In my opinion, the Grassmann number "apparatus" is one of the least intuitive things in modern physics.

I remember that it took a lot of effort when I was studying this. The problem was not in the algebraic manipulations themselves -- it was rather psychological: "Why would one want to consider such a crazy stuff!?"

Later one just gets used to it and have no more such strong feelings for the anticommuting numbers. But this psychological barrier still exists for newbies.

Is there a way to explain or motivate one to learn the Grassmann numbers? Maybe there is a simple yet practical problem that demonstrates the utility of those? It would be great if this problem would provide the connection to some "not-so-advanced" areas of physics, like non-relativistic QM and/or statistical physics.

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    $\begingroup$ Lubos is right - however non-relativistic QM still benefits from the path integral formulation, and how do you make path integrals for fermions? Likewise, the path integral formulation is the cornerstone of numerical equilibrium statistical mechanics. There would be no Determinant Monte Carlo for fermions without Grassmann numbers. The problem is that indistinguishable particle statistics are a complicated and unintuitive, but easily stated fact of nature -- Grassmann numbers are a math trick that helps with formal understanding; there's no way to make them more intuitive than the physics! $\endgroup$ – wsc Feb 11 '11 at 16:53
  • $\begingroup$ I wonder, how did you learn about Grassmann numbers? When I learned about them, their motivation was immediately clear as a computational tool for fermions (especially path integrals). $\endgroup$ – Raskolnikov Feb 11 '11 at 17:36
  • $\begingroup$ I was hoping someone would recommend a reference for this $\endgroup$ – user37343 Apr 2 '14 at 21:22
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    $\begingroup$ I personally just think of them as formal operators that happen to have particularly simple commutation relations - everything anticommutes - so you can throw away almost all of their ordering information. It's not ideal for intuition, but I find that calling them "numbers" makes it too tempting to mentally assign them "values" as if they were real variables. It's important to remember that they're inherently dummy variables that only ever get differentiated or integrated, and can never appear in any final physical answer. $\endgroup$ – tparker Jul 23 '17 at 4:15
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    $\begingroup$ I only found myself comfortable with them once I gave up on trying to understand them intuitively, and just started treated them as formal symbols to be manipulated without having any picture in your head of what you're doing. They're very Zen that way. As Yoda said, "You must unlearn what you have learned." $\endgroup$ – tparker Jul 23 '17 at 4:18

I don't have an answer to the question "why would one want to consider such crazy stuff in physics?" since I don't know much physics, but as a mathematics student I do have an answer to the question "why would one want to consider such crazy stuff in mathematics?"

What physicists call Grassmann numbers are what mathematicians call elements of the exterior algebra $\Lambda(V)$ over a vector space $V$. The exterior algebra naturally arises as the solution to the following geometric problem. Say that $V$ has dimension $n$ and let $v_1, ... v_n$ be a basis of it. We would like a nice natural definition of the $n$-dimensional volume of the paralleletope defined by the vectors $\epsilon_1 v_1 + ... + \epsilon_n v_n, e_i \in \{ 0, 1 \}$. When $n = 2$ this is the standard parallelogram defined by two linearly independent vectors, and when $n = 3$ this is the standard paralellepiped defined by three linearly independent vectors.

The thing about the naive definition of volume is that it is very close to having really nice mathematical properties: it is almost multilinear. That is, if we denote the volume we're looking at by $\text{Vol}(v_1, ... v_n)$, then it is almost true that $\text{Vol}(v_1, ... v_i + cw, ... v_n) = \text{Vol}(v_1, ... v_n) + c \text{Vol}(v_1, ... v_{i-1}, w, v_{i+1}, ... v_n)$. You can draw nice diagrams to see this readily. However, it isn't actually completely multilinear: depending on how you vary $w$ you will find that sometimes the volume shrinks to zero and then goes back up in a non-smooth way when really it ought to keep getting more negative. (You can see this even in two dimensions, by varying one of the vectors until it goes past the other.)

To fix that, we need to look instead at oriented volume, which can be negative, but which has the enormous advantage of being completely multilinear and smooth. The other major property it satisfies is that if any of the two vectors $v_i$ agree (that is, the vectors are linearly dependent) then the oriented volume is zero, which makes sense. It turns out (and this is a nice exercise) that this is equivalent to oriented volume coming from a "product" operation, the exterior product, which is anticommutative. Formally, these two conditions define an element of the top exterior power $\Lambda^n(V)$ defined by the exterior product $v_1 \wedge v_2 ... \wedge v_n$, and choosing an element of this top exterior power (a volume form) allows us to associate an actual number to an $n$-tuple of vectors which we can call its oriented volume in the more naive sense. If $V$ is equipped with an inner product, then there are two distinguished elements of $\Lambda^n(V)$ given by a wedge product of an orthonormal basis in some order, and it's natural to pick one of these as a volume form.

