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Assume $m\ne 0$, since the trivial $m=0$ theory cannot fix $\beta$ or the vacuum expectation value. A zero-energy vacuum must satisfy both $V=0$ and stationarity in both real scalar directions. With the notation from the preceding solution, these become $H=H_x=H_y=0$, because the prefactor $|m{|}^4{e}^{x^2+y^2}$ is positive. The derivatives areThese conditions also show that the zero-energy stationary point must be real. If $y\ne 0$, the second equation gives $P=(3-{\beta }^2)/2$; the first then gives $x=-({\beta }^2+3)/(2\beta )$. Substituting into the definition of $P$ gives $x^2+y^2=2$. Butwhich is impossible. Thus $y=0$, without assuming a real vacuum in advance.

Set $t=x+\beta$ and $F=1+xt$. Since $t=0$ would give $H=1$, it cannot occur. The zero-energy equation gives $F=s\sqrt{3}t$, $s\in \{1,-1\}$. Stationarity givesCombining these equations gives $1=(s\sqrt{3}-x)t=t^2$. Writing $t=\epsilon \in \{1,-1\}$ producesThe condition $\beta >0$ leaves exactly $(s,\epsilon )=(1,1)$ and $(-1,1)$. The second branch is a saddle point, as its real-direction curvature is negative; the stability calculation in the next solution verifies this explicitly. The stable zero-energy Polonyi vacuum therefore selectsZero energy alone, without stationarity and stability, would not imply this parameter value. Even zero energy plus stationarity also admits $\beta =2+\sqrt{3}$ on the unstable branch.

For the stable branch found above, $t=x+\beta =1$ and $s=1$. Hence the vacuum expectation value isin the stated Planck units. To verify that it is a vacuum rather than merely a zero-energy stationary point, evaluate the Hessian matrix. At either zero-energy stationary branch,Because $H$ and its first derivatives vanish there, the Hessian matrix of $V$ is just $|m{|}^4{e}^{x^2}$ times this Hessian matrix. For $s=1$, both eigenvalues are positive. This proves a strict local minimum in both real scalar directions. For $s=-1$, $H_{xx}<0$, so the alternative $\beta =2+\sqrt{3}$, $⟨z⟩=-\sqrt{3}-1$ is a saddle point and is excluded from the stable zero-energy Polonyi vacuum.

There is also a useful global check. Set $u=x-(\sqrt{3}-1)$ on the stable branch. Directly completing squares givesBoth remaining coefficients are positive. Thus $H\ge 0$ everywhere, with equality only at $u=y=0$. The positive exponential prefactor proves that this is the unique global minimum, not just a metastable vacuum.

Finally, at the stable vacuum $z+\beta =1$ and $D_{z}W=\sqrt{3}{m}^2$. Its supergravity auxiliary field hasThus the Minkowski vacuum breaks supersymmetry, even though its cosmological constant vanishes. If $m=0$, the potential is flat and the displayed tuned parameter and scalar value are not selected.

Take $D_{\mu }={\partial }_{\mu }+i{K}_{\mu }$, with the Pauli matrices acting on the Higgs doublet. For electroweak hypercharge $Y=1/2$,This specifies every component of the gauge covariant derivative for the electroweak interaction. Expanding the gauge-covariant kinetic term makes its derivative, trilinear and quartic interactions explicit:where the Pauli matrix multiplication law givesThe antisymmetric Pauli contribution vanishes because $W_{\mu }^a{W}^{b\mu }$ is symmetric in $a,b$. Reversing the sign convention for $D_{\mu }$ reverses the linear gauge interactions consistently, without changing the masses.

A nonzero vacuum expectation value requires ${\mu }^2<0$. Minimizing the Higgs potential gives $v^2=-{\mu }^2/\lambda$. By an gauge transformation choosein unitary gauge. The electroweak doublet gauge-boson mass matrix follows by inserting the vacuum expectation value in the gauge-covariant kinetic term:Define the charged electroweak gauge bosons and the neutral rotation through the Weinberg angle bywith inverse $W_{\mu }^3=c_W{Z}_{\mu }+s_W{A}_{\mu }$ and $B_{\mu }=-s_W{Z}_{\mu }+c_W{A}_{\mu }$. ThenThe factors differ because $W^{+}$ and $W^{-}$ are conjugate fields whereas $Z$ is real. The massless photon corresponds to the unbroken Lie algebra generator $Q=T_3+Y$, which annihilates $⟨\varphi ⟩$. Thus three of the four real gauge bosons acquire mass, with $m_W=m_Z{c}_W$. The three would-be Goldstone bosons provide their longitudinal polarizations; the remaining scalar $h$ is the Higgs boson.

