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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?

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  • $\begingroup$ is your singlet “located” only on site 2? $\endgroup$ – ZeroTheHero May 20 at 11:31
  • $\begingroup$ Yes, both spins are on site 2 or site 1 so the (0,2) or the (2,0) configurations $\endgroup$ – PhysicsMan May 20 at 11:39
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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 $S_{z}$.

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

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