Alright, so what about the rest of the exterior powers $\Lambda^p(V)$ that make up the exterior algebra? The point of these is that if $v_1, ... v_p, p < n$ is a tuple of vectors in $V$, we can consider the subspace they span and talk about the $p$-dimensional oriented volume of the paralleletope given by the $v_i$ in this subspace. But the result of this computation shouldn't just be a number: we need a way to do this that keeps track of what subspace we're in. It turns out that mathematically the most natural way to do this is to keep in mind the requirements we really want out of this computation (multilinearity and the fact that if the $v_i$ are not linearly independent then the answer should be zero), and then just define the result of the computation to be the universal thing that we get by imposing these requirements and nothing else, and this is nothing more than the exterior power $\Lambda^p(V)$.

This discussion hopefully motivated for you why the exterior algebra is a natural object from the perspective of geometry. Since Einstein, physicists have been aware that geometry has a lot to say about physics, so hopefully the concept makes a little more sense now.

Let me also say something about how modern mathematicians think about "space" in the abstract sense. The inspiration for the modern point of view actually derives at least partially from physics: the only thing you can really know about a space are observables defined on it. In classical physics, observables form a commutative ring, so one might say roughly speaking that the study of commutative rings is the study of "classical spaces." In mathematics this study, in the abstract, is called algebraic geometry. It is a very sophisticated theory that encompasses classical algebraic geometry, arithmetic geometry, and much more, and it is in large part because of the success of this theory and related commutative ring approaches to geometry (topological spaces, manifolds, measure spaces) that mathematicians have gotten used to the slogan that "commutative rings are rings of observables on some space."

Of course, quantum mechanics tells us that the actual universe around us doesn't work this way. The observables we care about don't commute, and this is a big issue. So mathematically what is needed is a way to think about noncommutative rings as "quantum spaces" in some sense. This subject is very broad, but roughly it goes by the name of noncommutative geometry. The idea is simple: if we want to take quantum mechanics completely seriously, our spaces shouldn't have "points" at all because points are classical phenomena that implicitly require a commutative ring of observables, which we know is not what we actually have. So our spaces should be more complicated things coming from noncommutative rings in some way.

Grassmann numbers satisfy one of the most tractable forms of noncommutativity (actually they are commutative if one alters the definition of "commutative" very slightly, but never mind that...), and even better it is a form of noncommutativity that is clearly related to something physicists care about (the properties of fermions), so anticommuting observables are a natural step up from commuting observables in order to get our mathematics to align more closely with reality while still being able to think in an approximately classical way.

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    $\begingroup$ Great answer, Qiaochu. I'd add that the exterior algebra is a natural object when we have no metric on a linear space, but we do have a completely antisymmetric structure, which in coordinate terms is the determinant. When there is also a metric structure, then the Clifford algebra is also natural. In the case of a Fermionic quantum field, there is a metric structure as well as the antisymmetric structure, so one can use either the exterior or Clifford algebraic structures, whichever seems most useful at the time. Physics papers and textbooks alike could housekeep more cleanly on this. $\endgroup$ – Peter Morgan Feb 11 '11 at 19:11
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    $\begingroup$ @Qiaochu Yuan, it's a good lesson on a subject that you chose to lecture about but one must be very careful before he learns about properties of Grassmann numbers from the structures you have described. The structures you have described are more linked to differential forms and exterior products - but in the normal physics edition, all their coordinates are still bosonic, Grassmann-even objects! That's also why you didn't encounter the Berezin integral for the Grassmann numbers - your story actually has almost nothing "detailed" to do with it, except for the name of Grassmann. $\endgroup$ – Luboš Motl Feb 11 '11 at 21:58
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    $\begingroup$ So it's really misleading to present physicists' Grassmann numbers as exterior products of vectors and p-forms. The vectors are multi-component objects with a lot of ordinary bosonic components, and the anticommutativity comes from an antisymmetric treatment of the bosonic components. But the Grassmann numbers used by physicists are literally one-component numbers, to be fully analogous to bosonic scalars and not bosonic vectors or forms, and their anticommutativity is inherent rather than a derived feature of a determinant-like formula that deals with many components. So it's not the same. $\endgroup$ – Luboš Motl Feb 11 '11 at 22:01
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    $\begingroup$ @Luboš: like I said, I don't know enough physics to say anything intelligent about the exact connection. But on the mathematical side, as far as I can tell all you're saying is that you're working in the category of super vector spaces (or supermanifolds, or whatever) instead of dealing with exterior powers in the category of ordinary vector spaces (ordinary manifolds, or whatever). As a mathematician I am well aware of this distinction, but I don't know how much it would have added to the discussion to talk about what the category of super vector spaces looks like. $\endgroup$ – Qiaochu Yuan Feb 11 '11 at 22:07
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    $\begingroup$ @Luboš: I am also wondering if you understood the second half of my answer. It precisely addresses a general state of mind where one thinks about "inherent anticommutativity" in the "scalars." $\endgroup$ – Qiaochu Yuan Feb 12 '11 at 0:41

