In this section, we reveal that the Hamiltonian described by the Equation (3) is non- hermitian if ϕ = − 1 . It is called Mandal Hamiltonian (i.e. H M ) and it takes the following form:

Taking account to the following identities:

let’s show that the Mandal Hamiltonian given by the above Equation (4) is non hermitian:

H M + = ( ε 2 σ z ) + + ( ω a + a ) + + [ ρ ( σ + a − σ − a + ) ] + ,

Comparing the expressions given by the Equations (4) and (6), we see that they are different (i.e. H M + ≠ H M ), as a consequence, we are allowed to conclude that the Mandal Hamiltonian H M is non-hermitian.

In this section, we prove that the Mandal Hamiltonian is non PT-symmetric [5] [6] . Recall that the parity operator is represented by the symbol P and the time-reversal operator is described by the symbol T.

The effect of the parity operator P implies the following changes [1] [2] :

Notice also the changes of the following quantities under the effect of the time reversal operator T:

Taking account to the relations (7) and (8), one can easily deduce the changes of the Mandal Hamiltonian under the effect of combined operators P et T as follows

,

This above relation (9) can be written as follows

Comparing the relations (4) and (10), we see that they are different (i.e. H M P T ≠ H M ), it means that the Mandal Hamiltonian H M is not invariant under the combined action of the parity operator P and the time-reversal operator T. In other words, the Mandal Hamiltonian H M is not PT-symmetric.

In this section, we first prove that the non PT-symmetric Mandal Hamiltonian is pseudo-hermitian with respect to third Pauli matrix σ z [1] [2] :

with σ z σ ± σ z − 1 = − σ ∓ and W n = e − ω x 2 2 ( P n − 1 ( x ) , P n ( x ) ) t .

Comparing the Equations (6) and (11), it is seen that the following relation is satisfied:

W n = e − ω x 2 2 ( P n − 1 ( x ) , P n ( x ) ) t (12)

Taking account to this above relation, we are allowed to conclude that the Mandal Hamiltonian is pseudo-hermitian with respect to σ z .

Finally, we reveal a pseudo-hermiticity of H M with respect to the parity operator P:

Here we have used the relations (7) in order to obtain this above equation (13). As a consequence, one can conclude that the Mandal Hamiltonian is pseudo-hermitian with respect to the parity operator P.

Note that even if H M is non hermitian and non PT-symmetric, its eigenvalues are entirely real due to the pseudo-hermiticity property [1] .

In this step, our purpose is to change the Mandal Hamiltonian given by the Equation (4) in appropriate differential operator (i.e. H M is expressed in the position operator x and in the impulsion operator H J C + = ε 2 σ z + ω a + a + [ ρ ( σ + a + σ − a + ) ] ). Thus, referring to the ideas of exactly and quasi-exactly solvable operators studied in the Refs. [7] [8] [9] [10] , we reveal that H M preserves a family of vector spaces of polynomials in the variable x.

With this aim, we use the usual representation of the creation and annihilation operators of the harmonic oscillator respectively a + and a [1] [2] :

where ω is the oscillation frequency, m denotes the mass, x refers to the position operator and the impulsion operator is, p 2 = − d 2 d x 2 .

Using appropriate units, we can assume m = ℏ = 1 and the operators a + and a take the following forms:

Replacing the operators a + and a by their expressions given by this above Equation (15) in the Equation (4), the Mandal Hamiltonian H M takes the following form:

In order to reveal the exact solvability of the above operator H M , we first perform the standard gauge transformation [2] :

After some algebraic manipulations, the new Hamiltonian H ˜ M (known as gauge Hamiltonian) is obtained

Replacing the Pauli matrices and by their respective expressions given by the relation (2), the final form of the gauge Hamiltonian is:

,

Note that one can easily check if this above gauge Hamiltonian H ˜ M preserves the vector spaces of polynomials with n ∈ Ν . As the integer n doesn’t have to be fixed (i.e. it is arbitrary), H ˜ M is exactly solvable. Indeed, its all eigenvalues can be computed algebraically. Even if the gauge Mandal Hamiltonian H ˜ M is non-hermitian and non PT-symmetric, its spectrum energy is entirely real due to the property of the pseudo-hermiticity [1] [2] .

