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The convention used here chooses units in which the numerical values of the reduced Planck constant, electron mass, elementary charge and $4\pi$ times the vacuum permittivity are one.

The fluctuation-dissipation theorem relates equilibrium fluctuations to dissipative linear response. The ordering and normalization of a quantum correlation function matter. For a Hermitian operator $A$ in thermal equilibrium, defineFor nonzero frequency, with spectral function $\rho (\omega )=-2\mathrm{Im}{G}_R(\omega )$ and $\beta =1/(k_{B}T)$,Here $k_B$ is the Boltzmann constant and $\hslash$ is the reduced Planck constant. In units $k_B=\hslash =1$, this reduces to $S^{>}(\omega )=\rho (\omega )/(1-e^{-\omega /T})$.

To see the origin, the Kubo--Martin--Schwinger condition gives $S^{<}(\omega )=e^{-\beta \hslash \omega }{S}^{>}(\omega )$ for the reversed ordering. The commutator spectral function obeys $\hslash \rho =S^{>}-S^{<}$, which proves the displayed relation. The symmetrized spectrum instead satisfiesAt $|\hslash \omega |\ll k_{B}T$, both forms approach $k_{B}T\rho (\omega )/\omega$. This connects a measured noise spectrum with the dissipative part of the corresponding response, after matching the experimental spectrum's ordering and Fourier conventions. ^[ref]

The harmonic oscillator transition kernel is the position-space unitary time evolution amplitudeFor $0<T<\pi /\omega$,Here $\hslash$ is the reduced Planck constant. The interval fixes the initial square-root branch; extension beyond focusing times requires its continued phase.

In the path integral, write a path as the classical oscillator path plus a fluctuation vanishing at both endpoints. The quadratic action splits exactly into the classical action and the endpoint-independent fluctuation action. The Dirichlet oscillator determinant ratio gives the prefactor, while the classical boundary action gives the exponent. ^[ref]

As $\omega \to 0$ with $T$ fixed, the expression reduces to the free-particle propagatorAt $T=n\pi /\omega$, the coordinate kernel is distributional because classical paths refocus. The Feynman i-epsilon prescription fixes the continuation through these caustics. Integrating the kernel against an initial wavefunction gives its later wavefunction.

Consider an explicitly stipulated dilute independent spin-one-half two-level system with zero applied magnetic field. Each constituent has $J_z=\pm \hslash /2$ and $m_z=\gamma J_z$, with constant signed gyromagnetic ratio $\gamma$. Ignore interactions and the specimen's own magnetic field. At positive temperature $T$, thermal equilibrium with the rotating lattice gives energy levels $E_{\pm }=\mp \hslash \Omega /2$. Define the dimensionless variable $x=\hslash \Omega /(2k_{B}T)$, where $\hslash$ is the reduced Planck constant and $k_B$ is the Boltzmann constant. The partition function and Boltzmann distribution giveTaking the expected value giveswhere $n$ is number density. This is the rotational version of the independent-moment magnetic response. The angular momentum aligns with positive rotation even when the magnetic moment points oppositely. The probability ratio $p_{+}/p_{-}=e^{2x}$ provides a direct population check of the sign.

For $|x|\ll 1$, expand the hyperbolic tangent to obtainThe slope against $B_{\Omega }$ is positive for either sign of $\gamma$, while the slope against $\Omega$ has the sign of $\gamma$. At fixed $T>0$, reversing $\Omega$ reverses both means, and $\Omega =0$ gives zero. At fixed nonzero $\Omega$, the formal $T\to 0^{+}$ limit saturates $|⟨J_z⟩|$ at $\hslash /2$; the linear approximation must not be extrapolated there. At $T\to \infty$, the response vanishes. These limits describe the toy canonical ensemble, whose equilibration is assumed.

For a numerical example, stipulate $n=10^{28}{\mathrm{m}}^{-3}$, $T=300\mathrm{K}$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and rotation frequency $f=1000\mathrm{Hz}$. With $\Omega =2\pi f$, $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}J/K$, the model givesThese are teaching inputs and calculated outputs, not measured material values. The corresponding internal angular momentum density is $M_z/\gamma =+4.22\times 10^{-17}\mathrm{J}s/{\mathrm{m}}^3$. Keeping a rotor at prescribed $\Omega$ while its internal angular momentum changes requires the drive to supply the appropriate angular impulse. In an isolated body the mechanical part instead changes oppositely, as in magnetomechanical angular momentum balance. There is no creation of net angular momentum by choosing a rotating frame.

