I would have put a "does not equal" symbol there in the title if I could because that is more my concern. I have recently been playing around with Weyl spinors a bit, but I have also been reading Penrose and Rindler (spinors and spacetime) and have begun to notice some things that concern me a little. Penrose and Rindler give me all the algebraic properties of spinors well enough, but it's not particularly consistent with the physics notation and so confuses me a little. Ok my main question is are the components of a spinor necessarily Grassmann numbers? Penrose and Rindler never mention this and it in fact seems to contradict some things in their book, which is leading me to the conclusion that perhaps in general spinors do NOT have Grassmann components, but that the fermionic fields in QFT DO have Grassmann components, making them some special class of spinors. Does anyone know the truth, or have some insight? None of the literature I have read seems to make any note of this which is somewhat concerning. Oh, on a side note, this came about because I had some strange idea in my head that it was possible to write out expressions containing spinors out totally in component form, such that all their anticommuting properties are contained in epsilon tensors and associated index contractions, but if the components themselves are Grassmann variables then this is impossible, I believe. And another thing, Penrose and Rindler say that the inner product of two spinors anticommutes, while the exact opposite seems to be the case in the QFT books! This one really bugs me because it seems like it is a fundamental property of spinors and not subject to differences in notation etc. I can only hope it has something to do with the aforementioned Grassmann components/not Grassmann components conundrum.
I've never really noticed that but having had a quick discussion with people in the office we reached the consensus that they are different. When working over a supermanifold, ie in superspace with coordinates \((x^{\mu},\theta^{\alpha},\theta_{\dot{\beta}})\) the \(\theta\) are true Grassman numbers. They are constant quantities in the same way '5' is. The spinors you see in field theory are (surprisingly enough) fields and thus can be regarded in terms of mode raising and lowering operators. These are not Grassmann in the sense the \(\theta\) are but instead get their anticommuting nature from the canonical commutation relations which define quantisation. The fact the resultant commutation relations are pretty much the same is not too much of a coincidence since we use Grassmann stuff precisely because it is of the same mathematical structure as required for spinor field theory.
Ok, before I begin---I bet you learned all you know about spinors from Peskin, right? ( Please Register or Log in to view the hidden image! ) The problem is, Peskin NEVER treats Weyl spinors (probably the most useful form of things for people interested in the SM/MSSM) until very late in his book. In my opinion, this is a terrible idea, and it took me forever to learn the formalism after Pesking chapter 3. What about their notations confuse you? I'd suggest a few different sources for Weyl Spinors: Wess and Bagger (App. A and B) is good, so is Ryder. Beware of Wess and Bagger, as it is more or less a list of equations. Spinors are definitely written in terms of anti-commuting numbers. Maybe there is some confusion because Penrose and Rindler may be working in a curved space-time, instead of flat. In that case, maybe there's some new difficulty? The metric tensor on super-space is anti-symmetric: it's the Levi-Cevita thing. This means that the coordinates on superspace are anti-commuting numbers, and the inner product on superspace is anti-commuting. So: what is confusing about Penrose and Rindler's text? Ummmm....it is? This is the only way to learn supersymmetry! When in doubt, put in all the spinor indeces. Let me demonstrate : So I'm a bit confused about what you're asking here, but I think you mean \( \psi^\alpha \chi_{\alpha} = -\chi_{\alpha}\psi^{\alpha}\) This can be seen by just putting in the appropriate epsilon tensors: \(\psi^{\alpha} \epsilon_{\alpha \beta}\chi^{\beta}\)
Hmm, ok, well that may help. So you say the anticommuting nature comes from the canonical commutation relations, and so is in some sense in purely a quantum mechanical phenomenon. The spinors in Penrose and Rindler are purely classical objects so perhaps this is why they don't have this property. Hmm.
Hmm thanks I'll check those out. And the notation in Penrose and Rindler is fine really, it is just that it's a little different and some conventions regarding which indices of the epsilon tensor one needs to contract spinors with to raise and lower the indices are the opposite to most physics books, possibly the epsilon itself is defined slightly differently too, I can't remember. I mean things like that though, which make me always a little unsure if I have my negative signs right, which is rather crucial for issue at hand. Well Penrose and Rindler aren't talking about supersymmetry, or even QFT at all, they are talking about spinors as defined by their relation to the Lorentz group, so the metric tensor is a perfectly normal minkowski metric. Perhaps they go to curved space later in the book, but I imagine one then has to bring in the vielbein formalism and I don't really want to go there just now. I am doing all this to learn supersymmetry though and so I think perhaps something happens to the classical spinors when we go to QFT, perhaps to do with the imposition of the commutation relations as AN mentioned. Ahh if only that were the case. This is how I have come to expect things to behave in the SUSY literature, however Penrose and Rindler make it clear that \( \psi^{\alpha} \chi_{\alpha} = -\chi^{\alpha}\psi_{\alpha}\) Although they write it explicitly as an inner product: \( \{\psi,\chi\} = -\{\chi,\psi\}\) (They use curly braces for inner products, its a little strange looking.) Also, your expression is true only when the spinor components themselves anticommute, which is in addition to the anticommuting properties of the epsilon tensor. I believe in Penrose and Rindler one can perfectly well write out expressions in terms of components and rearrange that however one likes, because the anticommuting properties are only contained in the epsilon tensor and index contractions. If the components themselves anticommute as they do in QFT then one can't do this anymore and has to be careful about order. Thus there seems to be a difference between classical and quantum spinor objects, which I never realised and doesn't seem to really be explained anywhere. Perhaps it is supposed to be obvious because we imposed extra commutivity properties on our fields when we made them quantum. I didn't think about that before, I thought I could get away with imagining all this classically first, but that may have been a bad idea so I'll go check this hypothesis out...
