I am reading Efetov's article on Anderson Localization, where some kind of supersymmetric formalism is used, and I am currently trying to make sense of the definitions. The most useful reference is this long review article, around equation 2.20.

Here are the useful definitions. I will only use 2x2 supermatrices, which is enough for my question, and I will write complex numbers with latin letters, and Grassmann numbers with greek letters.

The complex conjugate of a Grassmann number $\theta$ is $\theta^*$, and we use the definition $$(\theta^*)^*=-\theta.$$

A supermatrix $F$ takes the form

$$F=\begin{pmatrix}a & \theta \\ \eta & b\end{pmatrix},$$ and its transpose is defined as $$F^T=\begin{pmatrix}a & -\eta \\ \theta & b\end{pmatrix},$$ which implies the nice property that if we denote the hermitian conjugate by $F^\dagger=(F^T)^*$, then $(F^\dagger)^\dagger=F$. (NB: it is important that the Grassmann elements of $F$ are complex to have this property).

A supermatrix $U$ is said to be unitary if $$U U^\dagger = U^\dagger U =1.$$

Now here's the question. Writing $U$ in the form of $F$, I am trying to find the most general form of $U$ such that it is unitary. The problem is that I cannot find a consistent way to do that.

The relevant equations, using $U U^\dagger = U^\dagger U =1$, are $$ |a|^2-\theta^*\theta=1,\\ |b|^2+\eta^*\eta=1,\\ |a|^2-\eta^*\eta=1,\\ |b|^2+\theta^*\theta=1,\\ a\theta^*+b^*\eta=0,\\ a^*\eta+b\theta^*=0,\\ $$ as well as the complex conjugate of the last two equations.

Playing with this, one finds (without any assumptions) that $|a|^2+|b|^2=2$ and $\eta^*\eta=\theta^*\theta$. Now assuming that $a\neq 0$, one gets $\eta=-b/a^*\theta^*$, which implies that (if $\theta^*\neq 0$) $|a|^2=|b|^2=1$.

Here comes the problem : we also have that $|a|^2-\theta^*\theta=1$ which implies that $\theta^*\theta=0$, which is not possible, unless $\theta^*=\theta$, but this breaks our (important) assumption that $\theta$ is complex (to have $(U^\dagger)^\dagger=U$).

If instead we have $a=0$, this implies $\theta^*\theta=-1$, and I am not sure how to interpret this... (Yes, $\theta^*\theta$ is bosonic, but I don't think this equality makes sense, since if we integrate with respect to $\theta$ and $\theta^*$ both side, we get $1=0$...).

Otherwise, one could just impose $\theta=\eta=0$, but in that case, there is not really a point to define unitary supermatrices, since their effect is kind of trivial (i.e. change the phase of the bosons and fermions independently).

  • $\begingroup$ D0es multiplication of supermatrices differ from ordinary matrix multiplication? I am not versed in the subject, but just going from what you have written and using matrix multiplication I get different equations for being unitary. Namely, I get some (but not all) of your minus signs to be plus signs and vice versa. $\endgroup$ Oct 4, 2016 at 13:43
  • 1
    $\begingroup$ Yes, matrix multiplication is the same. But $\theta^*\theta=-\theta\theta^*$, which might explain some signs. $\endgroup$
    – Adam
    Oct 4, 2016 at 13:45

1 Answer 1


The problem was in assuming that because one has $$ (|b|^2-|a|^2)\theta^*=0,$$ then $|b|^2=|a|^2$. Indeed, the only thing it tells you is that $$|b|^2=|a|^2+\sigma \theta^*,$$ where $\sigma$ can be another Grassmanian (or a complex number). In fact, one can check that the two independent equations $|b|^2=1-\theta^*\theta$ and $|a|^2=1+\theta^*\theta$ are compatible with $ (|b|^2-|a|^2)\theta^*=0$.

The best way to solve the above equations is to solve the equations for $a$ and $b$ in terms of $\theta^*\theta$. One gets $$a=e^{i\alpha}\left(1+\frac{\theta^*\theta}{2}\right),\\ b=e^{i\beta}\left(1-\frac{\theta^*\theta}{2}\right), $$ with $\alpha$ and $\beta$ two arbitrary phases. Then the equation linking $\eta$ and $\theta^*$ is used to get $$\eta=-e^{i(\alpha+\beta)}\theta^*.$$ One then checks that these results are compatible with all the equations in the question.

Thus, a general 2x2,unitary supermatrix is parametrized by only three numbers : two real numbers $\alpha$ and $\beta$, and a Grassman number $\theta$, and it reads $$U=\begin{pmatrix} e^{i\alpha}\left(1+\frac{\theta^*\theta}{2}\right) && \theta \\ -e^{i(\alpha+\beta)}\theta^* && e^{i\beta}\left(1-\frac{\theta^*\theta}{2}\right). \end{pmatrix}$$

One then checks that it is indeed superunitary.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.