QuantumMob/QuantumPlaybook | Paclet Repository

Quantum Dot Systems

Here, we consider a system of quantum dots that are described by the Hubbard model.

Q	Particle number operator
Hop	Hopping operator
FermionBasis	Fermion basis with certain features

Some useful functions when studying quantum dot systems.

First, make sure the package is loaded.

In[112]:=

Needs["QuantumMob`Q3`"]

Choose a symbol to denote the dot operators.

In[113]:=

Let[Fermion,d]

It is also convenient to explicitly set the energy parameters to be real.

In[114]:=

Let[Real,ϵ,t,U]

Single Quantum Dot

Single-particle (quadratic or non-interacting) part

Let us start from the single-particle (quadratic) part of the Hamiltonian. There are several equivalent ways:

The most obvious method is using the fermion operators:

In[115]:=

Dagger[d[Up]]**d[Up]

Out[115]=

†

↑

The above has a short cut Q[...] as following:

In[6]:=

Q[d[Up]]

Out[6]=

†

↑

You need to sum over all spin (up and down):

In[7]:=

Sum[Q[d[s]],{s,{Up,Down}}]

Out[7]=

†

↓

†

↑

The above can be even simpler as the following two (equivalent) examples:

In[8]:=

Q[d[{Up,Down}]]Q[d[All]]

Out[8]=

†

↓

†

↑

Out[9]=

†

↓

†

↑

So, the non-interacting part of the QD Hamiltonian is given by

In[116]:=

H0=ϵ*Q[d[All]]

Out[116]=

ϵ

†

↓

†

↑



On-site interaction

Now, let us turn to the interaction (quartic) part:

In[117]:=

Hint=UQ[d[Up]]**Q[d[Down]]

Out[117]=

†

↓

†

↑

↓

Many-body eigenstates

Finally, the Hamiltonian for the quantum dot is given by

In[118]:=

HH=H0+Hint

Out[118]=

ϵ

†

↓

†

↑

+U

†

↓

†

↑

↓

The above Hamiltonian may be described in graph as follows. Here, the vertex

with dashed lines denotes the interaction.

In[81]:=

GraphForm[HH]

Out[81]=

We need to construct a basis for the fermionic Fock space. Recall that the Hamiltonian conserves both the charge (

) and spin (

). Therefore, it is a good idea to organize the basis accordingly. Here, note that the sector

{Q=1,S=1/2}

has only one state



↑



but not



↓



. This is to avoid redundancy because both gives the same matrix elements.

In[34]:=

bs=FermionBasis[d]

Out[34]=

{0,0}{|␣〉},1,



↑

,{2,0}-

↓

↑



Naturally, the Hamiltonian is block diagonal in this basis.

In[35]:=

big=MatrixIn[HH,bs,bs];big=Values@KeyGroupBy[big,First,Values];MatrixForm@Map[MatrixForm,big,{2}]

Out[37]//MatrixForm=

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

)

(

U+2ϵ

)

It is thus efficient to work on each diagonal block separately.

In[38]:=

mat=MatrixIn[HH,bs];MatrixForm/@mat

Out[39]=

{0,0}(

),1,

(

),{2,0}(

U+2ϵ

)

Double Quantum Dot

Let us now consider a double quantum dot (DQD). From the above example of single-orbital quantum dot, you now see that the following two statements are equivalent:

In[89]:=

Dagger[d[1,Up]]**d[1,Up]Q[d[1,Up]]

Out[89]=

†

1,↑

Out[90]=

†

1,↑

You should also be familiar with the following equivalent statements:

In[91]:=

Sum[Q[d[j,Up]],{j,1,2}]Q[d[{1,2},Up]]

Out[91]=

†

1,↑

†

2,↑

Out[92]=

†

1,↑

†

2,↑

What about the sum over spin? The following two statements are equivalent:

In[93]:=

Q[d[1,{Up,Down}]]Q[d[1,All]]

Out[93]=

†

1,↓

†

1,↑

Out[94]=

†

1,↓

†

1,↑

The single-particle part of the two isolated quantum dots is given by

In[95]:=

H0=ϵ*Q[d[{1,2},All]]

Out[95]=

ϵ

†

1,↓

†

1,↑

†

2,↓

†

2,↑



Now, let us turn to the hopping between the two quantum dots. Again one straightforward method is as following:

In[96]:=

Sum[d[1,s]^Dagger**d[2,s],{s,{Up,Down}}]

Out[96]=

†

1,↓

2,↓

†

1,↑

2,↑

The above has a short-cut as following:

In[97]:=

Hop[d[1,All],d[2,All]]Hop[{d[1,All],d[2,All]}]

Out[97]=

†

1,↓

2,↓

†

1,↓

2,↑

†

1,↑

2,↓

†

1,↑

2,↑

Out[98]=

†

1,↓

2,↓

†

1,↑

2,↑

Do not forget to make it Hermitian. (There is one important reason why Hop[...] or FockHopping[...] does not include the hopping terms for both directions.)

So, the hopping part is given by

Then, the overall single-particle part is given by

On-site interactions

Here is the onsite interaction terms.

Many-body eigenstates

Finally, the total Hamiltonian for the DQD is given by

Naturally, the Hamiltonian is block diagonal in this basis.

Therefore, it is efficient to consider each block separately.