Quantum Matter Reference and Data

This page helps you find a quantum-matter formula, identify its conventions, and compare numerical values from different sources. The worked examples show why a factor of $2\pi$, a sample thickness, or the sign of an energy axis can change the physical interpretation.

Helpful background. Conventions for Quantum Matter explains lattice coordinates, Fourier transforms, response tensors, and normalizations. You can also use the examples below independently.

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Your question	Start here
Which signs, bases, and normalizations define a formula?	Conventions for Quantum Matter
What does a term or symbol mean?	Glossary and Symbol Index
Which formula applies?	Quantum Matter Formulas
Which degrees of freedom and interactions define a model?	Condensed-Matter Models and Choosing a Model
What does an experiment measure?	How Quantum Matter Is Measured
How should I compare numerical calculations?	Computational Quantum Matter and Software, Notebooks, and Benchmarks
Where can I read more?	Condensed Matter References

A formula card is a convenient reminder once its assumptions are understood. For a first encounter, follow its link to the explanation. In particular, a model Hamiltonian describes chosen degrees of freedom and interactions; using that Hamiltonian for a material requires a separate physical justification.

Energy, frequency, and temperature scales

The same excitation energy can be expressed through ordinary frequency $f$, angular frequency $\omega$, or an equivalent temperature $T_E$:

$$\begin{aligned} E&=hf=\hbar\omega=k_{\mathrm B}T_E,\\ \omega&=2\pi f. \end{aligned}$$

Here $T_E$ is a way to express an energy scale. It is not a measurement of the specimen’s temperature. The spectroscopic wavenumber is $\widetilde{\nu}=E/(hc)$; it also differs from the angular wave number $2\pi/\lambda$.

For example, take the synthetic excitation energy

$$E=(18.00\pm0.12)\,\mathrm{meV}.$$

The resulting conversions are

$$\begin{aligned} f&=(4.352\pm0.029)\,\mathrm{THz},\\ \omega&=(27.35\pm0.18)\,\mathrm{ps}^{-1},\\ T_E&=(208.9\pm1.4)\,\mathrm K,\\ \widetilde{\nu}&=(145.18\pm0.97)\,\mathrm{cm}^{-1}. \end{aligned}$$

All four relative uncertainties equal $0.12/18.00$. They describe the same uncertain energy, so they are correlated conversions, not four independent measurements.

The SI definitions make $h$, $e$, $k_{\mathrm B}$, and $c$ exact. Thus these conversions introduce no uncertainty from those constants; numerical rounding is a separate issue. The BIPM SI Brochure gives their definitions, and the NIST constants database distinguishes exact constants from measured constants in the current CODATA adjustment.

Reciprocal coordinates and physical distances

A reciprocal coordinate such as $(1.25,0,0)$ specifies components in a reciprocal-lattice basis. It is not a magnitude in inverse ångströms. With the convention

$$\mathbf b_i\cdot\mathbf a_j=2\pi\delta_{ij},$$

the scattering wave vector is

$$\mathbf Q=H\mathbf b_1+K\mathbf b_2+L\mathbf b_3.$$

For a cubic crystal of lattice constant $a$, the basis vectors are orthogonal and have length $2\pi/a$. Take

$$\begin{aligned} a&=(3.900\pm0.004)\,\text{\AA},\\ \mathbf Q&=(1.25,0,0)\ \text{r.l.u.} \end{aligned}$$

where “r.l.u.” means reciprocal-lattice units. Relative to the reciprocal-lattice point $\mathbf G=(1,0,0)$, the reduced wave vector is $\mathbf q=\mathbf Q-\mathbf G=(0.25,0,0)$ r.l.u. Hence

$$\begin{aligned} |\mathbf q|&=\frac{0.25(2\pi)}a\\ &=(0.4028\pm0.0004)\,\text{\AA}^{-1},\\ |\mathbf Q|&=\frac{1.25(2\pi)}a\\ &=(2.0138\pm0.0021)\,\text{\AA}^{-1}. \end{aligned}$$

These uncertainties include only the stated uncertainty in $a$, treating the dimensionless coordinates as exact. Their ratio is exactly five under this assumption. For a nonorthogonal lattice, calculate the vector length using the reciprocal-basis dot products; the cubic formula does not apply.

The distinction between $\mathbf Q$ and $\mathbf q$ matters in neutron scattering: the reduced wave vector labels a crystal excitation, while form factors and scattering geometry depend on the full momentum transfer $\hbar\mathbf Q$. See Reciprocal Lattice for the general construction and Squires (2012) for scattering conventions.

From a Hall slope to a carrier estimate

A Hall measurement first produces a voltage-to-current ratio. Converting it into a bulk resistivity requires the specimen geometry; interpreting it as a carrier density requires a transport model.

