Find the value of A to normalize the wave function: {A, -L `

Thursday, June 10, 2010

Find the value of A to normalize the wave function: {A, -L

Hello!

There is no given $\displaystyle{L}$ and no given $\displaystyle{A}.$ But, this is not a problem, we can write expressions with variables.

To normalize a wave function means to find the expression for $\displaystyle{A}$ (depending on $\displaystyle{L}$ ) such that the probability of finding a particle somewhere on the entire axis is exactly $\displaystyle{1}.$ It is known that a probability density function $\displaystyle{p}{d}{\left({x}\right)}$ is the square of a wave function $\displaystyle\psi{\left({x}\right)}.$

Also, by the definition of a probability density function, the probability of a particle being (at least) somewhere is $\displaystyle{\int_{{-\infty}}^{{+\infty}}}{p}{d}{\left({x}\right)}{\left.{d}{x}\right.}={\int_{{-\infty}}^{{+\infty}}}{{\left|\psi{\left({x}\right)}\right|}}^{{2}}{\left.{d}{x}\right.}.$

In our case $\displaystyle\psi{\left({x}\right)}={0}$ for $\displaystyle{x}$ outside $\displaystyle{\left[-{L},{L}\right]},$ therefore the integral is equal to

$\displaystyle{\int_{{-{L}}}^{{+{L}}}}{{\left|\psi{\left({x}\right)}\right|}}^{{2}}{\left.{d}{x}\right.}={\int_{{-{L}}}^{{+{L}}}}{{\left|{A}\right|}}^{{2}}{\left.{d}{x}\right.}={2}{L}{{A}}^{{2}},$

and this must be $\displaystyle{1},$ so $\displaystyle{{A}}^{{2}}=\frac{{1}}{{{2}{L}}},$ $\displaystyle{A}=\frac{{1}}{\sqrt{{{2}{L}}}}.$ This is the answer.

Ninenotes

Thursday, June 10, 2010

Find the value of A to normalize the wave function: {A, -L

No comments:

Post a Comment

find square roots of -1+2i