The interaction picture divides quantum dynamics between an exactly handled reference Hamiltonian and an interaction. Write $H(t)=H_0+V(t)$ with time-independent $H_0$, and definewhere $\hslash$ is the reduced Planck constant and the subscript $S$ denotes the Schrödinger picture. A linear operator transforms as $A_I(t)=U_0^{†}{A}_S(t)U_0$. Differentiating the transformed quantum state and using the Schrödinger equation cancels the $H_0$ terms and givesThe resulting unitary time evolution isThe Dyson series expands this time-ordered expression in powers of the interaction. ^[ref]

This is useful when $H_0$ has a known eigenbasis of energy eigenstates and the interaction is weak enough for perturbation theory. The transformation itself is exact and preserves expectation values: $⟨{\psi }_S|A_S|{\psi }_S⟩=⟨{\psi }_I|A_I|{\psi }_I⟩$. For time-dependent $H_0(t)$, use its exact unitary time evolution in place of the simple matrix exponential.

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

Make the Wick rotation $t^{\prime }=-i\tau$, and write $T_E=\beta \hslash$ for the final imaginary time, where $\beta$ is inverse temperature and $\hslash$ is the reduced Planck constant. Then $d{t}^{\prime }=-id\tau$ and $dx/d{t}^{\prime }=idx/d\tau$. Thus the continued action isBoth the kinetic energy and the potential energy enter this Euclidean action with positive signs relative to their real-time definitions. Also $e^{-iH(-i\beta \hslash )/\hslash }=e^{-\beta H}$, so the Euclidean path integral isIts boundary conditions fix the two endpoint positions. They need not coincide: identifying and integrating over them would instead give the thermal trace.

The normalization of $\mathcal{D}x$ can be fixed without continuing a formal measure. Apply the Trotter product formula to $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide $T_E$ into $M$ intervals of size $ϵ=T_E/M$, and integrate the intermediate positions. The resulting time-sliced configuration-space path integral isThis also explains the upper limit $\beta \hslash$ rather than $\beta$ when the reduced Planck constant has not been set to one.

Solutions to all three questions of the Further quantum theory course of the University of Oxford Part B examination, held on 29 May 2023. Source: original released PDF, bundle pages 268–271; paper pages 1–4. In question 2 the paper defines $k=\sqrt{2mE}$, a momentum; the corresponding wavenumber is $q=k/\hslash$. Keeping these distinct retains the paper's convention without setting the reduced Planck constant to one.

In Fermi's golden rule,$\Gamma$ is a transition decay rate, measured in inverse time; $\hslash$ is the reduced Planck constant; $M=⟨f|H_{\mathrm{int}}|i⟩$ is the transition matrix element of the interaction part of the Hamiltonian; and $\rho$ is the final density of states per unit energy, evaluated subject to conservation of energy. Final spin and other discrete labels must be summed, with initial labels averaged when the preparation requires it. If $M$ varies across the final phase space, $|M{|}^2\rho$ denotes the corresponding integral rather than a product of constants.

A convention warning: many quantum field theory texts use $\Gamma$ for an energy decay width. That convention is ${\Gamma }_{\mathrm{width}}=\hslash {\Gamma }_{\mathrm{rate}}$; the formula as printed uses the rate convention.

Useful examples with no net quark flavor are the eta c meson ${\eta }_c=c\bar{c}$ and the eta b meson ${\eta }_b=b\bar{b}$. Examples with nonzero quark flavor are the kaon $K^{+}=u\bar{s}$ and the D meson $D^{+}=c\bar{d}$. These are pseudoscalar mesons: in the lowest quark model configuration the constituent spins combine to $S=0$, the relative orbital angular momentum is $L=0$, and the total angular momentum and parity are $J^P=0^{-}$. For the self-conjugate examples, meson parity and charge conjugation also gives $C=(-1{)}^{L+S}=+1$. Electrical neutrality alone would not imply zero net quark flavor; for example $K^0=d\bar{s}$ has strangeness.

