Eigenfunctions and Eigenvalues of Basic Quantum Operators

Welcome back. In the previous lesson, you used operators to calculate expectation values such as $⟨x⟩$ and $⟨p_x⟩$. An expectation value is an average over many measurements. This lesson addresses the sharper question: when does a state guarantee one particular measurement result?

The answer is given by the eigenvalue equation. You will learn to apply basic operators, decide whether a function is an eigenfunction, and identify the corresponding eigenvalue. This distinction is fundamental for interpreting momentum, position, and—soon—energy states of electrons.

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The eigenvalue equation

An operator is a mathematical instruction that acts on a function. For example, the momentum operator differentiates a wavefunction, while the position operator multiplies it by position.

A function $u(x)$ is an eigenfunction of an operator $\hat{A}$ if applying the operator returns the same function, multiplied only by a constant:

$$\boxed{\hat{A}u(x)=au(x)}$$

In this equation:

• $\hat{A}$ is the operator.
• $u(x)$ is an eigenfunction of that particular operator.
• $a$ is the eigenvalue, a constant numerical result associated with that eigenfunction.

The decisive word is constant. The factor $a$ may depend on a label such as $n$ or $k$, but it cannot depend on position $x$.

For a quantum observable, the physical interpretation is powerful:

If a system is in a normalized eigenfunction of an observable’s operator, a measurement of that observable gives the associated eigenvalue with certainty.

For instance, if

$${\hat{p}}_x\psi =p_0\psi ,$$

then the state has definite momentum $p_0$ in the $x$-direction.

Quantum observables are represented by Hermitian operators, with suitable boundary conditions. One essential consequence is that their eigenvalues—the possible outcomes of physical measurements—are real numbers.

A crucial qualification follows from the definition: “eigenfunction” is not a permanent property of a function by itself. It is always relative to an operator. The same wavefunction can be an eigenfunction of energy but not of momentum, as you will see below.

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The basic operators in one dimension

In the position representation, several operators will appear repeatedly throughout quantum chemistry.

Observable	Operator	What it does to $f(x)$
Position	$\hat{x}=x$	$\hat{x}f(x)=xf(x)$
Momentum	${\hat{p}}_x=-i\hslash \frac{d}{dx}$	Differentiates $f(x)$
Kinetic energy	$\hat{T}=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$	Takes a second derivative
Total energy	$\hat{H}=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}+V(x)$	Adds kinetic and potential contributions

The Hamiltonian $\hat{H}$ is the total-energy operator. Its eigenvalue equation is the time-independent Schrödinger equation:

$$\boxed{\hat{H}\psi (x)=E\psi (x)}$$

Here, $E$ is an allowed energy and $\psi (x)$ is an energy eigenfunction. Such states are called stationary states. You will construct this equation explicitly in the next module.

For this lesson, focus on the testing process:

1. Apply the operator to the candidate function.
2. Simplify the result.
3. Compare it with the original function.
4. If the result is a constant times that unchanged function, identify the constant as the eigenvalue.
5. If the shape changes, the function is not an eigenfunction of that operator.

Multiplying an eigenfunction by a nonzero normalization constant or an overall constant phase does not change its eigenvalue. For example, if $u(x)$ is an eigenfunction, so is $Nu(x)$, provided $N\ne 0$.

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Momentum eigenfunctions: complex traveling waves

Consider the complex exponential

$${\psi }_k(x)=N{e}^{ikx},$$

where $k$ is a real constant with units of inverse length. Apply the momentum operator:

$${\hat{p}}_x{\psi }_k(x)=-i\hslash \frac{d}{dx}\left(N{e}^{ikx}\right).$$

Since

$$\frac{d}{dx}{e}^{ikx}=ik{e}^{ikx},$$

we obtain

$${\hat{p}}_x{\psi }_k(x)=-i\hslash \left(ikN{e}^{ikx}\right)=\hslash k\left(N{e}^{ikx}\right).$$

Therefore,

$$\boxed{{\hat{p}}_x{\psi }_k=(\hslash k){\psi }_k}$$

The original function is unchanged except for multiplication by the constant $\hslash k$. Thus:

$$\boxed{{\psi }_k(x)=N{e}^{ikx}\text{ is a momentum eigenfunction, with }p=\hslash k.}$$

Similarly,

$${\psi }_{-k}(x)=N{e}^{-ikx}$$

satisfies

$${\hat{p}}_x{\psi }_{-k}=-\hslash k{\psi }_{-k}.$$

It has the same spatial wavelength but a momentum eigenvalue of opposite sign.

