Free Particle Hamiltonian

The free-particle Hamiltonian describes nonrelativistic motion without position-dependent potential energy or external fields. Its compact expression is

$$\hat H=\frac{\hat{\mathbf p}^{\,2}}{2m}, \qquad m>0.$$

This formula fixes the local kinetic energy, but not the complete operator. The configuration space, Hilbert space, domain, and boundary conditions determine whether the momentum is continuous or discrete, which symmetries survive, and whether the spectrum contains only scattering states.

The canonical solution, including wave packets, currents, and normalization, lives at Free Particle. This page is the compact Hamiltonian card.

Quick Reference

Property	Standard full-space realization
Degrees of freedom	One spinless particle in $d$ Euclidean dimensions
Hilbert space	$L^2(\mathbb R^d,d^dx)$
Hamiltonian	$\hat H=\hat{\mathbf p}^{\,2}/(2m)$
Position representation	$\hat H=-\hbar^2\nabla^2/(2m)$
Momentum representation	$(\hat H\widetilde\psi)(\mathbf p)=\mathbf p^2\widetilde\psi(\mathbf p)/(2m)$
Natural domain	Sobolev space $H^2(\mathbb R^d)$
Spectrum	$[0,\infty)$, purely continuous
Generalized eigenstates	Momentum plane waves $\lvert\mathbf p\rangle$
Dispersion relation	$E(\mathbf p)=\mathbf p^2/(2m)$
Solvability	Exactly diagonalized by the Fourier transform
Main symmetries	Translations, rotations, parity, and spinless time reversal

If an uncoupled spin degree of freedom is retained, the Hamiltonian is $\hat{\mathbf p}^{\,2}/(2m)\otimes I_{\mathrm{spin}}$. Spin then supplies a degeneracy but does not alter the spatial evolution.

Definition and Parameters

In one Cartesian dimension,

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

In $d$ Cartesian dimensions,

$$\hat H =-\frac{\hbar^2}{2m}\nabla^2, \qquad \nabla^2=\sum_{i=1}^{d}\frac{\partial^2}{\partial x_i^2}.$$

Symbol	Meaning	SI units	Constraint
$m$	Particle mass	$\mathrm{kg}$	$m>0$ in this nonrelativistic model
$\hat{\mathbf p}$	Canonical momentum	$\mathrm{kg\,m\,s^{-1}}$	Self-adjoint realization depends on geometry
$\hbar$	Reduced Planck constant	$\mathrm{J\,s}$	Exact SI defining constant through $h=2\pi\hbar$
$\nabla^2$	Cartesian Laplacian	$\mathrm{m^{-2}}$	Domain and boundary data are required

The displayed Hamiltonian assumes:

• a nonrelativistic dispersion relation;
• no position-dependent scalar potential;
• no electromagnetic vector potential;
• no spin-dependent coupling;
• a fixed background geometry;
• no interactions with other particles or an environment.

A spatially constant offset $E_0$ gives

$$\hat H'=\frac{\hat{\mathbf p}^{\,2}}{2m}+E_0I.$$

For one isolated fixed sector, it changes evolution only by the global phase $e^{-iE_0t/\hbar}$. It can still matter in thermodynamics, comparisons between sectors, and couplings for which the relative energy convention is physical.

Domain and Self-Adjointness

On $L^2(\mathbb R^d)$, the standard free Hamiltonian is the self-adjoint operator

$$\hat H=-\frac{\hbar^2}{2m}\nabla^2, \qquad \mathcal D(\hat H)=H^2(\mathbb R^d).$$

In momentum space this is multiplication by $\mathbf p^2/(2m)$, with domain

$$\mathcal D(\hat H) = \left\lbrace \widetilde\psi\in L^2(\mathbb R^d): \mathbf p^2\widetilde\psi\in L^2(\mathbb R^d) \right\rbrace.$$

The Fourier transform makes self-adjointness and nonnegativity transparent. For every state in the domain,

$$\langle\psi\vert\hat H\vert\psi\rangle = \int_{\mathbb R^d} \frac{\mathbf p^2}{2m} \lvert\widetilde\psi(\mathbf p)\rvert^2\,d^dp \geq0.$$

On an interval, half-line, ring, torus, or region with a boundary, the differential expression $-\hbar^2\nabla^2/(2m)$ does not by itself define a self-adjoint operator. Endpoint or boundary conditions are part of the Hamiltonian. The Boundary Conditions page develops that point.

