Free Particle

A free particle is a nonrelativistic particle with no potential energy. In one dimension its Hamiltonian is

$$\hat H=\frac{\hat p^2}{2m} =-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}.$$

The free particle is the canonical model for continuous spectra, momentum eigenstates, plane waves, wave packets, group velocity, and the local behavior of particles far from interactions.

What “Free” Specifies

On the full line, the model consists of the Hilbert space $L^2(\mathbb R)$ and the self-adjoint Hamiltonian

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

on an appropriate twice-differentiable domain. Saying only that $V(x)=0$ does not specify every global problem. The same differential expression on a ring, a half-line, or a finite interval has different boundary conditions and therefore a different spectrum.

A spatially constant potential $V_0$ is dynamically equivalent to the free problem up to an energy offset:

$$\hat H=\frac{\hat p^2}{2m}+V_0.$$

It leaves the spatial eigenfunctions and velocities unchanged and multiplies every evolving state by the common phase $e^{-iV_0t/\hbar}$. Potential differences, not an isolated additive constant, affect nongravitational wave mechanics.

Stationary Equation

The time-independent Schrödinger equation is

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} =E\psi.$$

For $E\gt0$, define

$$k=\frac{\sqrt{2mE}}{\hbar}.$$

Then the general stationary solution is

$$\psi(x)=Ae^{ikx}+Be^{-ikx}.$$

The two terms represent right-moving and left-moving momentum eigenstates. The corresponding momenta are

$$p=\hbar k, \qquad p=-\hbar k.$$

The energy-momentum relation is

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

The other energy cases explain the lower edge of the spectrum. At $E=0$,

$$\psi(x)=A+Bx,$$

and no nonzero solution is square-integrable on the full line. For $E\lt0$, writing $E=-\hbar^2\kappa^2/(2m)$ gives

$$\psi(x)=A e^{\kappa x}+B e^{-\kappa x}.$$

Decay at $+\infty$ requires $A=0$, whereas decay at $-\infty$ requires $B=0$. There is therefore no nonzero negative-energy bound state.

The operator gives the same conclusion without solving the equation. For a state in its domain,

$$\begin{aligned} \langle\psi|\hat H|\psi\rangle &=-\frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \psi^*\psi''\,dx\\ &=\frac{\hbar^2}{2m} \int_{-\infty}^{\infty} \lvert\psi'\rvert^2\,dx \ge0, \end{aligned}$$

where the boundary term vanishes on the domain. Thus the full-line free Hamiltonian is a nonnegative operator. Its spectrum is $[0,\infty)$ and is continuous; it has no square-normalizable energy eigenvectors.

Plane Waves

A right-moving plane wave has the time-dependent form

$$\psi_k(x,t)=A e^{i(kx-\omega t)}, \qquad \omega=\frac{\hbar k^2}{2m}.$$

It is an eigenfunction of momentum:

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

It is also an energy eigenfunction:

$$\hat H\psi_k =\frac{\hbar^2k^2}{2m}\psi_k.$$

Plane waves are useful because they diagonalize both $\hat p$ and the free Hamiltonian. Their normalization conventions are treated carefully in Plane Waves and Delta Normalization. They are not ordinary normalizable states on the full real line, because $\lvert\psi_k\rvert^2$ is constant and

$$\int_{-\infty}^{\infty}\lvert\psi_k(x,t)\rvert^2\,dx$$

diverges for nonzero amplitude.

Continuous Spectrum

The free particle on the full line has a continuous spectrum. The momentum label $p$ can take any real value, and the energy satisfies

$$E\ge 0.$$

Because $E=p^2/(2m)$, the same positive energy corresponds to two momenta, $p$ and $-p$, in one dimension. This degeneracy records the two possible directions of motion. For the formal distinction between discrete and continuous spectra, see Discrete and Continuous Spectra.

In the momentum representation, the spectral resolution is especially simple:

$$\hat H = \int_{-\infty}^{\infty}dp\, \frac{p^2}{2m} |p\rangle\langle p|.$$

If states are labeled by energy instead, a direction label must accompany every $E\gt0$. With $p_E=\sqrt{2mE}$, one may use $|E,+\rangle$ and $|E,-\rangle$ for the $+p_E$ and $-p_E$ branches. The Jacobian between momentum and energy normalization is discussed in Normalization Conventions.

