Does this vacuum preserve supersymmetry?

A supersymmetric action need not have a supersymmetric vacuum. Conversely, a field configuration with nonzero auxiliary fields need not be a vacuum at all. Before interpreting a positive potential as spontaneous symmetry breaking, we must find a minimum and test its fluctuations.

The required preparation is superspace, interactions, and supersymmetric vacua. We will finish its charged-pair example by finding the actual ground state, then examine a model whose potential approaches zero without attaining it.

Throughout, use four-dimensional Minkowski spacetime with signature $(+---)$, unitary global $\mathcal N=1$ supersymmetry, and canonical positive scalar kinetic terms. The action has four real supercharges. We study translation-invariant bosonic configurations with constant scalars, vanishing fermions and field strengths, and a gauge in which $A_\mu=0$. Unless stated otherwise, the potential and stability calculations are classical. Neither curved-space supersymmetry nor supergravity uses the same vacuum-energy test.

What the algebra tells us

In the four-dimensional convention already established,

$$\{Q_\alpha,Q^\dagger_{\dot\beta}\} =2\sigma^\mu_{\alpha\dot\beta}P_\mu, \qquad \sigma^0=I.$$

The spatial Pauli matrices have zero trace. Summing the two diagonal entries therefore gives

$$H=P_0=\frac14\sum_{\alpha=1}^{2} \{Q_\alpha,Q_\alpha^\dagger\}.$$

For a normalized state $|\Omega\rangle$ in the domain of these operators,

$$\langle H\rangle_\Omega =\frac14\sum_{\alpha=1}^{2} \left( \|Q_\alpha|\Omega\rangle\|^2+ \|Q_\alpha^\dagger|\Omega\rangle\|^2 \right)\geq0.$$

Thus zero energy is equivalent to annihilation by all supercharges. A ground state with positive energy breaks supersymmetry. The zero of energy here is the one fixed by the displayed algebra; an arbitrary subtraction cannot turn a broken-symmetry state into a supersymmetric one. Tong, section 3.4

For an infinite homogeneous system, interpret this argument with a supersymmetry-preserving volume regulator and then use the energy density. In a broken phase the integrated charge need not exist as an operator on the infinite-volume vacuum; the local supercurrent statement remains meaningful. This qualification does not change the classical potential test below. Intriligator–Seiberg v3, introduction

On our constant bosonic configurations the fermion transformations reduce to

$$\delta\psi_{i\alpha}=\sqrt2\,\epsilon_\alpha F_i, \qquad \delta\lambda_\alpha=i\epsilon_\alpha D,$$

and their Lorentzian conjugates. Consequently $F_i=D=0$ preserves all four real supercharges. A nonzero scalar $F_i$ or $D$ forces the preserved parameter to vanish. This conclusion depends on the stated background: a configuration with a field strength or spatial gradient has additional terms in these transformations.

A charged model and its auxiliary equations

Take a rank-one $U(1)$ gauge theory with two independent chiral multiplets $\Phi_+$ and $\Phi_-$ of charges $+1$ and $-1$. Their superpotential is

$$W=m\Phi_+\Phi_-.$$

Use a canonically normalized vector field and $D_\mu\phi_\pm=(\partial_\mu\mp igA_\mu)\phi_\pm$ with $g>0$. Add the Fayet–Iliopoulos, or FI, term $+\xi D$ with real $\xi$. The Abelian FI operator and this charged mass deformation are described in Tong, section 4.1.2, equation (4.16). We retain our preceding lesson’s canonical $D$ normalization and define the sign of $\xi$ by $+\xi D$.

The dimensions are $[\phi_\pm]=[m]=1$, $[F_\pm]=[D]=[\xi]=2$, and $[g]=0$. To analyze the nontrivial case, take $m\ne0$ and $\xi>0$; reversing $\xi$ exchanges the roles of the two charges. Write $a=|m|^2$, a positive mass squared.