Representations

Position representation

With $\hat{\mathbf p}=-i\hbar\nabla$,

$$(\hat H\psi)(\mathbf x) =-\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf x).$$

The time-dependent Schrödinger equation is

$$i\hbar\frac{\partial\psi}{\partial t} =-\frac{\hbar^2}{2m}\nabla^2\psi.$$

Momentum representation

The Hamiltonian is diagonal:

$$(\hat H\widetilde\psi)(\mathbf p) =\frac{\mathbf p^2}{2m}\widetilde\psi(\mathbf p).$$

Consequently, exact evolution multiplies each momentum component by a phase,

$$\widetilde\psi(\mathbf p,t) = \exp\left( -\frac{i\mathbf p^2(t-t_0)}{2m\hbar} \right) \widetilde\psi(\mathbf p,t_0).$$

The momentum probability density is constant in time. Position-space spreading comes from the momentum-dependent phase, not from any change in $\lvert\widetilde\psi(\mathbf p)\rvert^2$.

Generalized momentum basis

With the convention

$$\langle\mathbf x\vert\mathbf p\rangle =\frac{1}{(2\pi\hbar)^{d/2}} \exp\left(\frac{i\mathbf p\cdot\mathbf x}{\hbar}\right),$$

the generalized eigenstates satisfy

$$\langle\mathbf p\vert\mathbf p'\rangle =\delta^{(d)}(\mathbf p-\mathbf p'), \qquad \hat H\lvert\mathbf p\rangle =\frac{\mathbf p^2}{2m}\lvert\mathbf p\rangle.$$

They provide the spectral resolution

$$\hat H = \int_{\mathbb R^d} d^dp\, \frac{\mathbf p^2}{2m} \lvert\mathbf p\rangle\langle\mathbf p\rvert.$$

These plane waves are generalized eigenstates, not square-normalizable vectors in $L^2(\mathbb R^d)$. See Plane Waves and Delta Normalization for the rigged-Hilbert-space interpretation and normalization conventions.

Spectrum and Degeneracy

For the standard full-space realization,

$$\sigma(\hat H)=[0,\infty).$$

The spectrum is continuous and the Hamiltonian has no normalizable energy eigenvectors. In one dimension, every $E>0$ corresponds to two momentum branches,

$$p_\pm=\pm\sqrt{2mE}.$$

In $d>1$, fixed positive energy corresponds to the momentum-space sphere

$$\mathbf p^2=2mE.$$

The degeneracy is therefore associated with propagation direction. In three dimensions it is naturally resolved by momentum direction or by angular-momentum partial waves.

There are no negative-energy states in the standard full-space realization because $\hat H\geq0$. The lower spectral edge $E=0$ is not a normalizable eigenstate: a constant position-space wavefunction is not in $L^2(\mathbb R^d)$.

Geometry and Boundary Conditions

The word free describes the absence of a bulk potential, not necessarily an unbounded configuration space.

