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.

Helpful background. Plane Waves and Delta Normalization to handle generalized momentum eigenstates.

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

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

A generalized momentum eigenket is represented formally by the distributional amplitude below. This is not the distributional limit of unit-$L^2$ packet amplitudes: for normalized narrowing Gaussians, the momentum probability densities tend to a delta function while the amplitudes tend to zero as distributions.

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

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

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

Here the active convention is $(\hat T(a)\psi)(x)=\psi(x-a)$: a packet initially centered at zero moves to $a$. Momentum eigenstates therefore transform only 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 $dp=\hbar\,dk$.

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 the Dirichlet kinetic-energy operator $-\hbar^2d^2/dx^2$, but not of $\hat p$. Dirichlet conditions do not make $-i\hbar d/dx$ self-adjoint; see Expectation Values in Wave Mechanics.

3. A normalized state has momentum-space wavefunction $\phi(p)$ supported only on $p\gt0$. If its first momentum moment is finite, 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

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

References

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