Why Supersymmetry?

Why Are We Writing This Series?

Supersymmetry is one of the most debated and easily misunderstood subjects in theoretical physics. To someone hearing about it for the first time, it sounds rather strange: every boson should have a fermionic partner, and every fermion a bosonic partner. The electron should have a scalar partner, quarks should have scalar partners, the photon a fermionic partner, and the gluon a fermionic partner. At first glance, the natural question is: Why would we need such a thing? Do we not already have an extraordinarily successful theory in the Standard Model?

This is a very reasonable question. The best way to understand supersymmetry is not to begin immediately with mathematical equations. First, we need to understand where the Standard Model succeeds, where it falls short, and why we feel the need to seek a deeper theory. Without that motivation, supersymmetry looks like little more than a mathematical toy. Yet supersymmetry plays a fundamental role in high-energy physics, particle physics, quantum field theory, cosmology, and string theory.

That is precisely why we are writing this series. Our aim is not simply to line up the formulas of supersymmetry and say, “This is supersymmetry.” We aim to explain, step by step, which physical problems gave rise to supersymmetry, what mathematical structure it has, and what kind of theory it attempts to build beyond the Standard Model. This series will therefore try to answer not only “What is supersymmetry?” but also “Why was supersymmetry proposed?”, “Which problem does it try to solve?”, and “How do theoretical physicists construct a new model?”

Before we begin, one point must be stated clearly: supersymmetry has not yet been experimentally confirmed. The supersymmetric partners of Standard Model particles have not been directly observed at the Large Hadron Collider. It would therefore be wrong to discuss supersymmetry as though its existence in nature were certain. But it would be equally wrong to dismiss its importance simply because it has not yet been found experimentally. Supersymmetry provides a powerful theoretical framework for protecting scalar masses in quantum field theory, unifying gauge coupling constants, furnishing dark-matter candidates, and studying supergravity and string theory.

Throughout this series, we will discuss openly where supersymmetry is powerful, which problems it genuinely solves, which new problems it creates, and what its experimental status means. For that reason, we will not jump straight into the supersymmetry algebra in this first part. We will first see exactly where the Standard Model puts us under strain. To understand the road to supersymmetry, we must formulate the problem correctly.

What Will We Do in This Part?

Unfortunately, to follow this article we expect you to have a basic knowledge of quantum field theory and particle physics. I will not explain topics such as what a hadron is or how to draw a Feynman diagram here. If you do not know these things, do not worry: we will soon begin Yaren’s series on Quantum Field Theory. You can read that series and return to this one afterward.

For those continuing, let us now outline what we will do in this part. Our principal aim here is not to construct supersymmetry directly, but to understand why it is needed. We will begin with a brief review of the Standard Model. In particular, we will show clearly why the Higgs field is necessary, how fermions are given mass, and where the electroweak scale comes from.

Then we will reach the central problem: Why does the Higgs mass create a naturalness problem? That is the question we will seek to answer.

If you are ready, let us begin.

A Brief Look at the Standard Model

The fundamental gauge symmetry of the Standard Model is

$$SU(3)_C \times SU(2)_L \times U(1)_Y$$

This expression looks very compact, but it actually summarizes much of the theory.

$SU(3)_C$ describes the strong interaction. The $C$ denotes color charge. Quarks carry color charge, while gluons are the gauge bosons that mediate this interaction.

$SU(2)_L \times U(1)_Y$ describes the electroweak interaction. The letter $L$ matters because the weak interaction treats left-handed fermions specially. Left-handed fermions reside in $SU(2)_L$ doublets, whereas right-handed fermions are singlets.

Electric charge, weak isospin, and hypercharge are related by

$$Q = T_3 + \frac{Y}{2}.$$

Here $Q$ denotes electric charge, $T_3$ the third component of weak isospin, and $Y$ hypercharge. We will use this hypercharge convention throughout the article.

Let us see an immediate example. Left-handed leptons reside in the doublet

$$L_L = \begin{pmatrix} \nu_L \\ e_L \end{pmatrix}$$

Within this doublet, the weak isospin of the upper component is

$$T_3(\nu_L)=+\frac{1}{2}$$

while that of the lower component is

$$T_3(e_L)=-\frac{1}{2}$$

Neutrinos are electrically neutral:

$$Q(\nu_L)=0.$$

Write the electric-charge relation for the neutrino:

$$0 = \frac{1}{2}+\frac{Y_L}{2}.$$

It follows that

$$\frac{Y_L}{2}=-\frac{1}{2}$$

and

$$Y_L=-1$$

Now use the same relation for the electron. Because the electron is the lower component of the same doublet, its hypercharge is again

$$Y_L=-1$$

Therefore,

$$Q(e_L) = -\frac{1}{2}+\frac{-1}{2}$$

which gives

$$Q(e_L)=-1$$

This small calculation tells us something important. The charges of particles in the Standard Model are not assigned at random. How a field transforms under each symmetry determines its charge and interactions. The Standard Model is therefore not a catalog of particles; it is a quantum field theory built on symmetry principles. As you will appreciate, this is why a sound knowledge of quantum field theory is needed to understand the Standard Model.