Configuration space	Typical domain data	Consequence
$\mathbb R^d$	Sobolev domain $H^2(\mathbb R^d)$	Continuous spectrum $[0,\infty)$
Rectangular torus	Periodic boundary conditions	Discrete momentum lattice and traveling waves
Finite interval or bounded region	Dirichlet hard walls	Discrete standing-wave spectrum
Half-line	Dirichlet, Neumann, or Robin condition at the endpoint	Continuous spectrum; some Robin choices also support a boundary state
Ring with flux or twist	Quasiperiodic endpoint condition	Shifted discrete momenta

For a rectangular periodic cell with side lengths $L_i$,

$$p_i=\frac{2\pi\hbar n_i}{L_i}, \qquad n_i\in\mathbb Z,$$

and

$$E_{\mathbf n} =\sum_{i=1}^{d} \frac{2\pi^2\hbar^2n_i^2}{mL_i^2}.$$

This finite-volume regulator should not be confused with hard-wall confinement. Periodic boundaries preserve translations on a torus, whereas hard walls break continuous translation symmetry and select standing waves.

Symmetries and Conserved Quantities

On full Euclidean space, the Hamiltonian is a function only of $\hat{\mathbf p}^{\,2}$. Therefore

$$[\hat H,\hat p_i]=0, \qquad [\hat H,\hat L_i]=0.$$

Momentum and orbital angular momentum are conserved, although the components of angular momentum do not commute with one another. The model is also invariant under parity and, for a spinless particle with no fields, under time reversal.

The Heisenberg equations are exact:

$$\frac{d\hat{\mathbf p}_H}{dt}=0, \qquad \frac{d\hat{\mathbf x}_H}{dt} =\frac{\hat{\mathbf p}_H}{m}.$$

Hence

$$\hat{\mathbf p}_H(t)=\hat{\mathbf p}, \qquad \hat{\mathbf x}_H(t) =\hat{\mathbf x}+\frac{t-t_0}{m}\hat{\mathbf p}.$$

Boundaries can remove these symmetries even when the interior differential expression remains unchanged. For example, a hard-wall interval is not invariant under continuous translations.

Physical Interpretation

The dispersion relation

$$E(\mathbf p)=\frac{\mathbf p^2}{2m}$$

has group velocity

$$\mathbf v_g =\nabla_{\mathbf p}E =\frac{\mathbf p}{m}.$$

This is the classical particle velocity. A one-dimensional plane wave has phase velocity $v_{\mathrm{ph}}=p/(2m)$, which is not the packet velocity.

A normalizable free particle need not be delocalized. Localized states are wave packets built from a range of momenta. Because the dispersion is quadratic, different momentum components accumulate phases at different rates, and a generic packet spreads. The exact Gaussian example is developed at Gaussian Wave Packets.

Validity and Nearby Hamiltonians

Change in physics	Replacement
Position-dependent scalar potential	$\hat H=\hat{\mathbf p}^{\,2}/(2m)+V(\hat{\mathbf x},t)$
Electromagnetic field	$\hat H=(\hat{\mathbf p}-q\mathbf A)^2/(2m)+q\Phi$
Spin magnetic coupling	Use the Pauli Hamiltonian
Relativistic kinematics without spin	$E(\mathbf p)=\sqrt{\mathbf p^2c^2+m^2c^4}$ requires a different framework
Relativistic spin-$1/2$ particle	Use the Dirac Hamiltonian with its field-theory caveats
Coupling to an environment	Reduced dynamics generally requires a master equation rather than a state-vector Hamiltonian alone

The nonrelativistic approximation requires characteristic momenta $\lvert\mathbf p\rvert\ll mc$. Expanding the relativistic energy gives

$$\sqrt{\mathbf p^2c^2+m^2c^4} =mc^2+\frac{\mathbf p^2}{2m} -\frac{\mathbf p^4}{8m^3c^2}+\cdots.$$

The free Hamiltonian retains the leading momentum-dependent term after the rest energy is removed.

Canonical Links

• Free Particle is the canonical solution.
• Free Particle in Three Dimensions treats energy shells and three-dimensional propagation.
• Momentum Operator records the generator, domain, and representation conventions.
• Momentum Eigenstates develops the plane-wave basis.
• Gaussian Wave Packets gives a normalized evolving state.
• Wave-Packet Spreading derives the variance growth.
• Free-Particle Propagator: First Encounter gives the position-space kernel.
• Schrödinger Equation is the dynamics formula card.
• Probability Current gives the flux convention used in scattering.