Introduce the left-handed lepton doublet $L=({\nu }_{eL},e_L{)}^T$, with $Y_L=-1/2$, and the right-handed singlet $e_R$, with $Y_R=-1$. Use the chiral projectors $P_L=(1-{\gamma }^5)/2$ and $P_R=(1+{\gamma }^5)/2$ in the course's convention. The minimal Standard Model has no right-handed neutrino. The gauge-invariant fermion terms areThe doublet contraction in the Yukawa interaction is a singlet, and its total hypercharge is $+1/2+1/2-1=0$. A bare term ${\bar{e}}_L{e}_R$ would fail electroweak gauge invariance. The gauge covariant derivatives above already give all the requested fermion-gauge couplings. In terms of mass eigenstates, $e=g{s}_W=g^{\prime }{c}_W$ and they becomeThus the weak charged current is chiral and the neutrino has zero electric charge. The gauge-invariant electron Yukawa mass follows from the Yukawa interaction after electroweak symmetry breaking:Rephase $e_R$ to make $y_e$ real and positive. It givesThe electron Dirac mass is therefore compatible with the original gauge symmetry through the Higgs mechanism. The neutrino remains massless in this minimal renormalizable lepton sector.

Use natural units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real scalar field assigns a real variable $\varphi (t,\mathbf{x})$ to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation $\square \varphi +V^{\prime }(\varphi )=0$. With $V(\varphi )=m^2{\varphi }^2/2$, this is the Klein-Gordon equation. An additional nonlinear part of $V$ describes interactions.

The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -V^{\prime }(\varphi )$ recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.

To see its origin, first consider one coordinate with $H=p^2/(2M)+V(q)$. Split a time interval into $N$ steps of length $\epsilon$ and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of $q$ are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains $(M/(2\pi i\epsilon ){)}^{N/2}{\prod }_{j=1}^{N-1}d{q}_j$. This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring $\hslash$ replaces the weight by $e^{iS/\hslash }$; stationary phase explains the emergence of classical trajectories.

For the field, use scalar field configuration eigenstates $|\phi ⟩$, satisfying $\hat{\varphi }(\mathbf{x})|\phi ⟩=\phi (\mathbf{x})|\phi ⟩$. Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral $\mathcal{N}\int \mathcal{D}\varphi e^{iS[\varphi ]}$. The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.

For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: $e^{-TH}|n⟩=e^{-T{E}_n}|n⟩$, so after normalization only the lowest-energy component remains as $T\to \infty$. This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With $t=-i\tau$, the Euclidean path integral has the weight $e^{-S_E}$, whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.

Introduce a classical source $J(x)$ and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down $i\varphi (x)$. The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.

The free theory illustrates the method. Its quadratic kernel is $K=-\square -m^2$ with the vacuum pole prescription, and completing the square gives the Gaussian evaluation of a free scalar generating functionalTwo source derivatives give ${\Delta }_F=⟨\Omega |T\{\hat{\varphi }(x)\hat{\varphi }(y)\}|\Omega ⟩$. Higher derivatives give all pairings, the content of Wick theorem. For an interaction $S_{\mathrm{int}}$, one may use path-integral perturbation by source derivatives:Expanding this expression generates Feynman diagrams and their Wick contractions. The connected generating functional $W[J]=-i\log Z[J]$ retains connected contributions; in particular $\delta W/\delta J=⟨\varphi {⟩}_J$. These functionals turn the computation of field-operator expectations into source differentiation of an ordinary regulated integral.

