### Time-independent Schrodinger equation 02

** — 2. The infinite square well — **

Imagine a situation where a projectile in confined to a one dimensional movement and bounces off two infinitely rigid walls while conserving kinetic energy.

This situation can be modeled by the following potential:

$ \displaystyle f(x) = \begin{cases} 0 \quad 0\leq x \leq a\\ \infty \quad \mathrm{otherwise} \end{cases}$

Classically speaking the description is basically what we said in our initial paragraph and on what follows we'll derive the Quantum Mechanical description of the physics that result from this potential.

Outside the potential well it is $ {\psi=0}$ since the wave function cannot exist outside the well.

Inside the well the potential is $ {0}$, hence the particle is a free particle and its wave equation is

$ \displaystyle -\frac{\hbar ^2}{2m}\frac{d^2 \psi}{dx^2}=E \psi $

Since $ {E>0}$ we can define a new quantity $ {k}$

$ \displaystyle k=\frac{\sqrt{2mE}}{\hbar}$

and rewrite the wave equation in the following way

$ \displaystyle \frac{d^2 \psi}{dx^2}=-k^2\psi$

The previous equation is the equation of a harmonic oscillator whose solution in known to be of the form

$ \displaystyle \psi=A\sin kx+B\cos kx $

Were $ {A}$ and $ {B}$ are arbitrary constants whose value will be defined by normalization and boundary conditions.

The boundary conditions that have to be respected are

- $ {\psi(0)=A\sin k0+B\cos k0=0 }$
- $ {\psi(a)=A\sin ka+B\cos ka=0 }$

since one has to have continuity of the wave function and outside the potential well the wave function is vanishing.

The first condition implies $ {B=0}$ and hence the wave function simply is

$ \displaystyle \psi=A\sin kx$

For the second condition it is $ {A\sin ka=0}$. This implies that either $ {A=0}$ or $ {\sin ka=0}$. The first possibility can be discarded since with $ {A=B=0}$ one would have $ {\psi (x)=0}$ and that solution has no physical interest whatsoever. Hence one is left with $ {\sin ka=0}$.

The previous condition implies that

$ \displaystyle ka=0,\pm\pi,\pm 2\pi,\pm 3\pi,\cdots$

Since $ {k=0}$ again leads us to $ {\psi (x)=0}$ we'll discard this value.

The parity of the sine function ($ {\sin (-x)=-\sin x}$) allows one to absorb the sign of the negative solutions into the constant $ {A}$ (which remains undetermined) and we are left with

$ \displaystyle k_n=\frac{n\pi}{a} $

where $ {n}$ runs from $ {1}$ to infinity.

Since the second boundary condition determines the allowed values of $ {k}$ it also determines the allowed values of the energy of the system.

$ \displaystyle E_n=\frac{\hbar ^2 k_n^2}{2m}=\frac{n^2\pi^2\hbar^2}{2ma^2} $

Even though in a classical context a particle trapped in a infinite square well can have any value for its energy that doesn't happen in the quantum mechanical context. The allowed values for the energy arise from the fact that we imposed boundary conditions in a differential equation. Even in classical contexts one is sure to face quantization conditions as long as one imposes boundary conditions in differential equations.

The wave functions are

$ \displaystyle \psi_n(x)=A\sin\left(\frac{n\pi}{a}x\right) $

At this point one just have to find the value of $ {A}$ in order to determine the time-independent wave function.

In order to determine $ {A}$ one must normalize the wave function:

$ {\begin{aligned} 1&=\int_0^a|A|^2\sin^2(kx)dx\\ &=|A|^2\dfrac{a}{2} \end{aligned}}$

Hence

$ \displaystyle A=\sqrt{\frac{2}{a}}$

Where we have chosen the positive real root because the phase of $ {A}$ has no physical significance.

Finally one can write the wave functions that represent the particle inside of the infinite square well:

$ \displaystyle \psi_n(x)=\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right) $

Like we said previously the time-independent Schroedinger equation has an infinite set of solutions. As we can see in the expression for $ {E}$ the energy of the particle increases as $ {n}$ increases. For that reason the state $ {n=1}$ is said to be the ground state while the other values of $ {n}$ are said to be excited states.

** — 2.0.1. Properties of infinite square well solutions — **

The solutions we just found to the infinite square well have some interesting properties. As an example we'll sketch the first four solutions to the time-independent Schroedinger equation (here we set $ {a=1}$).

- The wave functions are alternatively even and odd relative to the center of square well.
- The wave functions of successive energy states have one more node than the wave function that precedes it. $ {\psi_11}$ has $ {0}$ nodes, $ {\psi_2}$ has one node, $ {psi_3}$ has two nodes, and so on and so forth.
- The wave functions are orthonormal
$ \displaystyle \int\psi_m^*(x)\psi_n(x)dx =\delta_{mn}$

- The set of wave functions is complete.
$ \displaystyle f(x)=\sum_{n=1}^{\infty}c_n\psi_n(x)=\sqrt{\frac{2}{a}}\sum_{n=1}^{\infty}c_n\sin\left(\frac{n\pi}{a}x\right) $

To evaluate the coefficients $ {c_n}$ one uses the expression $ {\int \psi_m^*(x)f(x)dx}$. The proof is

$ {\begin{aligned} \int \psi_m^*(x)f(x)dx&=\sum_{n=1}^{\infty}c_n\int \psi_m^*(x)\psi_n(x)dx\\ &=\sum_{n=1}^{\infty}c_n\delta_{mn}\\ &=c_m \end{aligned}}$

The first property is valid for all symmetric potentials. The second property is always valid. Orthonormality also is universally valid. The property of completeness is the more subtle one, but for all practical purposes we can consider that for every potential we'll encounter that the set of solutions is complete.

** — 2.1. Time-dependent solutions — **

The stationary states for the infinite square well are

For our job to be done we need to show that general solutions to the time-dependent Schroedinger equation can be written as linear combinations of stationary states.

In order to do that one must first write the general solution for $ {t=0}$

$ \displaystyle \Psi(x,0)=\sum_{n=1}^\infty c_n\psi _n(x) $

Since $ {\psi _n}$ form a complete set we know that $ {\Psi (x,0)}$ can be written in that way. Using the orthonormality condition we know that the coefficients are:

$ \displaystyle c_n=\sqrt{\frac{2}{a}}\int_0^a \sin\left(\frac{n\pi}{a}x\right)\Psi(x,0)dx$

With this we can write

$ \displaystyle \Psi(x,t)=\sum_{n=1}^\infty c_n\sqrt{\frac{2}{a}}\sin\left(\frac{n\pi}{a}x\right)e^{-in^2\pi^2\hbar/(2ma^2)t} $

which is the most general solution to the infinite square well potential.

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