Derivation

The de Broglie Wavelength of a Material Particle

D-431 Home PU-202 Threads matter · waves Depends on The Energy-Momentum Four-Vector, The Four-Wavevector and Phase Invariance

Statement

A free particle of rest mass \(m\), momentum \(\vec{p}\), and total energy \(E\) is associated with a quantum wave of angular frequency \(\omega\) and wave vector \(\vec{k}\) such that \(E = \hbar\omega\) and \(\vec{p} = \hbar\vec{k}\). The de Broglie wavelength is defined as \(\lambda = 2\pi/|\vec{k}| = h/p\), where \(p = |\vec{p}|\) and \(h = 2\pi\hbar\). For a non-relativistic electron of charge \(e\) accelerated from rest through a potential difference \(V\), this yields \(\lambda = h/\sqrt{2meV}\).

Why it matters

The de Broglie hypothesis unifies wave and particle descriptions of matter. It explains why electrons diffract, why atoms have discrete energy levels (standing waves), and why quantum mechanics is required at atomic scales. The formula \(\lambda = h/p\) is the single most used relation in electron microscopy, neutron scattering, and solid-state physics.

It also reveals a deep symmetry: the same Planck constant \(h\) that quantizes light's energy and momentum quantizes matter's energy and momentum. This universality is the foundation of the wave-particle duality that defines quantum theory.

Assumptions

The particle is free (no external potential).In a potential \(V(\vec{r})\), the local wavelength varies as \(\lambda(\vec{r}) = h/\sqrt{2m(E-V(\vec{r}))}\); there is no single global wavelength. The simple formula \(\lambda = h/p\) applies only where the kinetic energy is well-defined and constant.

The particle is in a momentum eigenstate (or a narrow momentum packet).A general wave packet has a spread of momenta \(\Delta p\), so the wavelength is not sharply defined. The relation \(\lambda = h/p\) is exact only for a plane wave with definite \(\vec{p}\).

The particle is non-relativistic (\(p \ll mc\)) for the simplified formula \(\lambda = h/\sqrt{2mE_k}\).At relativistic speeds, \(E^2 = p^2c^2 + m^2c^4\), and the correct relation is \(\lambda = h/p\) with \(p = \sqrt{E^2/c^2 - m^2c^2}\). The non-relativistic approximation fails when \(E_k \gtrsim mc^2\).

The particle is uncharged or the electromagnetic field is accounted for.For a charged particle in an electromagnetic field, the canonical momentum is \(\vec{p} = m\vec{v} + q\vec{A}\), and the phase of the wavefunction includes the vector potential. The simple \(\lambda = h/p\) uses the kinetic momentum \(m\vec{v}\), not the canonical one.

Derivation

\[ E = hf, \qquad p = \frac{E}{c} \quad (\text{massless photon}) \]

Planck's relation for photon energy and the massless dispersion relation \(E = pc\). A

\[ p = \frac{hf}{c} = \frac{h}{\lambda} \]

Since \(f = c/\lambda\) for light, the photon momentum is \(p = h/\lambda\). A

\[ E = \hbar\omega, \qquad \vec{p} = \hbar\vec{k} \quad (\text{de Broglie hypothesis}) \]

De Broglie postulated that the same relations hold for massive particles, with \(\omega = 2\pi f\) and \(\vec{k}\) pointing along the direction of propagation with magnitude \(|\vec{k}| = 2\pi/\lambda\). This is the core hypothesis, not a derivation. C

\[ \lambda = \frac{2\pi}{|\vec{k}|} = \frac{2\pi\hbar}{p} = \frac{h}{p} \]

From \(\vec{p} = \hbar\vec{k}\), the wavelength is \(2\pi/|\vec{k}|\), and since \(h = 2\pi\hbar\), we get \(\lambda = h/p\). A

\[ p = \sqrt{2mE_k} \quad (\text{non-relativistic}) \]

For a non-relativistic particle, kinetic energy \(E_k = p^2/2m\), so \(p = \sqrt{2mE_k}\). A

\[ \lambda = \frac{h}{\sqrt{2mE_k}} \]

Substitute Step 5 into Step 4. A

\[ E_k = eV \quad (\text{electron accelerated through } V) \]

