Momentum Eigenstates

A momentum eigenstate is a state with a sharp value of momentum. In one-dimensional coordinate representation, the momentum operator is

$$\hat p = -i\hbar\frac{d}{dx},$$

so the eigenvalue equation is

$$\hat p\,\psi_p(x) = p\,\psi_p(x).$$

Solving this differential equation gives a plane wave. On the full real line, that plane wave is not square-normalizable, so exact momentum eigenstates are generalized states. Physical states with finite norm are wave packets built from a distribution of momenta.

Solving The Eigenvalue Equation

The eigenvalue equation reads

$$-i\hbar\frac{d\psi_p}{dx} = p\,\psi_p.$$

For real $p$, the solution is

$$\psi_p(x)=C e^{ipx/\hbar}.$$

The constant $C$ is fixed by the chosen continuum normalization. With delta normalization in momentum,

$$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}.$$

These functions satisfy

$$\hat p\,\langle x\vert p\rangle = p\,\langle x\vert p\rangle$$

in the distributional sense.

Why The Eigenvalue Is Real

The momentum operator is an observable only after a suitable self-adjoint domain is specified. On the full line, with wavefunctions that decay appropriately or are treated distributionally, the formal integration-by-parts identity is

$$\int \phi^*(-i\hbar\psi')\,dx = \int (-i\hbar\phi')^*\psi\,dx$$

when the boundary term vanishes. Self-adjointness is what makes momentum measurement outcomes real.

On finite intervals, boundary conditions matter. Periodic boundary conditions admit momentum eigenstates. Hard-wall boundary conditions do not: sine standing waves are energy eigenstates, but they are not eigenstates of $\hat p$.

Periodic Box Momentum States

In a periodic box of length $L$,

$$\psi(x+L)=\psi(x).$$

For $\psi(x)=e^{ikx}$, periodicity gives

$$e^{ikL}=1,$$

so

$$k_n=\frac{2\pi n}{L}, \qquad p_n=\hbar k_n=\frac{2\pi\hbar n}{L}, \qquad n\in\mathbb Z.$$

The normalized momentum eigenfunctions are

$$\psi_n(x)=\frac{1}{\sqrt L}e^{ik_nx}.$$

They satisfy

$$\int_0^L \psi_m^*(x)\psi_n(x)\,dx = \delta_{mn}.$$

The periodic box is a useful regulator for the continuum, but it represents different boundary conditions from an infinite square well with hard walls.

The finite-volume construction is developed in Periodic Boundary Conditions.

Momentum Space Wavefunction

For a normalizable state $\lvert\psi\rangle$, the momentum-space wavefunction is

$$\phi(p)=\langle p\vert\psi\rangle.$$

Using the plane-wave kernel,

$$\phi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-ipx/\hbar}\psi(x)\,dx.$$

The inverse relation is

$$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{ipx/\hbar}\phi(p)\,dp.$$

Thus a momentum eigenstate is the limiting case in which the momentum-space amplitude is a delta distribution:

$$\phi_{p_0}(p)=\delta(p-p_0).$$

Substituting this into the inverse transform gives the plane wave with momentum $p_0$.

Momentum Probability Density

For a normalized wave packet,

$$\int_{-\infty}^{\infty} \lvert\phi(p)\rvert^2\,dp =1.$$

The probability of measuring momentum in an interval $\Delta$ is

$$\mathbb P(p\in\Delta) = \int_\Delta \lvert\phi(p)\rvert^2\,dp.$$

The expectation value of momentum can be computed in momentum space as

$$\langle p\rangle = \int_{-\infty}^{\infty} p\,\lvert\phi(p)\rvert^2\,dp.$$

In position space the same quantity is

$$\langle p\rangle = \int_{-\infty}^{\infty} \psi^*(x) \left( -i\hbar\frac{d}{dx} \right) \psi(x)\,dx.$$

These two formulas agree when the Fourier transform is used with consistent domains and boundary behavior.

