# Superspace Uncertainty Principle

Do the "operator for translations in superspace" and the "position in superspace operator" follow an uncertainty principle? How "real" is superspace? Aside from being weird (and possibly just a "mathematical trick") is superspace really just an extension of space on the same footing as real space (with off-shell degrees of freedom)?

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## 1 Answer

1) Well, no measuring device is going to measure a Grassmann-odd$^1$ number. A measuring device can only produce real outputs $\subseteq\mathbb{R}$.

However, one can measure a fermionic condensate, i.e. an expectation value $\subseteq\mathbb{R}$ of an composite bosonic Hermitian operator, where each term is built from an even number of elementary fermionic operators, typically two.

2) It also makes sense to e.g. consider canonical anticommutation relations (CAR) for fermionic operators in complete analogy with canonical commutation relations (CRR) for bosonic operators. As is well-known, CCRs can be viewed as the source/origin of Heisenberg's uncertainty principle (HUP).

It is natural to ponder if there exists a fermionic HUP? It turns out that we can only see a HUP effect of CARs indirectly. We first have to construct composite bosonic Hermitian operator observables, say $A$ and $B$, build out of some elementary fermionic (and bosonic) operators that obeys CARs (and CCRs). Again, the composite bosonic operators $A$ and $B$ must necessarily both consist of an even number of elementary fermionic operators in each of their terms. The uncertainty

$$\langle ( \Delta A )^{2} \rangle \langle ( \Delta B )^{2} \rangle ~\geq~ \frac{1}{4} | \langle [ A,B ] \rangle |^{2}$$

of $A$ and $B$ is in principle measurable, and can be linked to the pertinent CARs (and CCRs).

3) An Example. Consider Hermitian elementary fermions $c^i$, $i=1, \ldots, n$, with CAR

$$\tag{1} \{c^i, c^j\}~=~ \hbar g^{ij}{\bf 1} , \qquad g^{ji}~=~g^{ij}~\in~\mathbb{R},$$

and a fermionic condensate

$$\chi^{ij}~:=~\frac{\mathrm{i}}{2}\langle [c^i, c^j] \rangle, \qquad \chi^{ji}~=~-\chi^{ij}~\in~\mathbb{R}.$$

Let the two composite bosonic operators be

$$A~=~ \frac{\mathrm{i}}{4}\alpha_{ij} [c^i, c^j] , \qquad \alpha_{ji}=-\alpha_{ij}~\in~\mathbb{R},$$

and

$$B~=~ \frac{\mathrm{i}}{4}\beta_{ij} [c^i, c^j] , \qquad \beta_{ji}=-\beta_{ij}~\in~\mathbb{R}.$$

Then one may calculate that

$$\tag{2} \langle [ A,B ] \rangle~=~-\mathrm{i}\hbar\alpha_{ij}g^{jk}\beta_{k\ell}\chi^{\ell i} .$$

The presence of the metric $g^{jk}$ on the right-hand side of (2) is due to the CAR (1), which leads to uncertainty in measuring the observables $A$ and $B$ simultaneously.

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$^1$ In this answer, bosonic (fermionic) will mean Grassmann-even (Grassmann-odd), respectively.

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Ah, this is interesting (both the question and the answer); I never brooded about an uncertainty principle involving fermionic operators. Dear Qmechanic, could you write an additional line such that I can explicitely see how the anticommutators of the fermionic constituents of A and B are linked to the uncertainty? – Dilaton Jul 19 '12 at 16:32
I updated the answer. – Qmechanic♦ Jul 19 '12 at 19:51
Hey thanks Qmechanic, if I could I would +1 once again :-) – Dilaton Jul 19 '12 at 20:03
Thanks for the detailed answer! I only have time to skip it over at the moment so i'll comment if I have further questions about your answer. – tachyonicbrane Aug 9 '12 at 14:31
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