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How one can simultaneously represent fermionic operators (denoted with hats) and its corresponding Grassmann variables (denoted without hats), so that all the anticommutation relations between them and also states would take place?

$$ \hat{c}\hat{c}^\dagger+\hat{c}^\dagger\hat{c}=1 \\ c^2 = 0 \\ \overline{c}^2 = 0 \\ c\overline{c}+\overline{c}c = 0 \\ c\hat{c}+\hat{c} c = 0 \\ c\hat{c}^\dagger+\hat{c}^\dagger c = 0 \\ \overline{c}\hat{c}+\hat{c} \overline{c} = 0 \\ \overline{c}\hat{c}^\dagger+\hat{c}^\dagger \overline{c} = 0 \\ c\left|0\right>-\left|0\right>c = 0 \\ c\left|1\right>+\left|1\right>c = 0 \\ \overline{c}\left|0\right>-\left|0\right>\overline{c} = 0 \\ \overline{c}\left|1\right>+\left|1\right>\overline{c} = 0. \\ $$ It seems like it's impossible to represent both these operators and Grassmann numbers as matrices or am I wrong? The anticommutation with states requires numbers to be antihermitian but it should be triangular to satisfy nilpotency, this only works for zero matrix.

In my case I need this also to work for two fermions with four states. I can represent operators as matrices and states as vectors as follows

$$ \left|0\right>=\{1,0,0,0\}\\ \left|\downarrow\right>=\{0,1,0,0\}\\ \left|\uparrow\right>=\{0,0,1,0\}\\ \left|\downarrow\uparrow\right>=\{0,0,0,1\}\\ \hat{c_\sigma} = \left(\array{0&\delta_{\sigma\downarrow}&\delta_{\sigma\uparrow}&0\\0&0&0&-\delta_{\sigma\uparrow}\\0&0&0&\delta_{\sigma\downarrow}\\0&0&0&0}\right)\\ c_\sigma = ?\\ \overline{c_\sigma}=? $$ So how should I represent these Grassmann numbers?

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2 Answers 2

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I) Yes, if OP insists of the presence of Grassmann variables$^1$, it is possible to represent the fermionic operators as matrices with the caveat that the fermionic Fock space of states is a super vector space, and the matrices are super matrices.

If we have 2 creation operators $\hat{c}^{\dagger}_{\sigma}$, $\sigma\in \{\uparrow,\downarrow\}$, then there are:

  • 2 bosonic states (1 vacuum state $\left|0\right>$ and 1 two-particle state $\left|\uparrow\downarrow\right>$), and

  • 2 fermionic single-particle states, $\left|\uparrow\right>$ and $\left|\downarrow\right>$.

See also e.g. my Phys.SE answer here. Let us represent the 4 states

$$ \left|0\right> =\begin{pmatrix} 1\\ 0 \\ 0 \\ 0 \end{pmatrix}, \qquad \left|\uparrow\downarrow\right> =\begin{pmatrix} 0\\ 1 \\ 0 \\ 0 \end{pmatrix}, \qquad \left|\uparrow\right> =\begin{pmatrix} 0\\ 0 \\ 1 \\ 0 \end{pmatrix}, \qquad \left|\downarrow\right> =\begin{pmatrix} 0\\ 0 \\ 0 \\ 1 \end{pmatrix}.\tag{1}$$

as 4 basis vectors in the super vector space $\mathbb{C}^{2|2}$. In other words, the Fock space is isomorphic to $\mathbb{C}^{2|2}$.

The fermionic operators are represented by $(2+2)\times (2+2)$ supermatrices in ${\rm End}(\mathbb{C}^{2|2})=L(\mathbb{C}^{2|2},\mathbb{C}^{2|2})$. They will have four $2\times 2$ blocks. For instance:

$$ \hat{c}_{\uparrow} ~=~\begin{pmatrix} 0&0&1&0\\ 0&0&0&0 \\ 0&0&0&0 \\ 0&1&0&0 \end{pmatrix}, \qquad \hat{c}^{\dagger}_{\uparrow} ~=~\begin{pmatrix} 0&0&0&0\\ 0&0&0&1 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}. \tag{2} $$

Notice that both the above supermatrices are Grassmann-odd despite the fact that all non-zero matrix elements are Grassmann-even. This is because the non-zero matrix elements sit in the off-diagonal $2\times 2$ Bose-Fermi blocks.