Momentum Versus Energy

For a free particle,

$$\hat H=\frac{\hat p^2}{2m}.$$

Every momentum eigenstate is therefore also a free-particle energy eigenstate:

$$\hat H\lvert p\rangle = \frac{p^2}{2m}\lvert p\rangle.$$

But energy does not determine momentum uniquely in one dimension. The two momentum eigenstates $\lvert p\rangle$ and $\lvert -p\rangle$ have the same positive energy:

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

This is why free-particle energy eigenstates are often discussed as right-moving and left-moving components. A real standing wave can have definite energy while not having definite momentum.

Translation Interpretation

Momentum generates spatial translations. A finite translation by $a$ is represented by

$$\hat T(a) = \exp\left(-\frac{i}{\hbar}a\hat p\right).$$

Acting on a momentum eigenstate gives

$$\hat T(a)\lvert p\rangle = e^{-iap/\hbar}\lvert p\rangle,$$

up to the active/passive sign convention used for translations. The important point is that momentum eigenstates transform by phases under translations. This is the symmetry reason plane waves appear whenever the Hamiltonian is translation invariant.

Practical Use

Momentum eigenstates are useful when:

• the Hamiltonian is a function of $\hat p$, as for a free particle;
• the system is translation invariant;
• asymptotic scattering states are approximately free;
• a wave packet is specified by its momentum spread;
• a potential is easier to treat through momentum transfer.

They are less convenient for hard-wall boundary conditions and sharply localized potentials, where position-space matching or numerical methods may be simpler.

Common Mistakes

• Treating $\lvert p\rangle$ as a normalizable Hilbert-space vector.
• Forgetting that hard-wall box eigenstates are not momentum eigenstates.
• Confusing a state with definite energy and a state with definite momentum.
• Dropping the $\hbar$ in $e^{ipx/\hbar}$.
• Treating $\lvert\phi(p)\rvert^2$ as a probability rather than a probability density.
• Mixing the $p$ label with the wavenumber $k$ without the Jacobian $dp=\hbar\,dk$.

Where This Is Used

• Plane Waves and Delta Normalization explains the continuum normalization of $\lvert p\rangle$.
• Free Particle uses momentum eigenstates to diagonalize the Hamiltonian.
• Periodic Boundary Conditions explains why a finite periodic interval supports normalized momentum eigenstates.
• Momentum-Space Representation gives the abstract representation viewpoint.
• Momentum Representation gives the transform and operator-action machinery.
• Gaussian Wave Packets constructs localized states from momentum amplitudes.
• Reflection and Transmission Coefficients uses incoming and outgoing momentum components.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.

Exercises

1. Solve the momentum eigenvalue equation in position space.

Solution

Start from

$$-i\hbar\frac{d\psi}{dx}=p\psi.$$

Then

$$\frac{d\psi}{dx} = \frac{ip}{\hbar}\psi.$$

Integrating gives

$$\psi(x)=C e^{ipx/\hbar}.$$

The constant $C$ is fixed by the normalization convention, not by square normalization on the full line.

2. Why is $\sin(n\pi x/L)$ not a momentum eigenfunction on $0\lt x\lt L$?

Solution

Applying $\hat p=-i\hbar d/dx$ gives

$$\hat p\sin\frac{n\pi x}{L} = -i\hbar\frac{n\pi}{L} \cos\frac{n\pi x}{L}.$$

The result is proportional to a cosine, not to the original sine. The sine is an eigenfunction of $\hat p^2$ with hard-wall boundary conditions, but not an eigenfunction of $\hat p$.

3. A normalized state has momentum-space wavefunction $\phi(p)$ supported only on $p\gt0$. What can you say about $\langle p\rangle$?

Solution

The expectation value is

$$\langle p\rangle = \int_0^\infty p\,\lvert\phi(p)\rvert^2\,dp.$$

Since $p\gt0$ on the support and $\lvert\phi(p)\rvert^2$ is nonnegative, $\langle p\rangle$ is positive unless the state is zero almost everywhere.

4. Show that a periodic-box momentum eigenstate is normalized.

Solution

For

$$\psi_n(x)=\frac{1}{\sqrt L}e^{ik_nx},$$

one has

$$\int_0^L \lvert\psi_n(x)\rvert^2\,dx = \int_0^L \frac{1}{L}\,dx =1.$$