Is Hilbert space of a fermionic harmonic oscillator a Hilbert space over Grassmann algebra?

Question

I'm reading the book Basics of Thermal Field Theory also available on the authors webpage http://www.laine.itp.unibe.ch/basics.pdf

In the section 4.1 (Path integral for the partition function of a fermionic oscillator) they introduce Grassmann variables $c$ and $c^*$ that satisfy several properties, in particular

• $c$, $c^*$ are treated as independent variables, like $x$, $p$.
• $c^2=(c^*)^2$, $cc^*=-c^*c$
• $c$, $c^*$ are defined to anticommute with $\hat a$, $\hat a ^\dagger$ as well.

These are classical analogs of fermionic creation and annihilation operators $\hat a$ and $\hat a ^\dagger$. To define the path integral (using classical fields $c(\tau)$ and $c^*(\tau)$) $$ \int \mathcal{D}c(\tau) \mathcal{D} c^*(\tau)e^{-\frac{1}{h}S} $$ of the fermionic harmonic oscillator they introduce coherent states $$ |c\rangle = e^{-c\hat a^\dagger} |0>. $$

So here $c$ plays a role of a "number". The only way I see how to make sense of this definition is to extend scalars of the Hilbert space: instead of two dimensional Hilbert space $H$ we consider Hilbert space over Grassmann algebra $A=\mathbb{C}<c, c^*>$ i.e. tensor product $H_A=A \otimes_\mathbb{C} H$, then fermionic coherent state is an element of $H_A$, but not an element of $H$.

So, for quantisation of fermions we need to extend scalars of the Hilbert space from $\mathbb{C}$ to Grassmann numbers $A$. Is that correct?

Answer 1

I know nothing about supernumbers, but the construction that you mention for the Fermi oscillator looks easy to formalize.

On the wikipedia page, they say that the set of polynomials in $n$ Grassmann generators can be identified with the exterior algebra of a linear space, and in the case of complex Grassmann number we should take a space on the complex one. Actually, I think that this is a bit imprecise, since in quantum field theory we want the $\theta$'s and $\theta^*$ 's to be two independent set of Grassmann generators. In fact, everything works out if we identify the set of Grassmann polynomials with the covariant exterior algebra: $$\Lambda _{\bullet}(V\oplus V^*),$$ where $V$ and $V^*$ are two complex anti-isomorphic vector spaces (I have worked out this by myself, so please correct me if I'm wrong).

In any case, this is not the real point of the question. Suppose that you have a set of Grassmann numbers $\mathscr G$ generated by two elements $\lbrace \theta,\theta ^* \rbrace$. We want to make sense of an expression such as $P(\theta,\theta ^*) \vert \alpha \rangle$, and we can do this in the same way which allows us to multiply a real vector by a complex scalar: tensor product.

If the "physical" Hilbert $\mathcal H$ space consists of vectors: $$\vert \rangle = \alpha \vert 0 \rangle + \beta \vert 1\rangle,$$where $\alpha,\beta$ are ordinary complex numbers, we define a vector space $\mathcal H' =\mathscr G \otimes \mathcal H $. The multiplication by Grassmann numbers can be defined on decomposable tensors by: $$Q (P \otimes \vert \rangle)=(QP )\otimes \vert \rangle$$and extended by linearity on the whole space $\mathcal H '$. In particular, the old $\mathcal H$ can be seen as the subspace spanned by the elements $1\otimes \vert \rangle$, and the multiplication by the scalar $\lambda$ in $\mathcal H$ agrees with the multiplication by the "Grassmann number" $\lambda$ on $\mathcal H '$.

All other things are just formal manipulations, plus some ad-hoc definition, for example to make sense of the conjugation $P\vert \rangle \to \langle \vert P^*$. For example, you can introduce the "coherent state" $$\vert \theta \rangle = \vert 0\rangle + \theta \vert 1\rangle,$$ which is an eigenvector of $a\equiv\text {id}_{\mathscr G} \otimes a$ :$$a\vert \theta \rangle = \theta \vert \theta \rangle,$$ which may be the reason why your book is introducing Grassmann numbers.