Common Mistakes

• Writing $-\hbar^2\nabla^2/(2m)$ without specifying its domain or boundary conditions.
• Concluding that every system with $V=0$ has a continuous spectrum.
• Treating a full-space plane wave as a normalized physical state.
• Forgetting the two momentum directions associated with one positive energy in one dimension.
• Confusing a periodic simulation cell with an infinite square well.
• Interpreting the phase velocity as the particle velocity.
• Assuming that free motion prevents wave-packet spreading.
• Calling canonical momentum mechanical momentum after electromagnetic minimal coupling is introduced.
• Applying the quadratic dispersion when momenta are not small compared with $mc$.
• Treating an energy offset as irrelevant without checking which sectors or thermodynamic quantities are being compared.

Exercises

1. Periodic cell

A particle moves in a rectangular periodic cell with side lengths $L_x,L_y,L_z$. Find the normalized momentum eigenfunctions and their energies.

Solution

Periodicity requires

$$k_iL_i=2\pi n_i, \qquad n_i\in\mathbb Z.$$

Writing $\mathcal V=L_xL_yL_z$, the normalized modes are

$$\psi_{\mathbf n}(\mathbf x) =\frac{1}{\sqrt{\mathcal V}} \exp\left[ 2\pi i \left( \frac{n_xx}{L_x} +\frac{n_yy}{L_y} +\frac{n_zz}{L_z} \right) \right].$$

Their momenta and energies are

$$\begin{aligned} \mathbf p_{\mathbf n} &=2\pi\hbar \left( \frac{n_x}{L_x}, \frac{n_y}{L_y}, \frac{n_z}{L_z} \right),\\ E_{\mathbf n} &=\frac{\mathbf p_{\mathbf n}^{\,2}}{2m}. \end{aligned}$$

Degeneracies depend on the cell shape. A cubic cell has additional permutations and sign symmetries.

2. Exact Heisenberg motion

Use $[\hat x_i,\hat p_j]=i\hbar\delta_{ij}$ to derive the Heisenberg equations for $\hat{\mathbf x}$ and $\hat{\mathbf p}$.

Solution

For an operator with no explicit time dependence,

$$\frac{d\hat A_H}{dt} =\frac{i}{\hbar}[\hat H_H,\hat A_H].$$

Since the momentum components commute,

$$[\hat H,\hat p_i]=0,$$

so $d\hat p_{i,H}/dt=0$. Using

$$[\hat p_j^2,\hat x_i] =-2i\hbar\delta_{ij}\hat p_j,$$

gives

$$\frac{d\hat x_{i,H}}{dt} =\frac{\hat p_{i,H}}{m}.$$

Integrating yields

$$\hat{\mathbf x}_H(t) =\hat{\mathbf x}_H(t_0) +\frac{t-t_0}{m}\hat{\mathbf p}_H(t_0).$$

3. Global offset versus potential step

Explain why adding the same constant $E_0$ everywhere does not change position probabilities for an isolated particle, whereas adding $E_0$ only for $x>0$ changes scattering.

Solution

For a global shift,

$$\hat H'=\hat H+E_0I,$$

and the commuting terms give

$$U'(t)=e^{-iE_0t/\hbar}U(t).$$

Every state acquires the same global phase, so expectation values and outcome probabilities are unchanged.

If the added term is $E_0\Theta(x)$, it is not proportional to the identity. The kinetic energy differs between the two spatial regions, so the local wave numbers differ. Matching the wavefunction and its derivative then produces reflection and transmission. This is a physical Potential Step, not a change of energy origin.

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• B. C. Hall, Quantum Theory for Mathematicians, Springer, 2013.
• M. Reed and B. Simon, Methods of Modern Mathematical Physics II: Fourier Analysis, Self-Adjointness, Academic Press, 1975.