# Anticommutation of Fermions and the Lorentz irreps

1. Mar 11, 2015

### Trifis

Hi everyone!

I have two questions that arose during the path integral quantization of theories involving fermions.

First of all, when we prove the equivalence between the path integral formalism and the canonical quantization we make use of the eigenvalue defining equation: $\hat{φ}(x)|Φ>=φ(x)|Φ>$ in order to get rid of the operators and start working with numbers.

My first question is: What happens when we try to find the eigenvalue of a fermionic field? To my understanding, the eigenvalues of creation/annihilation operators for fermions, somehow, have to be decomposable into normal functions, i.e. numbers, and Grassman numbers.

And then a more general consideration came to mind: Does the anticommutation property of fermionic operators follow solely from the Lorentz group representation theory? I know that Dirac spinor part of the fermionic field transforms as the (1/2,0)⊕(0,1/2) representation of the Lorentz group but what about the creation/annihilation operators? As far as I can tell, anticommutation relations are not imposed on the spinors but on the operators! In other words, is the usage of Grassman numbers justified by the spacetime symmetry or there is some other kind of fundamental axiom?

2. Mar 11, 2015

### vanhees71

That there are only fermions and bosons in Nature is due to the topology of many-body configuration space. If you have a space dimension larger or equal to 3, this implies that there are only fermions and bosons, if you assume that two particles must not be at the same place at a given time, which makes sense since the fundamental interactions are singular for two particles meeting at the same space-time point. This is due to the famous paper,

Laidlaw, M. G. G., DeWitt, Cécile Morette: Feynman Functional Integrals for Systems of Indistinguishable Particles, Phys. Rev. D 3, 1375, 1970

Then in the relativistic case, you assume that relativistic QT is described by local, microcausal QFTs with a stable ground state. From the representation theory of the Poincare group then you can deduce that fields with integer spin have to be quantized as bosons (equal-time commutator relations) and such with half-integer spin as fermions (equal-time anticommutator relations). This is known as the spin-statistics theorem. Another conclusion is that despite symmetry under the proper orthochronous Lorentz group (i.e., the subgroup continuously connected with the group identity) also the discrete operation, CPT (charge conjugation, followed by space reflections (parity) and time reversal) is necessarily a symmetry. It is proven that the weak interaction breaks P, T, and CP (each measured independently now!).

You need Grassmann numbers for the path-integral of fermions, because to derive the path integral, you have to introduce coherent states for fermions, and you cannot have usual c-number eigenvalues for anticommuting fields, because this would lead to a contradiction. Also the path integral would not give the correct result for fermions, when you'd use usual (real or complex) "field eigenvalues".

3. Mar 12, 2015

### Trifis

This is a very nice answer, thank you!

I would appreciate it if you could clarify some points though.

So the spin-statistics theorem can be derived, using ONLY the Poincare group representation theory? Can you provide such a proof?

Naively, one could write $\hat{ψ}(x)|Ψ>=ψ(x)|Ψ>$ with a fermionic field $\hat{ψ}(x)$ acting on some state. Does it make any sense to express the eigenvalue ψ(x) as some kind of combination of complex and grassman numbers? From a mathematical point of view, I cannot imagine how this could be done in a rigorous way, but even physically ψ(x) cannot be a grassman number, since ψ(x) has to be an observable!

4. Mar 12, 2015

### Trifis

What is more, I have never seen a derivation of $\{\hat{a},\hat{a}^†\}=1$, coming from the representation theory. I was under the impression, that Poincare group does practically determine the spinor part ($u^s(k)$ for particles or $v^s(k)$ for antiparticles) of the fermionic field $$\hat{ψ}(x) \sim \hat{a}^su^s(k)e^{-ikx} + \hat{a}^{s†}v^s(k)e^{ikx}$$ and has nothing to do with the creation/annihilation operators.

