Check your starting knowledge

Supersymmetry adds a new kind of symmetry to field theory. You can begin without knowing superspace, complex geometry, localization, or conformal field theory. The useful starting tools are more familiar: varying an action, following a gauge transformation, evaluating a Gaussian integral, and multiplying matrices.

Try the four questions below before opening their solutions. A missed sign is a reason to revisit one calculation, not to postpone the entire course. Each solution includes a short refresher that you can use immediately. Afterward, begin with what a supercharge does.

A conserved current

Work in four-dimensional Minkowski space with metric $\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$ and units $\hbar=c=1$. A complex scalar field has Lagrangian density

$$\mathcal L=\partial_\mu\phi^*\partial^\mu\phi-m^2\phi^*\phi.$$

Under a constant phase rotation, $\delta\phi=i\alpha\phi$ and $\delta\phi^*=-i\alpha\phi^*$. Derive the Noether current in the convention

$$\delta\mathcal L=j^\mu\partial_\mu\alpha$$

when $\alpha$ is temporarily allowed to depend on position. Check conservation on the equations of motion. Is the current real?

Hint

The terms proportional to $\alpha$ cancel. Keep the terms in which the derivative acts on $\alpha$.

Complete solution

The differentiated field variation is

$$\delta(\partial_\mu\phi)=i\alpha\partial_\mu\phi +i(\partial_\mu\alpha)\phi,$$

with the complex-conjugate formula for $\phi^*$. Substitution gives

$$\delta\mathcal L =i\bigl(\phi\partial^\mu\phi^*-\phi^*\partial^\mu\phi\bigr) \partial_\mu\alpha.$$

Thus our specified current is

$$j^\mu=i\bigl(\phi\partial^\mu\phi^*-\phi^*\partial^\mu\phi\bigr).$$

Taking the divergence cancels the products of first derivatives:

$$\partial_\mu j^\mu =i\bigl(\phi\Box\phi^*-\phi^*\Box\phi\bigr)=0,$$

because $(\Box+m^2)\phi=0$ and its conjugate hold. Complex conjugation reverses the sign of both $i$ and the bracket, so $(j^\mu)^*=j^\mu$. In mass units $[\phi]=1$ and $[\partial_\mu]=1$, giving $[j^\mu]=3$ and a dimensionless integrated charge $\int d^3x\,j^0$.

Choosing $\delta\phi=-i\alpha\phi$ instead would reverse the current. A current sign cannot be compared between sources without also comparing the transformation convention.

Refresher on varying an action

For fields $\varphi_a$, integrating by parts gives

$$\delta S=\int d^dx\left[ \left(\frac{\partial\mathcal L}{\partial\varphi_a} -\partial_\mu\frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}\right) \delta\varphi_a +\partial_\mu\left( \frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)} \delta\varphi_a\right)\right].$$

Here repeated field labels are summed. Treat $\phi$ and $\phi^*$ as independent variables while varying, then impose their physical conjugation relation. Stationarity for arbitrary compactly supported variations gives the Euler–Lagrange equations. Localizing a symmetry parameter isolates the current without guessing it. Return to the current question and identify exactly where the equations of motion were needed.

A gauge-covariant derivative

Let a scalar of charge $q$ transform as $\phi'=e^{iq\alpha(x)}\phi$. Define

$$D_\mu\phi=(\partial_\mu-iqA_\mu)\phi, \qquad A_\mu'=A_\mu+\partial_\mu\alpha.$$

Show that $D_\mu'\phi'=e^{iq\alpha}D_\mu\phi$. Then calculate $[D_\mu,D_\nu]\phi$. We have absorbed the gauge coupling into $A_\mu$ here; it is not an additional missing factor.

Hint

Differentiate the exponential before using the transformation of $A_\mu$. In the commutator, terms with two derivatives and terms quadratic in the Abelian connection cancel.

Complete solution

The product rule gives

$$\begin{aligned} D_\mu'\phi' &=(\partial_\mu-iqA_\mu-iq\partial_\mu\alpha) (e^{iq\alpha}\phi)\\ &=e^{iq\alpha}\bigl[ iq(\partial_\mu\alpha)\phi+\partial_\mu\phi -iqA_\mu\phi-iq(\partial_\mu\alpha)\phi\bigr]\\ &=e^{iq\alpha}D_\mu\phi. \end{aligned}$$

Expanding both orders of covariant differentiation leaves

$$[D_\mu,D_\nu]\phi=-iqF_{\mu\nu}\phi, \qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.$$

The curvature $F_{\mu\nu}$ is unchanged by $A_\mu\mapsto A_\mu+\partial_\mu\alpha$ for smooth $\alpha$, since mixed partial derivatives commute. When $q=0$, both formulas reduce to ordinary differentiation. This check would catch an unpaired sign change in the definition of $D_\mu$.

