A particle is confined to a one-dimensional box (an infinite well) on the x-axis between x = 0 and x = L. The potential height of the walls of the...

Tuesday, June 17, 2014

A particle is confined to a one-dimensional box (an infinite well) on the x-axis between x = 0 and x = L. The potential height of the walls of the...

The wave functions of a particle confined to an infinite potential well between x = 0 and x = L are

$\displaystyle\psi_{{n}}{\left({x}\right)}=\sqrt{{\frac{{2}}{{L}}}}{\sin{{\left(\frac{{{n}\pi{x}}}{{L}}\right)}}}$ , where n is an integer (n = 1, 2, 3...). These wave functions are normalized so that the probability of finding the particle in the well is 1 and the probability of finding the particle outside of the well is 0.

Since the particle in this problem is in the ground state, n = 1 and its wave function is

$\displaystyle\psi_{{1}}{\left({x}\right)}=\sqrt{{\frac{{2}}{{L}}}}{\sin{{\left(\pi\frac{{x}}{{L}}\right)}}}$ .

b) The probability of finding the particle between x = 0 and x = L/3 is then

$\displaystyle{P}={\int_{{0}}^{{\frac{{L}}{{3}}}}}{{\left|\psi_{{1}}\right|}}^{{2}}{\left.{d}{x}\right.}$

Let's work with the integrand first and rewrite it using a trigonometric half-angle identity:

$\displaystyle{{\left|\psi_{{1}}\right|}}^{{2}}=\frac{{2}}{{L}}{{\sin}}^{{2}}{\left(\pi\frac{{x}}{{L}}\right)}=\frac{{2}}{{L}}\cdot\frac{{1}}{{2}}\cdot{\left({1}-{\cos{{\left({2}\pi\frac{{x}}{{L}}\right)}}}\right)}=\frac{{1}}{{L}}-\frac{{1}}{{L}}{\cos{{\left({2}\pi\frac{{x}}{{L}}\right)}}}$ .

Then, the original probability integral breaks up into the two integrals. The first one is

$\displaystyle{\int_{{0}}^{{\frac{{L}}{{3}}}}}\frac{{1}}{{L}}{\left.{d}{x}\right.}=\frac{{1}}{{L}}\cdot\frac{{L}}{{3}}=\frac{{1}}{{3}}$

and the second one is

$\displaystyle{\int_{{0}}^{{\frac{{L}}{{3}}}}}\frac{{1}}{{L}}{\cos{{\left({2}\pi\frac{{x}}{{L}}\right)}}}{\left.{d}{x}\right.}=\frac{{1}}{{L}}\cdot\frac{{L}}{{{2}\pi}}{\left({\sin{{\left({2}\pi\frac{{x}}{{L}}\right)}}}{{\mid}_{{0}}^{{\frac{{L}}{{3}}}}}=\frac{{1}}{{{2}\pi}}{\sin{{\left({2}\frac{\pi}{{3}}\right)}}}=\frac{{1}}{{{2}\pi}}\cdot\frac{\sqrt{{{3}}}}{{2}}=\frac{\sqrt{{{3}}}}{{{4}\pi}}\right.}$

So the probability will be $\displaystyle{P}=\frac{{1}}{{3}}-\frac{\sqrt{{{3}}}}{{{4}\pi}}={0.196}$ , which confirms your result.

Part a seems to be less straightforward. The probability density function $\displaystyle{{\left|\psi{\left({x}\right)}\right|}}^{{2}}$ describes the probability of finding a particle at a given point. I am not sure what is meant by probability per unit length. Just dividing the total probability (1) by the length L would result in 1/L, not 2/L.

Ninenotes

Tuesday, June 17, 2014

A particle is confined to a one-dimensional box (an infinite well) on the x-axis between x = 0 and x = L. The potential height of the walls of the...

No comments:

Post a Comment

find square roots of -1+2i