Insignificance of the anomalous magnetic moment of the quarks in presence of chiral imbalance

Abstract

We incorporate the anomalous magnetic moment (AMM) of quarks in the framework of PNJL model to study hot and dense magnetised matter with chiral imbalance. For this purpose, the eigen energy solution of the Dirac equation is obtained in presence of constant background magnetic field and chiral chemical potential (CCP) along with the minimal anomalous magnetic moment interaction of the fermion. Although there is a marginal enhancement in the IMC behaviour of the quark condensate due to the combined effects of AMM and CCP, we find that the overall behaviour of the Polyakov loop and the chiral charge density is dominated by the chiral chemical potential. It is further shown that the AMM effects in presence of CCP remains insignificant even after consideration of thermo-magnetically modified moments.

Appendix A: Determination of energy eigenvalues

In a cylindrical coordinate system, we have

$$\begin{aligned} \dfrac{\partial }{\partial x}= & {} \cos \phi \dfrac{\partial }{\partial \rho } - \sin \phi \frac{1}{\rho } \dfrac{\partial }{\partial \phi }, \end{aligned}$$

(A1)

$$\begin{aligned} \dfrac{\partial }{\partial y}= & {} \sin \phi \dfrac{\partial }{\partial \rho } + \cos \phi \frac{1}{\rho } \dfrac{\partial }{\partial \phi }. \end{aligned}$$

(A2)

Also noting that, $x\pm iy = e^{\pm i \phi }$, we can write

$$\begin{aligned} \Pi _x \pm i \Pi _y= & {} -i \left( {\dfrac{\partial }{\partial x}\pm i \dfrac{\partial }{\partial y}}\right) - e\left( {A_x \pm i A_y}\right) \nonumber \\= & {} -i e^{\pm i \phi } \left\{ {\left( {\dfrac{\partial }{\partial \rho }\pm \frac{i}{\rho } \dfrac{\partial }{\partial \phi }}\right) \pm \rho \frac{1}{2}eB }\right\} . \end{aligned}$$

(A3)

Substituting Eqs. (8) and (9) into Eq. (7) and using Eq. (A3), we arrive at

$$\begin{aligned} E \begin{pmatrix} e^{i(l-1) \phi } f_1 \\ e^{i l \phi } f_2 \\ e^{i(l-1) \phi } f_3 \\ e^{i l \phi } f_4 \end{pmatrix} =\hat{H} \begin{pmatrix} e^{i(l-1) \phi } f_1 \\ e^{i l \phi } f_2 \\ e^{i(l-1) \phi } f_3 \\ e^{i l \phi } f_4 \end{pmatrix} \end{aligned}$$

(A4)

where, $\hat{H}$ is given by,

$$\begin{aligned} \hat{H} = \begin{pmatrix} m-\mu -a B &{} 0 &{} \mu _5 +p_z &{} -i e^{- i \phi } \left( {{\frac{\partial }{\partial \rho } - \frac{i}{\rho } \frac{\partial }{\partial \phi }} - \frac{1}{2}eB \rho }\right) \\ 0 &{} m-\mu +a B &{} -i e^{ i \phi } \left( {{\frac{\partial }{\partial \rho } + \frac{i}{\rho } \frac{\partial }{\partial \phi }} + \frac{1}{2}eB \rho }\right) &{} \mu _5 -p_z \\ \mu _5+p_z &{} -i e^{- i \phi } \left( {{\frac{\partial }{\partial \rho } - \frac{i}{\rho } \frac{\partial }{\partial \phi }} - \frac{1}{2}eB\rho }\right) &{}-m-\mu +a B &{} 0 \\ -i e^{ i \phi } \left( {{\frac{\partial }{\partial \rho } + \frac{i}{\rho } \frac{\partial }{\partial \phi }} + \frac{1}{2}eB\rho }\right) &{} \mu _5 -p_z &{}0 &{} -m-\mu -a B \end{pmatrix}. \end{aligned}$$

Writing down all the rows of the matrix in Eq. (A4) separately, we arrive at the following four equations:

$$\begin{aligned} \left( {E +\mu - m_1}\right) f_1(\rho )= & {} \left( \mu _5 + p_z\right) f_3(\rho ) \nonumber \\&- i\left( \frac{d}{d \rho }+ \frac{l}{\rho } - \xi \rho \right) f_4(\rho ), \end{aligned}$$