Why Does the Question of Mass Matter?

Let us gradually approach the main issue. In everyday intuition, mass seems entirely natural. A particle has a mass; we write it into the equation and move on. In the Standard Model, however, we are not always free to write down a mass. Mass terms must respect gauge symmetry.

For a Dirac fermion, the mass term generally has the form

$$\mathcal{L}_m=-m\bar{\psi}\psi$$

Here the Dirac adjoint is defined as

$$\bar{\psi}=\psi^\dagger \gamma^0$$

Split the Dirac spinor into its left- and right-chiral components:

$$\psi=\psi_L+\psi_R.$$

Here,

$$\psi_L=P_L\psi, \qquad \psi_R=P_R\psi$$

and the projection operators are

$$P_L=\frac{1-\gamma^5}{2}, \qquad P_R=\frac{1+\gamma^5}{2}$$

Their fundamental properties are

$$P_L^2=P_L, \qquad P_R^2=P_R, \qquad P_LP_R=P_RP_L=0, \qquad P_L+P_R=1.$$

We now want to write the mass term in terms of chiral components. First, let us carefully examine the chiral components of the Dirac adjoint.

The Dirac adjoint of the left-chiral component is

$$\bar{\psi}_L=(\psi_L)^\dagger \gamma^0$$

But because

$$\psi_L=P_L\psi$$

we have

$$\bar{\psi}_L=(P_L\psi)^\dagger\gamma^0.$$

Taking the Hermitian conjugate gives

$$\bar{\psi}_L=\psi^\dagger P_L^\dagger\gamma^0.$$

Because the projection operators are Hermitian,

$$P_L^\dagger=P_L$$

and therefore

$$\bar{\psi}_L=\psi^\dagger P_L\gamma^0.$$

To put this expression into the form $\bar{\psi}=\psi^\dagger\gamma^0$, insert

$$\gamma^0\gamma^0=1$$ $$\bar{\psi}_L = \psi^\dagger\gamma^0(\gamma^0P_L\gamma^0).$$

Since

$$\psi^\dagger\gamma^0=\bar{\psi}$$

we have

$$\bar{\psi}_L = \bar{\psi}(\gamma^0P_L\gamma^0).$$

Now use

$$P_L=\frac{1-\gamma^5}{2}$$ $$\gamma^0P_L\gamma^0 = \gamma^0\left(\frac{1-\gamma^5}{2}\right)\gamma^0.$$

This becomes

$$\gamma^0P_L\gamma^0 = \frac{\gamma^0\gamma^0-\gamma^0\gamma^5\gamma^0}{2}$$

Furthermore,

$$(\gamma^0)^2=1$$

and

$$\gamma^0\gamma^5\gamma^0=-\gamma^5$$

so

$$\gamma^0P_L\gamma^0 = \frac{1+\gamma^5}{2}$$

But

$$\frac{1+\gamma^5}{2}=P_R.$$

Thus,

$$\boxed{\bar{\psi}_L=\bar{\psi}P_R}$$

Similarly, one obtains

$$\boxed{\bar{\psi}_R=\bar{\psi}P_L}$$

The important point to notice is that while

$$\psi_L=P_L\psi$$

we have

$$\bar{\psi}_L=\bar{\psi}P_R$$

In other words, taking the Dirac adjoint interchanges the left and right projectors.

Now expand the mass term:

$$\bar{\psi}\psi = (\bar{\psi}_L+\bar{\psi}_R)(\psi_L+\psi_R).$$

Distributing the product gives

$$\bar{\psi}\psi = \bar{\psi}_L\psi_L + \bar{\psi}_L\psi_R + \bar{\psi}_R\psi_L + \bar{\psi}_R\psi_R$$

Now let us show why the terms of equal chirality vanish. First consider

$$\bar{\psi}_L\psi_L$$

Use the results obtained above:

$$\bar{\psi}_L=\bar{\psi}P_R, \qquad \psi_L=P_L\psi.$$

Therefore,

$$\bar{\psi}_L\psi_L = \bar{\psi}P_RP_L\psi.$$

But for the projectors,

$$P_RP_L=0$$

so

$$\boxed{\bar{\psi}_L\psi_L=0}$$

Likewise,

$$\bar{\psi}_R\psi_R = \bar{\psi}P_LP_R\psi.$$

Because

$$P_LP_R=0$$

we obtain

$$\boxed{\bar{\psi}_R\psi_R=0}$$

Only the cross terms remain:

$$\bar{\psi}\psi = \bar{\psi}_L\psi_R+ \bar{\psi}_R\psi_L.$$

Thus the Dirac mass term

$$\mathcal{L}_m=-m\bar{\psi}\psi$$

can be written as

$$\boxed{ \mathcal{L}_m = -m\left(\bar{\psi}_L\psi_R+\bar{\psi}_R\psi_L\right) }$$

The physical meaning is that the Dirac mass term connects a left-chiral fermion with a right-chiral fermion:

$$\psi_L \leftrightarrow \psi_R.$$

A Dirac fermion therefore needs both left- and right-chiral components in order to acquire mass.

