# Low dimensional anticommutation example

jostpuur
Can somebody give an example of two matrices $a_1$ and $a_2$ which would satisfy the relation

$$a_i a_j^{\dagger} + a_j^{\dagger} a_i = \delta_{ij}$$

I know that

$$\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array}\right) \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}\right) + \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}\right) \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array}\right) = 1$$

but I found myself unable to modify this for the index $i\in\{1,2\}$.

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meopemuk
One obvious example is creation and annihilation operators for electrons. These operators act in the Fock space, which is a direct sum of N-electron Hilbert spaces, where N runs from 0 to infinity. So, the corresponding matrices are infinite-dimensional.

Eugene.

jostpuur
Raising and lowering operators for fermions were my original motivation for this problem. I didn't feel comfortable by the way books only give the properties of the operators, but I also want to see how these operators could be defined. I decided to make the problem simpler by allowing "only two Fourier modes" for the particles.

My first idea was that I should use four dimensional vector space, denote the basis vectors as $|00\rangle$, $|01\rangle$, $|10\rangle$, $|11\rangle$, and then define $a_1$ so that it maps $|01\rangle\mapsto |00\rangle$ and $|11\rangle\mapsto |10\rangle$, and $a_2$ so that it maps $|10\rangle\mapsto |00\rangle$ and $|11\rangle\mapsto |01\rangle$. With the following notation choice:

$$|00\rangle = \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \\ \end{array}\right),\quad |01\rangle = \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \\ \end{array}\right),\quad |10\rangle = \left(\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \\ \end{array}\right),\quad |11\rangle = \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \\ \end{array}\right)$$

the operators would be

$$a_1 = \left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array}\right),\quad\quad a_2 = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right)$$

But this does not work as wanted. These work like this:

$$a_1 a_2^{\dagger} + a_2^{\dagger} a_1 = \left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right)$$

Because $$a_1 a_2^{\dagger}$$ maps $|01\rangle\mapsto |11\rangle\mapsto |10\rangle$, and $$a_2^{\dagger} a_1$$ maps $|01\rangle\mapsto |00\rangle\mapsto |10\rangle$.

jostpuur
I was informed elsewhere, that matrices

$$a_1 = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array}\right),\quad\quad a_2 = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\right)$$

satisfy the property I was asking for. This did not solve my problem though, because it only made me realize that I should have also demanded that $a_1 a_1 = 0$ and $a_2 a_2 = 0$ in the beginning.

Well, using your 4-D vector space from post #3, I think the following matrices fit the bill:

$$\left(\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{array}\right),\quad\quad and \left(\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array}\right)$$

but they are adjoints of each other, so perhaps that is not what you are looking for. I guess it's just a block diagonal generalization of your original 2x2 case, now that I look at it.

jostpuur
That's right. I'm surprised to notice only now, that the matrices

$$a_1 = \left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array}\right),\quad\quad a_2 = \left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}\right)$$

But there still seems to be something that doesn't make sense in this. Perhaps I'll need to come up with a better formulation for the problem...

yuanyuan5220
this is one example of dirac algebra, and can be generalized to clifford algebra.

jostpuur
I have now understood that not every non-zero number has to be one.

So I want

$$a_1 |00\rangle = 0$$
$$a_1 |01\rangle \propto |00\rangle$$
$$a_1 |10\rangle = 0$$
$$a_1 |11\rangle \propto |10\rangle$$

$$a_2 |00\rangle = 0$$
$$a_2 |01\rangle = 0$$
$$a_2 |10\rangle \propto |00\rangle$$
$$a_2 |11\rangle \propto |01\rangle$$

and

$$a_ia_j^{\dagger} + a_j^{\dagger}a_i = \delta_{ij}$$

If I choose any four complex numbers $\alpha,\beta,\gamma,\delta$ such that $|\alpha|=|\beta|=|\gamma|=|\delta|=1$ and $\beta\delta^* + \gamma^*\alpha = 0$, then the desired operators can be defined as

$$a_1 = \left(\begin{array}{cccc} 0 & \alpha & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & \beta \\ 0 & 0 & 0 & 0 \\ \end{array}\right),\quad\quad\quad a_2 = \left(\begin{array}{cccc} 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & \delta \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array}\right)$$

For example $(\alpha,\beta,\gamma,\delta)=(1,1,1,-1)$ is one possibility.

But there is something inelegant about this. It seems laborous to try to generalise this for three Fourier modes next. Would I need to solve again even more equations then?

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jostpuur
Suppose we have a fermion system with N Fourier modes. Then for $i\neq j$ the raising operator $$a_j^{\dagger}$$ and lowering operator $$a_i$$ will commute modulo some coefficient, so that

$$a_i a_j^{\dagger} |\delta_1\delta_2 \cdots \underset{(i)}{1} \cdots \underset{(j)}{0} \cdots \delta_N\rangle \propto |\delta_1\delta_2 \cdots \underset{(i)}{0} \cdots \underset{(j)}{1} \cdots \delta_N\rangle$$

$$a_j^{\dagger} a_i|\delta_1\delta_2 \cdots \underset{(i)}{1} \cdots \underset{(j)}{0} \cdots \delta_N\rangle \propto |\delta_1\delta_2 \cdots \underset{(i)}{0} \cdots \underset{(j)}{1} \cdots \delta_N\rangle$$

It seems, that we can always choose these coefficients so that $$a_i a_j^{\dagger} + a_j^{\dagger} a_i = 0$$ (when $i\neq j$).

Why do physicist insist on getting the anticommutativity property holding? If the crucial property of fermion system is that each Fourier mode always has only one excitation state, the anticommutativity of raising and lowering operators does not seem to be inherent to the system. We might as well have fermion systems without anticommuting raising and lowering operators too, right?