With a bar denoting complex conjugation, the auxiliary density is

$$\begin{aligned} \mathcal L_{\rm aux} ={}&|F_+|^2+|F_-|^2 +m(\phi_-F_++\phi_+F_-)+\text{h.c.}\\ &+\frac12D^2+ \left[g(|\phi_+|^2-|\phi_-|^2)+\xi\right]D. \end{aligned}$$

Here “h.c.” applies only to the two terms proportional to $m$. Completing the squares, or varying $\bar F_+,\bar F_-,D$, gives

$$F_+=-\bar m\,\bar\phi_-, \qquad F_-=-\bar m\,\bar\phi_+, \qquad D=-g(|\phi_+|^2-|\phi_-|^2)-\xi.$$

The reduced auxiliary density is $-V$, where

$$V=a(|\phi_+|^2+|\phi_-|^2) +\frac12\left[g(|\phi_+|^2-|\phi_-|^2)+\xi\right]^2.$$

Since $m\ne0$, both F equations vanish only at $\phi_+=\phi_-=0$. At that point $D=-\xi\ne0$. There is no supersymmetric classical vacuum. We have not yet established where the theory settles.

Test the origin before calling it a vacuum

Define $x=|\phi_+|^2$, $y=|\phi_-|^2$, and $B=g(x-y)+\xi$. Treating each scalar and its conjugate as independent variables during differentiation,

$$\frac{\partial V}{\partial\bar\phi_+}=(a+gB)\phi_+, \qquad \frac{\partial V}{\partial\bar\phi_-}=(a-gB)\phi_-.$$

Both derivatives vanish at the origin. Its energy density is $\xi^2/2$, and its exact expansion is

$$V=\frac{\xi^2}{2} +(a+g\xi)|\phi_+|^2+(a-g\xi)|\phi_-|^2 +\frac{g^2}{2}(|\phi_+|^2-|\phi_-|^2)^2.$$

For a complex scalar $\phi=(u+iv)/\sqrt2$, the kinetic term is $[(\partial u)^2+(\partial v)^2]/2$. The coefficient of $|\phi|^2$ is therefore the mass squared of each of its two real components. At the origin these are

$$m_+^2=a+g\xi,\qquad m_-^2=a-g\xi.$$

If $a<g\xi$, the negative $m_-^2$ is a tachyonic fluctuation: the potential decreases under a small displacement, so the origin is unstable. It is not a long-lived false vacuum.

If $a\geq g\xi$, every term in $V-\xi^2/2$ is nonnegative because $x,y\geq0$. This proves that the origin is a global classical minimum, not merely a candidate with a favorable Hessian. At equality the quadratic test has a zero eigenvalue; the positive quartic term still fixes the minimum. Exercise 2 examines that boundary.

Find the minimum when a scalar condenses

Now assume $0<a<g\xi$. Setting $\phi_+=0$ and taking a nonzero $\phi_-$ in its stationary equation requires $B=a/g$. Hence

$$|\phi_-|^2=v^2=\frac{g\xi-a}{g^2}, \qquad V_{\min}=\frac{a\xi}{g}-\frac{a^2}{2g^2}>0.$$

The phase of $\phi_-$ is gauge equivalent, so choose a representative $\phi_-=v>0$. At this representative,

$$F_+=-\bar m v,\qquad F_-=0,\qquad D=-\frac ag.$$

Supersymmetry remains broken. Both an F contribution and a D contribution now enter the energy.

To check that no lower configuration was missed, complete a different square directly in the physical potential:

$$V-V_{\min} =2ax+\frac{g^2}{2}(x-y+v^2)^2\geq0.$$

Equality requires $x=0$ and $y=v^2$. This establishes the full set of minima modulo the gauge phase.

The transverse fluctuation check has independent value: it identifies which fields destabilized the origin and whether a zero mode is physical. Set

$$\phi_+=\frac{u_1+iu_2}{\sqrt2}, \qquad \phi_-=v+\frac{s+ip}{\sqrt2}.$$

Expanding the last square through quadratic order yields

$$V-V_{\min} =a(u_1^2+u_2^2)+g^2v^2s^2+O(\text{fluctuations}^3).$$

The real scalar Hessian therefore has eigenvalues

$$2a,\quad 2a,\quad 2(g\xi-a),\quad 0.$$

The zero eigenvector is the infinitesimal gauge phase $p$, which supplies the longitudinal mode of the massive vector. It is not an additional physical scalar modulus. All three transverse scalar masses squared are positive. The positive scalar and Maxwell kinetic terms then also penalize spatial gradients in the classical energy.

For a numerical check with a reference mass $M$, let $g=1/2$, $\xi=4M^2$, and $a=M^2$. The origin has $m_-^2=-M^2$. The true minimum has $v=2M$, $D=-2M^2$, and

$$V_{\min}=6M^4 =|F_+|^2+\frac12D^2 =4M^4+2M^4.$$

All three physical scalar Hessian eigenvalues are $2M^2$. If instead $a=3M^2$ at the same $g,\xi$, the origin is the minimum with scalar masses squared $5M^2$ and $M^2$ and energy $8M^4$.

