# The Infinite Square Well
###### tags: `Quantum Physics I`
###### language: `en`
---
## Solving equation
First, write down the **time-independent Schrödinger equation**
$-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}\psi(x) + V(x)\psi(x) = E\psi(x)$.

And then we introduce the potential $V(x) = \left\{\begin{matrix}
0,&0\leq x\leq a\\
\infty,&\text{otherwise}
\end{matrix}\right.$.

The particle is in such a potential that it traverses freely _inside an infinitely well_ and the well also restricts it from escaping. For a classical analogue, one may consider a ball, under frictionless conditions and with perfect elasticity, bouncing back and forth between walls.

Such potential is "artificially" simple, since you can never find a infinitely large force in nature. But anyway, we can imagine it and treat it as a simulation of reality. This is what we always do in physics.

Within the interval $0 \leq x\leq a$, the equation becomes
$-\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}\psi(x)= E\psi(x)$
or
$\dfrac{d^2}{dx^2}\psi(x)= -k^2\psi(x)$,
where $k=-\dfrac{\sqrt{2mE}}{\hbar}$, which we introduce for simplicity.

The general solotion is
$\psi(x) = A\sin kx+B \cos kx$
or
$\psi(x) =Ce^{ikx}+De^{-ikx}$.

where $A$ and $B$ or $C$ and $D$ are arbitrary constants, which are _constrained_ by the **boundary conditions** (B.C.s) of the problem. By the way, we shall use the first form of general solution; we'll see why later.

## Applying boundary conditions
What are the B.C.s for $\psi(x)$? Generally speaking, we assume that: (See Eisberg, p.187)
* $\psi$ is continuous, single-valued, and finite;
* so is $d\psi /dx$.

But here we've met an extreme case -- the potential goes to infinity, at which the continuity of $d\psi /dx$ doesn't apply, while that of $\psi$ does.

So the continuity of $\psi$ requires that
$\psi(0) = 0$ and $\psi(a)= 0$.

Substituting both of these into the first form of general solution, we have
$\psi(0)=B=0$ and $\psi(a)=A\sin ka=0$.
The former tells us the solution comprises only sine function; the latter implies either $A=0$ or else $\sin ka = 0$.

With $A=0$, we are led to a trivial solution $\psi(x)\equiv 0$, which is non-normalizable and also means the particle doesn't even exist!

On the other hand, if $\sin ka = 0$, then $ka$ is the integral multiples of $\pi$, i.e. $ka = n\pi,\, n\in\mathbb{Z}$.

Actually we can ignore the cases with negative $n$'s, for sine is an odd function and the minus sign indside can be extracted and absorbed into the constant $A$. Also, we shall exclude the case $n=0$, which is trivial again.
Therefore, we can write
$ka = n\pi,\, n\in\mathbb{N}$.
or
$k_n=\dfrac{n\pi}{a},\,n\in\mathbb{N}$,
from which we see something noticeable -- the number $k$ is _quantized_ by the B.C.s, in other words, only certain values of $k$ are allowed. Don't forget $k$ depends on $E$. So $E$ is _quantized_ as well.
$E_n = \dfrac{k_n^2\hbar^2}{2m}=\dfrac{n^2\pi^2\hbar^2}{2ma^2}$

Mathematically speaking, we can say the problem is solved because we are able to write a set of solution as $\psi_n(x) = A\sin (n\pi/a)$, where the arbitrary constant $A$ remains to be determined; it can be any value other than zero.

## Normalisation
However, the statistical interpretation of the wave function requires that $\int^\infty_{-\infty} |\Psi(x,t)|^2 dx = 1$. This identity requires the value of $A$ being determined.

We know that $\Psi(x,t)$ is just the linear combination of $\psi_n(x)$ multiplying its temporal factor $e^{-iE_nt/\hbar}$, of which the magnitude is unity. So if we can normalise $\psi(x)$, then we can normalise $\Psi(x,t)$ as well.

Putting $\psi(x)=A\sin kx$ into $\int^\infty_{-\infty} |\psi(x)|^2 dx = 1$,
we obtain
$\displaystyle \int_0^a |A|^2 \sin^2(kx)dx = |A|^2\frac{a}{2}=1$,
and thus $|A|^2 = 2/a$. Since $A$ is complex, this result only determines the magnitude of $A$. But its phase doesn't matter in regard to physical significance, so we simply pick the positive real root $A = \sqrt{2/a}$.

