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Λ-Matter in a Cavity
In this example, we consider a system of Λ-level (artificial) atoms in a single-mode cavity. Each atom has two ground-state levels and and a excited-state level , and no transition is allowed between 0 and 1; and hence Λ-level structure. Suppose that transition is induced resonantly by photon in the cavity whereas transition is driven classically by an external laser.
|0〉
|1〉
|2〉
0→2
1→2
Constructs the computational basis for a system of species. | |
Gives the matrix representation of an operator in the computational basis. | |
Zeroes computational basis states satisfying certain condition. | |
Modifies computational basis states according to certain rules. | |
Returns the Schur basis. | |
Is like GroupBy but applies to keys of an association. |
Q3 functions used in this documentation.
In[1]:=
Needs["QuantumMob`Q3`"]
The Model
Consider a system of Λ-level atoms. Choose symbol A (or any one you like) to designate a class of such atoms.
n
In[2]:=
Let[Qudit,A]
In[3]:=
$n=3;kk=Range[$n]
Out[4]=
{1,2,3}
The atoms are coupled to the cavity mode (we consider a single mode). Choose symbol c to refer to the photon.
In[5]:=
Let[Boson,c]
In[6]:=
Hg=g*Dagger[c]**Total[A[kk,20]];Hg=PlusDagger[Hg]
Out[7]=
*
g
(|2〉〈0|)
1
(|2〉〈0|)
2
(|2〉〈0|)
3
†
c
(|0〉〈2|)
1
†
c
(|0〉〈2|)
2
†
c
(|0〉〈2|)
3
Here g is the coupling constant. We will put g=1, that is, choose it as the energy scale of the system.
In[8]:=
g=1;
The transition is driven by a classical laser. Let be the Rabi transition amplitude.
1↔2
Ω
In[9]:=
Hw=Ω*Total[A[kk,12]];Hw=PlusDagger[Hw]
Out[10]=
Ω(++)+(++)
(|2〉〈1|)
1
(|2〉〈1|)
2
(|2〉〈1|)
3
(|1〉〈2|)
1
(|1〉〈2|)
2
(|1〉〈2|)
3
*
Ω
Construct the total Hamiltonian.
In[11]:=
HH=Hg+Hw
Out[11]=
Ω(++)+(++)+c+c+c+++
(|2〉〈1|)
1
(|2〉〈1|)
2
(|2〉〈1|)
3
(|1〉〈2|)
1
(|1〉〈2|)
2
(|1〉〈2|)
3
*
Ω
(|2〉〈0|)
1
(|2〉〈0|)
2
(|2〉〈0|)
3
†
c
(|0〉〈2|)
1
†
c
(|0〉〈2|)
2
†
c
(|0〉〈2|)
3
Examine the matrix representation of the Hamiltonian in the standard basis.
In[12]:=
mat=Matrix[HH]
Out[12]=
SparseArray
 |  |
|
In[13]:=
mat〚;;20,;;20〛//MatrixForm
Out[13]//MatrixForm=
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | * Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | * Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | * Ω | 0 | * Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | * Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | * Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | Ω | 0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | * Ω | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | * Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | * Ω |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | * Ω | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | * Ω | 0 | * Ω | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | * Ω | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | * Ω | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | Ω | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | Ω | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
We note that many matrix elements vanish. This is due to symmetries and consequent selection rules that we will investigate below.
Symmetries
In[14]:=
Needs["QuantumMob`Q3`"]
In[15]:=
Let[Qudit,A]
Now we exploit symmetries of the system. The symmetries drastically simplifies the study of the system.
Conservation of Total Excitations
Conservation of Total Excitations
Note that transition can occur only at the cost of a photon in the cavity. Therefore, the number of total excitations,
0→2
N:=c+
†
c
n
∑
k=1
(|1〉〈1|+|2〉〈2|)
k
In[16]:=
NN=Q[c]+Total[A[kk,11]]+Total[A[kk,22]]
Out[16]=
(|1〉〈1|)
1
(|2〉〈2|)
1
(|1〉〈1|)
2
(|2〉〈2|)
2
(|1〉〈1|)
3
(|2〉〈2|)
3
†
c
In[17]:=
Commutator[NN,HH]
Out[17]=
0
Since the total number of excitations is conserved, it is convenient to focus on a fixed excitation number .
p
In[18]:=
$p=2
Out[18]=
2
We need to choose a proper basis. We start with the standard tensor-product basis of atoms (not yet including photon).
In[21]:=
bs0=Basis[A@kk]
Out[21]=
,,,,,,,,,,,,,,,,,,,,,,,,,,
0
A
1
0
A
2
0
A
3
0
A
1
0
A
2
1
A
3
0
A
1
0
A
2
2
A
3
0
A
1
1
A
2
0
A
3
0
A
1
1
A
2
1
A
3
0
A
1
1
A
2
2
A
3
0
A
1
2
A
2
0
A
3
0
A
1
2
A
2
1
A
3
0
A
1
2
A
2
2
A
3
1
A
1
0
A
2
0
A
3
1
A
1
0
A
2
1
A
3
1
A
1
0
A
2
2
A
3
1
A
1
1
A
2
0
A
3
1
A
1
1
A
2
1
A
3
1
A
1
1
A
2
2
A
3
1
A
1
2
A
2
0
A
3
1
A
1
2
A
2
1
A
3
1
A
1
2
A
2
2
A
3
2
A
1
0
A
2
0
A
3
2
A
1
0
A
2
1
A
3
2
A
1
0
A
2
2
A
3
2
A
1
1
A
2
0
A
3
2
A
1
1
A
2
1
A
3
2
A
1
1
A
2
2
A
3
2
A
1
2
A
2
0
A
3
2
A
1
2
A
2
1
A
3
2
A
1
2
A
2
2
A
3
Remove the states with excitations more than $p.
