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Bose-Einstein statistics describes identical noninteracting bosons in equilibrium. Any occupation number is allowed in a one-particle state; the resulting Bose-Einstein distribution is $n(E)=[e^{\beta (E-\mu )}-1{]}^{-1}$.

For the internal rotation symmetry of two real scalar fields with $\delta {\varphi }_1=\alpha {\varphi }_2$ and $\delta {\varphi }_2=-\alpha {\varphi }_1$, the Noether charge is $Q=i\int (a_1^{†}{a}_2-a_2^{†}{a}_1)d^{3}p/(2\pi {)}^3$. Its commutators satisfy $[Q,b_{\pm }^{†}]=\pm b_{\pm }^{†}$, so these creation operators produce charge eigenstates in the one-particle state sector. Substitution gives $Q=N_{+}-N_{-}$, converting the rotation of real components into opposite charges in a complex basis.

The condensate fraction is the population $N_0$ of a macroscopically occupied one-particle state divided by the total particle number $N$. For an interacting system, use a macroscopic eigenvalue of the one-particle density matrix, rather than assume that the relevant mode is an undeformed single-particle trap ground state. The condensate fraction is not generally the fraction measured as superfluid by a flow or rotation response.

Bose-Einstein condensation concerns occupation of a one-particle state. Matter-wave coherence concerns field correlations between positions or times. Superfluidity concerns a flow response, including suppression of dissipation under appropriate conditions. A density peak diagnoses neither a relative quantum phase between two separately prepared clouds nor a critical flow speed. These observables require different experiments.

For example, in a single-mode fixed-number Fock state, $⟨\hat{a}⟩=0$ but $⟨{\hat{a}}^{†}\hat{a}⟩=N$. Writing $\hat{\psi }(\mathbf{r})=\varphi (\mathbf{r})\hat{a}$ for that occupied mode givesThus vanishing mean field does not imply vanishing first-order matter-wave coherence. For two independent fixed-number modes, $⟨{\hat{a}}^{†}\hat{b}⟩=0$: there is no ensemble mean interference term establishing a fixed relative quantum phase. Interpreting fringes in an individual detection record requires considering correlations and the measurement; assigning a pre-existing definite relative quantum phase is unnecessary.

Wolfgang Ketterle's lecture describes a separate coherence test: two condensates expanded into one another and produced fringes. Selectively removing one condensate made the fringes disappear on alternating preparation cycles, testing against diffraction from the confinement. This compared two-cloud and one-cloud signals; it did not measure dissipationless flow.^[ref]

Choose the one-particle states of a hydrogen atom, or a central-field description of a chemical atom, with labelsThe principal quantum number $n$ identifies the electron shell. The orbital angular momentum label $\ell$ identifies the electron subshell; the magnetic quantum number $m_{\ell }$ labels $2\ell +1$ independent spatial orbitals. Each has two spin one-half along an axis choices, hence $2(2\ell +1)$ spin orbitals. Applying the Pauli exclusion principle once per spin orbital givesThese quantum numbers label wavefunctions and spin projections, not electron trajectories. They need not resolve exact energy eigenstates in the presence of spin-orbit interaction; changing to coupled angular momentum eigenstates preserves the count of independent one-particle states within this shell model.

For example, $n=3$ allows $\ell =0,1,2$, conventionally $3s,3p,3d$. Their spatial orbital counts are $1,3,5$; their spin orbital counts are $2,6,10$. The sum is eighteen. The algebra behind “one per complete state” is short: two-particle antisymmetrization gives $u\wedge u=(u\otimes u-u\otimes u)/\sqrt{2}=0$, and the zero vector cannot be a normalized quantum state. Opposite-spin electrons in the same spatial orbital have distinct complete states, so their antisymmetric product survives.

This supplies a structural link to the periodic table: finite electron subshell capacities make repeated patterns of occupied electron configurations possible. It does not predict the exact ground state electron configuration or energy of a many-electron chemical atom. Different electron subshells overlap in energy, and electron interactions change their ordering. In particular, the periodic table period lengths are not simply successive $2n^2$ values: the third periodic table period has eight chemical elements, although the full $n=3$ electron shell has eighteen available states. A capacity counts available spin orbitals; an actual filling order requires a Quantum Hamiltonian.