Finally, the solution set is
$\psi_n(x)=\sqrt{\dfrac{2}{a}} \sin\left(\dfrac{n\pi x}{a}\right),\,n\in\mathbb{N}$,
each of which corresponds to an positive integer $n$.

For $n = 1$, the energy $E_1$ is the lowest of all, and thus $\psi_1$ is called **ground state**. For $n \geq 1$, the energies increase in proportion to $n^2$, and those $\psi$'s are called **excited states**.

## Properties
1. The functions $\psi_n$ are alternatively symmetric and antisymmetric, or equivalently speaking, even and odd, with respect to the center of the well.
2. The $n$th state $\psi_n$ has $n-1$ nodes (zero-crossings, endpoints not included) in total.
3. The functions $\psi_n$ are **orthonormal**, or we may say, $\psi_n$ form an orthonormal basis. The term "orthonormality" entails "orthogonality" and "normalisation". In terms of formula, it is
$\displaystyle\int\psi_m^*(x)\psi_n(x)dx = \delta_{mn}$,
where the asterisk denotes the complex conjugate and $\delta_{mn}:=\left\{\begin{matrix}0,&m\neq n\\ 1,&m=n\end{matrix}\right.$ is called the **Kronecker delta**.
Sine and cosine functions are well-known examples for orthonormal basis in the space of continuous functions. Check more on linear algebra textbooks.
4. The functions $\psi_n$ are **complete**, or we may say, $\psi_n$ form a complete set. Here we put the completeness of a basis not rigorously as follows: for any other continuous function $f(x)$ defined in the same interval as the member of this basis, $f(x)$ can be expressed as a linear combination of them, or expanded in terms of them.
$\displaystyle f(x)=\sum_{n=1}^\infty c_n \psi_n(x) = \sqrt{\dfrac{2}{a}}\sum_{n=1}^\infty c_n \sin \dfrac{n\pi x}{a}$

## Remarks
1. Qualitatively, the wider the well is, the lower the energies are. This is why the energy units for atomic, subatomic, and quark scale are eV, MeV, and GeV, respectively.

2. It is more sensible to choose $\sin kx$ and $\cos kx$ instead of $e^{ikx}$ and $e^{-ikx}$ as the fundamental set because we get $B=0$ immediately when we realize that cosine vanishes at $x=0$ as one of the B.C. requires. If we chose otherwise, the result would be similar, but not that straightforward. In that case, we would have $\psi(0)=C+D=0$ or $D=-C$, and that $\psi(a)=Ce^{ika}+De^{-ika} = 2C\sin(ka)=0$. The behaviour of the functions becomes explicit only when we use sine and cosine.

3. In some textbook, the infinite square well is defined by
$V(x) = \left\{\begin{matrix}
0,&-a/2\leq x\leq a/2\\
\infty,&\text{otherwise}
\end{matrix}\right.$
In this case, the B.C.s are $\psi(-a/2)=\psi(a/2)=0$. One can simply translate each $\psi_n(x)$ leftwards by $a/2$ and obtain
$\psi_n(x) = \sqrt{\dfrac{2}{a}} \sin\left[\dfrac{n\pi }{a}\left(x+\dfrac{a}{2}\right)\right],\,n\in\mathbb{N}$, which can be reduced to a combination of sine and cosine.
Or we separate the odd terms and the even terms:
$\psi_n(x) = \left\{\begin{matrix}\sqrt{\dfrac{2}{a}} \sin\left(\dfrac{n\pi x}{a}\right),&n \text{ odd}\\\sqrt{\dfrac{2}{a}} \cos\left(\dfrac{n\pi x}{a}\right),&n \text{ even} \end{matrix}\right.$
Note that the symmetry of potential (with respect to $x=0$) still corresponds to the symmetry of $\psi(x)$. Besides, their energies are, of course, the same.

## Reference
* Griffiths, David J. _Introduction to Quantum Mechanics_ (2nd ed.). Prentice Hall. 2004. ISBN 0-13-191175-9.
* Eisberg, R. & Resnick, R. _Quantum Physics of Atoms, Molecules, Solids, Nuclei, and Particles_ (2nd ed.). Wiley. 1985. ISBN 0-471-87373-X.
* 姚珩（2018）。量子力學。新北市：滄海圖書。ISBN 978-986-5647-91-9.