# 5.3: Linear Combinations of Eigenfunctions

- Page ID
- 4500

It is not necessary that an electron be described by an eigenfunction of the Hamiltonian operator. Many problems encountered by quantum chemists and computational chemists lead to wavefunctions that are not eigenfunctions of the Hamiltonian operator. Science is like that; interesting problems are not simple to solve. They require adaptation of current techniques, creative energy, and a good set of skills developed by studying solutions to previously solved interesting problems.

Consider a free electron in one dimension that is described by the wavefunction

\[ \psi (x) = C_1\psi _1 (x) + C_2 \psi _2 (x) \label {5-21}\]

with

\[ \begin{align} \psi _1(x) &= \left ( \dfrac {1}{2L} \right )^{1/2} e^{ik_1x} \label {5-22} \\[4pt] \psi _2(x) &= \left ( \dfrac {1}{2L} \right )^{1/2} e^{ik_2x} \label {5-23} \end{align} \]

where \(k_1\) and \(k_2\) have different magnitudes. Although such a function is not an eigenfunction of the momentum operator or the Hamiltonian operator, we can calculate the average momentum and average energy of an electron in this state from the expectation value integral. (Note: "in-this-state" means "described-by-this-wavefunction".)

The function shown in Equation \(\ref{5-21}\) belongs to a class of functions known as **superposition functions**, which are linear combinations of eigenfunctions. A linear combination of functions is a sum of functions, each multiplied by a weighting coefficient, which is a constant. The adjective linear is used because the coefficients are constants. The constants, e.g. \(C_1\) and \(C_2\) in Equation \(\ref{5-21}\), give the weight of each component (\(\psi_1\) and \(\psi_2\)) in the total wavefunction. Notice from the discussion previously that each component in Equation \(\ref{5-21}\) is an eigenfunction of the momentum operator and the Hamiltonian operator although the linear combination function (i.e., \(\psi(x)\)) is **not**.

The expectation value, i.e. average value, of the momentum operator is found as follows. First, write the integral for the expectation value and then substitute into this integral the superposition function and its complex conjugate as shown below. Since we are considering a free particle in one dimension, the limits on the integration are \(–L\) and \(+L\) with \(L\) going to infinity.

\[ \begin{align} \left \langle p \right \rangle &= \int \psi ^* (x) \left ( -i\hbar \dfrac {d}{dx} \right ) \psi (x) dx \\[4pt]&= \dfrac {-i\hbar}{2L} \int \limits _{-L}^{+L} \left ( C_1^* e^{-ik_1x} + C_2^* e^{-ik_2x} \right ) \dfrac {d}{dx} \left ( C_1 e^{ik_1x} + C_2 e^{ik_2x} \right ) dx \\[4pt] &= \dfrac {-i\hbar}{2L} \int \limits _{-L}^{+L} \left ( C_1^* e^{-ik_1x} + C_2^* e^{-ik_2x} \right ) \left ( (ik_1)C_1 e^{ik_1x} + (ik_2)C_2 e^{ik_2x} \right ) dx \label {5-24} \end{align} \]

Cross-multiplying the two factors in parentheses yields four terms.

\[\left \langle p \right \rangle = I_1 + I_2 + I_3 + I_4 \nonumber \]

with

\[ \begin{align} I_1 &= \dfrac {\hbar k_1}{2L} C^*_1 C_1 \int \limits ^{+L} _{-L} dx = C^*_1 C_1 \hbar k_1 \\[4pt] I_2 &= \dfrac {\hbar k_2}{2L} C^*_2 C_2 \int \limits ^{+L} _{-L} dx = C^*_2 C_2 \hbar k_2 \\[4pt] I_3 &= \dfrac {\hbar k_1}{2L} C^*_1 C_2 \int \limits ^{+L} _{-L} e^{i(k_2 - k_1)x} dx \\[4pt] I_4 &= \dfrac {\hbar k_1}{2L} C^*_2 C_1 \int \limits ^{+L} _{-L} e^{i(k_1 - k_2)x} dx \label {5-25} \end{align} \]

An integral of two different functions, e.g. \(\int \psi _1^* \psi _2 dx\), is called an **overlap integral** or **orthogonality integral**. When such an integral equals zero, the functions are said to be orthogonal. The integrals in \(I_3\) and \(I_4\) are zero because the functions \(\psi_1\) and \(\psi_2\) are orthogonal. We know \(\psi_1\) and \(\psi_2\) are orthogonal because of the Orthogonality Theorem, described previously, that states that eigenfunctions of any Hermitian operator, such as the momentum operator or the Hamiltonian operator, with different eigenvalues, which is the case here, are orthogonal. Also, by using Euler's formula and following Example \(\PageIndex{1}\) below, you can see why these integrals are zero.

## Contributors

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