Thus, the vector spaces preserved by the operator H M have the following form

where P n − 1 ( x ) and P n ( x ) denote respectively the polynomials of degree n − 1 and n.

As the gauge Mandal Hamiltonian H ˜ M , it is obvious that the original Mandal Hamiltonian H M is exactly solvable. Due to this property of exact solvability, the whole spectrum of H M can be computed exactly (i.e. by the algebraic methods) [1] [2] [3] .

In this section, considering ϕ = + 1 , the Hamiltonian given by the Equation (3) leads to the standard Jaynes-Cummings Hamiltonian of the following form

H J C = ε 2 σ z + ω a + a + ρ ( σ + a + σ − a + ) (21)

Our aim is now to prove that the above Hamiltonian H J C is hermitian.

Indeed, in order to reveal the hermiticity of the Jaynes-Cummings Hamiltonian given by the above relation (21), the following relation H J C + = H J C must be satisfied.

Consider now the following relation

H J C + = ( ε 2 σ z ) + + ( ω a + a ) + + [ ρ ( σ + a + σ − a + ) ] + , (22)

Taking account to the identities of the relation (5), this above equation leads the following expression:

H J C + = ε 2 σ z + ω a + a + [ ρ ( σ + a + σ − a + ) ] . (23)

Comparing the Equations (21) and (23), one can write that

H J C + = H J C . (24)

Referring to this equation (24), it is obvious that the standard Jaynes-Cummings Hamiltonian is hermitian. As a consequence, its eigenvalues are real due to the property of hermiticity.

Along the same lines as in the above section 3.4, our purpose is to change the Jaynes-Cummings Hamiltonian given by the Equation (21) in appropriate differential operator (i.e. H J C is expressed in the position operator x and in the impulsion operator p = − i d d x ).

With this purpose, we use the usual expressions of the creation and annihilation operators of the harmonic oscillator respectively a + and a given by the Equation (15).

Substituting (15) in the Equation (21), the Jaynes-Cummings Hamiltonian H J C is written now as follows

H J C = ε 2 σ z + p 2 − ω + ω 2 x 2 2 + ρ [ σ + ( p − i ω x ) + σ − ( p + i ω x ) ] 2 ω (25)

Operating on the above operator H J C the standard gauge transformation as

H ˜ J C = R − 1 H J C R ,       R = exp ( − ω x 2 2 ) , (26)

after some algebraic manipulations, the new Hamiltonian H ˜ J C (known as gauge Hamiltonian) is obtained

H ˜ M = ε 2 σ z − 1 2 d 2 d x 2 + ω x d d x + ρ [ σ + p + σ − ( p + 2 i ω x ) ] 2 ω = ε 2 σ z + p 2 2 + i ω x p + ρ [ σ + p + σ − ( p + 2 i ω x ) ] 2 ω (27)

Replacing the Pauli matrices σ z , σ + and σ − respectively by their matrix form given by the relation (2), the final form of the gauge Hamiltonian H ˜ J C is

H ˜ M = ε 2 ( 1 0 0 − 1 ) + ( p 2 2 + i ω x p 0 0 p 2 2 + i ω x p ) + ρ ( 0 p 2 ω p + 2 i ω x 2 ω 0 ) ,

H ˜ M = ( p 2 2 + i ω x p + ε 2 ρ p 2 ω ρ p + 2 i ω x 2 ω p 2 2 + i ω x p − ε 2 ) . (28)

Note that one can easily check if this above gauge Hamiltonian H ˜ J C preserves the finite dimensional vector spaces of polynomials namely V n = ( P n − 1 ( x ) , P n ( x ) ) t with n ∈ Ν . As the integer n is arbitrary, the gauge Jaynes-Cummings Hamiltonian H ˜ J C is exactly solvable.

As a consequence, its all eigenvalues can be computed algebraically. Indeed, the vector spaces preserved by the operator H J C have the following form

W n = e − ω x 2 2 ( P n − 1 ( x ) , P n ( x ) ) t (29)

where P n − 1 ( x ) and P n ( x ) denote respectively the polynomials of degree n − 1 and n.

As the gauge Jaynes-Cummings Hamiltonian H ˜ J C , it is obvious that the standard Jaynes-Cummings Hamiltonian H J C is exactly solvable. In other words, all eigenvalues associated to the Hamiltonian H J C can be calculated algebraically (i.e. by the algebraic methods) [1-3].