There is one mathematical subtlety worth keeping in view. On the entire infinite line, a plane wave $e^{ikx}$ cannot be normalized in the usual way because its magnitude is constant everywhere. It is an idealized, generalized eigenfunction. Physically realistic wave packets can be normalizable while being built from many such momentum eigenfunctions. For the present purpose, the plane wave clearly reveals the momentum-eigenfunction relationship.

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An energy eigenfunction need not have definite momentum

Now consider a real standing wave:

$$\psi (x)=N\cos (kx).$$

Apply the momentum operator:

$${\hat{p}}_x\psi (x)=-i\hslash \frac{d}{dx}\left[N\cos (kx)\right].$$

Because

$$\frac{d}{dx}\cos (kx)=-k\sin (kx),$$

we find

$${\hat{p}}_x\psi (x)=i\hslash kN\sin (kx).$$

This is not a constant times $N\cos (kx)$:

$$i\hslash kN\sin (kx)\ne pN\cos (kx).$$

So,

$$\boxed{\cos (kx)\text{ is not a momentum eigenfunction.}}$$

The expression $i\hslash k$ is not the momentum eigenvalue. It is merely part of the result of applying the operator. An eigenvalue can only be identified when the entire result has the required form: a constant multiplying the original function.

Now apply the kinetic-energy operator:

$$\hat{T}=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}.$$

Since

$$\frac{d^2}{d{x}^2}\cos (kx)=-k^2\cos (kx),$$

we get

$$\hat{T}\psi (x)=-\frac{{\hslash }^2}{2m}\left[-k^{2}N\cos (kx)\right].$$

Thus,

$$\boxed{\hat{T}\psi (x)=\frac{{\hslash }^2{k}^2}{2m}\psi (x)}$$

The standing wave is therefore a kinetic-energy eigenfunction, with eigenvalue

$$\boxed{T=\frac{{\hslash }^2{k}^2}{2m}.}$$

For a free particle, where $V(x)=0$, the Hamiltonian equals the kinetic-energy operator:

$$\hat{H}=\hat{T}.$$

Therefore $\cos (kx)$ is also an energy eigenfunction of a free particle.

This result becomes clearer if you use Euler’s relation:

$$\cos (kx)=\frac{1}{2}\left(e^{ikx}+e^{-ikx}\right).$$

The cosine is a combination of a right-moving momentum eigenfunction with momentum $+\hslash k$ and a left-moving momentum eigenfunction with momentum $-\hslash k$. Both momentum values give the same kinetic energy because kinetic energy depends on momentum squared:

$$T=\frac{p^2}{2m}.$$

The two opposite momentum states therefore share one energy eigenvalue.

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Position eigenfunctions and a continuous spectrum

The position operator is exceptionally simple:

$$\hat{x}f(x)=xf(x).$$

Suppose we try a typical extended function such as $e^{ikx}$:

$$\hat{x}{e}^{ikx}=x{e}^{ikx}.$$

The extra factor $x$ is not a constant, so $e^{ikx}$ is not a position eigenfunction.

An exact position eigenstate centered at $x_0$ is represented by the Dirac delta distribution:

$${\psi }_{x_0}(x)=\delta (x-x_0).$$

It satisfies

$$\hat{x}\delta (x-x_0)=x\delta (x-x_0)=x_0\delta (x-x_0).$$

Therefore,

$$\boxed{\hat{x}\delta (x-x_0)=x_0\delta (x-x_0).}$$

The eigenvalue is $x_0$: a position measurement on this ideal state returns $x_0$ with certainty.

