Pauli Lubanski Vector Operator, Eigenvalue Problem and Entanglement

TL;DR

This paper investigates the eigenvalue problem of the Pauli-Lubanski vector operator's spin matrix, exploring the entanglement properties of its eigenvectors and their relation to degeneracy.

Contribution

It introduces a detailed analysis of the eigenvalue problem for the Pauli-Lubanski vector operator and links entanglement characteristics to degeneracy phenomena.

Findings

Eigenvectors exhibit specific entanglement properties.

Degeneracy correlates with entanglement features.

New insights into the structure of the Pauli-Lubanski vector eigenstates.

Abstract

We study an eigenvalue problem for a spin matrix arising in the Pauli Lubanski vector operator. Entanglement of the eigenvectors and its connection with degeneracy is discussed.

Equations

$$[S_1,S_2]=i{S}_3,[S_2,S_3]=i{S}_1,[S_3,S_1]=i{S}_2.$$

$$S_1^2+S_2^2+S_3^2=s(s+1)I_{2s+1}$$

$$\text{mbox}tr(S_j^2)=\frac{1}{3}s(s+1)(2s+1)$$

$$\text{mbox}tr(S_j{S}_k)=0\text{mbox}forj\ne k\text{mbox}andj,k=1,2,3.$$

$$W^{\mu }=\sum _{\nu =0}^3\sum _{\rho =0}^3\sum _{\lambda =0}^3{ϵ}^{\mu \nu \rho \lambda }{S}_{\rho \lambda }{p}_{\nu }$$

$$g=-d{x}_0\otimes d{x}_0+d{x}_1\otimes d{x}_1+d{x}_2\otimes d{x}_2+d{x}_3\otimes d{x}_3.$$

$$\sum _{\mu =0}^3{W}_{\mu }{W}^{\mu }=s(s+1)(\sum _{\mu =0}^3{p}_{\mu }{p}^{\mu })I_{2s+1}$$

$$S=(S_{\rho\lambda})=\pmatrix{0_{2s+1}&S_{1}&S_{2}&S_{3}\cr-S_{1}&0_{2s+1}&-iS_{3}&iS_{2}\cr-S_{2}&iS_{3}&0_{2s+1}&-iS_{1}\cr-S_{3}&-iS_{2}&iS_{1}&0_{2s+1}}.$$

$$S^{-1}=\frac{1}{s(s+1)}\pmatrix{-I_{2s+1}&-S_{1}&-S_{2}&-S_{3}\cr S_{1}&I_{2s+1}&-iS_{3}&iS_{2}\cr S_{2}&iS_{3}&I_{2s+1}&-iS_{1}\cr S_{3}&-iS_{2}&iS_{1}&I_{2s+1}}.$$

$$T_{1}=\pmatrix{0_{2s+1}&S_{1}&S_{2}&S_{3}\cr-S_{1}&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr-S_{2}&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr-S_{3}&0_{2s+1}&0_{2s+1}&0_{2s+1}},\quad T_{2}=\pmatrix{0_{2s+1}&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr 0_{2s+1}&0_{2s+1}&-iS_{3}&iS_{2}\cr 0_{2s+1}&iS_{3}&0_{2s+1}&-iS_{1}\cr 0_{2s+1}&-iS_{2}&iS_{1}&0_{2s+1}}$$

$$[S,S^{*}]=SS^{*}-S^{*}S=2\pmatrix{0_{2s+1}&i[S_{2},S_{3}]&i[S_{3},S_{1}]&i[S_{1},S_{2}]\cr i[S_{2},S_{3}]&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr i[S_{3},S_{1}]&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr i[S_{1},S_{2}]&0_{2s+1}&0_{2s+1}&0_{2s+1}}.$$

$$\text{mbox}tr(S)$$

$$\text{mbox}tr(S^2)$$

$$\text{mbox}tr(S^3)$$

$$\text{mbox}tr(S^4)$$

$$\text{mbox}tr(S^5)$$

$$\text{mbox}tr(S^6)$$

$$\text{mbox}tr(S^7)$$

$$\text{mbox}tr(S^8)$$

$$\text{mbox}tr(S^N)$$

$$\frac{1}{2}(4\text{mbox}times),\frac{1}{2}(-1+\sqrt{2}i)(2\text{mbox}times),\frac{1}{2}(-1-\sqrt{2}i)(2\text{mbox}times)$$