Restoring reduced Planck constant, decay width and mean lifetime satisfy $\Gamma \tau =\hslash$. Thus the representative numbers correspond toThe mean lifetimes differ by at least $3.0\times 10^{10}$. Hidden-heavy-flavor quarkonium can undergo quark-antiquark annihilation into gluons, followed by hadronization. A $0^{-+}$ color-singlet state can annihilate into two gluons. A lightest open-flavor meson cannot simply annihilate through the strong interaction, which conserves net quark flavor. With no lighter accessible final state carrying that flavor, it must undergo a weak interaction changing a constituent's flavor. The small weak interaction amplitude, CKM matrix factors and available Lorentz-invariant phase space produce a much smaller decay width.

Qualification. A decay width of order tens of MeV is not a universal property of flavor-neutral pseudoscalar mesons. The neutral pion has $\Gamma \simeq 7.8\mathrm{eV}$ and decays predominantly by the electromagnetic interaction, a concrete counterexample. Its mass lies below the lightest allowed hadronic final state. The examples above illustrate the intended annihilation-versus-open-flavor comparison; the numerical scale depends on thresholds and selection rules.^[ref]

For fixed electron number $N$, the specified electron degeneracy pressure scales as $P\propto V^{-5/3}$. Differentiating at fixed low temperature gives the degeneracy-pressure bulk modulus,A monovalent sample has one conduction electron per atom. Its carrier density isusing the Avogadro constant $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$. Substitution with reduced Planck constant $\hslash =1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$ and electron mass $m_e=9.10938\times 10^{-31}\mathrm{kg}$ givesEquivalently, $P=2n{E}_F/5$ and $K_T=2n{E}_F/3$ for this free electron model. This is the assumed electronic contribution, not a calculation of the complete equation of state of real potassium: ionic cohesion and interaction corrections have been excluded as instructed. The supplied pressure is the zero-temperature form, appropriate to leading order for $T\ll T_F$.

For the supplied linear rigid rotor spectrum, sum over every nonnegative integer $J$ and include all $2J+1$ magnetic substates:It is convenient to introduce the rotational temperatureso that the first terms are $Z_{\mathrm{rot}}=1+3e^{-2{\theta }_r/T}+5e^{-6{\theta }_r/T}+\cdots$. Here $I$ is the moment of inertia and $\hslash$ is the reduced Planck constant, as confirmed in the original PDF. The stipulated degeneracies are used as given, without adding nuclear-spin weights or omitting selected $J$ values.

A cubic reflecting cavity supports standing waves with positive integer mode indices and wavevector components $k_i=\pi n_i/L$. A cell in mode space has volume $(\pi /L{)}^3$. In the continuum approximation, a spherical shell of radius $k$ and thickness $dk$ has one-eighth of its volume in the positive octant. There are two independent transverse photon polarizations for a generic wavevector. HenceThe photon energy is $ϵ=\hslash \omega =\hslash ck$, so $dk=dϵ/(\hslash c)$. ThereforeThis is the leading bulk density of states, valid when the cavity is large compared with the relevant wavelengths and many modes lie in the counting interval. A finite cavity has a discrete mode spectrum, with boundary corrections beyond this approximation. Modes with vanishing indices require separate boundary counting but do not change the leading $L^3$ term. Here $\hslash$ is the reduced Planck constant, as printed in the original PDF; replacing it by $h$ would introduce an erroneous factor of $(2\pi {)}^3$.

Use dimensions $\mathsf{M},\mathsf{L},\mathsf{T}$ for mass, length and time. The reduced Planck constant, gravitational constant and speed of light have dimensionsFor $l_p={\hslash }^{\alpha }{G}^{\beta }{c}^{\gamma }$ to have the dimensions of a length, dimensional analysis requiresThe first equation gives $\beta =\alpha$; the third gives $\gamma =-3\alpha$. Substitution in the second gives $2\alpha =1$. HenceWith $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ and $c=2.99792458\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$, the Planck length isThe proposed monomial fixes the dimensionless prefactor to one; dimensional analysis on its own cannot select an arbitrary dimensionless multiplier.

