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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.

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.

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.

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.

Use Minkowski spacetime with $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and units $\hslash =c=1$. The transformation rotates the two components in their internal plane. In particular,Thus both terms in the Lagrangian density are invariant: this is an internal rotation symmetry of two real scalar fields, rather than a transformation of spacetime. The equal masses are essential to this symmetry.

The Noether theorem gives the Noether currentIndeed, the cross terms in its divergence cancel, and the two Klein-Gordon equations implyFor fields decaying sufficiently at spatial infinity, there is no outward current flux. The conserved Noether charge is therefore

In canonical quantization, the canonical momenta are ${\pi }_k={\dot{\varphi }}_k$, with equal-time canonical commutation relationsConsequently $Q=\int d^{3}x({\pi }_1{\varphi }_2-{\pi }_2{\varphi }_1)$. Operators of different components commute, so this expression is Hermitian without an ordering correction. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$, $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $a_{k\mathbf{p}}=a_{\mathbf{p}}^k$. Substituting the oscillator expansions and using the Dirac delta function from the spatial integral givesThe terms with two annihilation operators cancel after $\mathbf{p}\mapsto -\mathbf{p}$; the terms with two creation operators cancel in the same way. In the remaining terms useThere is no commutator constant between the distinct components. The result is already in normal ordering:

For the charged oscillator basis of a scalar doublet, computeIt follows that $[Q,b_{\pm \mathbf{p}}^{†}]=\pm b_{\pm \mathbf{p}}^{†}$. The Fock vacuum satisfies $Q|0⟩=0$, so $b_{+\mathbf{p}}^{†}|0⟩$ has charge $+1$, and $b_{-\mathbf{p}}^{†}|0⟩$ has charge $-1$. To obtain a normalized one-particle state, replace the sharp momentum by ${\int }_{\mathbf{p}}f(\mathbf{p})b_{\pm \mathbf{p}}^{†}|0⟩$, where ${\int }_{\mathbf{p}}|f(\mathbf{p}){|}^2=1$. The charge is unchanged. Equivalently,The generator convention is $\delta {\varphi }_k=i\alpha [Q,{\varphi }_k]$, which reproduces the original rotation signs.

For stability of a two-scalar quartic potential, the relevant potential energy density isAlong either axis boundedness requires $\lambda \ge 0$; along ${\varphi }_1={\varphi }_2$ it requires $\lambda +\mu \ge 0$. Conversely the displayed sum is nonnegative whenever both conditions hold. With $m^2>0$ the origin is then the unique global minimum, including the equality cases. Hence the stable-vacuum conditions areStrict positivity of the quartic term away from the origin would instead require $\lambda >0$ and $\mu >-\lambda$. That stronger condition is unnecessary for a positive mass term. If $m=0$, the same non-strict conditions give boundedness, but the equality cases have flat directions and do not give an isolated classical vacuum.

Finally, the interaction changes the variation of the potential energy byThe interacting Euler-Lagrange equations consequently giveThus the same Noether charge is conserved for all field configurations precisely when $\mu =\lambda$. Then $V_4=\lambda ({\varphi }_1^2+{\varphi }_2^2{)}^2$ preserves the internal rotation symmetry of two real scalar fields; imposing stability additionally requires $\lambda \ge 0$.

The Pauli exclusion principle says that each complete one-particle state can have fermionic occupation number zero or one. “Complete” includes spin and any other internal labels, as well as the spatial wavefunction. Thus two Electrons can share a spatial orbital with opposite spin projections: these are two different spin orbitals. Equal energy alone does not mean equal quantum state, because degenerate energy levels contain several independent one-particle states.

This distinction explains how an electron shell can accommodate several Electrons while exclusion still allows only one per complete one-particle state. It also makes a finite model of fermionic state counting surprisingly simple: two spatial orbitals with two spin choices each give four complete one-particle states, and exactly six two-Electron basis states. The following diagram enumerates them; its boxes represent occupation numbers, not locations in space.

Pauli's Exclusion principle and quantum mechanics by Wolfgang Pauli is his retrospective account, delivered on 13 December 1946 for the 1945 Nobel Prize. On printed pp. 29–30, he describes proposing an extra two-valued Electron property in autumn 1924 and formulating exclusion in 1925. The crucial clue linked counts of resolved energy eigenstates in alkali spectra under a magnetic field with the populations of closed electron shells. Resolving every label turned a complicated shell capacity into one Electron per complete one-particle state. This historical route preceded the modern spin interpretation and the antisymmetric wavefunction formulation; it was not originally a deduction from a Slater determinant.

A Slater determinant extends two-particle antisymmetrization to $N$ fermions. Here $x_i$ contains position and spin or other internal variables. Distinct orthonormal one-particle states give the normalized wavefunction displayed in the title. Repeating a one-particle state repeats a column of the determinant, so the wavefunction vanishes. Exchanging particle slots exchanges rows and reverses its sign. A general fermionic quantum state can be a quantum superposition of Slater determinants; it need not be one Slater determinant.

When spatial and spin factors separate, a spin orbital combines a spatial orbital ${\varphi }_a$ with a normalized spin state ${\chi }_{\sigma }$. For an Electron, choose spin one-half along an axis with $\sigma =\uparrow ,\downarrow$. The two spin orbitals $u_{a\uparrow }$ and $u_{a\downarrow }$ are orthogonal even though their spatial orbital is identical, since their inner product contains $⟨{\chi }_{\uparrow }|{\chi }_{\downarrow }⟩=0$. When spatial and spin variables do not separate, the complete one-particle state is instead a multicomponent wavefunction $u(\mathbf{r},s)$; exclusion still applies to that complete quantum state.