At the closest physical-strip pole of $S_{2,1}$, the difference of the analytically continued external momenta has squared mass $m_1^2$. Using $m_2=2m_1\cos a$ makes the identity immediate. Thus the exchanged t-channel one-particle state is the lightest Sine-Gordon breather, not a new member of the spectrum.

The density of available one-particle states multiplied by their mean occupation. For equilibrium electrons, $Z(E)=D(E)f(E)$ with the Fermi-Dirac distribution $f$. The integral of $Z$ counts electrons; the integral of $D$ counts available states.

In a dilute gas, Bose-Einstein condensation makes a single one-particle state hold a substantial fraction of the bosons. This is a population effect: cooling reduces the number that the excited energy eigenstates can accommodate at thermal equilibrium. The surplus accumulates in a Bose-Einstein condensate. For an ideal gas the occupied mode is the trap ground state; interactions can deform that mode, as described by the Gross–Pitaevskii equation. Dilute does not mean classical: a small number density can still give overlapping thermal de Broglie wavelengths at sufficiently low temperature.

The appeal is that changing the trap changes a calculable Quantum Hamiltonian, while absorption imaging makes its population distribution visible. The Bose condensation in a three-dimensional harmonic trap below turns density of states into a prediction for a critical temperature and condensate fraction. These are separate questions from measuring matter-wave coherence or superfluidity.

Eric A. Cornell and Carl E. Wieman's joint lecture emphasizes the competition between useful elastic collisions, which restore thermal equilibrium during evaporative cooling, and collisions that remove or heat the trapped sample. A long trap lifetime alone is insufficient: reequilibration must keep up with evaporation.^[ref]

Exclusion constrains complete one-particle states, not a collection of little spatial boxes. In particular, two opposite-spin electrons in the same spatial orbital haveThe spatial factor is symmetric; the spin-one-half singlet state supplies the required antisymmetry. At ${\mathbf{r}}_1={\mathbf{r}}_2$, this wavefunction need not vanish for opposite spin labels. For equal spin labels it does vanish. In a continuous position description, an exact coincident point has zero measure anyway; the useful comparison concerns joint probability densities or detection in finite regions. Equal positions do not imply equal complete quantum states.

The fermionic occupation numbers $n_{\alpha }=0,1$ refer to a chosen orthonormal basis of one-particle states. A quantum superposition need not have definite occupation numbers in that basis, and an expectation value $⟨n_{\alpha }⟩$ can be fractional. For example, one electron in $(u+v)/\sqrt{2}$ has mean occupation number $1/2$ in each of two orthogonal spin orbitals, without splitting into two particles. The canonical anticommutation relations give $(a_{\alpha }^{†}{)}^2=0$ and make $a_{\alpha }^{†}{a}_{\alpha }$ a projector, so its expectation value stays between zero and one in every normalized quantum state.

The Fermi-Dirac distribution follows from Fermi-Dirac statistics and is $n(E)=[e^{\beta (E-\mu )}+1{]}^{-1}$.

The mean occupation of a fermionic one-particle state of energy $E$ isAt zero temperature this becomes the step function $1_{E<E_F}$.

Choosing $N$ different complete one-particle states from an orthonormal basis of size $M$ gives $\left(\genfrac{}{}{0ex}{}{M}{N}\right)$ independent Slater determinants. This is the binomial coefficient because only the occupied subset matters: permutations change the sign of the Slater determinant, not the physical quantum state. Summing over particle numbers gives $2^M$ states in the finite fermionic Fock space.

For the diagram's two spatial orbitals $A,B$, take energies $0,\Delta$ with $\Delta >0$, each independent of spin. In the complete-state order $(A\uparrow ,A\downarrow ,B\uparrow ,B\downarrow )$, the six $N=2$ occupation number strings are $1100,1010,1001,0110,0101,0011$. Their total energies are $0,\Delta ,\Delta ,\Delta ,\Delta ,2\Delta$, giving energy multiplicities $1,4,1$. For $N=3$, the ground state energy is $\Delta$ with two independent ground states, according to which spin orbital in $B$ is occupied. Across $N=0,1,2,3,4$, the counts are $1,4,6,4,1$, summing to $16$. These are exact counts for this model, not measured atomic energies.

The four-momentum operator collects the generators of spacetime translations in quantum field theory. With physical speed of light $c$,For the mostly-minus Minkowski metric,All components of $P^{\mu }$ therefore have units of momentum; $P\cdot a$ has units of action. The natural units convention $c=1$ gives the common abbreviation $P^{\mu }=(H,\mathbf{P})$.