The time-independent Schrödinger equation is the energy eigenvalue equation $\hat{H}\psi =E\psi$ for a particle of mass $m$. The Hamiltonian operator is the sum of the kinetic operator $-({\hslash }^2/2m)d^2/d{x}^2$ and the prescribed potential energy $V(x)$. Here $\hslash$ is the reduced Planck constant. A corresponding stationary state is $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$. The Born rule identifies $|\Psi {|}^2=|\psi {|}^2$ as probability density for a state with wavefunction normalization $\int |\Psi {|}^{2}dx=1$; integrating it over an interval gives a position probability. A plane wave scattering state is not a normalizable wavefunction on the whole line, but its probability currents give the reflection and transmission fractions of an incident wave packet.

There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses ${\hslash }^2$. The standard physical equation uses the reduced Planck constant $\hslash =h/(2\pi )$, where $h$ is the Planck constant; that convention is used below. Reading the printed $h$ literally as the Planck constant would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.

For $E>V_0$, both sides allow oscillatory wavefunctions. A classical particle always crosses the potential step and slows because its kinetic energy falls from $E$ to $E-V_0$. In quantum mechanics there is generally both reflection and transmission even though the particle is above the step. Define the positive wavenumbersWith no wave incident from the right, writeA finite potential step with constant mass makes both the wavefunction and its first derivative continuous. The two boundary conditions at $x=0$ areSolving gives the reflected and transmitted wavefunctionsFor the specified electron energies, $q/k=\sqrt{10/50}=1/\sqrt{5}$, so the relative probability amplitudes areThe transmitted probability amplitude exceeding one is compatible with conservation of quantum probability, since the outgoing speed is smaller. In fact the probability current $j=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$ givesThese are the quantum reflection probability and transmission probability, which must be distinguished from the requested relative amplitudes.

Solutions to all three questions of the University of Oxford Part A quantum theory paper, sat on 16 June 2021. Source: original released PDF, bundle pages 145–148; paper pages 1–4. The symbol $\hslash$ denotes the reduced Planck constant. All four measurement experiments in question 1(c) start independently from the wavefunction in question 1(b).

Solutions to all three questions of the University of Oxford Part A quantum theory paper, sat on 22 June 2023. Source: original released PDF, bundle pages 153–156; paper pages 1–4. Question statements are omitted. The symbol $\hslash$ denotes the reduced Planck constant.

Take room temperature as $T=300\mathrm{K}$. With the supplied gyromagnetic ratio, reduced Planck constant and Boltzmann constant,soAt $293\mathrm{K}$ the ratio is $1.05\times 10^{-5}$; the room-temperature convention makes only a small difference. The corresponding fractional nuclear spin polarization has magnitude about five parts per million. The source's negative sign convention assigns the label up to the higher-energy state; it should not be confused with the lower-energy magnetic moment aligned with a positive proton magnetic field.

The Planck constant $h$ connects the frequency $\nu$ of an electromagnetic field mode with a single photon's energy, $E=h\nu$. Its physical dimension is action, or energy multiplied by time. It sets the scale of quantum mechanics: the de Broglie wavelength of a particle is $\lambda =h/p$, and discrete energy levels often have spacings involving $h$.

In the International System of Units, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ exactly. Fixing this value defines the kilogram through already defined units of length and time; it does not remove experimental uncertainty from a practical mass measurement.^[ref] The reduced Planck constant $\hslash =h/(2\pi )$ is convenient when angular frequency replaces ordinary frequency.

The Planck mass combines the reduced Planck constant, speed of light and gravitational constant into a mass scale, $m_P=\sqrt{\hslash c/G}$. Its associated Compton wavelength, using the reduced convention $\hslash /(m_{P}c)$, equals the Planck length ${\ell }_P=\sqrt{\hslash G/c^3}$. It therefore identifies a scale at which these particular quantum and gravitational length estimates coincide.

This dimensional comparison motivates the use of Planck units in quantum gravity; it is not an experimental observation of a particle with that mass. The reduced Planck mass instead uses an additional factor $1/\sqrt{8\pi }$, so a gravitational equation written with either convention must state which one is meant.