Moreover, one may check that the anticommutator of the two above supermatrices (2) is the $4\times 4$ identity matrix, as it should be to mimic the CAR algebra.

The operators $\hat{c}_{\downarrow}$ and $\hat{c}^{\dagger}_{\downarrow}$ have similar representations in terms of supermatrices. We leave it as an exercise to the reader to work them out.

Be aware that the equal symbols '$=$' in eqs. (1) & (2) mean are represented by rather than are equal to. In particular, be aware that Grassmann numbers still commute/anticommute with the operators/states based on their Grassmann-parity.

II) If there are no Grassmann variables but only fermionic operators and states, then we can represent the fermionic Fock space as an exterior algebra $\bigwedge\!{}^{\bullet}V$ generated by the space of 1-particle states $$V~:=~{\rm span}_{\mathbb{C}}\{\left|\uparrow\right>,\left|\downarrow\right>\}~\cong~\mathbb{C}^2.\tag{3} $$ The vacuum state $\left|0\right>$ is implemented as $$ \bigwedge\!{}^{0}V~\cong~\mathbb{C}~\cong~\mathbb{C}\left|0\right>.\tag{4}$$ The 2 creation & 2 annihilation operators generate a Clifford algebra $Cl(W)\cong\mathbb{C}^{16}$, where $$ W~:=~{\rm span}_{\mathbb{C}}\{\hat{c}_{\uparrow},\hat{c}^{\dagger}_{\uparrow},\hat{c}_{\downarrow},\hat{c}^{\dagger}_{\downarrow}\} ~=~~{\rm span}_{\mathbb{C}}\{\hat{\gamma}_{\mu}| \mu=1,2,3,4\} ~\cong~\mathbb{C}^4,\tag{5} $$ where $$ \begin{align} \hat{\gamma}_{1}~=~& \hat{c}_{\uparrow}+\hat{c}^{\dagger}_{\uparrow},\qquad \hat{\gamma}_{2}~=~ \hat{c}_{\downarrow}+\hat{c}^{\dagger}_{\downarrow} \cr \hat{\gamma}_{3}~=~& \frac{\hat{c}_{\uparrow}-\hat{c}^{\dagger}_{\uparrow}}{i},\qquad \hat{\gamma}_{4}~=~ \frac{\hat{c}_{\downarrow}-\hat{c}^{\dagger}_{\downarrow}}{i}, \end{align}\tag{6} $$ so that $$\{\hat{\gamma}_{\mu},\hat{\gamma}_{\nu}\}_{+} ~=~2\delta_{\mu\nu}\hat{\bf 1},\tag{7} $$ cf. e.g. Ref. 1. It is well-known that the Clifford algebra $Cl(W)$ can be represented by $4 \times 4$ Dirac matrices.

References:

  1. M.B. Green, J.H. Schwarz & E. Witten, Superstring theory, Vol. 1, 1986; Appendix 5.A.

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$^1$ NB: In this answer we do not try to also make matrix representations of the Grassmann numbers.

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  • $\begingroup$ Definitely you can represent these Fermionic operators by matrices, which is standard quantum mechamics! But you can't do this for Grassmann variables. This is what I claimed in my answer. $\endgroup$
    – hyd
    Oct 22, 2014 at 2:24
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    $\begingroup$ This does not answer the question, because still Grassmann variables remain Grassmann variables, which are not any kind of operators and can't be reduced to a set of real numbers! $\endgroup$
    – hyd
    Oct 22, 2014 at 2:27
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    $\begingroup$ In quantum mechanics, in order to represent an entity by a matrix, the prerequisite is that this entity must act in the Hilbert space. A Grassmannn variable is not any operator and it does not operate in the Hilbert space. So, it is impossible to put it as a matrix. $\endgroup$
    – hyd
    Oct 22, 2014 at 3:04
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There is no way to represent Grassmann variables using matrices ! Actually, this is the big obstacle that hinders the use of the so-called quantum state diffusion approach for systems placed in Fermionic baths. You may find many papers on this by googling this topic. Also, individual Grassmann variable has no physical meaning. It is something invented mostly for 'bookkeeping' in path integral method! There is a nice but brief exposure in the book by X.G. Wen.

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