Refresher on connections

A position-dependent change of basis produces an extra derivative of that basis. A connection cancels this extra term so that the derivative transforms in the same way as the original field. The gauge field above is the simplest example. In GR, the Christoffel connection plays a related role for coordinate components; a spin connection does so for spinors in a local frame. You will learn the spinor construction when it is needed. For now, reproduce the cancellation above with a constant $\alpha$ and then with a varying one.

A Gaussian integral and its moment

Let $a>0$ be real, and define

$$Z(a,J)=\int_{-\infty}^{\infty}dx\, \exp\left(-\frac a2x^2+Jx\right)$$

for real $J$. Evaluate $Z(a,J)$ and the normalized moment

$$\langle x^2\rangle_a= \frac{\int dx\,x^2e^{-ax^2/2}}{\int dx\,e^{-ax^2/2}}.$$

Why does the condition $a>0$ matter?

Hint

Complete the square. Obtain the moment either by differentiating twice with respect to $J$, or once with respect to $a$.

Complete solution

Completing the square yields

$$-\frac a2x^2+Jx =-\frac a2\left(x-\frac Ja\right)^2+\frac{J^2}{2a}.$$

A real shift of the integration variable and the elementary Gaussian integral therefore give

$$Z(a,J)=\sqrt{\frac{2\pi}{a}}\, e^{J^2/(2a)}.$$

Since $\partial_J^2Z(a,J)|_{J=0}$ inserts $x^2$, division by $Z(a,0)$ gives $\langle x^2\rangle_a=1/a$. Independently,

$$-2\partial_a\log Z(a,0)=\frac1a$$

gives the same answer. Positivity of $a$ ensures decay at both ends of the real contour. At $a=0$ the zero-source integral has infinite volume; for $a<0$ its integrand grows exponentially. Neither case is covered by the formula as a convergent real integral.

If $x$ has units, the exponent requires $[a]=[x]^{-2}$ and $[J]=[x]^{-1}$. Thus $1/a$ has precisely the units of $x^2$.

Refresher on Gaussian normalization

To fix the normalization without memorization, square $I=\int_{\mathbb R}e^{-x^2/2}dx$. The resulting two-dimensional integral in polar coordinates is $I^2=2\pi\int_0^\infty r e^{-r^2/2}dr=2\pi$. Positivity gives $I=\sqrt{2\pi}$. Rescaling $x\mapsto x/\sqrt a$ supplies the factor $a^{-1/2}$. Return to the Gaussian question and check the power of $a$ before doing any differentiation.

Two spin matrices

Consider the Pauli matrices

$$\sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad \sigma_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix}, \quad \sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

Compute $\sigma_1\sigma_2$ and $\sigma_2\sigma_1$. For an ordinary complex two-component column $u=(a,b)^T$, compute $u^\dagger\sigma_3u$. No anticommuting variables are involved in this question.

Hint

Use row-by-column multiplication. The dagger combines complex conjugation and transpose.

Complete solution

Multiplication gives

$$\sigma_1\sigma_2 =\begin{pmatrix}i&0\\0&-i\end{pmatrix}=i\sigma_3, \qquad \sigma_2\sigma_1=-i\sigma_3.$$

Thus $\{\sigma_1,\sigma_2\}=0$, whereas $[\sigma_1,\sigma_2]=2i\sigma_3$. Braces mean $AB+BA$ and brackets mean $AB-BA$ here. Replacing one bracket with the other changes the algebra.

Since $u^\dagger=(a^*,b^*)$,

$$u^\dagger\sigma_3u=|a|^2-|b|^2,$$

which is real but can have either sign. The reality follows from $\sigma_3^\dagger=\sigma_3$; Hermiticity alone does not imply positivity. For $u=(1,0)^T$ the answer is $1$, while for $u=(0,1)^T$ it is $-1$.

Refresher on spinor notation

At this stage a two-component spinor is a column on which rotation matrices act; the details of its Lorentz transformation will be introduced separately for each dimension. Do not infer a conjugation rule merely from the number of components. Bilal’s Introduction to Supersymmetry, section 2.2 provides an optional four-dimensional introduction. Its dotted and undotted notation is not required to solve the matrix question.

Decide what to read next

If these calculations are comfortable, go directly to the first lesson. If one was unfamiliar, use its refresher and repeat that calculation with a changed input: reverse the phase convention, set $q=0$, replace $a$ by $2a$, or choose another column $u$. A useful self-check explains why the answer changes, rather than only producing a new number.

The next new tool is Grassmann calculus. It changes how odd quantities reorder and differentiate. Ordinary matrix noncommutativity and Grassmann anticommutation are different ideas; the course will keep them distinct.