# 3.7: The Average Momentum of a Particle in a Box is Zero

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Now that we have mathematical expressions for the wavefunctions and energies for the particle-in-a-box, we can answer a number of interesting questions. The answers to these questions use quantum mechanics to predict some important and general properties for electrons, atoms, molecules, gases, liquids, and solids. Key to addressing these questions is the formulation and use of expectation values. This is demonstrated below and used in the context of evaluating average properties (momentum of the particle in a box for the case below).

## Expectation Values

The expectation value is the probabilistic expected value of the result (measurement) of an experiment. It is not the most probable value of a measurement; indeed the expectation value may have zero probability of occurring. The expected value (or expectation, mathematical expectation, mean, or first moment) refers to the value of a variable one would "expect" to find if one could repeat the random variable process an infinite number of times and take the average of the values obtained. More formally, the expected value is a weighted average of all possible values.

The extension of the classical expectation (average) approach in Example \(\PageIndex{1}\) using Equation \ref{Cl2} to evaluating quantum mechanical expectation values requires three small changes:

- Switch from descretized to continuous variables
- Substitute the wavefunction squared for the probability weights (i.e., the probability distribution)
- Use an operator instead of the scalar

Hence, the quantum mechanical expectation value \(\langle o \rangle\) for an observable, \(o\), associated with an operator, \(\hat{O}\), is given by

\[ \langle o \rangle = \int _{-\infty}^{+\infty} \psi^* \hat{O} \psi \, dx \label{expect}\]

where \(x\) is the range of space that is integrated over (i.e., an integration over all possible probabilities). The expectation value changes as the wavefunction changes and the operator used (i.e, which observable you are averaging over).

In general, changing the wavefunction changes the expectation value for that operator for a state defined by that wavefunction.

## Average Energy of a Particle in a Box

If we generalize this conclusion, such integrals give the average value for any physical quantity by using the operator corresponding to that physical observable in the integral in Equation \(\ref{expect}\). In the equation below, the symbol \(\left \langle H \right \rangle\) is used to denote the average value for the total energy.

\[ \begin{align} \left \langle H \right \rangle &= \int \limits ^{\infty}_{-\infty} \psi ^* (x) \hat {H} \psi (x) dx \\[4pt] &= \int \limits ^{\infty}_{-\infty} \psi ^* (x) \hat {KE} \psi (x) dx + \int \limits ^{\infty}_{-\infty} \psi ^* (x) \hat {V} \psi (x) dx \\[4pt] &= \underset{\text {average kinetic energy} }{ \int \limits ^{\infty}_{-\infty} \psi ^* (x) \left ( \frac {-\hbar ^2}{2m} \right ) \frac {\partial ^2 }{ \partial x^2} \psi (x) dx} + \underset{ \text {average potential energy} }{\int \limits ^{\infty}_{-\infty} \psi ^* (x) \hat{V} (x) \psi (x) dx} \label{3-35} \end{align}\]

The Hamiltonian operator consists of a kinetic energy term and a potential energy term. The kinetic energy operator involves differentiation of the wavefunction to the right of it. This step must be completed before multiplying by the complex conjugate of the wavefunction. The potential energy, however, usually depends only on position and not momentum (i.e., it involves conservative forces). The potential energy operator therefore only involves the coordinates of a particle and does not involve differentiation. For this reason we do not need to use a caret over \(V\) in Equation \(\ref{3-35}\).

Equation \ref{3-35} can be simplified

\[ \langle H \rangle = \langle KE \rangle + \langle V \rangle \label{3-35 braket}\]

The potential energy integral then involves only products of functions, and the order of multiplication does not affect the result, e.g. 6×4 = 4×6 = 24. This property is called the **commutative property**. The average potential energy therefore can be written as

\[ \left \langle V \right \rangle = \int \limits ^{\infty}_{-\infty} V (x) \psi ^* (x) \psi (x) dx \label{3-36}\]

This integral is telling us to take the probability that the particle is in the interval \(dx\) at \(x\), which is \(ψ^*(x)ψ(x)dx\), multiply this probability by the potential energy at \(x\), and sum (i.e., integrate) over all possible values of \(x\). This procedure is just the way to calculate the average potential energy \(\left \langle V \right \rangle\) of the particle.

What is the lowest energy for a particle in a box? The lowest energy level is \(E_1\), and it is important to recognize that this lowest energy of a particle in a box is not zero. This finite energy is called **the zero-point energy**, and the motion associated with this energy is called the zero-point motion. Any system that is restricted to some region of space is said to be bound. The zero-point energy and motion are manifestations of the wave properties and the Heisenberg Uncertainty Principle, and are general properties of bound quantum mechanical systems.

