Particle Polarization, Spin Tensor, and the Wigner Distribution in Relativistic Systems

Abstract

Particle spin polarization is known to be linked both to rotation (angular momentum) and magnetization of many particle systems. However, in the most common formulation of relativistic kinetic theory, the spin degrees of freedom appear only as degeneracy factors multiplying phase-space distributions. Thus, it is important to develop theoretical tools that allow to make predictions regarding the spin polarization of particles, which can be directly confronted with experimental data. Herein, we discuss a link between the relativistic spin tensor and particle spin polarization, and elucidate the connections between the Wigner function and average polarization. Our results may be useful for the theoretical interpretation of heavy-ion data on spin polarization of the produced hadrons.

Appendix: Expectation Values of Creation and Destruction Operators

In this appendix, we show details of the calculations of the expectation values of creation and destruction operators. In particular, we find an interesting and intuitive link between the average (anti)particle number and the quantum fluctuations required for the mixed terms ($\langle a^\dagger b^\dagger \rangle$ and $\langle a b\rangle$) to be nonvanishing.

The starting point is the density matrix (5.2), which reads

$$\begin{aligned} \rho = \sum _i \mathsf{P}_i \left| \psi _i \right\rangle \left\langle \psi _i\right| . \end{aligned}$$

(5.86)

All $\mathsf{P}_i$’s are classical probabilities

$$\begin{aligned} \sum _i \mathsf{P}_i =1. \end{aligned}$$

(5.87)

The states $|\psi _i\rangle$ are proper quantum states, that is, they are normalized to one

$$\begin{aligned} \langle \psi _i|\psi _i\rangle =1, \qquad \forall i. \end{aligned}$$

(5.88)

The expectation value $\mathcal{O}$ of any quantum operator¹⁵ $\hat{\mathcal{O}}$ is a weighted average of the expectation values in the pure states, with the classical weights $\mathsf{P}_i$

$$\begin{aligned} \mathcal{O} = \mathrm{tr}\left( \phantom {\frac{}{}} \rho \, \hat{\mathcal{O}} \right) = \sum _i \mathsf{P}_i \, \mathrm{tr}\left( \phantom {\frac{}{}} \left| \psi _i \right\rangle \left\langle \psi _i\right| \hat{\mathcal{O}} \right) , \end{aligned}$$

(5.89)

therefore, our problem reduces to the expectation value in a generic pure state. The trace must be taken over a complete set of independent states (not necessarily quantum states that are normalized to one). Since we are interested in the expectation values of the creation and destruction operators of four-momentum and polarization eigenstates, the most convenient states are N-particle and $\bar{N}$-antiparticle ones. The trace is defined as an integration over the momentum degrees of freedom and a sum over discrete polarizations, namely

$$\begin{aligned} \begin{aligned} \mathrm{tr}\left( \cdots \phantom {\frac{}{}} \right)&= \sum _r\int \frac{d^3 p}{(2\pi )^3 2 E_\mathbf{p}} \langle p,r| \cdots |p,r\rangle +\\&\quad + \sum _{r,s}\int \frac{d^3 p}{(2\pi )^3 2 E_\mathbf{p}}\frac{d^3 q}{(2\pi )^3 2 E_\mathbf{q}} \langle p,r,q,s| \cdots |p,r,q,s\rangle + \cdots , \end{aligned} \end{aligned}$$

(5.90)

and so on, until exausting all the combinations of N particles and $\bar{N}$ antiparticles. In the last formula the standard definition is used

$$\begin{aligned} |p,r\rangle = \sqrt{2 E_{p}}a^\dagger _r(\mathbf{p})|0\rangle , \end{aligned}$$

(5.91)

along with the analogous expressions for antiparticles and multiparticle states. The anticommutation relations have the form

$$\begin{aligned} \{a_r(\mathbf{p}),a^\dagger _s(\mathbf{p}^\prime )\}=\{b_r(\mathbf{p}),b^\dagger _s(\mathbf{p}^\prime )\} = (2\pi )^3 \delta _{rs}\,\delta ^3(\mathbf{p}-\mathbf{p}^\prime ), \end{aligned}$$

(5.92)

with the normalization

$$\begin{aligned} \langle p,r|q,s\rangle = 2E_\mathbf{p} (2 \pi )^3 \delta _{rs}\, \delta ^3(\mathbf{p}-\mathbf{p}^\prime ). \end{aligned}$$

(5.93)

