The normalized solution to the Schrodinger equation for a particular potential is `psi` = 0 for x 0....

Sunday, October 7, 2012

The normalized solution to the Schrodinger equation for a particular potential is $\displaystyle\psi$ = 0 for x 0....

Hello!

The probability of being between some c and d is $\displaystyle{\int_{{c}}^{{d}}}{{\left|\psi{\left({x}\right)}\right|}}^{{2}}{\left.{d}{x}\right.}.$ Therefore the probability in question is

$\displaystyle{\int_{{{a}-{0.027}{a}}}^{{{a}+{0.027}{a}}}}{{\left|\psi{\left({x}\right)}\right|}}^{{2}}{\left.{d}{x}\right.}={\int_{{{a}-{0.027}{a}}}^{{{a}+{0.027}{a}}}}\frac{{4}}{{{a}}^{{3}}}{{x}}^{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}{\left.{d}{x}\right.}$

( $\displaystyle{a}$ must be positive, so bounds of integration are also positive).

To compute the indefinite integral of $\displaystyle{{x}}^{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}$ we can use integration by parts twice: differentiate $\displaystyle{{x}}^{{2}}$ and then $\displaystyle{x}$ and integrate the exponent. Let's perform this:

$\displaystyle\int{{x}}^{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}{\left.{d}{x}\right.}={\left|{u}={{x}}^{{2}},{d}{v}={{e}}^{{-\frac{{{2}{x}}}{{a}}}}{\left.{d}{x}\right.},{d}{u}={2}{x}{\left.{d}{x}\right.},{v}=-\frac{{a}}{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}\right|}=$

$\displaystyle=-\frac{{a}}{{2}}{{x}}^{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}+\frac{{a}}{{2}}\int{\left({2}{x}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}\right)}{\left.{d}{x}\right.}.$

Then $\displaystyle{u}={x},{d}{v}={{e}}^{{-\frac{{{2}{x}}}{{a}}}}{\left.{d}{x}\right.},{d}{u}={\left.{d}{x}\right.},{v}=-\frac{{a}}{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}},$ and the remaining integral is equal to

$\displaystyle-\frac{{a}}{{2}}{x}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}+\frac{{a}}{{2}}\int{{e}}^{{-\frac{{{2}{x}}}{{a}}}}{\left.{d}{x}\right.}=-{x}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}-\frac{{a}}{{2}}\cdot\frac{{a}}{{2}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}.$

So the total indefinite integral is equal to $\displaystyle-\frac{{a}}{{4}}{{e}}^{{-\frac{{{2}{x}}}{{a}}}}{\left({{a}}^{{2}}+{2}{a}{x}+{2}{{x}}^{{2}}\right)}+{C},$ and the probability is

$\displaystyle-\frac{{4}}{{{a}}^{{3}}}\cdot\frac{{a}}{{4}}{\left({{e}}^{{-{2}{\left({1}+{0.027}\right)}}}{\left({{a}}^{{2}}+{2}{{a}}^{{2}}{\left({1}+{0.027}\right)}+{2}{{a}}^{{2}}{{\left({1}+{0.027}\right)}}^{{2}}\right)}-\right.}$

$\displaystyle-{{e}}^{{-{2}{\left({1}-{0.027}\right)}}}{\left({{a}}^{{2}}+{2}{{a}}^{{2}}{\left({1}-{0.027}\right)}+{2}{{a}}^{{2}}{{\left({1}-{0.027}\right)}}^{{2}}\right)}{\)}{\)}.$

$\displaystyle{a}$ vanishes and remains

$\displaystyle{{e}}^{{-{2}{\left({1}-{0.027}\right)}}}{\left({1}+{2}{\left({1}-{0.027}\right)}+{2}{{\left({1}-{0.027}\right)}}^{{2}}\right)}-$

$\displaystyle-{{e}}^{{-{2}{\left({1}+{0.027}\right)}}}{\left({1}+{2}{\left({1}+{0.027}\right)}+{2}{{\left({1}+{0.027}\right)}}^{{2}}\right)}\approx{0.0292}.$

This is the answer.

Ninenotes

Sunday, October 7, 2012

The normalized solution to the Schrodinger equation for a particular potential is $\displaystyle\psi$ = 0 for x 0....

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find square roots of -1+2i

Sunday, October 7, 2012

The normalized solution to the Schrodinger equation for a particular potential is = 0 for x 0....

No comments:

Post a Comment

find square roots of -1+2i

The normalized solution to the Schrodinger equation for a particular potential is $\displaystyle\psi$ = 0 for x 0....