One of your questions was whether there exists a use of the Grassmann number apparatus in statistical physics. The answer is yes. This is not surprising since there are many deep connections between QFT and statistical physics (even classical one). For example, the Grassman technique can be used as a tool for calculating the partition function of the dimer model - this model originates from the following question: if diatomic molecules are adsorbed on a surface, forming a single layer, how many ways can they be arranged, what is the entropy of the system? For a work on this, see: "Grassmann Variables and Exact Solutions for Two-Dimensional Dimer Models" ( http://arxiv.org/abs/cond-mat/9711156 ). One can also use Grassmann techniques in the field of classical spin systems, a review of this can be found in "Grassmann techniques applied to classical spin systems " ( http://arxiv.org/abs/0905.1104 ).

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    $\begingroup$ Good point. There can be classical statistical mechanics models whose partition functions may be identified with the quantum partition sums of quantum field theories with fermions. However, this still doesn't give one a "classical intuition" because you still need the quantum field theories to justify the formulae. $\endgroup$ – Luboš Motl Feb 11 '11 at 17:25
  • $\begingroup$ yes, i agree that this doesn't give such a "classical intuition", but (as i suggested) they can be useful for these problems. $\endgroup$ – Zoltan Zimboras Feb 11 '11 at 17:29
  • $\begingroup$ That is exactly what I am looking for. Motivation for Grassmann numbers for advanced students. The only problem -- they are a bit complex. $\endgroup$ – Kostya Feb 11 '11 at 17:35
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    $\begingroup$ But Kostya, how is this any more motivation than the simple construction of fermionic path integrals by themselves? The fact that $d$ dimensional quantum mechanics corresponds to classical equilibrium stat. mech. in $d+1$ dimensions is already a complication on top of the story. $\endgroup$ – wsc Feb 11 '11 at 18:14
  • $\begingroup$ @wsc either that, or a radical simplification. The correspondence between $d$ dim QM and $d+1$ equilibrium stat mech, lies at the heart of Horava's work on what is now known as "Horava-Lifshitz" gravity. In the abstract of Membranes at Quantum Criticality he states: "We propose a quantum theory of membranes designed such that the ground- state wavefunction of the membrane with compact spatial topology $\Sigma_h$ reproduces the partition function of the bosonic string on worldsheet $\Sigma_h$." Then he goes on to show how gravity arises naturally in the $d+1$.. $\endgroup$ – user346 Feb 11 '11 at 19:14


There can't be any role for the Grassmann numbers outside quantum physics. The reason is simple: they can't take any "values" from a particular "set". Instead, they only become analogous to the regular commuting numbers once we use them as intermediate steps and we integrate or differentiate over them or set them equal to zero. Indeed, the integrals are algebraically analogous to the integral over commuting variables. However, the overall sign in the presence of the Grassmann numbers is always a bit undetermined - and one must square the result to get rid of the ambiguity. That's why the Berezin (Grassmann) integrals must always be interpreted as probability amplitudes and quantum mechanics is needed.

One may say that a commuting number may be a product of two (or another even number of) Grassmann numbers - so a commuting number has an even number of previously unknown building blocks. But the measurable number - the number that Nature offers as a part of Her "user interface" to the consumers i.e. observers - is always commuting. However, at the fundamental level, in the core of Her inner workings, the Grassmann numbers are as natural as the commuting numbers.