It is convenient to introduce the compact notation for multiparticle states

$$\begin{aligned} \begin{aligned} \!\!\!\! |{\underline{p}},{\underline{r}};\bar{\underline{q}},\bar{\underline{s}}\rangle&=|p_1,r_1,p_2,r_2,\cdots p_N, r_N; {\bar{q}}_1,{\bar{s}}_1,{\bar{q}}_2,{\bar{s}}_2,\cdots {\bar{q}}_{\bar{N}},{\bar{s}}_{\bar{N}}\rangle ,\\ \int [d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}}&= \int \!\!\!\frac{d^3 p_1}{(2\pi )^3 2 E_{\mathbf{p}_1}}\cdots \frac{d^3 p_N}{(2\pi )^3 2 E_{\mathbf{p}_N}}\frac{d^3 {\bar{q}}_1}{(2\pi )^3 2 E_{{\bar{\mathbf{q}}}_1}}\cdots \frac{d^3 {\bar{q}}_{\bar{N}}}{(2\pi )^3 2 E_{{\bar{\mathbf{q}}}_{\bar{N}}}}, \end{aligned} \end{aligned}$$

(5.94)

where the bar is used to distinguish antiparticle from particle variables. In this way the trace (5.90) can be written in a more compact form as

$$\begin{aligned} \mathrm{tr}\left( \phantom {\frac{(}{}{}}\cdots \right) = \sum _{N,{\bar{N}}}\sum _{{\underline{r}},{\underline{\bar{s}}}} \int \!\!\![d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}} \langle {\underline{p}},{\underline{r}};\bar{\underline{q}},\bar{\underline{s}}| \cdots |{\underline{p}},{\underline{r}};\bar{\underline{q}},\bar{\underline{s}}\rangle . \end{aligned}$$

(5.95)

This compact notation is useful to write the generic quantum state $|\psi \rangle$

$$\begin{aligned} |\psi \rangle = \sum _{N,{\bar{N}}} \sum _{{\underline{r}},{\bar{\underline{s}}}}\int \!\!\![d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}}\; \alpha _{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}})\, |{\underline{p}},{\underline{r}};\bar{\underline{q}},\bar{\underline{s}}\rangle , \end{aligned}$$

(5.96)

where the complex functions $\alpha _{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}})$ are partial N-particle-$\bar{N}$-antiparticle wave functions in momentum space. The normalization reads

$$\begin{aligned} \begin{aligned} 1= \langle \psi |\psi \rangle&= \sum _{N,{\bar{N}}} \sum _{{\underline{r}},{\bar{\underline{s}}}}\int \!\!\![d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}}\; \alpha ^*_{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}}) \alpha _{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}})=\\&= \sum _{N,{\bar{N}}} \Vert \alpha _{N,{\bar{N}}}\Vert ^2, \end{aligned} \end{aligned}$$

(5.97)

with$\Vert \alpha _{N,{\bar{N}}}\Vert ^2$ being a shorthand notation for the (non-negative¹⁶) sum of integrals

$$\begin{aligned} \Vert \alpha _{N,{\bar{N}}}\Vert ^2=\sum _{{\underline{r}},{\bar{\underline{s}}}}\int \!\!\![d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}}\; \alpha ^*_{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}}) \alpha _{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}}). \end{aligned}$$

(5.98)

The tensor product $|\psi \rangle \langle \psi |$, that is, the projector on the quantum state $|\psi \rangle$ reads

$$\begin{aligned} \begin{aligned} \!\!\!\! |\psi \rangle \langle \psi |&= \sum _{N,{\bar{N}}} \sum _{{\underline{r}},{\bar{\underline{s}}}} \sum _{N^\prime ,{\bar{N}}^\prime } \sum _{{\underline{r}}^\prime ,{\bar{\underline{s}}}^\prime }\int \!\!\![d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}} [d {\underline{p}}^\prime ]^N [d {\underline{\bar{q}}}^\prime ]^{\bar{N}}\times \\&\qquad \times \alpha ^*_{N^\prime ,{\bar{N}}^\prime }({\underline{p}}^\prime ,{\underline{r}}^\prime ;{\underline{\bar{q}}}^\prime ,{\underline{\bar{s}}}^\prime )\alpha _{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}})\, |{\underline{p}},{\underline{r}};\bar{\underline{q}},\bar{\underline{s}}\rangle \langle {\underline{p}}^\prime ,{\underline{r}}^\prime ;\bar{\underline{q}}^\prime ,\bar{\underline{s}}^\prime |. \end{aligned} \end{aligned}$$

(5.99)