Local stability, metastability, and the limits of a Hessian

The calculation suggests a practical order of tests: eliminate auxiliaries, solve the zero-potential equations, find stationary points, check physical fluctuations, and then compare with other regions of field space.

Finding	What it establishes
$F_i=D^a=0$ with positive kinetic metric	A zero-energy classical minimum preserving supersymmetry
$\partial V=0$ with a negative physical Hessian eigenvalue	An unstable stationary point
A positive Hessian on all physical scalar directions	A strict local minimum in those directions
A local minimum with lower-energy configurations elsewhere, separated by a barrier	A false vacuum that may be metastable after quantization
A lower bound attained at the candidate	A global minimum within the stated classical theory
An unattained infimum along fields tending to infinity	A runaway rather than a ground state at a finite field value

A nonnegative Hessian with physical zero modes is inconclusive: higher-order terms can stabilize, destabilize, or remain flat. Gauge directions must be identified separately. Positive energy alone says nothing about metastability: both minima of the charged model above are global within their respective parameter regimes. A false vacuum’s decay rate requires a tunneling calculation; local masses alone do not establish a long lifetime. The distinction between local and global nonsupersymmetric minima is emphasized in Intriligator–Seiberg, section 2.2.

A failed F equation can signal a runaway

Consider a separate, gauge-neutral two-chiral model with canonical kinetic terms,

$$W=fX+\frac h2X^2Y, \qquad f\ne0,\quad h\ne0.$$

The same letters $X,Y$ denote its scalar components in the following calculation. Their dimensions are one, while $[f]=2$ and $[h]=0$. This example appears in Intriligator–Seiberg v3, section 2.4, equations (2.18)–(2.20).

The potential and F-flatness equations are

$$V=|f+hXY|^2+\frac{|h|^2}{4}|X|^4, \qquad W_X=f+hXY=0,\quad W_Y=\frac h2X^2=0.$$

The second equation forces $X=0$, contradicting the first. Nevertheless, for any nonzero $Y$ choose the path

$$X=-\frac{f}{hY}.$$

Then

$$V=\frac{|f|^4}{4|h|^2|Y|^4}\longrightarrow0 \qquad\text{as }|Y|\longrightarrow\infty.$$

Because $V\geq0$, this path proves that the infimum is zero. Because the F equations have no finite solution, that value is not attained. This is a runaway, not a supersymmetric vacuum “located at infinity.” Nor have we found a stable finite state that spontaneously breaks supersymmetry.

The chosen path need not solve the stationary equations at any finite $Y$; it is sufficient to bound the infimum. Exercise 3 checks all finite stationary points. This is a statement about the displayed classical potential with canonical kinetic terms; extrapolating an effective theory beyond its range, or claiming an exact quantum runaway, would require additional information.

Exercises

1. Positive energy at a stationary Wess–Zumino configuration

Take one neutral chiral with canonical kinetic term and $W=m\Phi^2/2+h\Phi^3/3$, where $m,h>0$ are real. Test the configuration $\phi_0=-m/(2h)$. Is it supersymmetric? Is it stationary? Find the two real scalar Hessian eigenvalues and decide whether “metastable vacuum” is an appropriate description.

Hint 1

The derivative of $V=|W'|^2$ is $W''\,\overline{W'}$. A stationary point can have $W''=0$ without having $W'=0$.

Hint 2

Write $\phi=\phi_0+\eta$ with $\eta=(u+iv)/\sqrt2$. Keep the cross term between the constant and quadratic pieces of $W'$.

Solution

At the proposed point,

$$W'(\phi_0)=-\frac{m^2}{4h}\ne0, \qquad W''(\phi_0)=0.$$

It is stationary and nonsupersymmetric, with $V_0=m^4/(16h^2)$. Since $W'=-m^2/(4h)+h\eta^2$, its exact expansion is

$$V=\frac{m^4}{16h^2} -\frac{m^2}{4}(u^2-v^2) +\frac{h^2}{4}(u^2+v^2)^2.$$

The canonically normalized real Hessian eigenvalues are $-m^2/2$ and $+m^2/2$. Thus the point is a saddle with a downhill real direction. The two supersymmetric minima $\phi=0,-m/h$ lie on that real axis. There is no local barrier protecting the saddle, so it is not a metastable vacuum. Using only $|W''|^2$ would incorrectly assign two zero scalar masses: the nonzero $W'$ supplies the missing quadratic cross term.

