The wave function for a particle is `psi(x)=Axe^(-x^2/a^2)` where A and a are constants. Where is the particle most likely to be found? Assume that...

Thursday, March 27, 2014

The wave function for a particle is $\displaystyle\psi{\left({x}\right)}={A}{x}{{e}}^{{-\frac{{{x}}^{{2}}}{{{a}}^{{2}}}}}$ where A and a are constants. Where is the particle most likely to be found? Assume that...

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The probability of finding a particle within some set $\displaystyle{R}$ is $\displaystyle\int_{{R}}{{\left|\psi{\left({x}\right)}\right|}}^{{2}}{\left.{d}{x}\right.}.$ The probability to find a particle at a specific point is zero, but there is a correct question: "what is the point $\displaystyle{x}_{{0}}$ such that the probability of finding the particle within a small interval with the center in $\displaystyle{x}_{{0}}$ is maximal?"

Since our $\displaystyle\psi{\left({x}\right)}$ is continuous, the integral over a small interval is almost equal to $\displaystyle\delta{x}\cdot{{\left|\psi{\left({x}_{{0}}\right)}\right|}}^{{2}}.$ So, we have to find the point(s) $\displaystyle{x}_{{0}}$ where $\displaystyle{{\left|\psi{\left({x}_{{0}}\right)}\right|}}^{{2}}$ has its maximum. This is the same as the $\displaystyle{\left|\psi{\left({x}_{{0}}\right)}\right|}$ maximum.

The factor $\displaystyle{\left|{A}\right|}$ has no effect on $\displaystyle{x}_{{0}},$ thus it is sufficient to find the maximum of $\displaystyle{f{{\left({x}\right)}}}={x}{{e}}^{{-\frac{{{x}}^{{2}}}{{{a}}^{{2}}}}}$ for $\displaystyle{x}\geq{0}$ (for $\displaystyle{x}\lt{0}$ the values are the same). At $\displaystyle{x}={0}$ the value is zero, at $\displaystyle+\infty$ the limit is also zero, so the maximum is somewhere in between. The necessary condition is $\displaystyle{f{'}}{\left({x}_{{0}}\right)}={0},$ so the equation is:

$\displaystyle{f{'}}{\left({x}\right)}={\left({x}{{e}}^{{-\frac{{{x}}^{{2}}}{{{a}}^{{2}}}}}\right)}'={{e}}^{{-\frac{{{x}}^{{2}}}{{{a}}^{{2}}}}}-{x}\cdot\frac{{{2}{x}}}{{{a}}^{{2}}}{{e}}^{{-\frac{{{x}}^{{2}}}{{{a}}^{{2}}}}}={{e}}^{{-\frac{{{x}}^{{2}}}{{{a}}^{{2}}}}}{\left({1}-\frac{{{2}{{x}}^{{2}}}}{{{a}}^{{2}}}\right)}={0}.$

The only such $\displaystyle{x}_{{0}}=\frac{{\left|{a}\right|}}{\sqrt{{{2}}}}$ (so there are two points of a maximum, $\displaystyle\frac{{\left|{a}\right|}}{\sqrt{{{2}}}}$ and $\displaystyle-\frac{{\left|{a}\right|}}{\sqrt{{{2}}}}$ ). Numerically for $\displaystyle{a}={2.49}{n}{m}$ $\displaystyle{x}_{{0}}\approx\pm{1.76}{n}{m}.$