Polarization in Relativistic Fluids: A Quantum Field Theoretical Derivation

Abstract

We review the calculation of polarization in a relativistic fluid within the framework of statistical quantum field theory. We derive the expressions of the spin density matrix and the mean spin vector both for a single quantum relativistic particle and for a quantum-free field. After introducing the formalism of the covariant Wigner function for the scalar and the Dirac field, the relation between the spin density matrix and the covariant Wigner function is obtained. The formula is applied to the fluid produced in relativistic nuclear collisions by using the local thermodynamic equilibrium density operator and recovering previously known formulae. The dependence of these results on the spin tensor and pseudo-gauge transformations of the stress-energy tensor is addressed.

Appendix: Angular Momentum Decomposition

We shall prove that the orbital part of the angular momentum operator (2.79) can be written as an integral in momentum space of on-shell functions. We will confine ourselves to the proof for the particle term in (2.48), its extension to the antiparticle term and the proof of the vanishing of the mixed term being alike. By using (2.48) and choosing the hyperplane $t=0$ as integration hypersurface, for the particle term, we can write:

$$\begin{aligned} \begin{aligned} \int \mathrm{d}^3 \mathrm{x}\; x^\mu k^\nu&\mathrm{tr}_4 (\gamma ^0 {\widehat{W}}(x,k)) =\\&\int \mathrm{d}^3 \mathrm{x}\; x^\mu k^\nu \sum _{r,s} \frac{1}{(2\pi )^3} \int \frac{\mathrm{d}^3 \mathrm{p}}{2\varepsilon }\frac{\mathrm{d}^3 \mathrm{p}^\prime }{2\varepsilon ^\prime } \mathrm{e}^{-i(p - p^\prime ) \cdot x} \times \\&\times \delta ^4 \left( k - \frac{p+p^\prime }{2} \right) {\widehat{a}}^\dagger _s (p^\prime ) {\widehat{a}}_r (p) \bar{u}_s(p') \gamma ^0 u_r(p) \\ =&8 \int \mathrm{d}^3 \mathrm{x}\; x^\mu k^\nu \sum _{r,s} \frac{1}{(2\pi )^3} \int \frac{\mathrm{d}^3 \mathrm{p}}{4\varepsilon \, \varepsilon _{k,p}} \mathrm{e}^{i( 2 k -2 p )\cdot x}\times \\&\times \delta \left( k^0 - \frac{\varepsilon +\varepsilon ^\prime _{k,p}}{2} \right) {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p), \end{aligned} \end{aligned}$$

(2.96)

where

$$ \varepsilon _{k,p} = \sqrt{(2 \mathbf{k} - \mathbf{p})^2 + m^2}. $$

and it is understood that in the arguments of creation and destruction operators, as well as of spinors u, only the spatial part of the four-vector k, that is $\mathbf{k}$, enters.

For $\mu =0$, the integration is straightforward as $x^0$ is constant on the hyperplane and we get, after integrating in $\mathrm{d}^3 \mathrm{x}$:

$$\begin{aligned}&x^0 k^\nu \! \sum _{r,s}\! \int \!\! \frac{\mathrm{d}^3 \mathrm{p}}{4\varepsilon \, \varepsilon _{k,p}} \delta ^3 (\mathbf{p}-\mathbf{k}) \delta \left( k^0 - \frac{\varepsilon +\varepsilon _{k,p}}{2} \right) {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p) \\&= x^0 k^\nu \sum _{r,s} \frac{1}{(2\pi )^3} \frac{1}{4\varepsilon _k^2} \delta (k^0 -\varepsilon ) {\widehat{a}}^\dagger _s (k) {\widehat{a}}_r (k) \bar{u}_s(k) \gamma ^0 u_r(k) \\&= x^0 k^\nu \delta ( k^0 -\varepsilon _k) \sum _{r} \frac{1}{2\varepsilon _k} {\widehat{a}}^\dagger _r (k) {\widehat{a}}_r (k) \end{aligned}$$

because $p'=2k-p$ and $\mathbf{p}=\mathbf{k}$ implies in turn $k=p=p^\prime$, hence k is on-shell; we have also used the known spinor relations.