Spectrum and symmetry. Choose a vacuum expectation value ${\varphi }_0=(0,v/\sqrt{2}{)}^T$. In unitary gauge the scalar field is $\varphi =(0,(v+h)/\sqrt{2}{)}^T$, with real $h$. The Pauli matrix multiplication law gives $\{{\tau }^a,{\tau }^b\}={\delta }^{ab}I/2$. Since $A^a{A}^b$ is symmetric in gauge indices, only this symmetric product contributes. The kinetic and potential terms therefore becomeComparison with $\frac{1}{2}{m}_A^2{A}_{\mu }^a{A}^{a\mu }$ and $\frac{1}{2}{m}_h^2{h}^2$ givesThese are the scalar interactions after complete SU2 breaking. The scalar mass uses the specified potential normalization: the prefactor $\lambda /2$ is important.

A nonzero complex doublet has trivial stabilizer in $SU(2)$: a matrix fixing $(0,v{)}^T$ must have second column $(0,1{)}^T$, and unitarity and determinant one then fix the first column too. Thus all three gauge generators are broken. Three scalar angular directions are absorbed by the Higgs mechanism, giving each gauge boson a longitudinal polarization. The physical degrees of freedom count isThere is no surviving electromagnetic $U(1)$ in this theory: it contains only the gauged $SU(2)$, not the electroweak product group. The would-be Goldstone bosons are not additional physical massless particles. This is complete breaking by an SU2 scalar doublet.

Scalar interactions. In unitary gauge, all interaction terms containing the physical scalar areThese give $hAA$, $hhAA$, $hhh$, and $hhhh$ Feynman vertices. Their factors are respectively $i{g}^{2}v{\delta }^{ab}{g}_{\mu \nu }/2$, $i{g}^2{\delta }^{ab}{g}_{\mu \nu }/2$, $-3i\lambda v$, and $-3i\lambda$, including the identical-leg multiplicities. Gauge-dependent descriptions also contain the absorbed scalar modes; the above are all physical-scalar interactions in the chosen gauge.

Gauge self-interactions. Write $f_{\mu \nu }^a={\partial }_{\mu }{A}_{\nu }^a-{\partial }_{\nu }{A}_{\mu }^a$. Expanding the field-strength square gives the cubic and quartic piecesThe sign of the cubic term follows from the minus sign in the given nonlinear field strength. The quartic term can also be written using ${ϵ}^{abc}{ϵ}^{ade}={\delta }^{bd}{\delta }^{ce}-{\delta }^{be}{\delta }^{cd}$. The cubic Feynman vertex joins three wavy gauge lines, has one momentum factor, and is proportional to $g{ϵ}^{abc}$. The quartic Feynman vertex joins four wavy lines, is proportional to $g^2$, and has products of structure constants with metric contractions. There are no higher pure-gauge vertices in this Lagrangian. These are the SU2 gauge self-interaction vertices.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

Let $V(\varphi )=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, where ${\varphi }^2={\sum }_i{\varphi }_i^2$. Its stationary-point equation is $(m^2+\lambda {\varphi }^2){\varphi }_i=0$.

For $m^2>0$, the unique minimum is $\varphi =0$. The quadratic mass term is $m^2{\delta }_{ij}$, so there are $n$ real scalar fields, all of mass $\sqrt{m^2}$, and the global special orthogonal group $SO(n)$ remains unbroken.

For $m^2<0$, write $v^2=-m^2/\lambda$. The vacuum manifold is the sphere ${\varphi }^2=v^2$. Choose $⟨\varphi ⟩=v{e}_n$. The transformations preserving this vacuum expectation value rotate the first $n-1$ components, giving $SO(n)\to SO(n-1)$. At this vacuum the potential's Hessian matrix isWriting $\varphi =({\pi }_1,\dots ,{\pi }_{n-1},v+\eta )$, the radial mode $\eta$ therefore hasThese are one massive radial scalar and $n-1$ physical Goldstone bosons. The Goldstone theorem applies to the broken global generators; the flat angular directions of the vacuum manifold explain their zero masses. The spontaneous symmetry breaking is a choice of vacuum, not a change in the invariant Lagrangian. For $n=1$ there are no continuous angular directions: only the radial scalar remains, and the potential's discrete $\varphi \mapsto -\varphi$ symmetry is broken.

On shell, a massless $\mathcal{N}=1$ vector multiplet contains a gauge boson with two transverse polarizations and a gaugino with two fermionic states. A chiral superfield contains a complex scalar field, hence two real bosonic states, and a Weyl spinor with two fermionic states. Their total is $4$ bosonic and $4$ fermionic states.