In summary,

$$\boxed{ \bar{\psi}\psi = \bar{\psi}_L\psi_R+\bar{\psi}_R\psi_L }$$

and the Dirac mass term connects the left- and right-chiral components.

If the left- and right-handed fields transform differently under gauge symmetries, this term can break the symmetry. In the Standard Model, the left-handed electron is part of an $SU(2)_L$ doublet:

$$L_L= \begin{pmatrix} \nu_L\\ e_L \end{pmatrix}.$$

The right-handed electron, by contrast, is a singlet field on its own:

$$e_R$$

The left- and right-handed electrons thus do not transform in the same way under electroweak symmetry. Consequently, directly writing a mass term of the form

$$-m_e\bar e_L e_R+\text{h.c.}$$

is not compatible with the gauge symmetry of the Standard Model.

How, then, do fermions acquire mass? The answer will come from the Higgs field. The Higgs field spontaneously breaks the symmetry, and fermion masses appear after that breaking. The important point is that the theory we write initially must respect gauge symmetry. The masses emerge because the Higgs field takes a nonzero value in the vacuum.

The Higgs Field and Its Potential

In the Standard Model, the Higgs field is an $SU(2)_L$ doublet:

$$\Phi= \begin{pmatrix} \phi^+\\ \phi^0 \end{pmatrix}.$$

Under the convention used in this article, the hypercharge of the Higgs field is

$$Y_\Phi=+1$$

Then, with $Q=T_3+Y/2$, the upper component has charge $+1$ and the lower component charge $0$.

The Higgs potential in its simplest form is

$$V(\Phi) = -\mu^2\Phi^\dagger\Phi + \lambda(\Phi^\dagger\Phi)^2$$

Here we take $\mu^2>0$ and $\lambda>0$. The condition $\lambda>0$ is important because we do not want the potential to be unbounded below at large field values.

Let us find the minimum of this potential. First define

$$x=\Phi^\dagger\Phi$$

The potential is then

$$V(x)=-\mu^2x+\lambda x^2$$

At a minimum, the derivative must vanish:

$$\frac{dV}{dx}=0.$$

Taking the derivative gives

$$\frac{dV}{dx} = -\mu^2+2\lambda x$$

Setting it to zero,

$$-\mu^2+2\lambda x=0$$

so

$$2\lambda x=\mu^2$$

and

$$x=\frac{\mu^2}{2\lambda}$$

But $x=\Phi^\dagger\Phi$. If in the vacuum we choose the Higgs field as

$$\langle\Phi\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 0\\ v \end{pmatrix}$$

then

$$\langle\Phi^\dagger\Phi\rangle = \frac{v^2}{2}$$

Therefore,

$$\frac{v^2}{2} = \frac{\mu^2}{2\lambda}.$$

Multiplying both sides by $2$ gives

$$v^2=\frac{\mu^2}{\lambda}$$

This tells us that the vacuum expectation value of the Higgs field follows from the parameters in the potential. The value $v$ does not appear arbitrarily; it is determined by the minimum of the Higgs potential.

Deriving the Higgs Mass

Now let us find the mass of the physical Higgs boson. Expand the Higgs field around the vacuum. In unitary gauge,

$$\Phi(x) = \frac{1}{\sqrt{2}} \begin{pmatrix} 0\\ v+h(x) \end{pmatrix}.$$

Here $h(x)$ represents the physical Higgs boson.

First,

$$\Phi^\dagger\Phi = \frac{(v+h)^2}{2}$$

Substitution into the potential gives

$$V(h) = -\mu^2\frac{(v+h)^2}{2} + \lambda\frac{(v+h)^4}{4}$$

To find the mass, we must read off the $h^2$ term in the potential. Write the expansions:

$$(v+h)^2 = v^2+2vh+h^2$$

and

$$(v+h)^4 = v^4+4v^3h+6v^2h^2+4vh^3+h^4.$$

Now select only terms containing $h^2$. The first part contributes

$$-\mu^2\frac{(v+h)^2}{2} \supset -\frac{\mu^2}{2}h^2$$

and the second contributes

$$\lambda\frac{(v+h)^4}{4} \supset \lambda\frac{6v^2h^2}{4} = \frac{3}{2}\lambda v^2h^2$$

Thus the $h^2$ part of the potential is

$$V(h)\supset \left(-\frac{\mu^2}{2}+\frac{3}{2}\lambda v^2\right)h^2$$

From the result above, we can write

$$\mu^2=\lambda v^2$$

Substituting it,

$$-\frac{\mu^2}{2}+\frac{3}{2}\lambda v^2 = -\frac{\lambda v^2}{2}+\frac{3}{2}\lambda v^2$$