2. Distinguish a quartic minimum from a vacuum modulus

In the charged model keep $g,\xi>0$. First set $a=g\xi$: determine the leading potential along $\phi_+=0$ and decide whether the massless quadratic fluctuation is a modulus. Then change the assumption to $m=0$. Find the complete zero-potential locus and explain why its degeneracy survives the gauge identification.

Hint 1

A modulus parametrizes distinct configurations of exactly equal minimum energy, not just a zero eigenvalue of the Hessian.

Hint 2

At $m=0$ the two F equations vanish identically. Solve the remaining real D equation and use the invariant product $\phi_+\phi_-$ to label gauge orbits.

Solution

At $a=g\xi$,

$$V-\frac{\xi^2}{2} =2g\xi|\phi_+|^2+ \frac{g^2}{2}(|\phi_+|^2-|\phi_-|^2)^2.$$

For $\phi_+=0$ this is $g^2|\phi_-|^4/2$. It vanishes only at $\phi_-=0$, so the origin remains the unique scalar minimum and the zero quadratic mass is not a modulus.

For $m=0$, instead,

$$F_\pm=0,\qquad |\phi_-|^2=|\phi_+|^2+\frac{\xi}{g}, \qquad D=V=0.$$

All these configurations preserve supersymmetry. If $x=|\phi_+|^2$, the gauge invariant $M=\phi_+\phi_-$ obeys

$$|M|^2=x\left(x+\frac{\xi}{g}\right).$$

For every $|M|\geq0$, this equation has one nonnegative solution $x=[-\xi/g+\sqrt{(\xi/g)^2+4|M|^2}]/2$. The phase of $M$ is the sum of the scalar phases and is unchanged by the opposite gauge rotations. Equal $M$ gives the same orbit, including $M=0$, where only the nonzero $\phi_-$ phase needs to be removed. Thus the classical vacua have one complex modulus. No flat metric on this space is implied. The limit $m=0$ changes the vacuum equations and is not the same operation as setting $a=g\xi$.

3. Does the runaway model hide a local minimum?

For the neutral model $W=fX+hX^2Y/2$, find all finite stationary points with $f,h\ne0$. Then choose field phases so that $f,h>0$ and test the line $X=t$, $Y=-t$, with real $t$ near zero. Is the path $X=-f/(hY)$ at finite $Y$ a stationary valley?

Hint 1

Set $A=f+hXY$. The equation obtained by differentiating with respect to $\bar Y$ is especially restrictive.

Hint 2

Separate $X=0$ from $X\ne0$. In the second case the first squared term can vanish, but the derivative of $|X|^4$ remains.

Solution

The two equations are

$$\frac{\partial V}{\partial\bar Y} =A\bar h\,\bar X=0, \qquad \frac{\partial V}{\partial\bar X} =A\bar h\,\bar Y+\frac{|h|^2}{2}X^2\bar X=0.$$

If $X\ne0$, the first requires $A=0$, and the second is then nonzero. If $X=0$, the second becomes $f\bar h\,\bar Y=0$, forcing $Y=0$. The origin is the only finite stationary point.

After rephasing the fields to make $f,h$ positive, the specified line has

$$V(t,-t)=f^2-2fh\,t^2+\frac54h^2t^4.$$

Small nonzero $t$ lowers the potential, so the origin is unstable. On the runaway path at any finite nonzero $Y$, one has $A=0$ but $X\ne0$, and $\partial V/\partial\bar X=|h|^2X^2\bar X/2\ne0$. The path was a sequence approaching the infimum, not a line of extrema or an exact solution of the time-dependent field equations. The absence of any other stationary point excludes a finite local minimum in this smooth classical model.

What remains after the classical test

A vacuum calculation has two logically separate outputs: which supersymmetries the candidate preserves, and whether the candidate is a stable state. The charged example required both the F/D equations and a stability calculation; the neutral example additionally required examining infinity.

The next question is what happens when a stable classical direction remains flat but supersymmetry is broken. An O’Raifeartaigh model and its Goldstino identifies the massless fermion and calculates the one-loop lifting of its scalar partner. Those steps require more than the classical tests completed here.