Lecture Note-Mean Field

Partons

We can decompose physical fermion into bosonic parton and fermionic parton.

\[c_{i l \sigma}= b_{i l } f_{i l \sigma}\]

Bosonic parton carries the total charge symmetry and the inter-layer symmetry, fermionic parton carries the spin SU(2) symmetry.

Bosonic parton mean field

By using mean field theory , we can get the Hamiltonian of bosonic parton witch resemble the bose hubbard model.

\[H=-t_{b}\sum_{l=1}^{2} \sum_{\langle i, j\rangle}\left(b_{il}^{\dagger} b_{jl}+\text { h.c. }\right)-\sum_{l=1}^{2}\sum_{i} \mu_{i} n_{il}+J_{b} \sum_{i} n_{i1}n_{i2}\]

Parameter l means different layers . $J_{b}=J\langle f_{1}^{\dagger}f_{2}f_{3}^{\dagger}f_{4} \rangle$ means the inter layer interaction parameter. $t_{b}=t\langle f_{i}f_{j} \rangle$ means the hopping parameter.

\[H_{i}=\sum_{l=1}^{2} -2 z t_{b}\left(\psi\left(b_{il}^{\dagger}+b_{il}\right)-\psi^{2}\right)-\mu_{i} n_{il}+J_{b} n_{i1}n_{i2}\]

So the Hamiltonian on each site can be written as follow

\[H=J_{b} n_{1}n_{2} -\mu n_{1}-\mu n_{2}-t_{b} \psi\left(b_{1}^{\dagger}+b_{1}+b_{2}^{\dagger}+b_{2}\right)+2t_{b} \psi^{2}\]

The expectation value of energy is

\[E_{0}=\langle n_{1}n_{2}|H| n_{1}n_{2}\rangle=J_{b} n_{1}n_{2} -\mu n_{1}-\mu n_{2}\]

We can find the ground state by minimizing the free energy

\(\partial_{n_{1}} E_{0}=\partial_{n_{1}}\left(J_{b} n_{1}n_{2} -\mu n_{1}-\mu n_{2}\right)=0\) \(\partial_{n_{2}} E_{0}=\partial_{n_{2}}\left(J_{b} n_{1}n_{2} -\mu n_{1}-\mu n_{2}\right)=0\)

\[n_{1}=n_{2}=[\frac{\mu}{J_{b}}]\]

And $

n_{1}n_{2}\rangle$ is the ground state.

The free energy can be expanded in power series of $\psi$

\[E(\psi) \equiv\langle\Psi|H| \Psi\rangle=E_{0}+E_{1} \psi+\frac{1}{2} E_{2} \psi^{2}+\frac{1}{6} E_{3} \psi^{3}+\frac{1}{24} E_{4} \psi^{4}+\ldots\]

Because of the projective $\mathbb{Z}_{2}^{\mathcal{S}}$ action, we shuold have:

\[E_{1}=E_{3}=\ldots=0\]

So the second order coefficient can be calculated from

\[E_{2}=\partial_{\psi}^{2} E=\sum_{m \neq n} \frac{\langle n|\partial H| m\rangle\langle m|\partial H| n\rangle+\langle n|\partial H| m\rangle\langle m|\partial H| n\rangle}{E_{n}-E_{m}}\] \[E_{2}=-2t_{b}^2 \sum_{(m_{1},m_{2}) \neq (n_{1},n_{2})} \frac{\langle n_{1},n_{2}|\partial H| m_{1},m_{2}\rangle\langle m_{1},m_{2}|\partial H| n_{1},n_{2}\rangle}{E_{n_{1},n_{2}}-E_{m_{1},m_{2}}}\]

where $\partial H=-t_{b}\left(b_{1}^{\dagger}+b_{1}+b_{2}^{\dagger}+b_{2}\right) $. So, the result is:

\[E_{2}=\frac{2\mu-J_{b}(n_{1}n_{2})}{(\mu - J_{b}n_{1})(\mu - J_{b}n_{2}) }\] \[E=E_{0}+\frac{1}{2}\left(E_{2}+4 t_{b}\right) \psi^{2}\]

so that we can obtain the energy as a faction of $\psi$ , $\mu$ , $J_{b}$ and $t_{b}$ (n should be replaced by the function of the ground state)

\[E_{b}\left(t_{b}, \psi, J_{b}\right)\] \[t_{b}=t\left\langle f_{i}^{\dagger} f_{j}\right\rangle, \quad J_{b}=J\langle\text { ffff }\rangle\]