Combining the right-hand side gives

$$-\frac{1}{2}\lambda v^2+ \frac{3}{2}\lambda v^2 = \lambda v^2$$

Therefore,

$$V(h)\supset \lambda v^2 h^2.$$

For a real scalar field, the mass term in the potential has the form

$$V(h)\supset \frac{1}{2}m_h^2h^2$$

We must therefore have

$$\lambda v^2h^2 = \frac{1}{2}m_h^2h^2$$

which gives

$$\boxed{m_h^2=2\lambda v^2}$$

Equivalently, because $v^2=\mu^2/\lambda$, we can write

$$m_h^2=2\mu^2$$

This equation will be very important later. It shows that the Higgs mass is associated with the electroweak scale. But we will shortly see that there is nothing automatically obvious about this mass remaining small under quantum corrections.

How Do Fermion Masses Come from the Higgs?

Let us continue with the electron. Its Yukawa interaction is

$$\mathcal{L}_Y = -y_e\bar L_L\Phi e_R+\text{h.c.}$$

Here $y_e$ is the electron Yukawa coupling.

Let us check that this term respects gauge symmetry. The left-handed lepton doublet has hypercharge

$$Y(L_L)=-1$$

Therefore,

$$Y(\bar L_L)=+1$$

For the Higgs field,

$$Y(\Phi)=+1$$

and for the right-handed electron,

$$Y(e_R)=-2$$

The total hypercharge is consequently

$$Y(\bar L_L)+Y(\Phi)+Y(e_R) =+1+1-2=0$$

The Yukawa term is thus neutral under $U(1)_Y$. The $SU(2)_L$ indices are also contracted appropriately in the product $\bar L_L\Phi$.

The left-handed lepton doublet is

$$L_L= \begin{pmatrix} \nu_L\\ e_L \end{pmatrix}$$

After symmetry breaking, the Higgs field is

$$\Phi = \frac{1}{\sqrt{2}} \begin{pmatrix} 0\\ v+h \end{pmatrix}$$

Now write the product $\bar L_L\Phi$ explicitly:

$$\bar L_L\Phi = \begin{pmatrix} \bar\nu_L & \bar e_L \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 0\\ v+h \end{pmatrix}.$$

Multiplying the matrices gives

$$\bar L_L\Phi = \frac{1}{\sqrt{2}} \left[\bar\nu_L\cdot 0+\bar e_L(v+h)\right]$$

The first term vanishes. Therefore,

$$\bar L_L\Phi = \frac{v+h}{\sqrt{2}}\bar e_L$$

Substitute this into the Yukawa term:

$$\mathcal{L}_Y = -y_e\frac{v+h}{\sqrt{2}}\bar e_L e_R +\text{h.c.}$$

Expanding the parentheses,

$$\mathcal{L}_Y = -\frac{y_ev}{\sqrt{2}}\bar e_L e_R - \frac{y_e}{\sqrt{2}}h\bar e_L e_R +\text{h.c.}$$

The first term is the electron mass term. Comparing with the standard mass term, we find

$$\boxed{m_e=\frac{y_ev}{\sqrt{2}}}$$

The second term gives the interaction between the Higgs boson and the electron. Since

$$y_e=\frac{\sqrt{2}m_e}{v}$$

it can be written as

$$-\frac{y_e}{\sqrt{2}}h\bar e_L e_R+\text{h.c.} = -\frac{m_e}{v}h\bar e e$$

For charged fermions in general, the masses are

$$\boxed{m_f=\frac{y_fv}{\sqrt{2}}}$$

A small technical detail should be noted here: charged leptons and down-type quarks have Yukawa interactions with the Higgs doublet $\Phi$, whereas up-type quarks use

$$\tilde\Phi=i\sigma^2\Phi^*$$

In the minimal Standard Model, neutrino masses are not explained by this simple Yukawa structure; they require additional structure beyond the Standard Model.

This result says that fermion masses in the Standard Model come from the vacuum expectation value of the Higgs field and the Yukawa coupling. But there is a small problem: the equation does not explain why the $y_f$ values are what they are. The electron’s Yukawa coupling is very small, while the top quark’s is of order one. Does the Standard Model explain this? No. It simply takes these values as parameters.

This shows us once more that the Standard Model has undeniable successes, but we would be mistaken to call it a final theory that explains everything.

Where Does the Electroweak Scale Come From?

The Higgs vacuum expectation value is known to be approximately

$$v\simeq 246\,\text{GeV}$$

Let us now see where this number comes from.

At low energies, weak interactions are described by Fermi theory. In this theory, the coefficient of the four-fermion interaction is

$$\frac{G_F}{\sqrt{2}}$$

In the Standard Model, this interaction instead arises from $W$-boson exchange.

If at low energy the momentum transfer is smaller than the $W$ mass,

$$q^2\ll M_W^2$$

then the $W$ propagator behaves approximately as

$$\frac{1}{q^2-M_W^2} \simeq -\frac{1}{M_W^2}$$

Thus $W$ exchange looks like a pointlike interaction at low energy.