As with plane waves, the delta function is an idealized generalized eigenfunction rather than an ordinary normalizable wavefunction. It nevertheless captures an important physical distinction:

• Position can take a continuum of possible values $x_0$.
• Momentum can also take a continuum of values $p=\hslash k$.
• A confined system can have discrete allowed energies because its boundary conditions restrict the possible energy eigenfunctions.

That last point will become concrete when you solve the infinite-well Schrödinger equation.

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Eigenstates, expectation values, and a common misconception

The preceding lesson found that a real box-state wavefunction can have

$$⟨p_x⟩=0.$$

That does not mean the box state is a momentum eigenfunction with eigenvalue zero.

For a momentum eigenstate with zero momentum, the equation would have to be

$${\hat{p}}_x\psi =0.$$

But the box sine state does not satisfy this: differentiating it produces a cosine, not zero times the original sine function. It has no single definite momentum.

The distinction is:

Statement	Meaning
$⟨p_x⟩=0$	Momentum measurements average to zero over many identically prepared systems.
${\hat{p}}_x\psi =0$	Every ideal momentum measurement returns exactly zero.

An eigenstate always has an expectation value equal to its eigenvalue. If $u_a$ is normalized and satisfies

$$\hat{A}{u}_a=a{u}_a,$$

then

$$⟨A⟩=\int u_a^{\ast }(x)\hat{A}{u}_a(x)dx=\int u_a^{\ast }(x)a{u}_a(x)dx.$$

Since $a$ is constant and the state is normalized,

$$⟨A⟩=a\int |u_a(x){|}^{2}dx=a.$$

So,

$$\boxed{\text{Definite observable value}⟹⟨A⟩=a.}$$

The reverse is not generally true. A state may have an expectation value of zero while individual measurements yield a spread of positive and negative values.

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A reliable identification checklist

When you encounter a proposed eigenvalue problem, use this compact routine:

1. Write the operator correctly.
   For example, use ${\hat{p}}_x=-i\hslash d/dx$, not merely $d/dx$.

2. Apply the operator to the entire candidate function.
   Differentiate carefully, including constants and complex phases.

3. Compare the output with the original function.
   The function must retain exactly the same shape.

4. Check whether the multiplier is constant.
   A factor such as $2k$ is allowed when $k$ is a constant. A factor such as $x$, $\sin (kx)$, or $V(x)$ is not generally an eigenvalue.

5. State the conclusion in operator-specific language.
   Say “$\psi$ is a momentum eigenfunction with eigenvalue $p$,” not simply “$\psi$ has an eigenvalue.”

6. Check the units.
   Momentum eigenvalues have units of momentum, energy eigenvalues have units of energy, and position eigenvalues have units of length.

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Key takeaways

An eigenfunction $u(x)$ of an operator $\hat{A}$ satisfies

$$\boxed{\hat{A}u(x)=au(x),}$$

where $a$ is a position-independent constant called the eigenvalue.

For quantum observables, an eigenstate gives a definite measurement result. The eigenvalue is that result.

The basic examples are:

$${\hat{p}}_x{e}^{ikx}=\hslash k{e}^{ikx},$$

so $e^{ikx}$ is a momentum eigenfunction with momentum $\hslash k$, while

$${\hat{p}}_x\cos (kx)=i\hslash k\sin (kx),$$

so $\cos (kx)$ is not a momentum eigenfunction. Yet for a free particle,

$$\hat{H}\cos (kx)=\frac{{\hslash }^2{k}^2}{2m}\cos (kx),$$

so the same cosine function is an energy eigenfunction.

Most importantly, eigenfunction status depends on the operator. A wavefunction can have definite energy without definite momentum, and an expectation value does not by itself establish a definite measured value.

Next, you will use the Hamiltonian more directly by writing the time-independent Schrödinger equation for a particle in a specified potential.