Consider a uniform film with current along $+x$ and field along $+z$. Choose the transverse voltage leads so that $V_y=E_yw$, where $w$ is the film width. With thickness $t$, define

$$R_{yx}=\frac{V_y}{I_x}, \qquad \rho_{yx}=tR_{yx}.$$

Assume the Hall signal is linear in the stated field interval, with slope $s=dR_{yx}/dB$. A synthetic dataset gives

$$\begin{aligned} t&=(50.0\pm0.5)\,\mathrm{nm},\\ s&=(-3.200\pm0.040)\,\Omega/\mathrm T,\\ R_\square&=(16.00\pm0.20)\,\Omega. \end{aligned}$$

Here $R_\square$ is the longitudinal sheet resistance. Treat the three input uncertainties as independent standard uncertainties. The Hall coefficient and longitudinal resistivity are

$$\begin{aligned} R_{\mathrm H}&=ts\\ &=(-1.600\pm0.026)\times10^{-7}\, \mathrm{m^3/C},\\ \rho_{xx}&=tR_\square\\ &=(80.0\pm1.3)\,\mu\Omega\,\mathrm{cm}. \end{aligned}$$

For example, first-order propagation gives

$$\frac{u(R_{\mathrm H})}{|R_{\mathrm H}|} = \sqrt{\left(\frac{u(t)}t\right)^2+ \left(\frac{u(s)}s\right)^2}.$$

The notation $u(x)$ denotes a standard uncertainty. The same thickness contributes to both derived quantities, so their uncertainties are correlated; the uncertainty rules in JCGM 100:2008 include these covariance terms.

Within the single-carrier Drude model, the negative sign corresponds to electrons with charge $-e$, and

$$\begin{aligned} n_{\mathrm H}&=\frac1{e|R_{\mathrm H}|}\\ &=(3.90\pm0.06)\times10^{25}\,\mathrm{m^{-3}},\\ \mu_{\mathrm H}&=\frac{|R_{\mathrm H}|}{\rho_{xx}} =\frac{|s|}{R_\square}\\ &=(0.2000\pm0.0035)\,\mathrm{m^2/(V\,s)}. \end{aligned}$$

The last mobility is also $(2000\pm35)\,\mathrm{cm^2/(V\,s)}$. Thickness cancels before uncertainties are propagated:

$$\frac{u(\mu_{\mathrm H})}{\mu_{\mathrm H}} = \sqrt{\left(\frac{u(s)}s\right)^2+ \left(\frac{u(R_\square)}{R_\square}\right)^2}.$$

Treating $R_{\mathrm H}$ and $\rho_{xx}$ as independent would count the thickness uncertainty twice.

A linear Hall signal alone does not establish the single-carrier model. Multiple bands, anisotropy, and anomalous Hall contributions can change the interpretation. In those cases $n_{\mathrm H}$ remains a defined Hall number, but need not equal the total carrier density. Hall Effect develops the models; Hall Measurements explains the measurement and geometric corrections.

Comparing measured widths and quoted values

An observed peak width includes instrumental resolution. For a one-dimensional Gaussian peak convolved with a Gaussian resolution function, full widths at half maximum obey

$$w_{\mathrm{obs}}^2=w_{\mathrm{int}}^2+w_{\mathrm{res}}^2.$$

This relation follows because Gaussian variances add under convolution and every Gaussian FWHM equals the same constant times its standard deviation. A Lorentzian intrinsic peak convolved with Gaussian resolution gives a Voigt profile; subtracting squared FWHMs is then incorrect. In a dispersive neutron spectrum, momentum resolution can also broaden the energy profile. The neutron-scattering discussion explains the combined energy–momentum resolution.

When the observed width is consistent with the resolution alone, describe the intrinsic width as unresolved. A numerical confidence upper bound requires a specified statistical model and confidence level; it does not follow from that phrase alone. Near zero intrinsic width, the square root is strongly nonlinear, and a symmetric error bar from linear propagation can be misleading. JCGM’s guides discuss nonlinear propagation and Monte Carlo methods.

Before comparing two material values, check the sample composition and state, the definition of the measured quantity, and its units and normalization. A reported density of states per cell cannot be compared directly with one per volume. A gap extracted from transport need not be the same quantity as a gap in a photoemission spectrum. Cite the original article or dataset version, and keep statistical uncertainties distinct from corrections or modeling assumptions. These details make a numerical comparison physically meaningful.

Exercises

1. Which Hall tensor component has the measured sign?

Suppose an isotropic in-plane conductivity tensor is

$$\boldsymbol{\sigma} = \begin{pmatrix} \sigma_{xx}&\sigma_{xy}\\ -\sigma_{xy}&\sigma_{xx} \end{pmatrix}.$$

Use $\mathbf j=\boldsymbol{\sigma}\mathbf E$ and $\mathbf E=\boldsymbol{\rho}\mathbf j$ to find $\rho_{yx}$. Why must the indices accompany a signed Hall coefficient?

Solution

Matrix inversion gives

$$\boldsymbol{\rho} =\frac1{\sigma_{xx}^2+\sigma_{xy}^2} \begin{pmatrix} \sigma_{xx}&-\sigma_{xy}\\ \sigma_{xy}&\sigma_{xx} \end{pmatrix}.$$

Thus $\rho_{yx}=\sigma_{xy}/(\sigma_{xx}^2+\sigma_{xy}^2)$ and $\rho_{xy}=-\rho_{yx}$. For $j_y=0$, the measured transverse field is $E_y=\rho_{yx}j_x$. Exchanging the indices changes the sign; reversing the voltage leads changes the sign again. These conventions must be fixed before comparing electron-like or hole-like Hall responses.