For $\mu = i \ne 0$, we can replace $x^\mu$ with a derivative of the exponential and, integrating by parts,

$$\begin{aligned} \int \mathrm{d}^3 \mathrm{x}&\; x^i k^\nu \mathrm{tr}_4 (\gamma ^0 {\widehat{W}}(x,k)) \\ =&4 i\int \mathrm{d}^3 \mathrm{x}\; k^\nu \sum _{r,s} \frac{1}{(2\pi )^3} \int \frac{\mathrm{d}^3 \mathrm{p}}{4\varepsilon \, \varepsilon _{k,p}} \frac{\partial }{\partial p^i} \mathrm{e}^{i(2 k -2 p) \cdot x} \times \\&\times \delta \left( k^0 - \frac{\varepsilon +\varepsilon _{k,p}}{2} \right) {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p) \\ =&4 i\int \mathrm{d}^3 \mathrm{x}\; k^\nu \sum _{r,s} \frac{1}{(2\pi )^3} \int \frac{\mathrm{d}^3 \mathrm{p}}{4\varepsilon \, \varepsilon _{k,p}} \times \\ \times&\frac{\partial }{\partial p^i} \left[ \mathrm{e}^{i(2 k - 2 p) \cdot x} \delta \left( k^0 - \frac{\varepsilon +\varepsilon _{k,p}}{2} \right) {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p) \right] + \\&- 4 i\int \mathrm{d}^3 \mathrm{x}\; k^\nu \sum _{r,s} \frac{1}{(2\pi )^3} \int \frac{\mathrm{d}^3 \mathrm{p}}{4\varepsilon \, \varepsilon _{k,p}} \mathrm{e}^{i(2 k -2 p) \cdot x } \frac{\partial }{\partial p^i}\times \\&\,\,\times \left[ \delta \left( k^0 - \frac{\varepsilon +\varepsilon _{k,p}}{2} \right) {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p) \right] . \end{aligned}$$

The first term gives rise to a boundary integral which vanishes for fixed k and only the second term survives. We can now integrate in $\mathrm{d}^3 \mathrm{x}$ getting:

$$\begin{aligned} \begin{aligned} - \frac{i}{2}&k^\nu \sum _{r,s} \int \frac{\mathrm{d}^3 \mathrm{p}}{4\varepsilon \, \varepsilon _{k,p}} \delta ^3(\mathbf{p}-\mathbf{k}) \frac{\partial }{\partial p^i}\times \\&\times \left[ \delta \left( k^0 - \frac{\varepsilon +\varepsilon _{k,p}}{2} \right) {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p) \right] . \end{aligned} \end{aligned}$$

(2.97)

There appear two derivative terms in the above expression: the derivative of the $\delta$ can be written as

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial p^i} \delta \left( k^0 - \frac{\varepsilon + \varepsilon ^\prime _{k,p}}{2} \right) =&- \frac{1}{2} \frac{\partial }{\partial k^0} \delta \left( k^0 - \frac{\varepsilon + \varepsilon _{k,p}}{2} \right) \frac{\partial }{\partial p^i} (\varepsilon + \varepsilon _{k,p})\\ =&- \frac{1}{2} \frac{\partial }{\partial k^0} \delta \left( k^0 - \frac{\varepsilon + \varepsilon _{k,p}}{2} \right) \left( \frac{p^i}{\varepsilon } - \frac{2 k^i - p^i}{\varepsilon _{k,p}} \right) , \end{aligned} \end{aligned}$$

(2.98)

while the derivative of the factor including creation and destruction operators and spinors yields, taking into account the $\delta ^3(\mathbf{p}-\mathbf{k})$:

$$\begin{aligned}&\delta ^3(\mathbf{p}-\mathbf{k}) \frac{\partial }{\partial p^i} {\widehat{a}}^\dagger _s (2k-p) {\widehat{a}}_r (p) \bar{u}_s(2k - p) \gamma ^0 u_r(p) \\ \nonumber&= \delta ^3(\mathbf{p}-\mathbf{k}) \left[ \left( {\widehat{a}}^\dagger _s (p) {{\mathop {\frac{\partial }{\partial p^i}}\limits ^{\leftrightarrow }}}{\widehat{a}}_r (p) \right) \bar{u}_s(p) \gamma ^0 u_r(p)+ {\widehat{a}}^\dagger _s (p){\widehat{a}}_r (p) \left( \bar{u}_s(p) {{\mathop {\frac{\partial }{\partial p^i}}\limits ^{\leftrightarrow }}}\gamma ^0 u_r(p) \right) \right] . \end{aligned}$$

