condensed matter - Particle hole symmetry of single site?

Let's consider I have a system with equal number of spin up and spin down particles

Now I consider a single site of system,I have a state $c_{i\uparrow} ^{\dagger}\mid 0\rangle$ under particle hole transformation it goes

$$c_{i\uparrow} ^{\dagger}\mid0\rangle\to c_{i\uparrow}\mid0\rangle$$

can I write the tranformation in this form $c_{i\uparrow} ^{\dagger}\mid0\rangle$$=c_{i\downarrow}^{\dagger}\mid0\rangle$? I am little confused about this transformation

Because we normally use $c_{i\uparrow} ^{\dagger}$$=c_{i\downarrow}^{\dagger}$ this transformation when we have equal spin up and spin down particle in a system and creating a hole with spin up means creating a particle with spin down because it corresponds to net increase in spin down, but if I consider in a system only a single site with net spin up then under particle hole transformation how it can correspond to creation of spin down particle when there is no down spin contribution at the site?

Answer

No, you can not write this transformation as $c_{i\uparrow}^\dagger|0\rangle=c_{i\downarrow}^\dagger|0\rangle$, because $c_{i\uparrow}^\dagger|0\rangle$ and $c_{i\downarrow}^\dagger|0\rangle$ are two orthogonal quantum states: they can not be equal. The transformation $c_{i\uparrow}^\dagger|0\rangle\to c_{i\uparrow}|0\rangle$ you start with is also wrong, because the resulting state is $c_{i\uparrow}|0\rangle= 0$, which makes the transformation not unitary. What you normally use $c_{i\uparrow}^\dagger=c_{i\downarrow}^\dagger$ is still wrong, because $c_{i\uparrow}^\dagger$ and $c_{i\downarrow}^\dagger$ are different operators and can not be equal.

The correct way of writing everything is to start by defining a particle-hole transformation operator $\mathcal{P}$, such that its action on the fermion operator is

$$\mathcal{P}c_{i\sigma}^\dagger\mathcal{P}^{-1} = c_{i\sigma}, \tag{1}$$

and its action on the vacuum state is

$$\mathcal{P}|0\rangle = \prod_{i,\sigma} c_{i\sigma}^\dagger|0\rangle. \tag{2}$$

It is important that the vacuum state $|0\rangle$ must also transform under $\mathcal{P}$. Physically, it is because the vacuum state is the state with no particle occupation, which means it is a state that is fully occupied by holes. So under the particle-hole transformation, the vacuum state will become a fully occupied state of particles, as expressed in Eq.(2). Mathematically, Eq.(2) follows from Eq.(1) as a result of consistency. Because the vacuum state is defined as the state annihilated by all the annihilation operator, i.e. $c_{i\sigma}|0\rangle=0$. Now if we apply $\mathcal{P}$ to both sides of the equation, we have $\mathcal{P}c_{i\sigma}|0\rangle=\mathcal{P}0=0$ (because any linear operator acts on $0$ is still $0$). However the left-hand-side reads $\mathcal{P}c_{i\sigma}|0\rangle=\mathcal{P}c_{i\sigma}\mathcal{P}^{-1}\mathcal{P}|0\rangle=c_{i\sigma}^\dagger\mathcal{P}|0\rangle$, meaning that the state $\mathcal{P}|0\rangle$ is annihilated by creation operator $c_{i\sigma}^\dagger$ instead, so the state $\mathcal{P}|0\rangle$ has to be a fully occupied state.

Applying Eq.(1) and Eq.(2) to a single site (omitting the site index $i$), we have

$$\mathcal{P}c_{\uparrow}^\dagger|0\rangle = (\mathcal{P}c_{\uparrow}^\dagger\mathcal{P}^{-1})(\mathcal{P}|0\rangle)=c_{\uparrow}c_{\uparrow}^\dagger c_{\downarrow}^\dagger|0\rangle=c_{\downarrow}^\dagger|0\rangle.$$

So this means under the particle-hole transformation, the spin-up state $c_{\uparrow}^\dagger|0\rangle$ is transformed to a spin-down state $c_{\downarrow}^\dagger|0\rangle$.