# Constructing singlet state in second quantization formalism

I am trying to understand the second quantization formalism. Let's say we have a system of fermions (e.g. electrons) with spin in an array of quantum dots. The creation and annihilation operators $$c^{\dagger}_{i\sigma}$$ and $$c_{i\sigma}$$, where $$i$$ and $$\sigma$$ indicate the index of the dot and the spin respectively, obey the caconical fermion commutation rules.

\begin{align} c^{\dagger}_{i\alpha}c_{j\beta}+c_{j\beta}c^{\dagger}_{i\alpha}=\delta_{i,j}\delta_{\alpha,\beta}, \hspace{5mm}c^{\dagger}_{i\alpha}c^{\dagger}_{i\alpha}=0, \hspace{5mm} c^{\dagger}_{i\alpha}c_{j\beta}^{\dagger}=-c_{j\beta}^{\dagger}c^{\dagger}_{i\alpha} \end{align}

I will be looking in the case of just 2 fermions, $$i={1,2}$$ and $$\sigma={\uparrow,\downarrow}$$. So both dots contain 1 fermion (1,1) or either of the dots contains 2 fermions (0,2) and (2,0).

So for example in the the (1,1) configuration I can create a state of two spin up fermions or one spin up and one spin down: $$c^{\dagger}_{1\uparrow}c^{\dagger}_{2\uparrow}|0,0\rangle=|\uparrow,\uparrow\rangle, \hspace{4mm}c^{\dagger}_{1\uparrow}c^{\dagger}_{2\downarrow}|0,0\rangle=|\uparrow,\downarrow\rangle$$.

My question now is, how do you represent the singlet state S in this ladder operator formalism?

$$S=\frac{1}{\sqrt{2}}|0,\uparrow\downarrow-\downarrow\uparrow\rangle\text{ or } \frac{1}{\sqrt{2}}|\uparrow\downarrow-\downarrow\uparrow,0\rangle$$

My first guess is something like: $$c^{\dagger}_{2\uparrow}c^{\dagger}_{2\downarrow}|0,0\rangle$$ But this should obviously result in $$c^{\dagger}_{2\uparrow}c^{\dagger}_{2\downarrow}|0,0\rangle=|0,\uparrow\downarrow\rangle$$

Maybe more general formulation of my question is: how do you represent superposition states in second quantization formalism?

• is your singlet “located” only on site 2? May 20, 2019 at 11:31
• Yes, both spins are on site 2 or site 1 so the (0,2) or the (2,0) configurations May 20, 2019 at 11:39

Your guess is actually correct. The occupation number representation you use is not simply the tensor product of the single-site states. It is already (anti-) symmetrized for (fermions) bosons. Indeed, if you exchange the two fermions you get a minus. This is because

$$c^{\dagger}_{2\uparrow}c^{\dagger}_{2\downarrow}=-c^{\dagger}_{2\downarrow}c^{\dagger}_{2\uparrow}$$

Since the two fermions share the same spatial wavefunction, the spin part of the wavefunction would be the anti-symmetric part and is thus the singlet.

You can also try to look at the second-quantized version of the spin operators of site $$2$$ (omitting the site index)

$$S_{x}=\dfrac{1}{2}\left(c^{\dagger}_{\uparrow}c_{\downarrow}+c^{\dagger}_{\downarrow}c_{\uparrow}\right)$$

$$S_{y}=\dfrac{1}{2i}\left(c^{\dagger}_{\uparrow}c_{\downarrow}-c^{\dagger}_{\downarrow}c_{\uparrow}\right)$$

$$S_{z}=\dfrac{1}{2}\left(c^{\dagger}_{\uparrow}c_{\uparrow}-c^{\dagger}_{\downarrow}c_{\downarrow}\right)$$

and check the eigenvalues of $$S^{2}=S^{2}_{x}+S^{2}_{y}+S^{2}_{z}$$ and $$S_{z}$$.

• Then how do you distinguish between the triplet and singlet? How is the following state contstructed: $T=\frac{1}{\sqrt{2}}=|0,\uparrow\downarrow+\downarrow\uparrow\rangle$ May 20, 2019 at 11:39
• @PhysicsMan I am not sure what you are asking. You can't get the triplet state for two fermions at the same site. May 20, 2019 at 11:51
• Why isn't it possible to get the triplet state for two fermions at the same site? If their spatial wavefunction is antisymmetric this should be possible right? May 20, 2019 at 11:57
• @PhysicsMan Because you can't get an anti-symmetric combination of the same single-particle wavefunction. The spatial part must be $\psi_{2}\left(\boldsymbol{r}_{1}\right)\psi_{2}\left(\boldsymbol{r}_{2}\right)$. May 20, 2019 at 11:59