(A5)

$$\begin{aligned} \left( E +\mu - m_2\right) f_2(\rho )= & {} \left( {\mu _5 - p_z}\right) f_4(\rho ) \nonumber \\&- i\left( \frac{d}{d \rho }- \frac{l-1}{\rho } + \xi \rho \right) f_3(\rho ), \end{aligned}$$

(A6)

$$\begin{aligned} \left( {E +\mu + m_1}\right) f_3(\rho )= & {} \left( {\mu _5 + p_z}\right) f_1(\rho ) \nonumber \\&- i\left( \frac{d}{d \rho } + \frac{l}{\rho } - \xi \rho \right) f_2(\rho ), \end{aligned}$$

(A7)

$$\begin{aligned} \left( {E +\mu + m_2}\right) f_4(\rho )= & {} \left( {\mu _5 - p_z}\right) f_2(\rho ) \nonumber \\&- i\left( \frac{d}{d \rho }- \frac{l-1}{\rho } + \xi \rho \right) f_1(\rho )\nonumber \\ \end{aligned}$$

(A8)

where, $m_1 = (m - a B)$, $m_2 = (m+a B)$ and $\xi = eB/2$.

Now multiplying both sides of Eq. (A5) by $- i\left( {\frac{d}{d \rho } - \frac{l-1}{\rho } + \xi \rho }\right)$, and, using Eqs. (A6) and  (A8), we get after some simplifications

$$\begin{aligned}&\left( {\frac{d^2 {} }{d {\rho }^2} + \frac{1}{\rho } \frac{d {} }{d {\rho }} + 2\xi (l-1) - \xi ^2 \rho ^2 -\frac{l^2}{\rho ^2} + \mathfrak {B}_1^\prime }\right) f_4(\rho ) \nonumber \\&\qquad = \mathfrak {B}_2^\prime f_2(\rho ) \end{aligned}$$

(A9)

where,

$$\begin{aligned} \mathfrak {B}_1^\prime= & {} \left( {E+\mu }\right) ^2 \nonumber \\&- \left( {E+\mu }\right) \left( {m_1 - m_2 }\right) - m_1 m_2 + \mu _5^2 - p_z^2 , \end{aligned}$$

(A10)

$$\begin{aligned} \mathfrak {B}_2^\prime= & {} 2\left( {E+\mu }\right) \mu _5\nonumber \\&- \left( {m_1 + m_2 }\right) \mu _5 + \left( {m_1 - m_2}\right) p_z . \end{aligned}$$

(A11)

Similarly starting from Eq. (A7) one arrives at

$$\begin{aligned}&\left( {\frac{d^2 {} }{d {\rho }^2} + \frac{1}{\rho } \frac{d {} }{d {\rho }} + 2\xi (l-1) - \xi ^2 \rho ^2 -\frac{l^2}{\rho ^2} + \mathfrak {D}_1^\prime }\right) \nonumber \\&\quad \times f_2(\rho ) = \mathfrak {D}_2^\prime f_4(\rho ) \end{aligned}$$

(A12)

where

$$\begin{aligned} \mathfrak {D}_1^\prime= & {} \left( {E+\mu }\right) ^2 + \left( {E+\mu }\right) \left( {m_1 - m_2 }\right) \nonumber \\&- m_1 m_2 + \mu _5^2 - p_z^2 , \end{aligned}$$

(A13)

$$\begin{aligned} \mathfrak {D}_2^\prime= & {} 2\left( {E+\mu }\right) \mu _5 + \left( {m_1 + m_2 }\right) \mu _5\nonumber \\&- \left( {m_1 - m_2}\right) p_z . \end{aligned}$$

(A14)

Introducing a dimensionless variable $\lambda = \xi \rho ^2$, we have $\frac{d {} }{d {\rho }} \rightarrow 2\sqrt{\lambda \xi } \frac{d {} }{d {\lambda }}$, so that, Eqs. (A9) and (A12) become

$$\begin{aligned}&\left[ {\lambda \frac{d^2 {} }{d {\lambda }^2} + \frac{d {} }{d {\lambda }} - \frac{l^2}{4\lambda } - \frac{\lambda }{4} -\frac{1}{2} \left( {l -1}\right) + \mathfrak {B}_1}\right] f_4(\lambda )\nonumber \\&\quad = \mathfrak {B}_2 f_2(\lambda ) , \end{aligned}$$

