${\mathrm{v}}^{\text{'}}={\mathrm{x}}^2$we have:

$\begin{array}{rcl}\mathbf{P}& \mathbf{=}& {\left[\frac{1}{3}{e}^{-2\mathrm{ax}}{x}^3-\int -\frac{2{\mathrm{ae}}^{-2\mathrm{ax}}{x}^3}{3}\mathrm{dx}\right]}_{\mathbf{0}}^{\frac{\mathbf{1}}{\mathbf{a}}}\\ & \mathbf{=}& {\left[\frac{1}{3}{e}^{-2\mathrm{ax}}{x}^3-\left(\frac{1}{3}{e}^{-2\mathrm{ax}}{x}^3+\frac{e^{-2\mathrm{ax}}\left(2a^2{x}^2+3\mathrm{ax}+1\right)}{4a^3}\right)\right]}_{\mathbf{0}}^{\frac{\mathbf{1}}{\mathbf{a}}}\\ & \mathbf{=}& {\left[-\frac{e^{-2\mathrm{ax}}\left(2a^2{x}^2+2\mathrm{ax}+1\right)}{4a^3}\right]}_{\mathbf{0}}^{\frac{\mathbf{1}}{\mathbf{a}}}\\ & \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{2}}\mathbf{-}\mathbf{5}}{\mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{2}}}\end{array}$

Further solving the main integral, we have:

$\begin{array}{rcl}\mathbf{P}& \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}{\mathbf{\int }}_{\mathbf{0}}^{\frac{\mathbf{1}}{\mathbf{a}}}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ax}}\mathbf{dx}\\ & \mathbf{=}& \mathbf{4}{\mathbf{a}}^{\mathbf{3}}\left(\frac{e^2-5}{4a^3{e}^2}\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{323}\end{array}$

The required probability of the particle is 0.323.