Where $\langle\text { ffff }\rangle $ can be calculate by using Wick Theory \(\langle f_{1}f_{2}f_{3}f_{4} \rangle = \langle f_{1}f_{2} \rangle \langle f_{3}f_{4} \rangle+\langle f_{1}f_{3} \rangle \langle f_{2}f_{4} \rangle+ \langle f_{1}f_{4} \rangle\langle f_{2}f_{3} \rangle\)

Fermionic parton mean field

By using mean field theory , we can get the Hamiltonian of fermionic parton.

\[H=-t_{f}\left(f_{i}^{\dagger} f_{j}+\text { h.c. }\right)+J\left(f_{i, 1 \downarrow}^{\dagger} f_{i, 2 \downarrow} f_{i, 2 \uparrow}^{\dagger} f_{i, 1 \uparrow}+h.c.\right)\]

where $ t_{f}=t \psi^{2} $,according to the mean field result of bosonic parton.

We can use mean field again to the Hamiltonian of fermionic parton and obtain:

\[H=-t \psi^{2}\sum_{l}^{4}\left(f_{li}^{\dagger} f_{lj}+h . c .\right)+\lambda(-)^{i}\left(f_{i, 2 \uparrow}^{\dagger} f_{i, 1 \uparrow}+f_{i, 1 \downarrow}^{\dagger} f_{i, 2 \downarrow} -f_{i, 1 \downarrow}^{\dagger} f_{i, 1 \uparrow}- f_{i, 2 \uparrow}^{\dagger}f_{i, 2 \downarrow} +h.c.\right)\]

where $\lambda$ is the mean field term

For the honeycomb lattice, we can write the Hamiltonian as follow

\(H_{r}=-t \psi^{2}\sum_{l}^{4}\left(f_{l(r+\delta_{1})}^{\dagger} f_{lr}+f_{l(r+\delta_{2})}^{\dagger} f_{lr}+f_{l(r+\delta_{3})}^{\dagger} f_{lr}+f_{l(r-\delta_{1})}^{\dagger} f_{lr+\delta_{1}}+f_{l(r-\delta_{2})}^{\dagger} f_{lr+\delta_{1}}+f_{l(r-\delta_{3})}^{\dagger} f_{lr+\delta_{1}}\right)\) \(+\lambda( f_{r, 2 \uparrow}^{\dagger} f_{r, 1 \uparrow}+f_{r, 1 \downarrow}^{\dagger} f_{r, 2 \downarrow} -f_{r, 1 \downarrow}^{\dagger} f_{r, 1 \uparrow}- f_{r, 2 \uparrow}^{\dagger}f_{r, 2 \downarrow}\) \(-f_{r+\delta_{1}, 2 \uparrow}^{\dagger} f_{r+\delta_{1}, 1 \uparrow}-f_{r+\delta_{1}, 1 \downarrow}^{\dagger} f_{r+\delta_{1}, 2 \downarrow} +f_{r+\delta_{1}, 1 \downarrow}^{\dagger} f_{r+\delta_{1}, 1 \uparrow}+ f_{r+\delta_{1}, 2 \uparrow}^{\dagger}f_{r+\delta_{1}, 2 \downarrow}+h.c.)\)

\(\boldsymbol{\delta}_{1}=(0,-1)\) \(\boldsymbol{\delta}_{2}=\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\) \(\boldsymbol{\delta}_{3}=\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)\)

r is the label of the unicellular which contain two atoms :A and B.

We can use fourier transformation and get the Hamiltonian in momentum space.