Making use of the normalization relations between the states, it is possible to write the trace in a pure state $|\psi \rangle$ of an operator $\hat{\mathcal{O}}$ in the compact form

$$\begin{aligned} \begin{aligned}&\!\!\!\!\mathrm{tr}\left( \phantom {\frac{}{}} \left| \psi \right\rangle \left\langle \psi \right| \hat{\mathcal{O}} \right) = \sum _{N,{\bar{N}}} \sum _{{\underline{r}},{\bar{\underline{s}}}} \sum _{N^\prime ,{\bar{N}}^\prime } \sum _{{\underline{r}}^\prime ,{\bar{\underline{s}}}^\prime }\int \!\!\![d {\underline{p}}]^N [d {\underline{\bar{q}}}]^{\bar{N}} [d {\underline{p}}^\prime ]^N [d {\underline{\bar{q}}}^\prime ]^{\bar{N}} \times \\&\times \alpha ^*_{N^\prime ,{\bar{N}}^\prime }({\underline{p}}^\prime ,{\underline{r}}^\prime ;{\underline{\bar{q}}}^\prime ,{\underline{\bar{s}}}^\prime )\alpha _{N,{\bar{N}}}({\underline{p}},{\underline{r}};{\underline{\bar{q}}},{\underline{\bar{s}}})\, \langle {\underline{p}}^\prime ,{\underline{r}}^\prime ;\bar{\underline{q}}^\prime ,\bar{\underline{s}}^\prime |\hat{\mathcal{O}} |{\underline{p}},{\underline{r}};\bar{\underline{q}},\bar{\underline{s}}\rangle . \end{aligned} \end{aligned}$$

(5.100)

There is a couple of results that can be immediately inferred from the last formula. The first one is that the expectation values of $a^\dagger b^\dagger$ and ba, for any momentum and polarization combination, can be nonvanishing if and only if the quantum state of the system is in a superposition of states with different particle content. More precisely, only the quantum interference between states that differ exactly by a particle-antiparticle pair can give a nonvanishing contribution ( understanding that the integral over the partial wave functions can still simplify and give a vanishing result).

The second observation is that the expectation value of $a^\dagger a$ and $b^\dagger b$ can be simplified. The only combinations that can give a contribution are the ones between states with exactly the same number of particles and the same number of antiparticles. In the following computations, we consider only the term $a^\dagger a$, understanding that the very same transformations hold for antiparticles.

As a particular case of (5.100) one can write the expectation value of $a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime )$

$$\begin{aligned} \begin{aligned}&\!\!\!\!\mathrm{tr}\left( \phantom {\frac{}{}} \left| \psi \right\rangle \left\langle \psi \right| a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) \right) = \sum _{N,{\bar{N}}}\sum _{{\underline{t}},{\underline{t}}^{\prime }}\sum _{{\bar{\underline{u}}},{\bar{\underline{u}}}^\prime }\int \!\!\![d {\underline{k}}]^N[d {\underline{k}}^{\prime }]^N [d {\underline{\bar{q}}}]^{\bar{N}}[d {\underline{\bar{q}}}^\prime ]^{\bar{N}} \times \\&\!\!\!\! \times \alpha ^*_{N,{\bar{N}}}({\underline{k}}^{\prime },{\underline{t}}^{\prime };{\underline{\bar{q}}}^\prime ,{\underline{\bar{u}}}^\prime )\alpha _{N,{\bar{N}}}({\underline{k}},{\underline{t}};{\underline{\bar{q}}},{\underline{\bar{u}}}) \langle {\underline{k}}^{\prime },{\underline{t}}^{\prime };{\underline{\bar{q}}}^{\prime },{\underline{\bar{u}}}^{\prime }|a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) |{\underline{k}},{\underline{t}};{\underline{\bar{q}}},{\underline{\bar{u}}}\rangle . \end{aligned} \end{aligned}$$

(5.101)

It is relatively simple to obtain the final formula by making use of the standard anticommutation relations

$$\begin{aligned} \begin{aligned} \!\!\!\!&\cdots a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) \sqrt{2E_{\mathbf{k}_j}}a^\dagger _{t_j}(\mathbf{k}_j)\cdots = \\ \!\!\!\!&\qquad \cdots a^\dagger _r(\mathbf{p}) \sqrt{2E_{\mathbf{k}_j}} \left( \{ a_s(\mathbf{p}^\prime ), a^\dagger _{t_j}(\mathbf{k}_j)\} - a^\dagger _{t_j}(\mathbf{k}_j) a_s(\mathbf{p}^\prime ) \right) \cdots = \\ \!\!\!\!&\cdots \sqrt{2E_{\mathbf{k}_j}}\left( a^\dagger _{t_j}(\mathbf{k}_j) a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) + a^\dagger _r(\mathbf{p}) (2\pi )^3\delta _{s t_j}\delta ^3(\mathbf{k}_j -\mathbf{p}^\prime ) \right) \cdots = \\ & \cdots \left[ \sqrt{2E_{\mathbf{k}_j}} a^\dagger _{t_j}(\mathbf{k}_j) \left( a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) \right) + \sqrt{2E_{\mathbf{p}^\prime }} (2\pi )^3\delta _{s t_j}\delta ^3(\mathbf{k}_j -\mathbf{p}^\prime ) a^\dagger _r(\mathbf{p})\right] \cdots =\\ \\ & =\cdots \left[ \sqrt{2E_{\mathbf{k}_j}} a^\dagger _{t_j}(\mathbf{k}_j) \left( a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) \right) + \right. \\&\qquad \left. + \sqrt{\frac{2E_{\mathbf{p}^\prime }}{2E_{\mathbf{p}}}} (2\pi )^3\delta _{s t_j}\delta ^3(\mathbf{k}_j -\mathbf{p}^\prime ) \sqrt{2E_{\mathbf{p}}}a^\dagger _r(\mathbf{p})\right] \cdots \end{aligned} \end{aligned}$$

