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I'm running into an annoying problem I am unable to resolve, although a friend has given me some guidance as to how the resolution might come about. Hopefully someone on here knows the answer.

It is known that a superfunction (as a function of space-time and Grassmann coordinates) is to be viewed as an analytic series in the Grassmann variables which terminates. e.g. with two Grassmann coordinates $\theta$ and $\theta^*$, the expansion for the superfunction $F(x,\theta,\theta^*)$ is

$$F(x,\theta)=f(x)+g(x)\theta+h(x)\theta^*+q(x)\theta^*\theta.$$

The product of two Grassmann-valued quatities is a commuting number e.g. $\theta^*\theta$ is a commuting object. One confusion my friend cleared up for me is that this product need not be real or complex-valued, but rather, some element of a 'ring' (I don't know what that really means, but whatever). Otherwise, from $(\theta^*\theta)(\theta^*\theta)=0$, I would conclude necessarily $\theta^*\theta=0$ unless that product is in that ring.

But now I'm superconfused (excuse the pun). If Dirac fields $\psi$ and $\bar\psi$ appearing the QED Lagrangian $$\mathcal{L}=\bar\psi(i\gamma^\mu D_\mu-m)\psi-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ are anticommuting (Grassmann-valued) objects, whose product need not be real/complex-valued, then is the Lagrangian no longer a real-valued quantity, but rather takes a value which belongs in my friend's ring??? I refuse to believe that!!

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    $\begingroup$ My friend followed up with saying that these Grassmann variables are like differential forms... how do I write the QED Lagrangian in that language? $\endgroup$
    – QuantumDot
    Commented Oct 14, 2012 at 1:44
  • $\begingroup$ In mathematics, a "ring" (derived from "number ring") is really a number system. You aren't given division in a ring (if you have it, you have a "division ring", or if additionally multiplication is commutative a "field")...so your thinking $(\theta\theta^*)=0$ is invalid. $\endgroup$
    – Alex Nelson
    Commented Oct 14, 2012 at 2:03
  • $\begingroup$ Also note differentiating and integrating Grassmann variables are quite different! They are the same! This is how you get rid of these "Grassmannian mysteries" in practice... $\endgroup$
    – Alex Nelson
    Commented Oct 14, 2012 at 2:05
  • $\begingroup$ @AlexNelson so what does that mean about the QED Lagrangian? Is it no longer a real-valued object? $\endgroup$
    – QuantumDot
    Commented Oct 14, 2012 at 3:12
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    $\begingroup$ In classical field theory $\psi$'s are of course usual complex valued fields. When you quantize you are bound to require that they satisfy anticommutation relations (for otherwise some things go wrong in quantum theory). So the "ring" is ring of operators on Hilbert space. However if you take path integral approach to quantization then you have to treat $\psi$'s as anticommuting variables so that your results agree with results of Hilbert space formalism. $\endgroup$
    – user10001
    Commented Oct 14, 2012 at 8:52

3 Answers 3

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A supernumber $z=z_B+z_S$ consists of a body $z_B$ (which always belongs to $\mathbb{C}$) and a soul $z_S$ (which only belongs to $\mathbb{C}$ if it is zero), cf. Refs. 1 and 2.

A supernumber can carry definite Grassmann parity. In that case, it is either $$\text{Grassmann-even/bosonic/a $c$-number},$$ or $$\text{Grassmann-odd/fermionic/an $a$-number},$$ cf. Refs. 1 and 2.$^{\dagger}$ The letters $c$ and $a$ stand for commutative and anticommutative, respectively.

One can define complex conjugation of supernumbers, and one can impose a reality condition on a supernumber, cf. Refs. 1-4. Hence one can talk about complex, real and imaginary supernumbers. Note that that does not mean that supernumbers belong to the set of ordinary complex numbers $\mathbb{C}$. E.g. a real Grassmann-even supernumber can still contain a non-zero soul.