Comparing the Standard Model with Fermi theory yields

$$\frac{G_F}{\sqrt{2}} = \frac{g^2}{8M_W^2}$$

Here $g$ is the $SU(2)_L$ gauge coupling.

The Higgs mechanism gives the $W$ boson the mass

$$M_W=\frac{gv}{2}$$

Squaring it gives

$$M_W^2=\frac{g^2v^2}{4}$$

Substitute this into the Fermi relation above:

$$\frac{G_F}{\sqrt{2}} = \frac{g^2}{8(g^2v^2/4)}.$$

In the denominator,

$$8\left(\frac{g^2v^2}{4}\right) = 2g^2v^2$$

so

$$\frac{G_F}{\sqrt{2}} = \frac{g^2}{2g^2v^2}$$

Canceling $g^2$ gives

$$\frac{G_F}{\sqrt{2}} = \frac{1}{2v^2}$$

It follows that

$$G_F=\frac{1}{\sqrt{2}v^2}$$

and

$$\boxed{ v=(\sqrt{2}G_F)^{-1/2}\simeq 246\,\text{GeV} }$$

The critical question is now this: Why is this scale so small? The Planck scale, one of the fundamental scales of the universe, is approximately

$$M_{\text{Pl}}\sim 10^{19}\,\text{GeV}$$

The scale expected in grand unified theories is often around

$$M_{\text{GUT}}\sim 10^{16}\,\text{GeV}$$

The electroweak scale, by contrast, is only of order

$$v\sim 10^2\,\text{GeV}$$

There is an astonishingly large gap between them.

That gap alone need not have been a problem. The real problem is that scalar fields such as the Higgs are extremely sensitive to high-energy scales at the quantum level. We can now examine this.

The Effective-Theory View: Heavy Physics and Light Physics

We have said that the Standard Model is not nature’s final theory. A better way to characterize its place is this: the Standard Model is an effective theory valid over a particular energy range. It gives correct results at low energy, but at very high energy it may give way to a more fundamental theory.

Let us call the scale at which this new physics begins

$$\Lambda$$

If it is the Planck scale, then

$$\Lambda\sim 10^{19}\,\text{GeV}$$

if it is the grand-unification scale, then

$$\Lambda\sim 10^{16}\,\text{GeV}$$

The important point is that even if we do not produce heavy particles directly, they can appear virtually in quantum loops and affect low-energy parameters. Scalar masses are especially vulnerable to these effects.

A typical loop correction to a scalar field’s mass can be written

$$\delta m^2 \sim \lambda \int^{\Lambda} \frac{d^4k_E}{(2\pi)^4} \frac{1}{k_E^2+m^2}.$$

Here $k_E$ is the Euclidean loop momentum, $\lambda$ represents the interaction constant, and $\Lambda$ is the cutoff scale. A brief technical clarification: a Wick rotation takes the propagator in Minkowski space into Euclidean form, which is why the integral contains $k_E^2+m^2$.

Let us evaluate the integral explicitly:

$$I(\Lambda) = \int_{|k_E|<\Lambda} \frac{d^4k_E}{(2\pi)^4} \frac{1}{k_E^2+m^2}.$$

In four-dimensional Euclidean momentum space, the volume element in spherical coordinates is

$$d^4k_E=2\pi^2k^3dk$$

Therefore,

$$I(\Lambda) = \frac{2\pi^2}{(2\pi)^4} \int_0^\Lambda dk\, \frac{k^3}{k^2+m^2}$$

Simplifying the coefficient,

$$\frac{2\pi^2}{(2\pi)^4} = \frac{2\pi^2}{16\pi^4} = \frac{1}{8\pi^2}$$

and hence

$$I(\Lambda) = \frac{1}{8\pi^2} \int_0^\Lambda dk\, \frac{k^3}{k^2+m^2}$$

For convenience, rewrite the integrand as

$$\frac{k^3}{k^2+m^2} = k-\frac{m^2k}{k^2+m^2}.$$

The integral is then

$$I(\Lambda) = \frac{1}{8\pi^2} \left[ \int_0^\Lambda k\,dk - m^2\int_0^\Lambda\frac{k\,dk}{k^2+m^2} \right]$$

The first integral gives

$$\int_0^\Lambda k\,dk=\frac{\Lambda^2}{2}$$

For the second integral, choose the variable

$$u=k^2+m^2$$

Then

$$du=2k\,dk$$

and

$$k\,dk=\frac{du}{2}$$

Change the limits as well. For $k=0$,

$$u=m^2$$

and for $k=\Lambda$,

$$u=\Lambda^2+m^2$$

Thus,

$$\int_0^\Lambda \frac{k\,dk}{k^2+m^2} = \frac{1}{2} \int_{m^2}^{\Lambda^2+m^2}\frac{du}{u}$$

which gives

$$\frac{1}{2} \ln\left(\frac{\Lambda^2+m^2}{m^2}\right)$$

Combining the result,

$$I(\Lambda) = \frac{1}{8\pi^2} \left[ \frac{\Lambda^2}{2} - \frac{m^2}{2} \ln\left(\frac{\Lambda^2+m^2}{m^2}\right) \right].$$