(5.102)

In other words, even if $a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime )$ doesn’t commute with the creation operators, it is possible to “move it to the right”. However, each time we do that we have to add a new state, with a delta between the j’th degrees of freedom and the destruction operator $a_s(\mathbf{p}^\prime )$, a numerical factor $(2\pi )^3\sqrt{E_{\mathbf{p}^\prime }/E_\mathbf{p}}$ and a substitution of the momentum and polarization at the j’th place with the ones related to the creation operator $a^\dagger _r(\mathbf{p})$. After moving to the right all the particle creation operators, $a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime )$ commutes with the creation operators of the antiparticles (if present).

In the end, after making use of the normalization of the eigenstates we find

$$\begin{aligned} \begin{aligned}&\!\!\!\!\mathrm{tr}\left( \phantom {\frac{}{}} \left| \psi \right\rangle \left\langle \psi \right| a^\dagger _r(\mathbf{p})a_s(\mathbf{p}^\prime ) \right) = 0 +\\&\quad + \frac{1}{\sqrt{2E_\mathbf{p}2E_{\mathbf{p}^\prime }}} \sum _{{\bar{N}},N>0}\sum _{j+1}^N\sum _{{\underline{t}}-t_j}\sum _{{\bar{\underline{u}}}}\int \!\!\![d {\underline{k}}]^{(N-j)} [d {\underline{\bar{q}}}]^{\bar{N}}\times \\&\!\!\!\! \times \alpha ^*_{N,{\bar{N}}}({\underline{k}}-\mathbf{k}_j,\mathbf{p}, {\underline{t}}-t_j, r;{\underline{\bar{q}}},{\underline{\bar{u}}})\alpha _{N,{\bar{N}}}({\underline{k}}-\mathbf{k}_j,\mathbf{p}^\prime , {\underline{t}}-t_j, s;{\underline{\bar{q}}},{\underline{\bar{u}}}). \end{aligned} \end{aligned}$$

(5.103)

The notation $\sum _{{\underline{t}}-t_j}\int d {[\underline{k}}]^{(N-j)}$ means that the integral and the sum is over all the particle degrees of freedom except for the j’th. In the similar way $\alpha _{N,{\bar{N}}}({\underline{k}}-\mathbf{k}_j,\mathbf{p}, {\underline{t}}-t_j, r;{\underline{\bar{q}}},{\underline{\bar{u}}})$ is a shorthand notation for the (partial) wave function with the j’th degrees of freedom fixed to the momentum $\mathbf{p}$ and polarization r.

The formula (5.103) has many interesting consequences. Besides the expected vanishing expectation value for purely antiparticle states, one can immediately check that the expectation value of $a^\dagger _r(\mathbf{p})a_r(\mathbf{p})$ is nonnegative, since it is a series of integrals and sums of squares. Moreover, as one could expect, it is linked to the average number of particles. Indeed, the expression

$$\begin{aligned} \sum _r \int \frac{d^3 p}{(2 \pi )^3} \mathrm{tr}\left( |\psi \rangle \langle \psi | a^\dagger _r(\mathbf{p})a_r(\mathbf{p}) \right) = \sum _N N \sum _{\bar{N}}\Vert \alpha _{N,{\bar{N}}}\Vert ^2, \end{aligned}$$

(5.104)

exactly gives the average number of particles in the state $|\psi \rangle$ because of the normalization (5.97). More interestingly, the expectation value of $a^\dagger _r(\mathbf{p})a_s(\mathbf{p})$ (same momentum, different polarization) performs the role of a momentum-dependent spin density matrix. The momentum integral of the trace is proportional to the average number of particles, but the matrix itself is sensitive to polarization in the r, s indices and can be used to obtain the average number of particles, per momentum cell, for some polarization states.

All these arguments do not change if one reinserts the classical probabilities $\mathsf{P}_i$ from (5.86) and deals with mixed states. The classical fluctuations do not change the properties of the spin density matrix, like the non-negative diagonal elements and normalization of the trace (after dividing by $(2\pi )^3$ and integrating over momentum, like for the pure states) does not change the average number of particles.