5. Mar 13, 2015

### samalkhaiat

Quantization of any system (of fields) is based upon Schrodinger equation $$i \partial_{t} | \Omega \rangle = \int d^{3} x \ \mathcal{H} ( \Psi , \Pi ) | \Omega \rangle ,$$ together with an equal-time operator algebra which, for fermion fields, is given by the following anti-commutation relations $$\{ \Psi_{\alpha} ( t , \vec{x} ) , \Pi^{\beta} ( t , \vec{y} ) \} = i \delta_{\alpha}^{\beta} \delta^{3} ( \vec{x} - \vec{y} ) ,$$ $$\{ \Psi_{\alpha} ( t , \vec{x} ) , \Psi_{\beta} ( t , \vec{y} ) \} = \{ \Pi^{\alpha} ( t , \vec{x} ) , \Pi^{\beta} ( t , \vec{y} ) \} = 0 .$$ The coordinate Schrodinger representation of the equal-time operator algebra is given by $$\Psi_{\alpha} ( 0 , \vec{x} ) | \psi \rangle = \psi_{\alpha} ( \vec{x} ) | \psi \rangle ,$$ $$\langle \psi | \Pi^{\alpha} ( 0 , \vec{x} ) | \tilde{\psi} \rangle = i \frac{\delta}{\delta \psi_{\alpha} ( \vec{x} )} \delta ( \psi - \tilde{\psi} ) .$$ Note that $\Psi ( 0 , \vec{x} )$ is an operator (the time-independent field/coordinate operator), while $\psi ( \vec{x} )$ is a time-independent field function (i.e., a “classical” field). Since the field operator $\Psi ( 0 , \vec{x} )$ squares to zero, its eigen-values, $\psi ( \vec{x} )$ must be spinors of anti-commuting components, $\psi^{2}_{\alpha} ( \vec{x} ) = 0$, i.e., functions taking values in the Grassmann algebra. Thus, quantization requires the “classical” spinor fields to be Grassmann-valued functions on space-time. However, like any other classical field on space-time, the classical Dirac field is defined to be a cross-section of the vector bundle $E^{\omega}$ (over $\mathbb{R}^{(1,3)}$) associated with the Dirac representation, $\omega ( r , \vec{p} ) \in ( 0 , 1/2 ) \oplus ( 1/2 , 0 )$, of $SL ( 2 , \mathbb{C} )$ over $\mathbb{C}^{4}$. In other words, the “classical” Dirac field has the following expansion (at $t = 0$) $$\psi_{\alpha} ( \vec{x} ) \sim \sum_{r = 1}^{4} \int d^{3} p \ b ( \vec{p} , r ) \ \omega_{\alpha} ( \vec{p} , r ) \ e^{ i \vec{p} \cdot \vec{x} } .$$
Some remarks are now in order.
(i) All factors in the expansion are numbers (not operators). It is in this sense we call $\psi ( \vec{x} )$ a classical field function (not operator).
(ii) In order for $\psi ( \vec{x} )$ to be the eigen-value of the field operator $\Psi ( 0 , \vec{x} )$, the number $b ( \vec{p} , r )$ must be a Grassmann number, $b^{2} = 0$. Notice that this Grassmannian nature is not implied by the above definition of classical fields for the ordinary Lorentz group on the ordinary (not super) space-time. However, the classical Dirac field is Grassmannian in the representation theory of super-Lorentz group on super-space.
(iii) The plane wave expansion of the Schrodinger field operator is obtained by promoting $b \to \hat{b} , \ \mbox{and} \ \psi_{\alpha} ( \vec{x} ) \to \Psi_{\alpha} ( 0 , \vec{x} )$: $$\Psi_{\alpha} ( 0 , \vec{x} ) \sim \sum_{r = 1}^{4} \int d^{3} p \ \hat{b} ( \vec{p} , r ) \ \omega_{\alpha} ( \vec{p} , r ) \ e^{ i \vec{p} \cdot \vec{x} } .$$
(iv) If we expand the time-dependent state vector $|\Omega (t) \rangle$ in the coordinate basis $|\psi \rangle$, the component of $|\Omega \rangle$ in the $|\psi \rangle$ direction is $\langle \psi | \Omega \rangle \equiv \Omega [ \psi ]$, which is a number, is called the wave functional. It is the probability amplitude for the field to be in the configuration $\psi ( \vec{x} )$ at time $t$.
(v) The quantization of the free Dirac field can, therefore, be reduced to solving the following Grassmann functional differential equation $$i \partial_{t} \Omega [ \psi ] = \int d^{3} x \int \mathcal{D} \ \tilde{\psi} \ \langle \psi | \mathcal{H} ( \Psi , \Pi ) | \tilde{\psi} \rangle \ \Omega [ \tilde{\psi} ] = \int d^{3} x \ \frac{\delta}{\delta \psi ( \vec{x} )} \left( - i \vec{\alpha} \cdot \vec{\nabla} \right) \psi ( \vec{x} ) \ \Omega [ \psi ] .$$

