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Schrodinger equation (view source)
Revision as of 12:40, September 19, 2016
, 12:40, September 19, 2016Added derivation for infinite square well example and a couple of other things
===Particle in a box===
===Particle in a box===
Consider a one-dimensional box of width a, where the potential energy is 0 inside the box and infinite outside of it. This means that <math>\psi</math> must be zero outside the box. One can verify (by substituting into the Schrodinger equation) that
Consider a one-dimensional box of width a, where the potential energy is 0 inside the box and infinite outside of it. We can describe this potential by:
<math>U(x) =
\begin{cases}
\infty &\quad\text{if } x < 0\\
0 &\quad\text{if } 0 \le x \le a \\
\infty &\quad\text{if } x > a \
\end{cases}
</math>
This is also known as "the infinite square well".
This means that <math>\psi</math> must be zero outside the box. One can verify (by substituting into the Schrodinger equation) that
<math>\psi=\sin(kx)</math>
<math>\psi=\sin(kx)</math>
<math>E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}</math>
<math>E_n=\frac{\hbar^2 k^2}{2m}=\frac{\hbar^2n^2\pi^2}{2m}</math>
===Derivation===
The particle cannot exits outside the box where the potential is infinite. Hence the wavefunction, <math>\psi</math>, here must be zero outside the box. Now let us consider inside the box. The Schrodinger equation inside the potential well becomes:
<math>- \frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi</math>
Rearranging this equation give
<math>\frac{d^2 \psi}{dx^2} = - \frac{2mE}{\hbar^2} \psi</math>
This has the same form as the basic equation for simple harmonic motion. Hence we can give the solution as
<math>\psi (x) = A \cos{\omega x} + B \sin{\omega x}</math>
where <math>A</math> and <math>B</math> are unknown constants and <math>\omega</math> is equal to <math>\sqrt{\frac{2mE}{\hbar^2}}</math>. To ind the value of the the constants, we substitute in boundary conditions. An additional constraint is that wavefunctions must be continious (they cannot suddenly change value). Hence
<math>\psi (0) = 0 = A \cos{\omega 0} + B \sin{\omega 0} </math>
From this we can see that <math>A</math> must be zero. Using the continuous constraint again, we see that
<math>\psi (a) = 0 = B \sin{\omega a} </math>
The continuous constraint is only satisfied when <math>\omega a = n \pi</math> where <math>n</math> is an [[integer]]. We also note that <math>n=0</math> is not a solution as the wavefunction would be zero everywhere and so the probability of finding the particle is 0. We also discard solutions for <math>n < 0 </math> since these are not "new" solutions. As <math> \sin{-n x} =-\sin{x}</math>, these are just the negative of solutions with positive <math>n</math>. By setting <math>\omega a = n \pi</math>, we can find the energy <math>E_n</math> as:
<math>\sqrt{\frac{2m E_n}{\hbar^2}} = n \pi</math>
And so:
<math>E_n=\frac{\hbar^2n^2\pi^2}{2m}</math>
You may wonder what happens about the constant <math>B</math>. Since the Schrodinger equation is linear, <math>B</math> can take any value and the resulting wavefunction would be a solution. Since <math>|\psi (x)|^2</math> represents the probability of finding the particle at position <math>x</math> and across all space, this probability must equal 1, we can solve for <math>B</math>. We do this by integrating <math>|\psi (x)|^2</math> from <math>-\infty</math> to <math> + \infty</math>. Therefore the full solution is:
<math>U(x) =
\begin{cases}
\sqrt{\frac{2}{a}} \sin{\omega x} &\quad\text{if } x < 0\\
0 &\quad\text{otherwise} \\
\end{cases}
</math>
where <math>\omega</math> is as above.
[[Category:Physics]]
[[Category:Physics]]
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