2. Convert a density of states

A calculation reports a density of states of $1.80$ states per eV per cell, including two identical spin contributions. The cell volume is $160\,\text{\AA}^3$. Find the density of states per cubic metre and the contribution of one spin per cell.

Solution

Since $1\,\text{\AA}^3=10^{-30}\,\mathrm{m^3}$,

$$\begin{aligned} D_V&=\frac{1.80}{160\times10^{-30}}\\ &=1.125\times10^{28}\, \mathrm{states/(eV\,m^3)}. \end{aligned}$$

One spin contributes $0.900$ states per eV per cell because the problem explicitly assumes two identical spin contributions. Dividing by two would be unjustified for unequal spin-resolved spectra. Volume normalization and spin counting are separate choices; see Density of States.

3. Convert a binding-energy slope to velocity

An isolated ARPES band is plotted against positive binding energy $E_B=E_{\mathrm F}-E$. Along increasing $k_x$, its fitted slope is

$$\frac{dE_B}{dk_x} =(2.40\pm0.08)\,\mathrm{eV\,\text{\AA}}.$$

Find the component of group velocity along the cut. Does the fit alone determine the Fermi velocity?

Solution

For a dispersion interpreted as an electron or quasiparticle energy,

$$v_x=\frac1\hbar\frac{dE}{dk_x} =-\frac1\hbar\frac{dE_B}{dk_x}.$$

Using $1\,\mathrm{eV\,\text{\AA}}/\hbar =1.51927\times10^5\,\mathrm{m/s}$ gives

$$v_x=-(3.65\pm0.12)\times10^5\,\mathrm{m/s}.$$

The sign is negative because increasing binding energy decreases the electron energy. The fitted slope gives a local velocity component over its fitting interval. Calling it a Fermi velocity additionally requires the derivative at a Fermi crossing; extrapolating a finite binding-energy window needs justification. Interpreting an intensity ridge as a dispersion also requires attention to overlapping bands and experimental resolution. See ARPES and Sobota, He, and Shen (2021).

4. Resolve a Gaussian width

A Gaussian fit gives $w_{\mathrm{obs}}=(1.10\pm0.10)\,\mathrm{meV}$, while an independent resolution measurement gives $w_{\mathrm{res}}=(0.80\pm0.05)\,\mathrm{meV}$. Under the one-dimensional Gaussian model, estimate the intrinsic FWHM and its first-order standard uncertainty. Explain why the calculation should not be reused unchanged for $w_{\mathrm{obs}}\approx w_{\mathrm{res}}$.

Solution

The central estimate is

$$\begin{aligned} w_{\mathrm{int}} &=\sqrt{1.10^2-0.80^2}\,\mathrm{meV}\\ &=0.755\,\mathrm{meV}. \end{aligned}$$

Differentiating the square root gives

$$u(w_{\mathrm{int}})^2 =\frac{w_{\mathrm{obs}}^2u(w_{\mathrm{obs}})^2}{w_{\mathrm{int}}^2} +\frac{w_{\mathrm{res}}^2u(w_{\mathrm{res}})^2}{w_{\mathrm{int}}^2}.$$

Therefore $u(w_{\mathrm{int}})=0.155\,\mathrm{meV}$, giving the approximate result $(0.75\pm0.16)\,\mathrm{meV}$. This is a Gaussian-model width estimate, not automatically a lifetime.

The derivatives diverge as $w_{\mathrm{int}}\to0$. A fit to the convolved profile with a nonnegative intrinsic width, together with a specified uncertainty procedure, is more appropriate near that boundary. A negative squared-width difference indicates that direct square-root subtraction cannot supply a real intrinsic-width estimate; it does not imply a negative physical width.

References

• Bureau International des Poids et Mesures. The International System of Units (SI), 9th edition, updated 2026. doi:10.59161/AUEZ1291.
• Joint Committee for Guides in Metrology. Evaluation of Measurement Data—Guide to the Expression of Uncertainty in Measurement, JCGM 100:2008. doi:10.59161/JCGM100-2008E. For nonlinear models, see Amendment 1 (2026) and Propagation of Distributions Using a Monte Carlo Method, JCGM 101:2008, doi:10.59161/JCGM101-2008.
• National Institute of Standards and Technology. The NIST Reference on Constants, Units, and Uncertainty: Fundamental Physical Constants, Standard Reference Database 121, CODATA 2022 values, data updated May 2024. Database.
• Sobota, Jonathan A., Yu He, and Zhi-Xun Shen. “Angle-resolved photoemission studies of quantum materials.” Reviews of Modern Physics 93, 025006 (2021). doi:10.1103/RevModPhys.93.025006.
• Squires, G. L. Introduction to the Theory of Thermal Neutron Scattering, 3rd edition. Cambridge University Press, 2012. doi:10.1017/CBO9781139107808.