(A15)

$$\begin{aligned}&\left[ {\lambda \frac{d^2 {} }{d {\lambda }^2} + \frac{d {} }{d {\lambda }} - \frac{l^2}{4\lambda } - \frac{\lambda }{4} -\frac{1}{2} \left( {l -1}\right) + \mathfrak {D}_1}\right] f_2(\lambda ) \nonumber \\&\quad = \mathfrak {D}_2 f_4(\lambda ) \end{aligned}$$

(A16)

where, $\mathfrak {B}_i = \mathfrak {B}_i^\prime / 4\xi ,\mathfrak {D}_i = \mathfrak {D}_i^\prime / 4\xi$. The functions $f_1$ and $f_3$ obey similar kind of equations but they will not concern us here as we are only interested to obtain the energy eigenvalue. The above two differential equations have regular singularities at $\lambda = 0$. So we can solve them using Frobenius method in which we assume

$$\begin{aligned} f_4(\lambda ) = e^{-\lambda /2} \lambda ^s \sum _{N = 0}^{\infty } c_N \lambda ^N ,\quad f_2(\lambda ) = e^{-\lambda /2} \lambda ^s \sum _{N = 0}^{\infty }d_N \lambda ^N. \end{aligned}$$

(A17)

Substituting Eq. (A17) into Eqs. (A15) and (A16)and equating the coefficient of $e^{-\lambda /2} \lambda ^{s+N-1}$, we get after some simplifications the following recursion relations

$$\begin{aligned}&\left\{ { \mathfrak {B}_1-N+\frac{1}{2} - \frac{1}{2}(l-1) -s }\right\} c_{N-1}\nonumber \\&\quad + \left\{ {(N+ s)^2 -\frac{l^2}{4} }\right\} c_N - \mathfrak {B}_2~ d_{N-1 }= 0 , \end{aligned}$$

(A18)

$$\begin{aligned}&\left\{ { \mathfrak {D}_1-N+\frac{1}{2} - \frac{1}{2}(l-1) -s }\right\} d_{N-1}\nonumber \\&\quad + \left\{ {(N+ s)^2 -\frac{l^2}{4} }\right\} ~d_N - \mathfrak {D}_2 ~c_{N-1 }= 0 . \end{aligned}$$

(A19)

If we take $N = 0$, we get $2s = \pm l$. But we will discard the solutions which diverge at $\lambda = 0$. Hence, considering $2s = l$, we get from Eqs. (A18) and (A19)

$$\begin{aligned} \left( { \mathfrak {B}_1-N+1-l }\right) c_{N-1} + N (N+l) c_N - \mathfrak {B}_2~ d_{N-1 }= & {} 0, \end{aligned}$$

(A20)

$$\begin{aligned} \left( { \mathfrak {D}_1-N+1-l }\right) d_{N-1} + N (N+l) d_N - \mathfrak {D}_2~ c_{N-1 }= & {} 0 .\nonumber \\ \end{aligned}$$

(A21)

To obtain well behaved wave function, we assume that the series must terminate at some $N = N^\prime$ (this ensures that, we get polynomial solution in $\lambda$ and since we already have an $e^{-\lambda /2}$ term, the solution must vanish as $\lambda \rightarrow \infty$ ) so that, $c_{N^\prime + 1} = 0$ and $d_{N^\prime +1 } = 0$. Thus from Eqs. (A20) and (A21), we get

$$\begin{aligned} \left( { \mathfrak {B}_1-N^\prime -l }\right) c_{N^\prime } - \mathfrak {B}_2 ~d_{N^\prime }= & {} 0 , \end{aligned}$$

(A22)

$$\begin{aligned} \left( { \mathfrak {D}_1-N^\prime -l }\right) ~d_{N^\prime } - \mathfrak {D}_2` c_{N^\prime }= & {} 0 . \end{aligned}$$

(A23)

For Eqs. (A22) and (A23) to have non-trivial solutions, we must have

$$\begin{aligned} \text {det} \begin{pmatrix} \mathfrak {B}_1-(N^\prime +l) &{}\quad - \mathfrak {B}_2 \\ - \mathfrak {D}_2 &{}\quad \mathfrak {D}_1-(N^\prime +l) \end{pmatrix} = 0 \end{aligned}$$