An electron of charge \(e\) gains kinetic energy \(eV\) when accelerated through potential difference \(V\). A

\[ \lambda = \frac{h}{\sqrt{2meV}} \]

Substitute Step 7 into Step 6. This is the standard formula for electron diffraction. A

\[ p^\mu = \left(\frac{E}{c},\, \vec{p}\right), \qquad k^\mu = (\omega/c,\, \vec{k}) \]

In relativistic notation, the four-momentum and four-wave-vector are defined. B

\[ p^\mu = \hbar k^\mu \quad \text{or} \quad \frac{E}{c} = \hbar\frac{\omega}{c},\ \vec{p} = \hbar\vec{k} \]

The de Broglie relations extend to four-vectors, ensuring Lorentz covariance. This is the relativistic consistency condition. C

\[ v_{\text{group}} = \frac{d\omega}{dk} = \frac{d(E/\hbar)}{d(p/\hbar)} = \frac{dE}{dp} \]

The group velocity of the matter wave is the derivative of energy with respect to momentum. B

\[ v_{\text{group}} = \frac{p}{m} = v_{\text{particle}} \quad (\text{non-relativistic}) \]

For \(E = p^2/2m\), \(dE/dp = p/m\), which is the classical particle velocity. A

\[ v_{\text{group}} = \frac{dE}{dp} = \frac{pc^2}{E} = v_{\text{particle}} \quad (\text{relativistic}) \]

For \(E = \sqrt{p^2c^2 + m^2c^4}\), \(dE/dp = pc^2/E\), which equals the relativistic particle velocity \(v = pc^2/E\). C

Result

\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}} = \frac{h}{\sqrt{2meV}} \quad (\text{non-relativistic electron}) \]

Reading. The de Broglie wavelength is inversely proportional to momentum. Heavier particles or faster particles have shorter wavelengths. For electrons, the wavelength depends on the accelerating voltage: higher voltage means higher momentum and shorter wavelength.

Units check. \([h/p] = \mathrm{J\cdot s / (kg\cdot m/s) = (kg\cdot m^2/s^2)\cdot s / (kg\cdot m/s) = m}\). Correct. For \(h/\sqrt{2meV}\): \([\sqrt{2meV}] = \sqrt{\mathrm{kg\cdot C\cdot V}} = \sqrt{\mathrm{kg\cdot J}} = \sqrt{\mathrm{kg^2\cdot m^2/s^2}} = \mathrm{kg\cdot m/s}\), so \([h/\sqrt{2meV}] = \mathrm{m}\). Consistent.

Limiting cases

Thermal neutrons. For a neutron in thermal equilibrium at temperature \(T\), the average kinetic energy is \(\tfrac{3}{2}k_BT\), so \(\lambda = h/\sqrt{3mk_BT}\). At \(T = 300\ \mathrm{K}\), \(\lambda \approx 0.15\ \mathrm{nm}\), comparable to atomic spacings.
Bohr quantization. Standing wave condition \(2\pi r = n\lambda\) gives \(2\pi r = n h/p\), so \(L = pr = nh/2\pi = n\hbar\). This recovers Bohr's angular momentum quantization.
Macroscopic objects. A \(0.1\ \mathrm{kg}\) ball at \(10\ \mathrm{m/s}\) has \(\lambda = 6.626\times10^{-34}/(0.1\times10) = 6.6\times10^{-34}\ \mathrm{m}\), far smaller than any measurable length, so wave effects are unobservable.
Relativistic electrons. Once \(eV\) is no longer negligible against \(m_ec^2 = 511\ \mathrm{keV}\) (already a 2% correction at \(V = 50\ \mathrm{kV}\)), use the exact \(\lambda = h/\sqrt{2m_e eV\,(1 + eV/2m_ec^2)}\).