Physical localized free-particle states are wave packets, built by superposing plane waves:

$$\psi(x,t) =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} dk\, a(k)e^{i(kx-\omega(k)t)}.$$

The function $a(k)$ controls the momentum distribution. A sharply localized packet requires a broad range of $k$ values.

With the displayed Fourier convention, a normalized packet obeys

$$\int_{-\infty}^{\infty}\lvert a(k)\rvert^2\,dk=1.$$

Free evolution changes only the phase of each momentum component:

$$a(k,t) =a(k,0) \exp\left(-i\frac{\hbar k^2}{2m}t\right).$$

Consequently, $\lvert a(k,t)\rvert^2$ and every moment of momentum are constant in time. Position-space interference among components still changes because their phases advance at different rates. This dispersive dephasing moves and generally spreads the packet even though its momentum distribution is fixed.

No normalizable free-particle state is exactly stationary. Energy eigenfunctions are generalized plane waves, while a normalizable packet necessarily contains a range of energies and changes shape or position under evolution.

Box Normalization

A common regulator is to place the particle in a box of length $L$ with periodic boundary conditions:

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

Then

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

and normalized plane waves are

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

The box discretizes momenta. After computing physical quantities, one often takes $L\to\infty$ and replaces sums by integrals:

$$\sum_n \longrightarrow \frac{L}{2\pi}\int dk.$$

This method is especially useful for density-of-states calculations and numerical approximations.

This “box” is a periodic cell, not an infinite square well. Hard walls select standing sine waves and different allowed wave numbers. Periodic boundaries preserve translation invariance on the circle and retain traveling-wave momentum eigenstates.

The finite-volume basis and sum-to-integral conversion are explained in Periodic Boundary Conditions.

Delta Normalization

On the full line, momentum eigenstates are often delta normalized:

$$\langle p\vert p'\rangle=\delta(p-p').$$

In position representation,

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

This convention is natural for Fourier transforms and continuum completeness:

$$\int_{-\infty}^{\infty} dp\, \lvert p\rangle\langle p\rvert=\hat I.$$

Delta-normalized plane waves are not physical localized particles by themselves. They are basis states used to construct normalizable packets.

Geometry Changes the Spectrum

The phrase “free particle” identifies the interior Hamiltonian, but global geometry and endpoint domains complete the model:

Configuration space	Typical condition	Spectrum and modes
Full line	square-integrable packets; generalized plane-wave basis	continuous $E\ge0$
Ring of circumference $L$	periodic endpoints	discrete traveling waves $k_n=2\pi n/L$
Finite interval	Dirichlet hard walls	discrete standing waves
Half-line	self-adjoint condition at the origin	continuous sector; some Robin domains also admit a boundary-bound state

Thus a particle can have $V=0$ throughout its allowed region and nevertheless possess a discrete spectrum. The quantization then comes from topology or boundary conditions, not from a force in the interior. Boundary Conditions develops this distinction.

Free Solutions Inside Constant-Potential Regions

Many scattering problems are assembled from regions where $V(x)=V_j$ is constant. The stationary equation there is

$$-\frac{\hbar^2}{2m}\psi'' =(E-V_j)\psi.$$

If $E\gt V_j$, the local solutions are traveling or standing waves with

$$k_j=\frac{\sqrt{2m(E-V_j)}}{\hbar}.$$

If $E\lt V_j$, the local wave number is imaginary and the solutions are exponential, with

$$\kappa_j=\frac{\sqrt{2m(V_j-E)}}{\hbar}.$$

The additive constant is unobservable only when it shifts the potential everywhere. Differences among regions change the available kinetic energy and therefore the wavelength, current, reflection, and tunneling behavior. The Potential Step is the first complete matching example.

Probability Current

For a one-dimensional wavefunction, the probability current is

$$j =\frac{\hbar}{2mi} \left(\psi^*\frac{d\psi}{dx} -\psi\frac{d\psi^*}{dx}\right).$$

For $\psi=Ae^{ikx}$,

$$j=\frac{\hbar k}{m}\lvert A\rvert^2.$$

Thus $e^{ikx}$ with $k\gt0$ carries probability to the right, while $e^{-ikx}$ carries probability to the left. This current interpretation is essential for scattering, where reflection and transmission are current ratios.

For a same-energy superposition

$$\psi(x,t) = \left(Ae^{ikx}+Be^{-ikx}\right)e^{-iEt/\hbar},$$

the cross terms cancel in the current, giving

$$j =\frac{\hbar k}{m} \left(\lvert A\rvert^2-\lvert B\rvert^2\right).$$

Equal right- and left-moving intensities form a standing wave with zero net current, even though its probability density is spatially modulated. Conversely, the constant modulus of one exact plane wave should not be interpreted as a normalizable uniform probability distribution on the infinite line; it is a property of a generalized eigenfunction.