(A24)

Simplification of Eq. (A24) yields,

$$\begin{aligned} \left( E+\mu \right) ^4 + \mathfrak {B} \left( E+\mu \right) ^2 + \mathfrak {C} = 0 \end{aligned}$$

(A25)

where

$$\begin{aligned} \mathfrak {B}= & {} - 2\left\{ p_z^2 + m^2 + \mu _5^2 + (a B)^2 \right\} - 8\xi (N^\prime + l) , \end{aligned}$$

(A26)

$$\begin{aligned} \mathfrak {C}= & {} \left\{ p_z^2 + m^2 -( \mu _5^2 + (a B)^2) \right\} ^2 + 4\left( m\mu _5 + p_z a B \right) \nonumber \\&+~ 8\xi (N^\prime + l ) \left\{ p_z^2 + m^2 -( \mu _5^2 + (a B)^2) \right\} \nonumber \\&+ 64 \xi ^2 (N^\prime + l)^2 , \end{aligned}$$

(A27)

and the discriminant is

$$\begin{aligned} (\mathfrak {B}^2 - 4\mathfrak {C})= & {} 16 \left[ \left\{ m^2 + 4\xi (N^\prime + l) \right\} \left( a B\right) ^2 \nonumber \right. \\&\left. + \left\{ p_z^2 + 4\xi (N^\prime + l) \right\} \mu _5^2 - 2 mp_z \mu _5 a B \right] . \nonumber \\ \end{aligned}$$

(A28)

Hence, the solution of Eq. (A25) is given by,

$$\begin{aligned}&(E+ \mu )^2 = \frac{-\mathfrak {B} \pm \sqrt{\mathfrak {B}^2- 4\mathfrak {C}}}{2}\nonumber \\&\quad = p_z^2 + m^2 + \mu _5^2 + \left( a B\right) ^2 \nonumber \\&\qquad + 2 (N^\prime + l) eB \pm 2 \left[ \left\{ m^2 + 2 (N^\prime + l)eB \right\} (a B)^2 \right. \nonumber \\&\qquad \left. + \left\{ p_z^2 + 2 (N^\prime + l)eB \right\} \mu _5^2 - 2 m p_z \mu _5 a B \right] ^{1/2} \end{aligned}$$

(A29)

Following Ref. [81], we replace $\pm \rightarrow -s {~\mathrm sign}(e B)$ where s is the helicity in the massless case [46, 82]. This will ensure that we get back the correct result when the non-relativistic limit is taken. Also, we identify $(N^\prime +l) =n$ as the Landau level. With these replacements, Eq. (A29) becomes

$$\begin{aligned} (E+ \mu )^2= & {} p_z^2 + m^2 + \mu _5^2 + (a B)^2 + 2 n eB \nonumber \\&- 2 s {~\mathrm sign} (eB) \left[ ( m^2 + 2 n eB ) (a B)^2 \nonumber \right. \\&\left. + ( p_z^2 + 2 n eB ) \mu _5^2 - 2 m p_z \mu _5 a B \right] ^{1/2}. \end{aligned}$$

(A30)

For ground state, we have $n = 0$ corresponding to the lowest Landau level (LLL), so that the energy eigenvalue becomes

$$\begin{aligned} (E_0 +\mu )^2 = (p_z - \mu _5)^2 + (m - \kappa \big |{e B}\big |)^2. \end{aligned}$$

(A31)

It is to be noted that, for a positively (negatively) charged fermion, the ground sate contribution comes from spin down (up) state. On the other hand, for higher Landau levels, $n \ge 1$ we get,

$$\begin{aligned} (E_{ns}+ \mu )^2= & {} p_z^2 + m^2 + \mu _5^2 + (\kappa e B)^2 + 2 n \big |{eB}\big | \nonumber \\&- 2 s {~\mathrm sign} (e B) \left[ ( m^2 + 2 n \big |{eB}\big | ) (\kappa e B)^2 \nonumber \right. \\&\left. + ( p_z^2 + 2 n \big |{eB}\big | ) \mu _5^2 - 2 m p_z \mu _5 \kappa e B \right] ^{1/2} . \nonumber \\ \end{aligned}$$

(A32)