Breaks when

Relativistic speeds. When \(v \gtrsim 0.1c\), the non-relativistic formula \(\lambda = h/\sqrt{2meV}\) underestimates the momentum and so overestimates the wavelength. Use \(\lambda = h/p\) with \(p = \sqrt{2meV + (eV)^2/c^2}\) or the full relativistic expression.
Bound particles. In a bound state (e.g., electron in an atom), the particle does not have a definite momentum, so there is no single de Broglie wavelength. The wavefunction is a superposition of many plane waves with different \(\vec{k}\).
Strong electromagnetic fields. In a magnetic field, the canonical momentum \(\vec{p} = m\vec{v} + q\vec{A}\) differs from the kinetic momentum \(m\vec{v}\). The phase of the wavefunction depends on the vector potential, and the simple \(\lambda = h/p\) must be modified.
Variable mass or effective mass. In a crystal, electrons have an effective mass \(m^*\) that depends on the band structure. Use \(\lambda = h/p\) with \(p\) related to \(m^*\) and the group velocity, not the bare electron mass.

Failure modes

Using the wrong mass. For an electron, use \(m_e = 9.109\times10^{-31}\ \mathrm{kg}\), not the proton mass or atomic mass unit. Confusing \(m_e\) with \(m_p\) gives a wavelength off by a factor of 1836.
Forgetting the square root. Writing \(\lambda = h/(2meV)\) instead of \(\lambda = h/\sqrt{2meV}\). The momentum is \(\sqrt{2meV}\), not \(2meV\).
Mixing up phase and group velocity. The phase velocity of a matter wave is \(v_{\text{phase}} = \omega/k = E/p = c^2/v\), which exceeds \(c\). The group velocity \(v_{\text{group}} = d\omega/dk = v\) is the particle velocity. Confusing them leads to incorrect energy-momentum relations.
Applying the formula to bound states. In an atom, the electron does not have a single wavelength. The de Broglie relation applies to free particles or local wave packets, not to stationary bound states.

Discussion

The de Broglie hypothesis is remarkable for its simplicity and universality. The same Planck constant \(h\) that appears in the photoelectric effect and blackbody radiation also sets the scale for matter waves. This universality is not a coincidence; it reflects the deep unity of quantum mechanics, where all particles are described by wavefunctions whose phase evolves according to \(e^{-i\omega t + i\vec{k}\cdot\vec{r}}\), with \(\omega = E/\hbar\) and \(\vec{k} = \vec{p}/\hbar\).

The relation \(\lambda = h/p\) explains why quantum effects are visible at atomic scales but not at macroscopic scales. For a \(1\ \mathrm{kg}\) object moving at \(1\ \mathrm{m/s}\), \(\lambda \approx 7\times10^{-34}\ \mathrm{m}\), some nineteen orders of magnitude smaller than a proton and utterly unmeasurable. For an electron at \(1\ \mathrm{eV}\), \(\lambda \sim 1.2\ \mathrm{nm}\), comparable to atomic spacings, so diffraction and interference are observable. The transition from quantum to classical behavior is a matter of scale, not of principle.

The group velocity of a matter wave equals the particle velocity, which is a non-trivial result. For a free particle, \(E = p^2/2m\), so \(v_{\text{group}} = dE/dp = p/m = v\). This ensures that the wave packet moves with the classical particle, preserving the correspondence principle. The phase velocity, however, is \(v_{\text{phase}} = \omega/k = E/p = c^2/v\), which is greater than \(c\) for non-relativistic particles. This does not violate relativity because phase velocity does not carry information; only the group velocity does.

At a deeper level, the de Broglie relations are a consequence of the symmetry between spacetime and phase space in quantum mechanics. The four-momentum \(p^\mu\) and four-wave-vector \(k^\mu\) are related by \(p^\mu = \hbar k^\mu\), which is Lorentz covariant. This means that the de Broglie relations hold in all inertial frames, and the wavelength and frequency transform according to the Lorentz transformation. This covariance is a strong argument for the correctness of the hypothesis, as it is consistent with special relativity.

Common misconceptions. The de Broglie wavelength is not the physical size of the particle; it is the wavelength of the quantum wave associated with the particle. A particle is not a classical wave; it is a quantum object whose behavior is described by a wavefunction. The wavelength is a property of the wavefunction, not of the particle itself. Also, the de Broglie relation applies to free particles or local wave packets, not to bound states, where the particle does not have a definite momentum.