A simultaneous momentum eigenstate satisfies $P^{\mu }|p⟩=p^{\mu }|p⟩$, where $p^{\mu }=(E/c,\mathbf{p})$ is its four-momentum. On a one-particle state of rest mass $m$, the mass-shell relation is $p_{\mu }{p}^{\mu }=m^2{c}^2$.

In a translation-invariant theory the four translation generators commute. A translation-invariant vacuum state has zero spatial momentum; after subtracting the additive vacuum energy, it also has zero Hamiltonian eigenvalue. This fixes the vacuum origin of the four-momentum spectrum.

A left-handed massless Weyl spinor has a positive-energy plane wave transforming by $e^{i\alpha /2}$ under a rotation about its momentum. Its particle has helicity $-1/2$. The negative-frequency term multiplies an antiparticle fermionic creation operator, so its corresponding one-particle state carries the conjugate rotation phase and helicity $+1/2$. Thus a left-handed field has one particle and one antiparticle helicity; the right-handed field interchanges the signs. A full massless Dirac field contains both chiral fields and hence both particle helicities.

The Källén–Lehmann spectral density is the nonnegative invariant-mass measure in a scalar vacuum two-point function. With complete intermediate states it satisfies $2\pi \theta (p^0)\rho (p^2)=(2\pi {)}^4{\sum }_{\alpha }|⟨\alpha |\varphi (0)|0⟩{|}^2{\delta }^4(p-p_{\alpha })$, with continuous state measures included in the sum. A stable one-particle state of mass $m$, with covariant normalization and squared field overlap $Z$, contributes $Z\delta (\sigma -m^2)$. Other particles and multiparticle states contribute additional nonnegative weight. The isolated pole of the scalar propagator then has residue $Z$.

For distinguishable particles with one-particle Hilbert spaces ${\mathcal{H}}_i$, the multiparticle state space is ${⨂}_i{\mathcal{H}}_i$. A product state is a tensor product of one-particle states, while a general state is a linear combination of product states.

A one-particle state belongs to the one-excitation sector of a Fock space. In a momentum basis with $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, the wavepacket $\int d^{3}pf(\mathbf{p})a_{\mathbf{p}}^{†}|0⟩/(2\pi {)}^3$ has norm squared $\int d^{3}p|f(\mathbf{p}){|}^2/(2\pi {)}^3$. Sharp-momentum states have distributional rather than unit normalization; factors of $\sqrt{2E_{\mathbf{p}}}$ convert to relativistic normalization of a one-particle state.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

For physical finite-spin one-particle states, a positive-energy Poincaré group representation is labelled by mass and spin $(m,j)$ when $m>0$, or mass and helicity $(0,h)$ when massless. The individual states also carry their four-momentum and a spin projection. For four-dimensional N=1 supersymmetry, the irreducible labels become $(m,j_{msuper})$, with $j_{msuper}$ the superspin, or $(0,h_{msuper})$, with a chosen superhelicity convention. The constituent representation labels arewith the last entry omitted for $j=0$, andA required CPT completion of a supermultiplet adds the conjugate labels $(0,-h),(0,-h-\frac{1}{2})$ and conjugates any internal charges.

Choose the extended supersymmetry normalizationThe central charge in supersymmetry is antisymmetric in $A,B$, because the full anticommutator is symmetric and ${ϵ}_{\alpha \beta }$ is antisymmetric. In a massive one-particle state rest frame, $P_{\mu }=(M,0,0,0)$, and the first anticommutator becomes $2M{\delta }_{\alpha \beta }{\delta }^A{}_B$.

A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$, with a remaining zero direction if $\mathcal{N}$ is odd. For one block write $z=|z|e^{i\phi }$, and label its two supersymmetries $1,2$. The combinationshaveFor example the mixed contribution to the first expression is $\pm (e^{-i\phi }z+e^{i\phi }\bar{z})/2=\pm |z|$. This is the phase-aligned version of taking a supercharge minus a unitary combination of its adjoints, as suggested by the hint.

For every operator $X$ and normalized state $|v⟩$,Applying this to $A_{-},B_{-}$ in every block proves the BPS boundIf the algebra is instead written with $2Z$ on its right side, the same result is $M\ge {\max }_r|Z_r|$; the factor is purely a central-charge convention.