An observable/measurable quantity can only consist of ordinary numbers (belonging to $\mathbb{C}$). It does not make sense to measure a soul-valued output in an actual physical experiment. A soul is an indeterminate/variable, i.e. a placeholder, except it cannot be replaced by a number to give it a value. A value can only be achieved by integrating it out!

In detail, a supernumber (that appears in a physics theory) is eventually (Berezin) integrated over the Grassmann-odd (fermionic) variables, say $\theta_1$, $\theta_2$, $\ldots$, $\theta_N$, and the coefficient of the fermionic top monomial $\theta_1\theta_2\cdots\theta_N$ is extracted to produce an ordinary number (in $\mathbb{C}$), which in principle can be measured.

E.g. the Grassmann-odd (fermionic) variables $\psi(x,t)$ in the QED Lagrangian should eventually be integrated over in the path integral.

References:

  1. planetmath.org/supernumber.

  2. Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.

  3. Pierre Deligne and John W. Morgan, Notes on Supersymmetry (following Joseph Bernstein). In Quantum Fields and Strings: A Course for Mathematicians, Vol. 1, American Mathematical Society (1999) 41–97.

  4. V.S. Varadarajan, Supersymmetry for Mathematicians: An Introduction, Courant Lecture Notes 11, 2004.

--

$^{\dagger}$ In this answer, the words bosonic (fermionic) will mean Grassmann-even (Grassmann-odd), respectively.

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  • $\begingroup$ Ok! So the Lagrangian is a real supernumber! Is that correct? $\endgroup$
    – QuantumDot
    Commented Oct 14, 2012 at 19:40
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    $\begingroup$ Yes, the QED Lagrangian is a real Grassmann-even supernumber with a non-zero soul. $\endgroup$
    – Qmechanic ♦
    Commented Oct 14, 2012 at 19:45
  • $\begingroup$ Awesome! Now I have a question about SUSY: The supercoordinates have the following supersymmetry transformation law: $(x^\mu,\theta,\bar\theta) \rightarrow \big((x+a)^\mu-i(\theta\sigma^\mu\bar\alpha-\alpha\sigma^\mu\bar\theta),\,\theta+\alpha,\,\bar\theta+\bar\alpha\big)$. Does this mean the space-time part is transformed into a supernumber (with a non-zero soul)? seems strange!!! $\endgroup$
    – QuantumDot
    Commented Oct 14, 2012 at 19:45
  • $\begingroup$ I have a follow up question you can find here: physics.stackexchange.com/q/40957 $\endgroup$
    – QuantumDot
    Commented Oct 16, 2012 at 15:31
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The Lagrangian may be proved to be real but the individual factors in its terms, such as $\psi$, are neither real nor complex. They're anticommuting. There are no "particular" elements of this set of anticommuting numbers that one could "enumerate" (except for zero) and they can't appear as final predictions for observable quantities but it still makes a perfect sense to do algebra with them. A product of an even number of anticommuting variables is commuting which means that it may take particular values that may be measured and compared with theoretical predictions.

I think that I am not the only one who doesn't really understand what you're asking about but there is a chance that the answer is either in the previous paragraph or the text below:

http://motls.blogspot.com/2011/11/celebrating-grassmann-numbers.html?m=1

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    $\begingroup$ This is in every intro to QFT that cares about such trivialities at all. You don't need a reference because it fits here. Taking $\psi$ as Grassmann-odd operators, it's known that the Hermitean conjugate to $\bar\psi \psi$ is the same thing backwards, so the mass term is Hermitian. The same with the cubic interaction term which has a real coefficient. The kinetic term needs an $i$ to be real because the Hermitian conjugation exchanges $\psi$ with $\partial_\mu\psi$ and to exchange them back, one pays a minus sign for Fermi statistics which implies $-1$ from $i^\dagger=-i$. $\endgroup$
    – Luboš Motl
    Commented Oct 14, 2012 at 7:13
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    $\begingroup$ Let me say that I was really proving that the object, when operators are substituted, is Hermitian, which implies that its eigenvalues are real. This working with the "full operators" - with a quantized theory in the operator approach - is a convenient and reliable way to decide how to complex-conjugate Grassmann numbers etc.: one Hermitian conjugates the operators using $(AB)^\dagger=B^\dagger A^\dagger$ and uses the same results for the Grassmann numbers, too. $\endgroup$
    – Luboš Motl
    Commented Oct 14, 2012 at 7:14
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    $\begingroup$ Dear QuantumDot, yes, $\theta$ times $\phi$ is a complex number, so the product has a value you can "name", but $\theta$ and $\phi$ separately don't have values you can "name". Here I am assuming you are talking about the most elementary Grassmann-odd variables. One could perhaps also talk about the quaternion-based Grassmann variabels, whose product would be a general quaternion, or - in QFT - Grassmann-odd operators, i.e. fermionic operators, whose product is a Grassmann-even i.e. bosonic ordinary operator. $\endgroup$
    – Luboš Motl
    Commented Oct 14, 2012 at 7:16
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    $\begingroup$ Thanks Lubos, but if $\theta\phi\in\mathbb{C}$ and $(\theta\phi)^2=0$ (by Grassmann algebra), then that would seem to suggest $\theta\phi$ must necessarily be zero. .. something has gone wrong in my reasoning...? (edit: yes, I am talking about the elementary Grassmann-odd variables) $\endgroup$
    – QuantumDot
    Commented Oct 14, 2012 at 7:18
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    $\begingroup$ @Luboš Motl: If something appears in an exponent, this does not necessarily mean that this "something" is "a real commuting c-number" - we know how to define exponents of, e.g., supernumbers. Your caveat "at least formally" can make pretty much any statement correct, "at least formally", but it strips any content from any statement. I insist that, strictly speaking, the Lagrangian in question is not a c-number, as it contains nonzero "soul". E.g., in a Grassmann algebra with a finite number of generators the fermionic part of the Lagrangian would vanish if multiplied by all of the generators. $\endgroup$
    – akhmeteli
    Commented Oct 14, 2012 at 13:50
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Let us sort out some terminology issues first. If the fermionic fields in your Lagrangian are Grassmanian, that means that the Lagrangian is classical, i.e. second quantization has not been performed yet. You can write a classical Lagrangian using c-number fermionic fields, but, as far as I understand, it is generally recognized now that one should use the classical Lagrangian with Grassmanian fermionic fields.

I also ran into the issue that you describe some time ago. I may be mistaken, but my conclusion was that indeed, the Lagrangian is not real, for the reasons that you give in your question. On the other hand, it is not obvious why this is necessarily bad.

EDIT: Maybe, to avoid ambiguity, I should have written that the Lagrangian is not real-valued

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  • $\begingroup$ Your conclusion is in line with my friend's conclusion. I feel very disconcerted by this. If I compute the stress-energy tensor by following Noether's theorem, doesn't this mean the energy is also not real-valued? What's going on??? $\endgroup$
    – QuantumDot
    Commented Oct 14, 2012 at 7:22
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    $\begingroup$ Yes, I may be mistaken, but it looks like the energy is not real-valued either. But again, why is this necessarily bad? We are not used to it, but there a lot of things in physics we are not used to. Remember, the classical Lagrangian (and the classical energy) are only some interim structures in QED. $\endgroup$
    – akhmeteli
    Commented Oct 14, 2012 at 7:51
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    $\begingroup$ Don't get freaked out--- the energy is not real valued, but the expectation value in any state is real valued. You have to remember that these grassman things are classical limits of quantum fields, and the states are constructed by using the ring elements, and the end result, when you compute some amplitude or expection value is always real. $\endgroup$
    – Ron Maimon
    Commented Oct 15, 2012 at 5:24

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