Hi kurros--- I don't think this is true: There is a difference between Wess and Bagger's conventions and, say, Penrose and Rindler's. Also, note the difference in your expressions and my expressions, specifically the up and down indeces, which (as you point out) differ by a minus sign coming from the epsilon tensor. Anyway, Penrose and Rindler may be working in some wacky conventions. I would highly suggest that you look up the book by Binetruy, and turn to page 449 and look at Problem 1. I can't believe I forgot to recommend this to you first! The problem starts with and ends with a table relating the conventions of several popular references.
I guess I must be a bit confused in all of this, too. Why can I write a chiral superfield as \(\Phi = \phi + \sqrt{2} \theta^{\alpha}\psi_{\alpha} + \theta^{\alpha}\theta_{\alpha} F\). If \(\phi\) is a c number, shouldn't \(\theta^{\alpha}\psi_{\alpha}\) also be a c number, as well? If \(\theta\) is Grassmanian, and \(\psi\) is not Grassmanian, how can the product of the two be (magically) a scalar? Something sounds fishy in all of this.
Yes, from the \(\{ q_{\psi} , p_{\psi} \} = i\delta\) anticommutation relations (q and p are the field and its conjugate etc). The field decomposes into the Fourier modes and you then get anticommutation relations on them. I'm sure you're aware of this, I'm more just speaking out loud for the benefit of anyone else. I haven't looked through Wess & Bagger in a few years but I half get the feeling that the spinor-spinor combinations either become c numbers or bosonic, in that they satisfy commutation relations and not anticommutation relations. Certainly if you start putting together pairs of spinors and they all individually anticommute with one another (ie there's no \(i\hbar \delta\) on the right hand side of the anticommutation relations, everything is classical) then you are doing to pick up pairs of -1's whenever you move things through one another. Being fast and loose with indices and tildes just mean 'another independent field' its something like this I think : \((\psi^{a}\tilde{\psi}^{b})(\phi^{c}\tilde{\phi}^{d}) = -\psi^{a}\phi^{c}\tilde{\psi}^{b}\tilde{\phi}^{d} = \psi^{a}\phi^{c}\tilde{\phi}^{d}\tilde{\psi}^{b} = -\phi^{c}\psi^{a}\tilde{\phi}^{d}\tilde{\psi}^{b} = (\phi^{c}\tilde{\phi}^{d}) (\psi^{a}\tilde{\psi}^{b})\) Certainly in the absense of a quantisation in the anticommutation relations you have \( (\psi^{a}\tilde{\psi}^{b})\) and the other pairing behaving as if they satisfy classical boson commutation relations. Putting in quantisation conditions is going to make it a bit messier but probably you'll find they still have the bosonic properties you'd expect of any integer spin object, as obtained by adding their non-integer spins. Certainly its possible to build, in experiments, bosonic objects from pairings of fermionic ones. The difference in how Helium-4 and Helium-3 go superfluidic (He-4 is much easier as it happens at a higher temperature) is related to this as the Helium-3 is not naturally bosonic in terms of its nucleon spin combinations. I'll ask the guy in the office who reads W&B a lot when he gets back from a seminar. /edit Thinking about it further (in relation to my first post) and having asked people in the office Grassman numbers are numbers, in the sense that their definition is a mathematical statement, not a physical one. Spinors in QFT are looked at in terms of physical meaning, so they perfectly anticommute when everything is classical and get quantum corrections when you make things quantum (no, really?!). The kind of 'bridging' between them is the notion of Clifford algebras. These have sets of objects \(\gamma^{a}\) which satisfy \(\{ \gamma^{a} , \gamma^{b} \} = 2g^{ab}\) for some metric g. If g=0 its not really a Clifford algebra because g needs to be non-degenerate etc and 0 is degenerate. The modes of a quantised spinor are not strictly speaking the basis objects in a C.A. because they are infinite dimensional but much of the structures are the same. So when you use Grassman numbers you are working with a bunch of objects which satisfy the properties you'd expect of a non-quantised spinor field's mode operator (the \(a^{\dag}\) etc) so they might make your workings simpler or allow you to get your head around certain spinor related properties. However, upon quantisation a Grassman number is just a number and you can no more quantise it than you would '5' and so its not appropriate to view spinors are Grassmanian.
