How one can simultaneously represent fermionic operators and its corresponding Grassmann variables, 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 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?

  • $\begingroup$ Have you looked at Wikipedia's comment on this? $\endgroup$ – ACuriousMind Oct 13 '14 at 12:39
  • $\begingroup$ Comment to the question (v1): Are hatted and unhatted c's different operators? Does an overline denote complex conjugate? $\endgroup$ – Qmechanic Oct 13 '14 at 12:39
  • $\begingroup$ @ACuriousMind Yes, I saw these, these matrices don't satisfy all relations. $\endgroup$ – swish Oct 13 '14 at 12:41
  • $\begingroup$ @Qmechanic Hatted c's are operators, and other c's are two Grassmann numbers. $\endgroup$ – swish Oct 13 '14 at 12:42
  • $\begingroup$ Comment to the question (v2): The equation $c\left|0\right>+\left|0\right>c = 0$ should be $c\left|0\right>-\left|0\right>c = 0$. Similarly for the equation with $\overline{c}$. $\endgroup$ – Qmechanic Oct 20 '14 at 19:33

Yes, 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 two 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 order the 4 states as follows:

$$ \left|0\right>, \qquad \left|\uparrow\downarrow\right>, \qquad\left|\uparrow\right>, \qquad\left|\downarrow\right>.$$

In other words, the Fock space is isomorphic to $\mathbb{C}^{2|2}$. Hence the $(2+2)\times (2+2)$ supermatrices will have four $2\times 2$ blocks. For instance:

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

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 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.

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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 '14 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 '14 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 '14 at 3:04

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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