Suppose charged scalar fields acquire vacuum expectation values at a supersymmetric vacuum, where F-flatness and D-flatness hold. For each broken generator of the gauge symmetry, the supersymmetric Higgs mechanism combines one massless vector multiplet and one chiral superfield into a massive N=1 vector multiplet. The gauge boson absorbs one real Goldstone boson, acquiring its third polarization. The other real scalar field remains physical. The gaugino and the relevant higgsino mix into the four fermionic states. Thus the spin content isUnbroken supersymmetry gives all these states the same mass through supersymmetric mass degeneracy. Remaining massive chiral superfields instead have spin content $(1/2,0,0)$, with two fermionic and two bosonic states of a common mass. The auxiliary fields ensure matching off shell but are not additional propagating particles in these counts.

The super-Higgs mechanism occurs when local supersymmetry is spontaneously broken in supergravity. Its spin-$3/2$ gauge field, the gravitino, absorbs the spin-$1/2$ goldstino:The extra states have helicities $\pm 1/2$, supplementing $\pm 3/2$. The ordinary supersymmetric Higgs mechanism can preserve supersymmetry while breaking the internal gauge symmetry; the super-Higgs mechanism breaks the local supersymmetry itself. Spontaneously broken global supersymmetry leaves a physical massless goldstino, because there is no dynamical gravitino to absorb it.

With the mode normalization given, the oscillator canonical commutation relations are $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$, with the other two commutators zero. Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p^0=E_{\mathbf{p}}$. The convention compatible with the requested numerator iswithout an additional factor of $i$ outside the vacuum expectation value. Put $\tau =x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$. The Fock vacuum is annihilated by $a_{\mathbf{p}}$, so only $a{a}^{†}$ contributes to the two Wightman functions. Consequently,In the second term we changed $\mathbf{p}\mapsto -\mathbf{p}$ to give the same spatial exponential.

Now perform the energy contour integralHere the residue step uses $E>0$; the massless zero-momentum point is interpreted through the smeared distribution limit, not as an isolated normalized oscillator. The Feynman i-epsilon prescription puts the positive-energy pole below the real axis and the negative-energy pole above it: $p^0=E-i0$ and $p^0=-E+i0$. For $\tau >0$, close in the lower half-plane, clockwise; the residue theorem gives $-i$ times the residue $i{e}^{-iE\tau }/(2E)$, hence $e^{-iE\tau }/(2E)$. For $\tau <0$, close in the upper half-plane, counterclockwise; the negative-energy residue is $-i{e}^{iE\tau }/(2E)$, again giving a positive $e^{iE\tau }/(2E)$. These are exactly the two time-ordered terms. Restoring the spatial integral proves the scalar Feynman propagator pole prescription:The prescription in this formula supplies the pole convention left unspecified in the printed display; an unprescribed ordinary real-axis integral would not be well-defined. It is a distribution limit after smearing, not an absolutely convergent four-dimensional integral. With this normalization the derivative jump of a free scalar time-ordered two-point function gives $(\square +m^2){\Delta }_F=-i{\delta }^4(x-y)$, an independent check of both the numerator and the sign.

The pole displacement is exaggerated in this original schematic. The contour orientation and selected pole reproduce the time ordering of the Feynman propagator.

The exponent in the original PDF is $e^{-\alpha S}$ with real $\alpha >0$, not the $e^{-aS}$ produced by the local TeX. Use that authoritative expression. PutThe domain conditions ensure a real Kähler potential and positive Kähler metric. There are no gauge multiplets specified, so the scalar potential is the supergravity F-term potentialThe Kähler covariant derivative of a superpotential and metric entries areState the inverse with its indices explicitly, since transposing the off-diagonal complex entries would change the answer:Here ${\sum }_j{K}_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, so the displayed upper-index array is the transpose of the ordinary matrix inverse of the displayed lower-index array.

Substitution shows that the cross terms involving $W$ cancel and that the $T,C$ sector givesThe first term cancels the universal negative term. Thus the no-scale supergravity potential isThis positivity is an algebraic cancellation, not an assumption that the individual Kähler derivatives vanish.