The first derivative of a function is the rate of change of the function, and the second derivative is the rate of change in the rate of change, also known as the curvature. A function with a large second derivative is changing very rapidly. Since the second derivative of the wavefunction occurs in the Hamiltonian operator that is used to calculate the energy by using the Schrödinger equation, a wavefunction that has sharper curvatures than another, i.e. larger second derivatives, should correspond to a state having a higher energy. A wavefunction with more nodes than another over the same region of space must have sharper curvatures and larger second derivatives, and therefore should correspond to a higher energy state.

## Average Position of a Particle in a Box

We can calculate the most probable position of the particle from knowledge of probability distribution, \(ψ^* ψ\). For the ground-state particle in a box wavefunction with \(n=1\) (Figure \(\PageIndex{1a}\))

\[\psi_{n=1} = \sqrt{\dfrac{2}{L}} \sin \left(\dfrac{\pi x}{L} \right) \label{PIB}\]

This state has the following probability distribution (Figure \(\PageIndex{1b}\)):

\[\psi^*_{n=1} \psi_{n=1} = \dfrac{2}{L} \sin^2 \left(\dfrac{\pi x}{L} \right)\]

The expectation value for position with the \(\hat{x} = x\) operation for any wavefunction (Equation \(\ref{expect}\)) is

\[ \langle x \rangle = \int _{-\infty}^{+\infty} \psi^* x \psi \, dx \]

which for the ground-state wavefunction (Equation \(\ref{PIB}\)) shown in Figure \(\PageIndex{1}\) is

\[ \begin{align} \langle x \rangle &= \int _{-\infty}^{+\infty} \sqrt{\dfrac{2}{L}} \sin \left(\dfrac{\pi x}{L} \right) x \sqrt{\dfrac{2}{L}} \sin \left(\dfrac{\pi x}{L} \right) \, dx \\[4pt] &= \dfrac{2}{L} \int _{-\infty}^{+\infty} x \sin^2 \left(\dfrac{\pi x}{L} \right) \label{GSExpect} \end{align}\]

## Average Momentum of a Particle in a Box

What is the average momentum of a particle in the box? We start with Equation \(\ref{expect}\) and use the momentum operator

\[\hat{p}_{x}=-i\hbar\dfrac{\partial}{\partial x}\label{3.2.3a}\]

We note that the particle-in-a-box wavefunctions are **not **eigenfunctions of the momentum operator (Exercise \(\PageIndex{4}\)). However, this does not mean that Equation \(\ref{expect}\) is inapplicable as Example \(\PageIndex{2}\) demonstrates.

It must be equally likely for the particle-in-a-box to have a momentum \(-p\) as \(+p\). The average of \(+p\) and \(–p\) is zero, yet \(p^2\) and the average of \(p^2\) are not zero. The information that the particle is equally likely to have a momentum of \(+p\) or \(–p\) is contained in the wavefunction. In fact, the sine function is a representation of the two momentum eigenfunctions \(e^{+ikx}\) and \(e^{-ikx}\) (Figure \(\PageIndex{2}\)).

The interpretation of the results of Exercise \(\PageIndex{6}\) is physically interesting. The exponential wavefunctions in the linear combination for the sine function represent the two opposite directions in which the electron can move. One exponential term represents movement to the left and the other term represents movement to the right (Figure \(\PageIndex{2}\)).

The electrons are moving, they have kinetic energy and momentum, yet the average momentum is zero.

## Orthogonality

In vector calculus, *orthogonality *is the relation of two lines at right angles to one another (i.e., perpendicularity), but is generalized into \(n\) dimensions via zero amplitude "dot products" or "inner products." Hence, orthogonality is thought of as describing **non-overlapping**, **uncorrelated**, or** independent** objects of some kind. The concept of oethogonality extends to functions (wavefunctions or otherwise) too. Two functions \(\psi_A\) and \(\psi_B\) are said to be orthogonal if

\[ \int \limits _{all space} \psi _A^* \psi _B d\tau = 0 \label{3.7.3}\]

In general, eigenfunctions of a quantum mechanical operator with different eigenvalues are orthogonal. Are the eigenfunctions of the particle-in-a-box Hamiltonian orthogonal?

## Contributors

David M. Hanson, Erica Harvey, Robert Sweeney, Theresa Julia Zielinski ("Quantum States of Atoms and Molecules")