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# Archive for May 31st, 2010

## Infinite square well wavefunction.

Posted by peeterjoot on May 31, 2010

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

Work problem 4.1 from [1], calculation of the eigensolution for an infinite square well, with boundaries $[-a/2, a/2]$. It’s actually a bit tidier seeming to generalize this slightly to boundaries $[a,b]$, which also implicitly solves the problem. This is surely a problem that is done in 700 other QM texts, but I liked the way I did it this time so am writing it down.

# Guts

Our equation to solve is $i \hbar \Psi_t = -(\hbar^2/2m) \Psi_{xx}$. Separation of variables $\Psi = T \phi$ gives us

\begin{aligned}T &\propto e^{-i E t/\hbar } \\ \phi'' &= -\frac{2 m E }{\hbar^2} \phi\end{aligned} \hspace{\stretch{1}}(2.1)

With $k^2 = 2 m E/\hbar^2$, we have

\begin{aligned}\phi = A e^{i k x } + B e^{-i k x},\end{aligned} \hspace{\stretch{1}}(2.3)

and the usual $\phi(a) = \phi(b) = 0$ boundary conditions give us

\begin{aligned}0 = \begin{bmatrix}e^{i k a } & e^{-i k a} \\ e^{i k b } & e^{-i k b}\end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.4)

We must have a zero determinant, which gives us the constraints on $k$ immediately

\begin{aligned}0 &= e^{i k (a - b)} - e^{i k (b-a)} \\ &= 2 i \sin( k (a - b) ).\end{aligned}

So our constraint on $k$ in terms of integers $n$, and the corresponding integration constant $E$

\begin{aligned}k &= \frac{n \pi}{b - a} \\ E &= \frac{\hbar^2 n^2 \pi^2 }{2 m (b-a)^2}.\end{aligned} \hspace{\stretch{1}}(2.5)

One of the constants $A,B$ can be eliminated directly by picking any one of the two zeros from 2.4

\begin{aligned}&A e ^{i k a } + B e^{-i k a} = 0 \\ &\implies \\ &B = -A e ^{2 i k a } \end{aligned}

So we have

\begin{aligned}\phi = A \left( e^{i k x } - e^{ ik (2a - x) } \right).\end{aligned} \hspace{\stretch{1}}(2.7)

Or,

\begin{aligned}\phi = 2 A i e^{i k a} \sin( k (x-a )) \end{aligned} \hspace{\stretch{1}}(2.8)

Because probability densities, currents and the expectations of any operators will always have paired $\phi$ and $\phi^{*}$ factors, any constant phase factors like $i e^{i k a}$ above can be dropped, or absorbed into the constant $A$, and we can write

\begin{aligned}\phi = 2 A \sin( k (x-a )) \end{aligned} \hspace{\stretch{1}}(2.9)

The only thing left is to fix $A$ by integrating ${\left\lvert{\phi}\right\rvert}^2$, for which we have

\begin{aligned}1 &= \int_a^b \phi \phi^{*} dx \\ &= A^2 \int_a^b dx \left( e^{i k x } - e^{ ik (2a - x) } \right) \left( e^{-i k x } - e^{ -ik (2a - x) } \right) \\ &= A^2 \int_a^b dx \left( 2 - e^{ik(2a - 2x)} - e^{ik(-2a + 2x)} \right) \\ &= 2 A^2 \int_a^b dx \left( 1 - \cos (2 k (a - x)) \right)\end{aligned}

This last trig term vanishes over the integration region and we are left with

\begin{aligned}A = \frac{1}{{ \sqrt{2 (b-a)}}},\end{aligned} \hspace{\stretch{1}}(2.10)

which essentially completes the problem. A final substitution back into 2.8 allows for a final tidy up

\begin{aligned}\phi = \sqrt{\frac{2}{b-a}} \sin( k (x-a )).\end{aligned} \hspace{\stretch{1}}(2.11)

# References

[1] R. Liboff. Introductory quantum mechanics. Cambridge: Addison-Wesley Press, Inc, 2003.