A BPS state saturates at least one of these inequalities. The corresponding zero-norm combinations of supercharges and their adjoints annihilate the representation, so part of supersymmetry is preserved. Its massive supermultiplet is shortened: each saturated block removes two complex fermionic oscillators, giving $(2j+1)2^{2\mathcal{N}-2k}$ states from a spin-$j$ Clifford vacuum when $k$ blocks saturate, before any necessary CPT completion of a supermultiplet. It preserves $4k$ of the $4\mathcal{N}$ real supercharges. Its mass is tied to the central charges in supersymmetry and remains protected while it stays on the same short branch. Stability is not unconditional: a wall of marginal stability can permit a charged BPS state to decay into other BPS states with the same total charge.

Use signature $(+---)$ and covariant one-particle state normalization $⟨p^{\prime }|p⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$. The future scattering condition uses the outgoing asymptotic scalar field:As with the incoming condition, this is understood in suitably smeared scattering matrix elements, not as unrestricted pointwise strong convergence of interacting and free local operators.

The free incoming field has expansionThe creation operator defines $|p⟩=a_{\mathrm{in}}^{†}(p)|0⟩$, with $a_{\mathrm{in}}|0⟩=0$, so $⟨p|{\varphi }_{\mathrm{in}}(x)|0⟩=e^{ipx}$. On the other hand, translation invariance givesMatching the stable one-particle state component in the asymptotic condition fixes the constant overlap. Choose its state phase so that it is real and nonnegative. ThenThe same stable particle appears in the outgoing description, with compatible one-particle phases. Thus $Z$ is the squared overlap measured by wave-function renormalization.

Insert a complete set of four-momentum eigenstates between the fields. For $x^0>0$, translation invariance and Hermiticity giveFor $x^0<0$, reversing the time-ordered product gives the same sum with $e^{i{p}_{\alpha }x}$. Therefore, with continuous-state integrals implicit in the completeness sum,Here $\theta$ is the Heaviside step function.

For the usual spectral representation, take a Lorentz-invariant vacuum, a positive-norm Hilbert space and the future-cone spectrum condition. Also take $⟨0|\varphi |0⟩=0$, as for the unbroken real scalar field, or subtract this constant and use the connected correlation function. Otherwise the vacuum intermediate state adds a disconnected constant not represented by the ordinary mass-shell measure below.

Let $W(x)=⟨0|\varphi (x)\varphi (0)|0⟩$ for that centered field, and use the Fourier transform convention $W(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}\tilde{W}(p)$. Define the Källén–Lehmann spectral density by the distributional equalityLorentz invariance makes the weight on each future mass shell a function or measure of $p^2$; the spectrum condition gives $p^2\ge 0$. Its nonnegativity follows from the displayed squared matrix elements. ThusApplying time ordering to this expression mass shell by mass shell proves the Källén–Lehmann spectral representation:The factor $i$ is outside ${\Delta }_F$ under the convention in the question.

The stable one-particle state part of completeness is $\int d^3\mathbf{p}[(2\pi {)}^{3}2E_p{]}^{-1}|p⟩⟨p|$. Its overlap is $\sqrt{Z}$, so its contribution to $W$ is $Z$ times the free mass-$m$ two-point function. HenceOther isolated particles as well as multiparticle states can belong to ${\rho }_{\mathrm{rest}}$. The full scalar propagator accordingly has isolated pole $Z/(p^2-m^2+i0)$, making $Z$ its pole residue.

Finally use canonical field normalization: the canonical momentum is $\dot{\varphi }$ and the equal-time canonical commutation relation is $[\dot{\varphi }(t,\mathbf{x}),\varphi (t,\mathbf{y})]=-i{\delta }^3(\mathbf{x}-\mathbf{y})$. The vacuum commutator $C(x)=W(x)-W(-x)$ has the same spectral decomposition. For each free mass shell,independently of $\sigma$. Comparing with the interacting equal-time commutator gives the canonical scalar spectral sum rule:For the particle genuinely interpolated by the stated asymptotic field, $Z>0$; the general nonnegative bound allows a field with zero overlap. The total weight and the upper bound require this canonical normalization and a well-defined distributional equal-time limit. They do not follow from the asymptotic condition alone: rescaling $\varphi$ rescales both $Z$ and the entire spectral measure.