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  How are "Classical Fermions" and Grassmann numbers related with each other?

+ 2 like - 0 dislike

In the theory of relativistic wave equations, we derive the Dirac equation and the Klein-Gordon equation by using representation theory of the Poincare algebra. 

For example, in this paper 


the Dirac equation in momentum space (equation [52], [57] and [58]) can be derived from the 1-particle state of irreducible unitary representation of the Poincare algebra (equation [18] and [19]). The ordinary wave function in position space is its Fourier transform (equation [53], [62] and [65]).  

Note that at this stage, this Dirac equation is simply a classical wave equation. i.e. its solutions are classical Dirac 4-spinors, which take values in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$.

If we regard the Dirac waves $\psi(x)$ and $\bar{\psi}(x)$ as 'classical fields', then the quantized Dirac fields are obtained by promoting them into fermionic harmonic oscillators. 

What I do not understand is that when we are doing the path-integral quantization of the Dirac fields, we are, in fact, treating $\psi$ and $\bar{\psi}$ as Grassmann numbers, which are counter-intuitive for me. As far as I understand, we do path-integral by summing over all 'classical fields'. While the 'classical Dirac wave $\psi(x)$' we derived in the beginning are simply 4-spinors living in $\Bbb{C}^{2}\oplus\Bbb{C}^{2}$. How can they be treated as Grassmann numbers instead? 

As I see it, physicists are trying to construct a 'classical analogue' of Fermions that are purely quantum objects. For instance, if we start from a quantum anti-commutators

$[\psi,\psi^{\dagger}]_{+}=i\hbar1$ and $[\psi,\psi]_{+}=[\psi^{\dagger},\psi^{\dagger}]_{+}=0$, 

then we can obtain the Grassmann numbers in the classical limit $\hbar\rightarrow0$. This is how I used to understand the Grassmann numbers. The problem is that if the Grassmann numbers are indeed a sort of classical limit of anticommuting operators in Hilbert space, then the limit $\hbar\rightarrow0$ itself does not make any sense from a physical point of view since in this limit $\hbar\rightarrow0$, the spin observables vanish totally and what we obtain then would be a $0$, which is a trivial theory.

Please tell me how exactly the quantum Fermions are related to Grassmann numbers.  

asked Jan 28, 2016 in Theoretical Physics by XIaoyiJing (50 points) [ revision history ]

To everyone who has interest in my question, I suddenly realized that in the path integral formalism of Fermions, the derivation process goes as follows. We start from considering the transition probability between two fermionic coherent states, say $<\bar{\psi}_{f}|\psi_{i}>$. Then we add a bunch of identity operators for the fermionic coherent states. We then encounter the grassmann numbers. Considering the above procedure, it seems to me that these grassmann numbers are not exactly the classical limit of fermionic fields. Instead, they are still 'semi-quantum'.

1 Answer

+ 4 like - 0 dislike

The Dirac equation is a classical field equation (PDE) describing the possible 1-particle states of a free fermionic quantum field theory. On the 1-particle level the fermionic nature is not yet visible - one just sees the spin 1/2 representation. The fermi statistics only enters through the spin-statistcs theorem, which holds for relativistic quantum field theories (only).

On the other hand, Grassmann numbers appear as limits of fermionic fields when $\hbar\to 0$ (usually called the classical limit), and hence figure in the functional integral. However, they are classical only in a generalized sense (what you call semiquantum), as the square of any Grassmann variable vanishes. Thus they cannot represent classical observables, they are just fomal classical elements.

The classical observables are the quadratic forms in Grassmann variables, which mutually commute and span the even part of the Grassmann algebra. And indeed, the functional integral is nonzero only on this even part, and the classical action figuring in the exponential part of the functional integral is also even.

answered Jan 28, 2016 by Arnold Neumaier (15,787 points) [ no revision ]

Thank you very much for your answer! 

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