That is,

$$\boxed{ I(\Lambda) = \frac{1}{16\pi^2} \left[ \Lambda^2 - m^2\ln\left(\frac{\Lambda^2+m^2}{m^2}\right) \right] }$$

If

$$\Lambda\gg m$$

the dominant term is

$$I(\Lambda)\simeq\frac{\Lambda^2}{16\pi^2}$$

The scalar-mass correction therefore behaves schematically as

$$\boxed{ \delta m^2 \sim \frac{\lambda}{16\pi^2}\Lambda^2 }$$

This is a very important result. It says that the correction grows rapidly as $\Lambda$ increases. This is a cutoff calculation, intended to display the quadratic sensitivity intuitively. Put more physically, if a heavy particle couples to the Higgs, the Higgs mass generally receives threshold corrections proportional to the square of that heavy scale.

What Is the Hierarchy Problem, Really?

Now write the situation for the Higgs mass. The physical Higgs mass can be viewed as the sum of a bare parameter and quantum corrections:

$$m_{h,\text{fiz}}^2 = m_{h,0}^2+ \delta m_h^2.$$

Here $m_{h,0}^2$ is the bare mass parameter and $\delta m_h^2$ the quantum correction. This notation is schematic because the separation between the bare parameter and the correction depends on the chosen renormalization scheme. The physical issue, however, is that the Higgs mass is sensitive to scales of heavy physics.

If the cutoff scale is near the Planck scale,

$$\Lambda\sim 10^{19}\,\text{GeV}$$

then

$$\Lambda^2\sim 10^{38}\,\text{GeV}^2$$

The Higgs mass, by contrast, is of order

$$m_h\sim 10^2\,\text{GeV}$$

and hence

$$m_h^2\sim 10^4\,\text{GeV}^2$$

Looking at the mass-renormalization equation above makes the problem plain. The left-hand side is small. On the right are two contributions, one the bare parameter and the other an enormous quantum correction:

$$\text{küçük fiziksel değer} = \text{çıplak değer} + \text{çok büyük kuantum düzeltmesi}.$$

For the physical Higgs mass to remain small, $m_{h,0}^2$ and $\delta m_h^2$ must cancel with extraordinary precision. Large contributions must cancel one another extremely accurately if the Higgs mass is to stay at the electroweak scale.

This is the hierarchy problem. The electroweak scale is

$$v\sim 10^2\,\text{GeV}$$

whereas the Planck scale is of order

$$M_{\text{Pl}}\sim 10^{19}\,\text{GeV}$$

There is an enormous separation between these scales. If quantum corrections make the Higgs mass sensitive to the large scale, why does it remain at the electroweak scale rather than rising to that high scale?

Without a satisfying answer, the Standard Model does not look natural.

What Does Naturalness Mean?

Let us clarify the word “naturalness.” A small parameter is not a problem simply because it is small. Small numbers are possible. The real question is: Is there a reason that protects this small value?

If taking a parameter to zero increases the symmetry of the theory, then a small value of that parameter is considered natural. The symmetry prevents quantum corrections from making it large. This idea is called technical naturalness.

Fermion masses are a good example. When the mass of a fermion is taken to zero, the theory gains chiral symmetry. Corrections to a fermion mass therefore generally take the form

$$\delta m_f \sim \frac{g^2}{16\pi^2}m_f \ln\left(\frac{\Lambda}{m_f}\right)$$

Notice that the correction is proportional to $m_f$. If the fermion mass is small, the correction is small as well. If

$$m_f=0$$

then

$$\delta m_f=0$$

The massless-fermion limit is therefore protected by symmetry.

A similar situation holds for gauge bosons. The photon’s masslessness is not an accident. Electromagnetic gauge symmetry forbids a photon mass. One might imagine writing a photon mass term as

$$\frac{1}{2}m_\gamma^2 A_\mu A^\mu$$

but this term is incompatible with $U(1)$ gauge symmetry. The photon’s masslessness is therefore protected by gauge symmetry.

What about the Higgs? The Higgs is a scalar field. Taking its mass to zero does not produce a new symmetry within the Standard Model that robustly protects it. The Higgs mass is therefore defenseless against high-energy scales.

This is exactly where supersymmetry enters. Supersymmetry places scalar fields such as the Higgs in the same symmetry structure as fermions. A protection similar to the mechanism that protects fermion masses can then emerge for scalar masses as well.

And Finally, Supersymmetry

We can now take the first step toward the idea of supersymmetry. Supersymmetry is a symmetry that relates bosons and fermions. Bosons have integer spin:

$$s=0,1,2,\ldots$$

Fermions have half-integer spin:

$$s=\frac{1}{2},\frac{3}{2},\ldots$$

In the Standard Model, the matter particles—quarks and leptons—are fermions. Force carriers are bosons. The Higgs boson is also a spin-$0$ scalar boson.

