states. The color-singlet projection with unimodular-like constraint is imposed in the all preceding Hagedorn states. It is reduced to α = 1 2 for color non-singlet SU (3) quark-gluon bags. The values of the mass spectral exponent α with various symmetries are displayed in Table (I). The variation of the chiral Hagedorn’s grand potential density with respect to σ is reduced to The scalar mean field, namely, σ = < σ > is found by extremizing Ω$_{σ}$ ( T, µ, V, σ ) which is presented in Eq.(149). The value of mean field σ = σ$_{0}$ can be approximated to The consequence of Eq.(160) when the temperature reaches the deconfinement one is im- portant. The integral term on the right hand side of Eq.(160) diverges for the mass spectral exponent α ≤ 5 2 + 1. It implies that the only possible solution is that ∂ b ( σ ) ∂σ = 0 and σ = 0. This means that the chiral field is certainly restored before the deconfinement phase transition point is reached. Hence, the chiral symmetry is restored far away below the de- confinement point for nuclear matter that is dominated by the orthogonal Hagedorn states (or the Hagedorn states with mass spectral exponent α ≤ 5 2 + 1). On the other hand, it is possible to have a finite value solution for σ for Hagedorn states with mass spectral exponent α > 5 2 + 1. Furthermore, when the exponent exceeds the limit α > 7 2 + 1, then it is pos- sible to have the first order chiral phase transition simultaneously with the deconfinement phase transition point. This support the existence of the tri-critical point at finite µ$_{B}$ and T where the nuclear matter at moderate baryonic densities is supposed to be dominated by the unitary Hagedorn states prior the color deconfinement phase transition. In order to study the deconfinement phase transition diagram in QCD, the hadronic density of states must be known for the entire energy domain below the borderline of the VI. DISCUSSIONS 43 m 2 σ σ ∝ ( ∂ b ( σ ) ∂σ ) × ∫ ∞ m$_{0}$ m − α +5 / $^{2}$e( b ( σ ) − T ) m d m. (160) 1 V ∂ ∂σ Ω$_{Hσ}$ ( T, µ, V, σ ) = − 2(2 J + 1) 1 (2 π ) 3 / 2 T 5 / 2 c ( ∂ b ( σ ) ∂σ ) × ∫ ∞ m$_{0}$ dm m − α +5 / $^{2}$e( b ( σ ) − T ) $^{m}$, ∼ − ( . . . ) ∫ ∞ m$_{0}$ dm m − α +5 / $^{2}$e( b ( σ ) − T ) $^{m}$. (159)