The Grassmann numbers are the $c$-numbers for fermionic creation or annihilation operators in the same way as the ordinary commuting $c$-numbers are the numbers for the ordinary bosonic creation or annihilation operators (and many other operators). That's the closest analogy. It's natural for two operators to "almost commute" with one another or "almost anticommute" with one another. It must be one of the two options because if one repeats the interchange twice, we must get back to the same object, so the factor we pick by interchanging them must satisfy $S^2=+1$, leaving $S=\pm 1$ as the only possibilities.

A person must first understand why the Fermi-Dirac statistics is as natural as the Bose-Einstein statistics; then he can understand that at a deeper level, the Grassmann numbers are exactly as natural as the commuting numbers. However, everyone must also be ready that he doesn't have any "material experience" with the Grassmann numbers because they can't play any role in the classical limit - they can't appear in the "user interface" of Nature.

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    $\begingroup$ That statement seems far too strong. "Grassmann numbers" are very useful in pure mathematics (geometry and representation theory) in ways which are, but don't have to be, related to quantum physics. $\endgroup$ – Qiaochu Yuan Feb 11 '11 at 17:00
  • $\begingroup$ I suspect that you are too "negative" too. But, anyway, thank you for some nice explanations and suggestions. +1 $\endgroup$ – Kostya Feb 11 '11 at 17:11
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    $\begingroup$ Dear @Qiaochu Yuan and @Kostya, I only meant that in physics, the Grassmann numbers require quantum physics. That was primarily an answer to the question whether they may be presented in statistical mechanics. Well, I actually haven't addressed whether they're relevant for non-relativistic quantum mechanics. My answer would still be "no". One needs to formulate fermions in terms of field-theoretical variables to really use Grassmann numbers. Well, in condensed matter physics, one may use non-relativistic field theories, so this could be an example. But it's not the normal non-rel. QM. $\endgroup$ – Luboš Motl Feb 11 '11 at 17:22
  • $\begingroup$ Well, maybe I didn't make it clear. I didn't request the "classical intuition" for Grassmann numbers -- suppose that person is familiar with basic courses in QM an statphysics. $\endgroup$ – Kostya Feb 11 '11 at 17:31
  • $\begingroup$ +1. But I want to make the same point as @Kostya and @Qiaochu. Your statement "There can't be any role for the Grassmann numbers outside quantum physics" is much too strong and also somewhat misleading. $\endgroup$ – user346 Feb 11 '11 at 17:41

You may also utilize Grassmann-odd numbers while considering generalization of some simple classical models to supersymmetric version of theirs. So for example in the paper


there're considered various supersymmetrizations of Landau model - that is the model describing charged particle moving in the plane orthogonal to uniform magnetic field. The supersymmetric version roughly speaking replaces commuting coordinates of this plane by anticommuting Grassmann-odd coordinates. Subsequent quantization gives quantum model with Landau levels being solutions of stationary Schroedinger equation.

  • $\begingroup$ Dear Mikhail, the classical theories with Grassmann variables are the "simplified model" of the quantum field theories, but their only conceivable classical interpretation emerges if all the Grassmann fields are simply set to zero - erased. The additional Grassmann fields only produce a "new physical theory" if the theory is treated quantum mechanically. Take e.g. the 11D supergravity. It has the gravitino fields, they may be seen in the Lagrangian, but if you want a classical theory, they have to be set to zero so they're just a redundancy. $\endgroup$ – Luboš Motl Feb 11 '11 at 17:28
  • $\begingroup$ Agreed. Your example with 11D SUGRA is illustrative. $\endgroup$ – Mikhail Feb 12 '11 at 16:12

I think a lot of the ideas discussed here are great. The mathematical motivation by @Qiaochu Yuan is spot on. Also Lubos is correct. I have personally struggled(somewhat still struggling) with this and I just thought I would say something here.

Essentially, a good way to perhaps proceed would be to start with http://www.theory.caltech.edu/~kapustin/Ph205/wp1.pdf, for a physical motivation may be after perhaps revisiting QM to show why we would make such arguments, then convince them of the geometric connection first with some naive arguments in classical space, showing how to compute volumes. http://en.wikipedia.org/wiki/Exterior_algebra, articles on Wikipedia are great, you just have to do it in an appropriate language depending on the student background,

If you like you may then fully fill Yuan's details. By this I mean recommend appropriate algebra texts. I happen to share the sentiment that there is a lot to explore here, but well

All in all the answers and comments here seem pretty awesome. My answer is not a substitute, I'm just a guy who has thought about these things, not any expert. So in all if you adopt the strategy I mention people like me might even understand it . . . .


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