Supersymmetry says that bosonic and fermionic degrees of freedom need not be entirely separate. They may be different parts of the same symmetric structure. Symbolically, we write

$$Q\,|\text{bozon}\rangle \sim |\text{fermiyon}\rangle$$

and

$$Q\,|\text{fermiyon}\rangle \sim |\text{bozon}\rangle$$

Here $Q$ is the supersymmetry generator.

We will not examine these expressions in great technical detail in this part; that comes later. For now, what matters is this: supersymmetry is a symmetry that changes spin. It is therefore not an ordinary internal symmetry. Color symmetry, for example, can change a quark’s color, but it does not turn a quark into a boson. Supersymmetry can place a scalar field and a fermionic field in the same multiplet.

This is crucial to the naturalness problem. Left on their own, scalar fields are sensitive to quantum corrections. But if a scalar field is paired supersymmetrically with a fermion, contributions from fermion loops can cancel the dangerous contributions from scalar loops.

How Do Bosonic and Fermionic Contributions Cancel?

The contribution of a bosonic loop to a scalar field’s mass has the schematic form

$$\delta m_{\text{bozon}}^2 \sim + \frac{g_B^2}{16\pi^2}\Lambda^2$$

The plus sign represents the sign of the bosonic contribution.

A fermion loop carries the opposite sign because of the minus sign associated with closed fermion loops:

$$\delta m_{\text{fermiyon}}^2 \sim - \frac{g_F^2}{16\pi^2}\Lambda^2.$$

In a generic theory, $g_B$ and $g_F$ are independent, so we would not expect these two contributions to cancel. In a supersymmetric theory, however, the bosonic and fermionic couplings are related by the same symmetry. The bosonic and fermionic degrees of freedom are also matched appropriately. With the proper coefficients, therefore,

$$\delta m_{\text{toplam}}^2 = \delta m_{\text{bozon}}^2 + \delta m_{\text{fermiyon}}^2 = 0$$

The point to note is that this cancellation is not fine-tuning performed by hand. We are not arbitrarily choosing two independent large numbers so that they cancel. The cancellation follows from symmetry. Good explanations in physics often work this way: what protects a small number is symmetry, not coincidence.

An Example of Cancellation in a Simple Supersymmetric Model

To make this cancellation more concrete, consider a simple Wess–Zumino-type model with a complex scalar field

$$\phi=\frac{1}{\sqrt{2}}(A+iB)$$

and a fermion field. Here $A$ and $B$ are real scalar fields.

In supersymmetric theories, interactions are often derived from a function called the superpotential. Take the simple example

$$W(\phi) = \frac{1}{2}m\phi^2 + \frac{1}{3}\lambda\phi^3$$

Here $m$ is a mass parameter and $\lambda$ an interaction constant.

The scalar potential is

$$V(\phi) = \left|\frac{dW}{d\phi}\right|^2$$

Now take the derivative explicitly:

$$\frac{dW}{d\phi} = \frac{d}{d\phi} \left( \frac{1}{2}m\phi^2 + \frac{1}{3}\lambda\phi^3 \right).$$

For the first term,

$$\frac{d}{d\phi}\left(\frac{1}{2}m\phi^2\right) = m\phi$$

and for the second,

$$\frac{d}{d\phi}\left(\frac{1}{3}\lambda\phi^3\right) = \lambda\phi^2$$

Therefore,

$$\frac{dW}{d\phi} = m\phi+ \lambda\phi^2$$

and

$$V(\phi) = |m\phi+ \lambda\phi^2|^2$$

The important point in this model is that interactions among the scalars and their interactions with fermions are governed by the same parameter $\lambda$. The bosonic and fermionic sectors are not chosen independently; supersymmetry ties them together.

Contributions to the scalar-mass correction therefore organize schematically as

$$\delta m_A^2\big|_{A} \sim +3\lambda^2I(\Lambda),$$ $$\delta m_A^2\big|_{B} \sim +\lambda^2I(\Lambda),$$

and

$$\delta m_A^2\big|_{\psi} \sim -4\lambda^2I(\Lambda)$$

Here $I(\Lambda)$ denotes the common quadratically divergent integral. The numerical coefficients follow from the interaction structure and the counting of degrees of freedom in the model.

The total contribution is

$$\delta m_A^2 \sim (3+1-4)\lambda^2I(\Lambda)$$

Adding the terms in parentheses gives

$$3+1-4=0$$

Therefore,

$$\boxed{ \delta m_A^2\big|_{\text{karesel}}=0 }$$

This simple example illustrates the central idea of supersymmetry very well. The most dangerous $\Lambda^2$ contributions to the scalar mass cancel between boson and fermion loops. The scalar mass is thereby protected from the high-energy scale.

Does Supersymmetry Really Exist in Nature?

