# 8: Topical Overview of PIB and Postulates QM (Lecture)

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Lecture 7

Conceptually, applying operator \(\hat{A}\) to its *eigenfunction *such as \(\psi (x)\) gives an *eigenvalue *\(a\) multiplied by the same eigenfunction. Mathematically, this looks like

\[\hat{A}\psi (x) = a\psi (x) \nonumber\]

Example of operators:

Wavefunctions have a probabilistic interpretation. More specifically, the wavefunction squared (or to be more exact, the \(\Psi^*\Psi\) is a probability density). To convert to a probability, we have to integrate \(\psi^*\psi\) over an interval. We can never discuss the probability at one point, except qualitatively (e.g., if we are referring to a node where the probability is zero). The probabilistic interpretation comes with key features of \(\psi^*\psi\) including finite, nonnegative and not infinite. A key aspect of this is that the wavefunctions must be **normalized**.

We introduced the particle-in-the-box model, which is arguably the most important "trapped" particle model. This system is relatively "easy" to solve the **Schrödinger Equation **to get oscillatory wavefunctions with corresponding eigenenergies. Other models are harder to solve as we will see.

## The Particle in a one Dimensional box (a Trapped Particle)

The *particle in the box *model system is the simplest non-trivial application of the Schrödinger equation, but one which illustrates many of the fundamental concepts of quantum mechanics. For a particle moving in one dimension (again along the x- axis), the Schrödinger equation can be written

\[-\dfrac{\hbar^2}{2m}\psi {}''(x)+ V (x)\psi (x) = E \psi (x)\label{3.5.1}\]

Assume that the particle can move freely between two endpoints \(x = 0\) and \(x = L\), but cannot penetrate past either end. This is equivalent to a potential energy dependent on *x* with

\[V(x)=\begin{cases}

0 & 0\leq x\leq L \\

\infty & x< 0 \; \text{and}\; x> L \end{cases}\label{3.5.2}\]

This potential is represented by the dark lines in Figure \(\ref{3.5.1}\). The infinite potential energy constitutes an impenetrable barrier since the particle would have an infinite potential energy if found there, which is clearly impossible.

The particle in a one-dimensional box is not a real system. As seen before with vibrating strings in the classical wave function, \( X(x) = A\sin(kx)\psi (x)\). In this system, solving the Schrödinger equation (Equation \ref{3.5.1}) results in this solution:

\[ \color{red} {\psi (x) = N\sin \left(\dfrac{n\pi x}{L}\right)} \label{1a}\]

for

\[0 \leq x \leq L \label{1b}\]

with \(N\) is an arbitrary Normalization constant.

Since a particle cannot have infinite energy, it cannot exist outside the box (Equation \ref{3.5.2}) and furthermore, two Boundary conditions are therefore expected:

\[\begin{align} \psi (x=0) = 0 \label{3.5.3A1} \\[4pt] \psi (x=L) = 0 \label{3.5.3A2} \end{align}\]

This results in the quantization in allowed levels. For the one dimensional box, the eigenvalue is

\[ \color{red} {E_n = \dfrac{h^2 n^2 }{8m_e L^2}} \label{3.5.11}\]

These are the only values of the energy which allows solutions of the Schrödinger Equation \(\ref{3.5.1}\) consistent with the boundary conditions in Equations \ref{3.5.3A1} and \ref{3.5.3A2}. The integer \(n\) in Equation \ref{1a} is called a *quantum number*, is appended as a subscript on \(E\) to label the allowed energy levels. Negative values of \(n\) add nothing new because the energies in Equation \ref{3.5.11} depends on \(n^2\).

Transition from one Eigenstate to another

Like the hydrogen atom, you can figure out the spectrum for transitions from \(n_1\) to \(n_2\) by the standard spectroscopy equation (i.e., conservation of energy)

\[\Delta E = E_{n_2} - E_{n_1} = hc/\lambda \nonumber\]

## Normalize the Wavefunction

To normalize the wave function \(\psi (x)\) it is important to recognize that \(\int_{-\infty}^{\infty} \psi_n^* \psi_n dx\) is the probability of finding a particle in all space, which is 1, or 100%. The integral can be broken up into three parts, defined by the system's boundary conditions:

\[\int_{-\infty}^{0} \psi_n^* \psi_n dx+ \int_{0}^{L} \psi_n^* \psi_n dx + \int_{L}^{\infty} \psi_n^* \psi_n dx = 1 \label{PIB2}\]

since \(\psi (x) = 0\) when \(x<0\) and \(x>a\), the first and last integral of the summation of integrals are zero. That leaves \(N^2\int_{0}^{L} \psi_n^* \psi_n dx\ = 1\), where \(N\) is the normalization constant. Thus,

\[1=N^2 \int_0^L {\sin}^2(n\pi x/L)dx = N^2(L/2) \label{PIB3}.\]

Solving for \(N\) = \(\sqrt{2/L}\). Going back to \(\psi (x) = B\sin(\frac {n\pi x}{a})\), the new equation becomes

\[ \color{red} {\psi (x) = \sqrt{\frac{2}{L}}\sin(\dfrac{n\pi x}{L})} \label{PIB4}\]

Java simulation of particles in boxes :https://phet.colorado.edu/en/simulation/bound-states