# Fermionic oscillator and reducible representation

Consider the fermionic oscillators $$\{a, a^\dagger\} = 1$$, $$\{a, a\} = \{a^\dagger, a^\dagger\} = 0$$. The commonly used irreducible representation is given by $$|0\rangle$$, $$a^\dagger |0\rangle$$ where $$a$$ annihilates $$|0\rangle$$.

I am curious about what happens if we consider $$|0\rangle$$ not annihilated by $$a$$? In this case, I think there are four states, $$|0\rangle, a|0\rangle, a^\dagger |0\rangle, a^\dagger a |0\rangle$$, in the game. Of course, the representation is reducible: for example, $$a|0\rangle, \ a^\dagger a|0\rangle$$ forms an invariant subspace, since $$aa^\dagger a |0\rangle = a|0\rangle$$.

But is this representation a direct sum of two irreducible representation? Note that the naive complement $$|0\rangle, a^\dagger |0\rangle$$ does not form an invariant subspace. Maybe it is decomposable, but I fail to see how to reorganize the states to make it explicit.

In case that it is not decomposable into two irreps, is it interesting to consider such representation as the Hilbert space?

• what about $a(a\vert 0\rangle)$? Commented Apr 20, 2022 at 14:56
• @ZeroTheHero Well I would say it's zero, given that $\{a, a\}= 0$ Commented Apr 20, 2022 at 15:05

To complement @Ipz 's answer, I will answer the question by considering the finite dimensional representation theory of the anti-commutation relations.

Define $$P = a^\dagger a$$ and $$Q = aa^\dagger$$. Then, using the anticommutation relation, we see that $$P^2 = P$$, $$Q^2 = Q$$, $$PQ = QP = 0$$ and $$1 = P + Q$$. Also $$P^\dagger = P$$ and $$Q^\dagger =Q$$.

Consider a finite dimensional representation $$V$$. Let $$V_0$$ be the kernel of $$P$$ and $$V_1$$ the kernel of $$Q$$. Then $$V = V_0 \oplus V_1$$ and $$V_0$$,$$V_1$$ are orthogonal subspaces. Therefore, $$a$$ is one-to-one from $$V_1$$ to $$V_0$$ and $$a^\dagger$$ is one-to-one from $$V_0$$ to $$V_1$$. If $$|0,i\rangle$$ for $$i=1,\ldots,k$$ are an orthogonal basis of $$V_0$$, then we define $$|1,i\rangle = a^\dagger |0,i\rangle$$. We have : $$\langle 1,j|1,i\rangle = \langle 0,j| aa^\dagger|0,i\rangle = \langle 0,j|0,i\rangle = \delta_{ij}$$

We see that $$V$$ is a direct sum of $$k$$ subrepresentations isomorphic to the $$2$$-dimensional one.

Now back to the question at hand. From the analysis above, we know that the representation are even dimensional. From one vector $$|\psi\rangle$$, we can generate at most a space of dimension $$4$$.

• If we have a representation of dimension $$4$$, then we know that it is of the form $$|0\rangle,|0'\rangle,|1\rangle,|1'\rangle$$. A generic vector is of the form : $$|\phi\rangle = a|0\rangle + b|0'\rangle + c |1\rangle + d |1'\rangle$$ with $$|a|^2 + |b|^2 + |c|^2 + |d|^2 + 1$$. It generates the whole $$4$$ dimensional space under the action of $$a$$ and $$a^\dagger$$ if, and only if : $$\det \begin{pmatrix} a & b \\ c&d\end{pmatrix} \neq 0$$

• If we have a vector space of dimension $$2$$, then we have a basis $$|0\rangle,|1\rangle$$, and any unit vector $$|\psi\rangle=a|0\rangle + b|1\rangle$$ with $$|a|^2 + |b|^2 = 1$$ generates the whole vector space.

the general method of finding the irreducible representations is to look at the kernel of $$a$$, orthogonal kets in this set will generate orthogonal irreducible representations. In your case, any ket orthogonal to $$a|0\rangle$$ in the kernel will do the trick.

Actually, you need more assumptions to study your representation. For example, you can have the case $$a^\dagger|0\rangle = 0$$, which gives the usual representation (switching $$a$$ and $$a^\dagger$$) and even if it not null, you could have for example $$|0\rangle = \frac{1}{\sqrt 2}(|0\rangle’+a^\dagger|0\rangle’)$$ with $$|0\rangle'$$ in the kernel of $$a$$.

I think you are interested in the case that the four kets $$|0\rangle,a|0\rangle,a^\dagger|0\rangle,a^\dagger a|0\rangle$$ are assumed to be independent, and thus generate a representation of dimension 4. In this case, the kernel of a is of dimension 2, generated by $$a|0\rangle,aa^\dagger|0\rangle$$, so the orthogonal representation would be generated by the ground state $$aa^\dagger|0\rangle-\frac{\langle 0|a|0\rangle}{\langle 0|a^\dagger a|0\rangle} a|0\rangle$$.

I hope it helps and please tell me if you find a mistake.

• I don’t see the link; I think I have a counter-example. Take $|0\rangle, |1\rangle=a^\dagger |0\rangle, |0’\rangle, |1’\rangle=a^\dagger |0’\rangle$ with $|0\rangle, |0’\rangle$ orthogonal ground states. By construction, N is diagonalizable in this space. Taking (before normalization) $|\psi\rangle=|1\rangle+ |0’\rangle$, the four kets become $|1\rangle+ |0’\rangle, |0\rangle, |1’\rangle, |1\rangle$ which is a basis.
– LPZ
Commented Apr 20, 2022 at 15:57
• Nevermind, I was mistaken Commented Apr 20, 2022 at 16:08