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  Is the classical action for fermions grassman valued or real valued?

+ 5 like - 0 dislike

Consider the classical action which appears in the exponent when calculating the amplitude of some scattering process using the path integral approach. For a theory with fermions, the fields are treated as classical anti-commuting variables: All fermionic terms in the action are constructed as the product of an even number of grassman variables in such a way that the action is a hermitian commuting number.

My question is the following: The product of an even number of grassman variables is never valued in the reals $\mathbb{R}$ (unless of course it is identically zero). It is valued in the grassman algebra. Does this mean that the action itself is valued in the grassman algebra? If so, then it seems to cause a problem in any theory which couples fermions to bosons: When extreemizing the action one would obtain equations of motion which involve both real valued terms as well as terms valued in the grassman algebra, and these terms must surely decouple from one another. As an example, consider a theory of gravity coupled to Dirac spinors. Varying the classical action with respect to the vierbein, one would obtain an Einstein equation with a stress energy tensor constructed from spinor bilinears. In other words, the left hand side of the Einstein equation would be real valued, while the right hand side would be valued in the grassman algebra. This doesn't seem to make sense so I must be misunderstanding something, but I don't quite see where.

1. Is my understanding correct that the classical action is constructed using grassman valued numbers, and is therefore valued in the grassman algebra and not the reals?

2. If the action is grassman valued, then what about the classical equations of motion? Are these grassman valued as well, and if so do the `grassman' and `real' parts necessarily decouple from one another? 

Any help you could give me would be greatly appreciated as I am pretty confused by this!

asked Jul 17, 2017 in Theoretical Physics by anonymous [ no revision ]
recategorized Jul 17, 2017 by Dilaton

1 Answer

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I don't fully understand Grassmann variables myself, but I think I can answer your question:

1. Yes, the action itself is Grassmann-valued. This is not as bad as it sounds, because in the path integral, we integrate over them, and we know how to do that with Grassmann variables. The action does not have to be real for the path integral to make sense.

In fact, bosonic fields take values in the Grassmann algebra as well. Bosonic values are always even elements of the algebra. For instance, when doing a Hubbard-Stratonovich transformation, we shift e.g. $\phi \to \phi + \overline{\psi}\psi$ where $\phi$ is a bosonic field and $\overline{\psi},\psi$ are fermionic fields. Clearly, this only makes sense when the field $\phi$ takes values in the Grassmann algebra, and is not just a real number.

That said, the action is always an even element of the Grassmann algebra, so it commutes with all other elements. In that way, it behaves more like a "number".

For a mathematically precise review of Grassmann variables and integration, I found Terence Tao's blog post "Supercommutative gaussian integration, and the gaussian unitary ensemble" extremely helpful. This question is answered there as well.

2. In a path integral setting, the classical equations of motion are obtained by looking at the stationary points of the exponent. This makes sense when the exponent is real-valued, because we can argue that the integral is dominated by the contributions from the stationary points. Unfortunately, I don't know whether this argument still makes sense for Grassmann fields! After all, we cannot argue that the integrand has a maximum, because that assumes that the integrand is a real number which can be compared with other real numbers. I don't know how to justify it.

Of course, we can still formally write down the Euler-Lagrange equations. For this, we only need to know how to perform derivatives with respect to fields. These equations now take values in the Grassmann algebra as well, and would need to be solved with this in mind. This has been done to obtain a classical variant of spin, e.g. in F.A. Berezin and S. Marinov, "Particle spin dynamics as the grassmann variant of classical mechanics" (1977). I am not sure how far this has been developed, I am unaware of any "serious" applications (and would love to be corrected on this).

answered Jul 17, 2017 by Greg Graviton (775 points) [ revision history ]
edited Jul 24, 2017 by Greg Graviton

Just one small remark. The article about spin dynamics was written by Berezin and Marinov.

@DavidBarMoshe Corrected, thanks! I had copied the author information from the Springer page, but the latter was apparently not complete.

Two small comments:

1. The Lagrangian density and the action are even Grassmann valued, hence live in a commutative subalgebra of the noncommutative Grassmann algebra. This is conceptually important.

2. The concept of a stationary point still makes sense mathematically as it only requires a notion of differentiation with respect to a field, and not an order relation.

@ArnoldNeumaier Good points, I will edit the answer. Concerning 2, while it's possible to define a stationary point is by differentiation, I don't see how it can related to the value of the integral (in the Grassmann case). Is there a way to do that?

@GregGraviton: What are you going to do with the integral value if it is a real number, for example?

I think I disagree with this answer. In the Hubbard-Stratonovich example, $\psi$ is valued in the Grassmann algebra and $\bar\psi$ in its dual, with $\bar \psi \psi$ short-hand for the pairing between them to the complex numbers.

@RyanThorngren Well, Terence Tao presents it this way, and I found it very convincing, as this way has the virtue of being mathematically rigorous. On the other hand, if $\overline{\psi}$ were the dual of $\psi$, then the expression $\psi\overline{\psi}$ would not make sense as a complex number, it would at best be a rank-2 tensor, and I guess it would be difficult to have $\psi\overline{\psi} = -\overline{\psi}\psi$ then.

I don't see why $\bar \psi \psi$ has to be a rank 2 tensor? Perhaps you should take the trace? In spinor variables, we define $\bar \psi \psi$ by contracting spinor indices with an anti-symmetric tensor, which is where $\psi \bar \psi = - \bar \psi \psi$ comes from.

@RyanThorngren According to the best of my knowledge $\psi$ and $\bar{\psi}$ are elements of a complex Grassmann algebra. Thus I think that Greg's answer is correct. Sometimes the quantization of a Grassmann algebra is performed such that $\psi$ is quantized as the multiplication operator $\psi$ on the Grassmann algebra and $\bar{\psi}$ is quantized as the differential operator $\frac{\partial}{\partial \psi}$. But in this case it is an operators on a Hilbert space expressed as differential operator of some supernanifold.

@GregGraviton: There is a formal power series expansion of integrals of exponentials around critical points of the exponent, called Wick's theorem. This has a Grassmann analogue.

@ArnoldNeumaier Wick's theorem tells us what happens in the vicinity of a stationary point, but I don't think that it tells us why expanding around the stationary points is a good idea (as opposed to expanding around other points).

@GregGraviton: Wick's theorem can be used to expand around anything - pick a quadratic  action and consider the difference between the intended and the picked action as your perturbation. Expanding around a stationary point gets rid of the linear terms, which is the usual starting point for formal (unrenormalized) perturbation theory. That's why stationary points are the relevant ones.

I am curious: If every term of the perturbation expansion is grassmann-valued, the what is the meaning of such an expansion? What should we compare - numerical coefficients at different terms or what? In the usual perturbation theory we compare entire different terms numerically, not just their coefficients.

@VladimirKalitvianski: One always compares entire terms with each other - or, indeed, groups of terms of the same order, since individual terms may diverge when removing the regularization.

@ArnoldNeumaier: I just wanted to know what is "larger", $\theta$ or $3\theta$, for example, where $\theta$ is a Grassmann variable. Another example, $\theta+0.1\cdot\bar{\theta}\theta$? Or one obtains something number-like solely in the perturbation theory above?

There is no order. One expands in a coupling constant or a power of $\hbar$, and the terms with higher powers are smaller. All terms of a fixed order in this constant and all lower order terms must be taken into account.

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