The creation and annihilation operator obey following equation

\(c_{i} =\frac{1}{\sqrt{N}} \sum_{k} c_{k} e^{i k x}\) \(c_{k} =\frac{1}{\sqrt{N}} \sum_{i} c_{i} e^{-i k x}\)

\(\frac{1}{N} \sum_{k} e^{i k j} e^{-i k j^{\prime}}=\delta_{j j^{\prime}}\) \(\frac{1}{N} \sum_{j} e^{i k j} e^{-i k^{\prime} j}=\delta_{k k^{\prime}}\)

Then we can obtain the Hamiltonian as a 8 x 8 matrix and we can diagonalize it and calculate the ground state and the expectation value of energy.Energy is a function of $t_{f}$,$J$and$\lambda$

\[E_{f}\left(t_{f}, J, \lambda\right)=E_{f}\left(t \psi^{2}, J, \lambda\right)\]

Conclusion

The total energy is equal to $E_{b}+E_{f}$, witch is a function of t,J, $\lambda$ and $\psi $

t and J are the parameter of the model. We can set them initially and find the value of $\lambda$ and $\psi $ witch make the energy minimum. As we tuning the ratio of t and J, $\lambda$ and $\psi $ will turn to be zero of not. There will be four situations and each one means a phenomenon.

\[\begin{array}{cccccc} ~ & \psi=\langle b\rangle &\lambda & a & \left\langle c_{2}^{\dagger} c_{1}\right\rangle & U(1)_{\text {layer }} \\ t>>\mathrm{J} & O(1) & 0 & \text { Higgs } & 0 & \text { symm } \\ J>>\mathrm{t} & 0 & O(1) & \text { confine } & 0 & \text { symm } \\ ? & 0 & 0 & \text { deconfine }(\mathrm{QED}) & 0 & \text { symm } \\ ? & O(1) & O(1) & \text { Higgs } & O(1) & \text { broken } \end{array}\]

When the hopping term is dominant, we guess $\lambda$ will be zero and fermion will act like a free fermion. When the interaction term is dominant, we guess $\psi$ will be zero because the bosonic parton will condense on each site and transform into Mott state.

During the tuning, there must be two probably intermediate state.

If $\lambda$and $\psi $ are both zero, that means the gauge field is deconfined, this system emergent the QED. If $\lambda$and $\psi $ are neither zero, that means the layer U(1) symmetry is broken.

Questions

I find some questions and confusions when I was summing up the note.

• how can I transform the initial Hamiltonian into partons’ Hamiltonian?

The Hamiltonian of SMG is \(H=-t \sum_{\langle i j\rangle} \sum_{a=1}^{4}\left(c_{i a}^{\dagger} c_{j a}+\text { h.c. }\right)-V \sum_{i}\left(c_{i 1}^{\dagger} c_{i 2}^{\dagger} c_{i 3}^{\dagger} c_{i 4}^{\dagger}+\text { h.c. }\right)\)

I find I still don’t understand what do the four flavours(a) of fermion mean. How can I transform four annihilation operators $c_{i 1}^{\dagger} c_{i 2}^{\dagger} c_{i 3}^{\dagger} c_{i 4}^{\dagger}$ into two annihilation operators and two creation operators $c_{i 1}^{\dagger} c_{i 2} c_{i 3}^{\dagger} c_{i 4}$ , so that we can get the form of $ J\left(f_{i, 1 \downarrow}^{\dagger} f_{i, 2 \downarrow} f_{i, 2 \uparrow}^{\dagger} f_{i, 1 \uparrow}+\ldots\right) $ and $J_{b} \sum_{i} n_{1i}n_{2i}$ ?

• What does the flavour of parton mean?

I find how to transform c into partons in your paper \(c_{i a}=\sum_{b=1}^{4} B_{i a b} f_{i b}\)

but I don’t understand why bosonic parton has two kinds of flavours and fermionic parton has four different flavours with fermion.How can I correct the flavour form of the Hamiltonian I wrote?

• How can I prove what will happen in the intermediate state?

I can only retell the conclusion of the two different probably intermediate state. How can I explain it more rigorously?