A serious question immediately arises. If supersymmetry were exactly preserved in nature, every Standard Model particle would have a superpartner of the same mass. The scalar superpartner of the electron, for example, would have the same mass as the electron. The scalar partners of quarks and the fermionic partners of gauge bosons would likewise have been observed.

But we do not see such a particle spectrum. If supersymmetry exists in nature, therefore, it cannot be exactly preserved. It must be broken.

The breaking must be done carefully, however. If supersymmetry is broken arbitrarily and hard, the elegant cancellation above is spoiled and the hierarchy problem returns. Realistic supersymmetric models therefore use a structure called soft supersymmetry breaking.

The idea behind soft breaking can be understood through a simple propagator expansion. Suppose the scalar mass receives an additional contribution from supersymmetry breaking:

$$m^2\rightarrow m^2+\Delta m^2.$$

The propagator in Euclidean momentum space is then

$$\frac{1}{k_E^2+m^2+\Delta m^2}$$

Expand it by treating $\Delta m^2$ as a small addition:

$$\frac{1}{k_E^2+m^2+\Delta m^2} = \frac{1}{k_E^2+m^2} \frac{1}{1+\frac{\Delta m^2}{k_E^2+m^2}}.$$

For small $x$,

$$\frac{1}{1+x}=1-x+x^2-\cdots$$

so

$$\frac{1}{k_E^2+m^2+\Delta m^2} = \frac{1}{k_E^2+m^2} - \frac{\Delta m^2}{(k_E^2+m^2)^2} + \cdots$$

The first term is the part canceled by the fermionic contribution in the supersymmetric case. At high momentum, the second behaves as

$$\frac{\Delta m^2}{k_E^4}$$

Since in four dimensions

$$d^4k_E\sim k^3dk$$

we have

$$\int^\Lambda d^4k_E\frac{\Delta m^2}{k_E^4} \sim \Delta m^2\int^\Lambda\frac{k^3dk}{k^4} = \Delta m^2\int^\Lambda\frac{dk}{k} \sim \Delta m^2\ln\Lambda$$

Soft breaking therefore does not reintroduce the quadratic divergence; it leaves only logarithmic sensitivity. Schematically, one obtains a structure such as

$$\delta m^2 \sim \frac{g^2}{16\pi^2}m_{\text{soft}}^2 \ln\left(\frac{\Lambda}{m_{\text{soft}}}\right)$$

Superpartners in supersymmetric models therefore need not have the same masses as Standard Model particles. They may be heavier. But if they become too heavy, the naturalness advantage weakens. This is one of the central debates in supersymmetry phenomenology.

Conclusion

We have not yet constructed supersymmetry mathematically in this part. That was deliberate: we first wanted to understand why supersymmetry is needed.

We reviewed the symmetry structure of the Standard Model. We then saw why fermion masses cannot be written directly and why the Higgs field is therefore necessary. By minimizing the Higgs potential, we obtained

$$v^2=\frac{\mu^2}{\lambda}$$

We then derived the Higgs mass as

$$m_h^2=2\lambda v^2$$

We saw that fermion masses arise from Yukawa couplings in the form

$$m_f=\frac{y_fv}{\sqrt{2}}$$

Next, we related the electroweak scale

$$v\simeq 246\,\text{GeV}$$

to the Fermi constant. Then we turned to the main problem: quantum corrections to scalar masses grow with the square of the high-energy scale,

$$\delta m^2 \sim \frac{\lambda}{16\pi^2}\Lambda^2$$

This explains why the Higgs mass is not natural.

Finally, we saw the idea by which supersymmetry addresses this problem. Supersymmetry relates bosonic and fermionic contributions. Because fermion loops carry a minus sign, quadratic corrections to scalar masses can cancel. This cancellation is not accidental; it is enforced by symmetry.

We can therefore summarize this part as follows:

One of the strongest roads to supersymmetry is the need to protect the Higgs mass naturally against quantum corrections.

What Will We Do in the Next Part?

In this part, we concentrated on establishing the physical motivation. In the next part, we will begin exploring the mathematical structure of supersymmetry.

First, we will ask what the supersymmetry generator is. If a symmetry transformation turns a boson into a fermion, what kind of object must generate that transformation? What is the difference between ordinary Lie algebras and the supersymmetry algebra? Why do anticommutators appear in place of commutators?

We will then write the fundamental relation of the supersymmetry algebra:

$$\{Q_\alpha,\bar Q_{\dot\beta}\} = 2(\sigma^\mu)_{\alpha\dot\beta}P_\mu.$$

In the next part, we will not merely write down this equation; we will unpack its meaning. In particular, we will see why two supersymmetry transformations correspond to a spacetime translation. This point is crucial to understanding why supersymmetry is not an ordinary internal symmetry.

We will then turn to supermultiplets. We will meet structures such as chiral and vector supermultiplets. We will address questions such as which fields can belong to the same multiplet, how bosonic and fermionic degrees of freedom are matched, and why auxiliary fields are needed.

See you in the next part. Wishing you good health.