Symmetries and Conserved Quantities

The full-line Hamiltonian is translation invariant:

$$[\hat H,\hat p]=0.$$

Momentum is therefore conserved. The Hamiltonian also commutes with parity, which sends $p$ to $-p$. This symmetry explains why the two propagation directions have equal energy. One may combine them into parity-even and parity-odd generalized eigenfunctions proportional to $\cos(kx)$ and $\sin(kx)$, respectively.

For a spinless particle with no fields, time reversal acts by complex conjugation in position space. It changes $e^{ikx}$ into $e^{-ikx}$ while leaving the energy unchanged. Translation, parity, and time-reversal statements depend on the full domain: a hard wall, for example, breaks continuous translation symmetry even where the interior potential vanishes.

Classical Correspondence

The dispersion relation is

$$\omega(k)=\frac{\hbar k^2}{2m}.$$

The group velocity of a wave packet centered at $k_0$ is

$$v_g=\left.\frac{d\omega}{dk}\right\rvert_{k_0} =\frac{\hbar k_0}{m} =\frac{p_0}{m}.$$

This matches the classical velocity. The phase velocity is

$$v_{\text{ph}}=\frac{\omega}{k} =\frac{\hbar k}{2m} =\frac{p}{2m},$$

which is not the particle velocity. The group velocity carries the packet envelope.

The free-particle Ehrenfest relations are exact, not merely narrow-packet approximations. In the Heisenberg picture,

$$\frac{d\hat p_H}{dt}=0, \qquad \frac{d\hat x_H}{dt}=\frac{\hat p_H}{m},$$

so

$$\hat p_H(t)=\hat p, \qquad \hat x_H(t)=\hat x+\frac{t}{m}\hat p.$$

For any state with the required moments,

$$\langle x\rangle_t =\langle x\rangle_0 +\frac{t}{m}\langle p\rangle_0.$$

The position variance obeys

$$\begin{aligned} (\Delta x)^2_t &=(\Delta x)^2_0 +\frac{t}{m} \left\langle \Delta x\,\Delta p +\Delta p\,\Delta x \right\rangle_0\\ &\quad +\frac{t^2}{m^2}(\Delta p)^2_0. \end{aligned}$$

An initially position-momentum-correlated packet can contract for a while, but the positive quadratic term dominates at large times whenever $\Delta p\ne0$. A square-normalizable packet of finite position width cannot have $\Delta p=0$, so asymptotic spreading is unavoidable. Wave Packet Spreading develops this result, while Gaussian Wave Packets gives the standard analytic example.

The Fourier integral is one representation of the full initial-value solution. The same unitary evolution can be written with a position-space kernel. Free-Particle Propagator: First Encounter states that kernel and explains its composition law; the general propagator theory remains in the Dynamics and Formulations volume.

Common Mistakes

• Treating a plane wave as a normalizable state on the full line.
• Claiming that a normalization constant can make an exact full-line plane wave square-integrable.
• Forgetting that positive energy in one dimension corresponds to left-moving and right-moving momentum states.
• Identifying the sign of energy with the direction of motion; both momentum signs have the same positive energy.
• Confusing phase velocity with particle velocity.
• Ignoring the normalization convention when comparing plane-wave amplitudes.
• Thinking a free particle must be perfectly delocalized; localized free particles are wave packets.
• Calling a normalizable packet an energy eigenstate.
• Forgetting that wave packets spread because $\omega(k)$ is nonlinear.
• Assuming every free packet broadens immediately; a correlated packet can initially contract.
• Concluding that $V=0$ guarantees a continuous spectrum without specifying the configuration space and boundary conditions.
• Treating a constant potential in one region as a globally irrelevant energy shift.

Where This Is Used

• Coordinate Representation explains how plane waves represent momentum eigenstates.
• Free Particle in Three Dimensions generalizes plane waves to momentum vectors and energy shells.
• Plane Waves and Delta Normalization explains continuum, box, and wave-packet normalization for plane waves.
• Periodic Boundary Conditions gives the finite-box momentum spectrum and density-of-states limit.
• Momentum Eigenstates connects the differential momentum operator to plane waves and momentum-space probabilities.
• Normalization Conventions compares box and delta normalization.
• Fourier Transform gives the transform machinery.
• Gaussian Wave Packets constructs localized free-particle states.
• Group Velocity and Phase Velocity develops the two velocities and their regimes of interpretation.
• Free-Particle Propagator: First Encounter gives the position-space evolution kernel without replacing the general propagator treatment.
• Potential Step and barrier pages use free-particle waves in each constant-potential region.