Sam

6. Mar 13, 2015

### vanhees71

Concerning the spin-statistics theorem: That's a very famous and well-known result by Pauli and Lüders. You find it in any textbook on relativistic QFT. The most comprehensive treatment for particles of any spin can be found in

Weinberg, The Quantum Theory of Fields, vol. 1

For the usual special low-spin cases, you can also find it in my lecture notes:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Further, the field eigenvalues are in general not observables. Particularly the Grassmann-valued eigenvalues of fermion fields do not represent observable values. What's observable are the probabilities encoded in the S-matrix elements and expectation values of observables like energy, momentum, and angular momentum. For free fields the latter are bilinear forms, obeying not anticommutators but commutators. Particularly the operators representing local densities commute at space-like separated arguments (principle of microcausality).

7. Mar 13, 2015

### Trifis

I think, this makes it clear for me. So, the Grassmanian nature of the fermionic eigenvalues IS NOT an inherent characteristic of the Lorentz group. This, alternatively, means that the Spin-Statistics theorem cannot be derived solely by arguments about the Lorentz spacetime symmetry.

EDIT : It seems that the anticommutation of the fermionic creation/annihilation operators can be derived by Lorentzian spacetime symmetry considerations and thus the anticommutating nature of their eigenvalues, namely the Grassman numbers, IS in a way a consequence of the ordinary Lorentz group.

Moreover, I did enjoy your presentation of the Dirac field eigenvalues as having both spinor and grassmanian parts in their expansion. It was more or less, what I pictured in post #4, but I had my doubts how this could be expressed in a rigorous mathematical way.

Last edited: Mar 13, 2015
8. Mar 13, 2015

### Trifis

Ok, so field operators after the second quantization do not follow the basic axioms of 1-particle QM. There must be a lot of mathematical subtleties in this one, but let us not go off-topic.

9. Mar 13, 2015

### vanhees71

There's no second quantization. Quantum field theory is just many-body quantum theory in Fock-space representation. The spin-statistics theorem is not based on fermionic coherent state representations but follows under the assumption of a microcausal local relativistic QFT with stable ground state. These assumptions also imply that in addition to the symmetry under proper isochronous Poincare transformations the discrete CPT transformation is also a symmetry (Pauli-Lüders theorem).

10. Mar 13, 2015

### Trifis

Yes, and the fact that operators are no longer observables in this many-body QM context, makes me think that the basic Dirac-von Neumann axioms do no longer have to hold.

I've done some reading on the proof of the spin-statistics theorem. It seems that the basic assumption is indeed the Lorentz invariance. Of course, there are some additional prerequisites, such as a lower energy bound which is needed to prove the CPT theorem, an essential tool of the proof, propagation with finite mass, localized excitations (microcasuality) etc.

All in all, we are allowed to say that the spinor part of the fermionic eigenvalues comes from the representation theory of the Lorentz group and the grassmanian part comes from the general spacetime stracture, i.e. the spin-statistics theorem.

Last edited: Mar 13, 2015
11. Mar 13, 2015

### vanhees71

Of course, spin follows from representation theory of the special orthochronous Poincare group, which is the symmetry group of special-relativistic spacetime. That there is half-integer spin is due to the fact that in quantum theory we look for unitary ray rather than unitary representations. In the case of the Poincare group all these are induced by the unitary representations of the covering group, where the proper orthochronous Lorentz subgroup is substituted by its covering group, which is SL(2,C). There are no non-trivial central charges (as in the case of the Galilei group, where mass is a non-trivial central charge).

The fact that in theories with $\geq 3$ spatial dimensions one has only fermions and bosons (and not more complicated representations of the symmetric group representing symmetry under indistinguishable-particle interchanges, called "anyons", which can occur in 2 spatial dimensions) follows from quite general assumptions on the configuration space of many-particle systems (Laidlav, de Witt Morette). That the integer-spin particles are bosons and half-integer-spin particles fermions follows then from the assumption of a local, microcausal QFT with a stable ground state.

12. Mar 13, 2015

### atyy

Even in basic QM the operators do not have to be observables. For example, the raising and lowering operators in the harmonic oscillator are not observables. Observanbles are operators, but operators don't have to be observables.