Sure they do, if the indices are contracted away (the indices are SU(2)xSU(2) indices) they are Lorentz invariant, and therefore bosonic. Of course this means that \(\theta^{\alpha}\) is also the component of a spinor, and \(\theta^{\alpha}\epsilon_{\alpha\beta}\psi^{\beta}\) is the antisymmetric part of \(1/2 \otimes 1/2\). So both \(\theta\) and \(\psi\) are definitely spinors, in that they are both spinor reps of SU(2). The contraction \(\theta^{\alpha}\psi_{\alpha}\) is definitely a boson (i.e. spin 0 rep of Lorentz grou)p. Do you agree? If this is the case, why should one spinor be written in Grassman numbers, while the other spinor be written in terms of some other objects? My hunch is that the construction/deconstruction operators, which you alluded to earlier, are also written in terms of Grassman numbers. Do you understand my confusion?
Check out theorem 8.3 in this paper, which is a nice survey anyway of current physics problems in relativity. http://dx.doi.org/10.1155/ijmms
I think I get what you mean. If \(\theta^{\alpha}\) are Grassman then they anticommute with themselves but commute with everything else so that raises the question of what it means then you consider the (anti)commutation relations between them and the modes of a spinor. In that case I am tempted to make the claim that if you have a spinor whose modes anticommute then you can write each anticommuting mode operator as a Grassman number combined with a bosonic commuting mode operator. That way you combine your quantisation of fermionic and bosonic modes since the bosonic quantisation condition feeds into the spinorial quantisation condition. Objects \(\{ \theta, \varphi \} = 0\) and \([a,a^{\dag}] = h\) combine into \(\theta a\) and \(\varphi a^{\dag}\) such that if we take the Grassman to commute with the operators : \((\theta a)(\varphi a^{\dag}) = \theta (\varphi a + [a,\varphi]) a^{\dag} = (-\varphi\theta + \{\varphi,\theta\}) a a^{\dag} = -\varphi\theta ( a^{\dag} a + [a,a^{\dag}] ) = -(\varphi a^{\dag}) (\theta a) - h\varphi\theta\) And so \(\{ (\theta a) , (\varphi a^{\dag}) \} = h\theta\varphi\) or equivalently \(\{ \theta a , \varphi a^{\dag} \} = \theta\varphi [a,a^{\dag}]\) How generally you can apply this I'm not sure.
This is the picture that I had, but my office mate has changed my mind. First of all, I apologize for misleading everyone with my confusions! Here's what we've arrived at after a half hour of heated discussion. I've also decided that it doesn't matter for kurros's original question, and that the minus sign is a matter of convention. This is verified by checking the Binetruy book and doing his Problem 1, that I recommended Please Register or Log in to view the hidden image! I think kurros is right now: What's the difference? Classically, you can write the fermion field in terms of grassman numbers to ``absorb'' the anti-commutation properties. Quantum mechanically, you write the field in terms of ladder operators a and a^+, which have anti-commutation properties. The point is that these guys are operators, and you don't care that they anti-commute, because that's just what happens sometimes. In the space of operators, some commute and some anti-commute. We say that the operators which describe the fermion field must have some relationship, inherited from the classical treatment, {a,a^+} = i ... Perhaps you CAN write a and a^+ in terms of Grassman numbers---we didn't prove or disprove that statement---but it doesn't matter. Of course, this discussion changes when you talk about the path integral. In some sense, you integrate over all paths between points A and B, weighting each path by its (classical) action. In this case, the Grassman numbers are just a mnemonic for getting your signs right. kurros: I would (again) encourage you to check out Binetruy, and figure out how Penrose/Rindler's conventions compare to those conventions you've dealt with before. Now, back to kurros's original question. The minus sign that you're confused about is convention, and Penrose and Rindler use dumb conventions. They still write a chiral superfield as \(\Phi = \phi + \theta^{\alpha}\psi_{\alpha} + \theta^2 F\). In order to do anything with this object, you have to write down a lagrangian, which involves integrating out the grassman numbers. Once you've integrated out the grassman numbers, then you can quantize the theory using your favorite method (I recommend path integrals).
The people in my office raised the same point, that you don't need to define how the field and its congugate momentum relate to one another when you do second quantisation because you're implicitly including it in the path integral in a natural way.
Hmm ok I'll get back to you once I process all this and figure out what we have decided here. I know Penrose and Rindler have SOME dumb conventions but I couldn't imagine how the anticommutivity of the inner product of spinors could be one of them. I'll probably need a couple of days here Please Register or Log in to view the hidden image!.