Worked examples

Example 1. An electron is accelerated from rest through a potential difference of \(50\ \mathrm{V}\). Find its de Broglie wavelength. Use \(m_e = 9.109\times10^{-31}\ \mathrm{kg}\), \(e = 1.602\times10^{-19}\ \mathrm{C}\), \(h = 6.626\times10^{-34}\ \mathrm{J\cdot s}\).

\[ E_k = eV = (1.602\times10^{-19}\ \mathrm{C})(50\ \mathrm{V}) = 8.01\times10^{-18}\ \mathrm{J} \]

Kinetic energy gained by the electron. A

\[ p = \sqrt{2m_e E_k} = \sqrt{2(9.109\times10^{-31}\ \mathrm{kg})(8.01\times10^{-18}\ \mathrm{J})} \]

Momentum from kinetic energy. A

\[ p = \sqrt{1.459\times10^{-47}} = 3.82\times10^{-24}\ \mathrm{kg\cdot m/s} \]

Arithmetic. A

\[ \lambda = \frac{h}{p} = \frac{6.626\times10^{-34}\ \mathrm{J\cdot s}}{3.82\times10^{-24}\ \mathrm{kg\cdot m/s}} \]

De Broglie wavelength. A

\[ \lambda = 1.73\times10^{-10}\ \mathrm{m} = 0.173\ \mathrm{nm} \]

Convert to nanometers. A

\[ \lambda = 0.173\ \mathrm{nm} \]

Reading. The wavelength is comparable to atomic spacings (\(\sim 0.1\ \mathrm{nm}\)), so electron diffraction is observable. This is why electron microscopes can resolve atomic structures.

Units check. \([h/p] = \mathrm{J\cdot s / (kg\cdot m/s) = m}\). Correct.

Example 2. A neutron has a kinetic energy of \(0.025\ \mathrm{eV}\) (thermal neutron at \(300\ \mathrm{K}\)). Find its de Broglie wavelength. Use \(m_n = 1.675\times10^{-27}\ \mathrm{kg}\), \(h = 6.626\times10^{-34}\ \mathrm{J\cdot s}\), \(1\ \mathrm{eV} = 1.602\times10^{-19}\ \mathrm{J}\).

\[ E_k = 0.025\ \mathrm{eV} = 0.025 \times 1.602\times10^{-19}\ \mathrm{J} = 4.005\times10^{-21}\ \mathrm{J} \]

Convert energy to joules. A

\[ p = \sqrt{2m_n E_k} = \sqrt{2(1.675\times10^{-27}\ \mathrm{kg})(4.005\times10^{-21}\ \mathrm{J})} \]

Momentum from kinetic energy. A

\[ p = \sqrt{1.342\times10^{-47}} = 3.66\times10^{-24}\ \mathrm{kg\cdot m/s} \]

Arithmetic. A

\[ \lambda = \frac{h}{p} = \frac{6.626\times10^{-34}\ \mathrm{J\cdot s}}{3.66\times10^{-24}\ \mathrm{kg\cdot m/s}} \]

De Broglie wavelength. A

\[ \lambda = 1.81\times10^{-10}\ \mathrm{m} = 0.181\ \mathrm{nm} \]

Convert to nanometers. A

\[ \lambda = 0.181\ \mathrm{nm} \]

Reading. The thermal neutron wavelength is comparable to atomic spacings, making thermal neutrons ideal for neutron diffraction studies of crystal structures.

Units check. \([h/p] = \mathrm{J\cdot s / (kg\cdot m/s) = m}\). Correct.