In[22]:=
bs1=KetPurge[bs0,Hold[Count[A@kk,1|2]>$p]]
Out[22]=
,,,,,,,,,,,,,,,,,,
0
A
1
0
A
2
0
A
3
0
A
1
0
A
2
1
A
3
0
A
1
0
A
2
2
A
3
0
A
1
1
A
2
0
A
3
0
A
1
1
A
2
1
A
3
0
A
1
1
A
2
2
A
3
0
A
1
2
A
2
0
A
3
0
A
1
2
A
2
1
A
3
0
A
1
2
A
2
2
A
3
1
A
1
0
A
2
0
A
3
1
A
1
0
A
2
1
A
3
1
A
1
0
A
2
2
A
3
1
A
1
1
A
2
0
A
3
1
A
1
2
A
2
0
A
3
2
A
1
0
A
2
0
A
3
2
A
1
0
A
2
1
A
3
2
A
1
0
A
2
2
A
3
2
A
1
1
A
2
0
A
3
2
A
1
2
A
2
0
A
3
In[23]:=
bs2=DeleteCases[bs1,0]
Out[23]=
,,,,,,,,,,,,,,,,,,
0
A
1
0
A
2
0
A
3
0
A
1
0
A
2
1
A
3
0
A
1
0
A
2
2
A
3
0
A
1
1
A
2
0
A
3
0
A
1
1
A
2
1
A
3
0
A
1
1
A
2
2
A
3
0
A
1
2
A
2
0
A
3
0
A
1
2
A
2
1
A
3
0
A
1
2
A
2
2
A
3
1
A
1
0
A
2
0
A
3
1
A
1
0
A
2
1
A
3
1
A
1
0
A
2
2
A
3
1
A
1
1
A
2
0
A
3
1
A
1
2
A
2
0
A
3
2
A
1
0
A
2
0
A
3
2
A
1
0
A
2
1
A
3
2
A
1
0
A
2
2
A
3
2
A
1
1
A
2
0
A
3
2
A
1
2
A
2
0
A
3
Note that the number of photons has just been fixed by the number of atoms in level 1 or 2. Specify the number of photons in each state vector.
In[24]:=
bs=KetUpdate[bs2,cHold[$p-Count[A@kk,1|2]]]
Out[24]=
,,,,,,,,,,,,,,,,,,
2
c
0
A
1
0
A
2
0
A
3
1
c
0
A
1
0
A
2
1
A
3
1
c
0
A
1
0
A
2
2
A
3
1
c
0
A
1
1
A
2
0
A
3
0
c
0
A
1
1
A
2
1
A
3
0
c
0
A
1
1
A
2
2
A
3
1
c
0
A
1
2
A
2
0
A
3
0
c
0
A
1
2
A
2
1
A
3
0
c
0
A
1
2
A
2
2
A
3
1
c
1
A
1
0
A
2
0
A
3
0
c
1
A
1
0
A
2
1
A
3
0
c
1
A
1
0
A
2
2
A
3
0
c
1
A
1
1
A
2
0
A
3
0
c
1
A
1
2
A
2
0
A
3
1
c
2
A
1
0
A
2
0
A
3
0
c
2
A
1
0
A
2
1
A
3
0
c
2
A
1
0
A
2
2
A
3
0
c
2
A
1
1
A
2
0
A
3
0
c
2
A
1
2
A
2
0
A
3
Note the dimension of the subspace.
In[25]:=
Length[bs]
Out[25]=
19
Now examine the matrix representation within the subspace.
There are still many vanishing elements of the above matrix. It implies there are other symmetries to exploit.
Permutation Symmetry
Permutation Symmetry
Before going further, here is a small utility function.
Note that in the above is shown a partial list.
As before, we take into account the conservation of excitations. First, remove state vectors with excitations more than $p.
Then, specify the number of photons according to the occupation number of level 1 and 2.
The matrix is block diagonalized.
It is thus convenient to examine each diagonal block separately.
Chiral Symmetry
Chiral Symmetry
The system has another unconventional symmetry. To see this symmetry, note that any atomic transition is through level 2. This implies that a many-body state with an odd number of atoms in level 2 can only be transformed to another many-body state with an even number of atoms in the same level. This leads to a chiral symmetry of the following operator.
Unlike conventional symmetry, chiral symmetry operator anti-commutes with the Hamiltonian.
Let us define a function testing if a given state is a “left” eigenstate of the chiral symmetry operator.
Examine each element and check its chirality to construct a chiral basis.
In the following basis, the elements are arranged according to the chirality eigenvalues.
In the chiral basis, the matrix representation of the Hamiltonian is off diagonal.
The block-off-diagonal form of the matrix representation in each sector indicates the zero-energy subspace (null space). For example, in the first sector, the lower off-diagonal block is 2×4. This means that the null space is two dimensional regardless of Ω.