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.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• J. R. Taylor, Scattering Theory: The Quantum Theory of Nonrelativistic Collisions, Dover, 2006.

Exercises

1. Verify that $e^{ikx}$ is an eigenfunction of both $\hat p$ and $\hat H=\hat p^2/(2m)$, and find the eigenvalues.

Solution

For momentum,

$$\hat p e^{ikx} =-i\hbar\frac{d}{dx}e^{ikx} =\hbar k e^{ikx}.$$

For energy,

$$\hat H e^{ikx} =-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}e^{ikx} =\frac{\hbar^2k^2}{2m}e^{ikx}.$$

2. A periodic box has length $L$. Show that periodicity implies $k_n=2\pi n/L$, and derive the one-dimensional continuum replacement for $\sum_n f(k_n)$.

Solution

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

$$e^{ik(x+L)}=e^{ikx}.$$

Thus $e^{ikL}=1$, so $kL=2\pi n$ for $n\in\mathbb Z$. Therefore $k_n=2\pi n/L$. Adjacent values have spacing $\Delta k=2\pi/L$. When $f$ varies slowly on that spacing,

$$\sum_n f(k_n) \approx \frac{1}{\Delta k}\int_{-\infty}^{\infty}f(k)\,dk =\frac{L}{2\pi} \int_{-\infty}^{\infty}f(k)\,dk.$$

3. Use positivity of the free Hamiltonian to show that it has no nonzero square-integrable eigenstate with $E\le0$ on the full line.

Solution

For an eigenstate in the Hamiltonian domain, integration by parts gives

$$E\langle\psi|\psi\rangle =\langle\psi|\hat H|\psi\rangle =\frac{\hbar^2}{2m} \int_{-\infty}^{\infty}\lvert\psi'(x)\rvert^2\,dx \ge0.$$

Therefore a nonzero eigenstate cannot have $E\lt0$. If $E=0$, the integral of $\lvert\psi'\rvert^2$ must vanish, so $\psi$ is constant almost everywhere. The only constant function in $L^2(\mathbb R)$ is the zero function. Hence there is no nonzero square-integrable eigenstate at zero energy either.

4. For

$$\psi(x,t) = \left(Ae^{ikx}+Be^{-ikx}\right)e^{-iEt/\hbar},$$

calculate the probability current. Under what condition is this state a standing wave with zero net current?

Solution

Insert the wavefunction into

$$j=\frac{\hbar}{2mi} \left(\psi^*\psi'-\psi\psi'^*\right).$$

Terms proportional to $A^*B e^{-2ikx}$ and its conjugate cancel. The remaining terms give

$$j=\frac{\hbar k}{m} \left(\lvert A\rvert^2-\lvert B\rvert^2\right).$$

The net current vanishes when $\lvert A\rvert=\lvert B\rvert$. The relative phase controls where the standing-wave nodes lie but does not change the zero-current condition.

5. Let

$$C_0 = \left\langle \Delta x\,\Delta p +\Delta p\,\Delta x \right\rangle_0.$$

Use free Heisenberg evolution to find the time at which the position variance is smallest when $C_0\lt0$. Find that minimum variance and explain why it cannot be negative.

Solution

The exact variance is

$$(\Delta x)^2_t =(\Delta x)^2_0 +\frac{C_0}{m}t +\frac{(\Delta p)^2_0}{m^2}t^2.$$

Differentiating with respect to time gives

$$t_{\min} =-\frac{mC_0}{2(\Delta p)^2_0},$$

which is positive when $C_0\lt0$. Substitution yields

$$(\Delta x)^2_{\min} =(\Delta x)^2_0 -\frac{C_0^2}{4(\Delta p)^2_0}.$$

The Schrödinger–Robertson uncertainty relation implies

$$(\Delta x)^2_0(\Delta p)^2_0 -\frac{C_0^2}{4} \ge\frac{\hbar^2}{4},$$

so

$$(\Delta x)^2_{\min} \ge \frac{\hbar^2}{4(\Delta p)^2_0} \gt0.$$