(2.99)

We can now plug Eqs. (2.98) and (2.99) into (2.97). The term (2.98) vanishes because

$$\begin{aligned}&\delta ^3(\mathbf{p}-\mathbf{k}) \left( \frac{p^i}{\varepsilon } - \frac{2 k^i - p^i}{\varepsilon _{k,p}} \right) k^\nu \frac{\partial }{\partial k^0} \delta \left( k^0 - \frac{\varepsilon + \varepsilon _{k,p}}{2} \right) \\&= - \delta ^3(\mathbf{p}-\mathbf{k}) \left( \frac{p^i}{\varepsilon } - \frac{2 k^i - p^i}{\varepsilon _{k,p}} \right) \delta ^\nu _0 \delta \left( k^0 - \frac{\varepsilon + \varepsilon _{k,p}}{2} \right) \\&= - \delta ^3(\mathbf{p}-\mathbf{k}) \delta ^\nu _0 \delta \left( k^0 - \varepsilon \right) \left( \frac{k^i}{\varepsilon _k} - \frac{k^i}{\varepsilon _k} \right) = 0, \end{aligned}$$

and we are just left with the term from (2.99).

We can now integrate in $\mathrm{d}^4 k$ according to Eq. (2.79). For $\mu =i$,

$$\begin{aligned} \begin{aligned}&\int \mathrm{d}^4 k \int \mathrm{d}^3 \mathrm{x}\; x^i k^\nu \mathrm{tr}_4 (\gamma ^0 {\widehat{W}}(x,k)) = - \frac{i}{2} \int \mathrm{d}^4 k \; k^\nu \sum _{r,s} \frac{1}{4\varepsilon _k^2} \delta (k_0 - \varepsilon _k ) \times \\&\times \left[ \left( {\widehat{a}}^\dagger _s (k) {{\mathop {\frac{\partial }{\partial k^i}}\limits ^{\leftrightarrow }}}{\widehat{a}}_r (k) \right) \bar{u}_s(k) \gamma ^0 u_r(k) + {\widehat{a}}^\dagger _s (k){\widehat{a}}_r (k) \left( \bar{u}_s(k) {{\mathop {\frac{\partial }{\partial k^i}}\limits ^{\leftrightarrow }}}\gamma ^0 u_r(k) \right) \right] \\ =&- \frac{i}{2} \int \mathrm{d}^3 \text {k}\; k^\nu \sum _{r,s} \frac{1}{4\varepsilon _k^2} \left[ \left( {\widehat{a}}^\dagger _s (k) {{\mathop {\frac{\partial }{\partial k^i}}\limits ^{\leftrightarrow }}}{\widehat{a}}_r (k) \right) \bar{u}_s(k) \gamma ^0 u_r(k)\right. \\&\left. + {\widehat{a}}^\dagger _s (k){\widehat{a}}_r (k) \left( \bar{u}_s(k) {{\mathop {\frac{\partial }{\partial k^i}}\limits ^{\leftrightarrow }}}\gamma ^0 u_r(k) \right) \right] \end{aligned} \end{aligned}$$

(2.100)

with $k^\nu$ again on-shell. We can then conclude that

$$ \int \mathrm{d}^4 k \int _{t=0} \mathrm{d}^3 \mathrm{x}\; x^\mu k^\nu \mathrm{tr}_4 (\gamma ^0 {\widehat{W}}(x,k)) = \int \frac{\mathrm{d}^3 \text {k}}{2 \varepsilon _k} \; \widehat{G}^\mu (k) k^\nu $$

with k on-shell and $\widehat{G}^0(k) = 0$ if $x^0 = 0$ is chosen.

Finally, we briefly address the calculation of the angular momentum tensor by using the Belinfante stress-energy tensor (2.82) where only the orbital part is involved. The calculation is very similar to the one just described, with the important difference that the second term of (2.82), obtained by swapping the indices of the first term in (2.82), leads to a term akin to the left hand side of Eq. (2.96) with exchanged indices:

$$ \int \mathrm{d}^3 \mathrm{x}\; x^\mu k^0 \mathrm{tr}_4 (\gamma ^\nu {\widehat{W}}(x,k)). $$

However, the final result is not proportional to $k^\nu$ and a double derivative term appears just like in Eq. (2.100); therefore, this term is not cancelled by the Levi-Civita tensor in the calculation of the mean spin, unlike in the canonical case.