Problems

An electron is accelerated through \(100\ \mathrm{V}\). Find its de Broglie wavelength. (Use \(m_e = 9.109\times10^{-31}\ \mathrm{kg}\), \(e = 1.602\times10^{-19}\ \mathrm{C}\), \(h = 6.626\times10^{-34}\ \mathrm{J\cdot s}\).)
Solution
\(E_k = eV = 1.602\times10^{-17}\ \mathrm{J}\). \(p = \sqrt{2m_e E_k} = \sqrt{2(9.109\times10^{-31})(1.602\times10^{-17})} = \sqrt{2.918\times10^{-47}} = 5.40\times10^{-24}\ \mathrm{kg\cdot m/s}\). \(\lambda = h/p = 6.626\times10^{-34}/5.40\times10^{-24} = 1.23\times10^{-10}\ \mathrm{m} = 0.123\ \mathrm{nm}\).
A proton has a kinetic energy of \(1.0\ \mathrm{keV}\). Find its de Broglie wavelength. (Use \(m_p = 1.673\times10^{-27}\ \mathrm{kg}\), \(h = 6.626\times10^{-34}\ \mathrm{J\cdot s}\).)
Solution
\(E_k = 1.0\times10^3 \times 1.602\times10^{-19} = 1.602\times10^{-16}\ \mathrm{J}\). \(p = \sqrt{2m_p E_k} = \sqrt{2(1.673\times10^{-27})(1.602\times10^{-16})} = \sqrt{5.360\times10^{-43}} = 7.32\times10^{-22}\ \mathrm{kg\cdot m/s}\). \(\lambda = h/p = 6.626\times10^{-34}/7.32\times10^{-22} = 9.05\times10^{-13}\ \mathrm{m} = 0.905\ \mathrm{pm}\).
An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength? By what factor?
Solution
\(\lambda = h/\sqrt{2mE_k}\). Since \(E_k\) is the same, \(\lambda \propto 1/\sqrt{m}\). The electron has a smaller mass, so it has a longer wavelength. The ratio is \(\lambda_e/\lambda_p = \sqrt{m_p/m_e} = \sqrt{1.673\times10^{-27}/9.109\times10^{-31}} = \sqrt{1836} = 42.8\). The electron's wavelength is 42.8 times longer.
A neutron has a de Broglie wavelength of \(0.15\ \mathrm{nm}\). Find its kinetic energy in eV. (Use \(m_n = 1.675\times10^{-27}\ \mathrm{kg}\), \(h = 6.626\times10^{-34}\ \mathrm{J\cdot s}\).)
Solution
\(p = h/\lambda = 6.626\times10^{-34}/1.5\times10^{-10} = 4.42\times10^{-24}\ \mathrm{kg\cdot m/s}\). \(E_k = p^2/2m_n = (4.42\times10^{-24})^2/(2\times1.675\times10^{-27}) = 1.954\times10^{-47}/3.35\times10^{-27} = 5.83\times10^{-21}\ \mathrm{J}\). Convert to eV: \(5.83\times10^{-21}/1.602\times10^{-19} = 0.0364\ \mathrm{eV}\).
An electron is accelerated through \(50\ \mathrm{kV}\). Find its de Broglie wavelength, including relativistic corrections. (Use \(m_e = 9.109\times10^{-31}\ \mathrm{kg}\), \(e = 1.602\times10^{-19}\ \mathrm{C}\), \(h = 6.626\times10^{-34}\ \mathrm{J\cdot s}\), \(c = 3.00\times10^8\ \mathrm{m/s}\).)
Solution
\(E_k = eV = 50\times10^3 \times 1.602\times10^{-19} = 8.01\times10^{-15}\ \mathrm{J}\). Relativistic: \(E_{\text{total}} = E_k + m_ec^2 = 8.01\times10^{-15} + 9.109\times10^{-31}\times(3.00\times10^8)^2 = 8.01\times10^{-15} + 8.20\times10^{-14} = 9.00\times10^{-14}\ \mathrm{J}\). \(p = \sqrt{E_{\text{total}}^2/c^2 - m_e^2c^2} = \sqrt{(9.00\times10^{-14})^2/(3.00\times10^8)^2 - (9.109\times10^{-31}\times3.00\times10^8)^2} \), which is equivalent to \(p = \sqrt{2m_e E_k + E_k^2/c^2}\) (expand \(E_{\text{total}}^2 = (E_k + m_ec^2)^2\)). \(p = \sqrt{2(9.109\times10^{-31})(8.01\times10^{-15}) + (8.01\times10^{-15})^2/(3.00\times10^8)^2} = \sqrt{1.459\times10^{-44} + 7.13\times10^{-46}} = \sqrt{1.530\times10^{-44}} = 1.237\times10^{-22}\ \mathrm{kg\cdot m/s}\). \(\lambda = h/p = 6.626\times10^{-34}/1.237\times10^{-22} = 5.36\times10^{-12}\ \mathrm{m} = 5.36\ \mathrm{pm}\), about 2% shorter than the non-relativistic \(5.49\ \mathrm{pm}\).