$$1(5\text{mbox}times),\frac{1}{2}(-1+\sqrt{7}i)(3\text{mbox}times),\frac{1}{2}(-1-\sqrt{7}i)(3\text{mbox}times),-2(1\text{mbox}times)$$

$$\frac{3}{2}(6\text{mbox}times),\frac{1}{2}(-1+\sqrt{14}i)(4\text{mbox}times),\frac{1}{2}(-1-\sqrt{14}i)(4\text{mbox}times),-\frac{5}{2}(2\text{mbox}times)$$

$$2(7\text{mbox}times),\frac{1}{2}(-1+\sqrt{23}i)(5\text{mbox}times),\frac{1}{2}(-1-\sqrt{23}i)(5\text{mbox}times),-3(3\text{mbox}times).$$

$$\frac{5}{2}(8\text{mbox}times),\frac{1}{2}(-1+\sqrt{34}i)(6\text{mbox}times),\frac{1}{2}(-1-\sqrt{34}i)(6\text{mbox}times),-\frac{7}{2}(4\text{mbox}times).$$

$$\frac{1}{2}(-1+i\sqrt{4s(s+1)-1}),\frac{1}{2}(-1-i\sqrt{4s(s+1)-1})$$

$$T_{3}:=[T_{1},T_{2}]=\pmatrix{0_{2s+1}&-S_{1}&-S_{2}&-S_{3}\cr-S_{1}&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr-S_{2}&0_{2s+1}&0_{2s+1}&0_{2s+1}\cr-S_{3}&0_{2s+1}&0_{2s+1}&0_{2s+1}}.$$

$$[T_3,T_1]$$

$$[T_3,T_2]$$

$$[[T_3,T_1],T_1]$$

$$[[T_3,T_1],T_2]$$

$$[[T_3,T_2],T_1]$$

$$[T_3,T_2],T_2]$$

$${\mathbf{v}}_1=\frac{1}{\sqrt{2}}\text{pmatrix}0\text{cr}0\text{cr}1\text{cr}0\text{cr}0\text{cr}0\text{cr}0\text{cr}1=\frac{1}{\sqrt{2}}\left(\text{pmatrix}1\text{cr}0\otimes \text{pmatrix}0\text{cr}1\otimes \text{pmatrix}1\text{cr}0+\text{pmatrix}0\text{cr}1\otimes \text{pmatrix}0\text{cr}1\otimes \text{pmatrix}0\text{cr}1\right)$$

$${\mathbf{v}}_2=\frac{1}{\sqrt{2}}\text{pmatrix}0\text{cr}0\text{cr}0\text{cr}0\text{cr}1\text{cr}0\text{cr}0\text{cr}i=\text{pmatrix}0\text{cr}1\otimes \frac{1}{\sqrt{2}}\text{pmatrix}1\text{cr}0\text{cr}0\text{cr}i$$

$${\mathbf{v}}_3=\frac{1}{\sqrt{2}}\text{pmatrix}0\text{cr}0\text{cr}0\text{cr}1\text{cr}0\text{cr}0\text{cr}-1\text{cr}0=\frac{1}{\sqrt{2}}\left(\text{pmatrix}1\text{cr}0\otimes \text{pmatrix}0\text{cr}1\otimes \text{pmatrix}0\text{cr}1-\text{pmatrix}0\text{cr}1\otimes \text{pmatrix}0\text{cr}1\otimes \text{pmatrix}1\text{cr}0\right)$$

Full text

Pauli Lubanski Vector Operator, Eigenvalue Problem and Entanglement

Garreth Kemp† and Willi-Hans Steeb*†***

† International School for Scientific Computing,

University of Johannesburg, Auckland Park 2006, South Africa,

e-mail: garrethkemp@gmail.com

e-mail: steebwilli@gmail.com

Abstract. We study an eigenvalue problem for a spin matrix arising in the Pauli Lubanski vector operator. Entanglement of the eigenvectors and its connection with degeneracy is discussed.

Let $S_{1}$, $S_{2}$, $S_{3}$ be the spin matrices [1] for spin $s=\frac{1}{2},\,1,\,\frac{3}{2},\,2,\dots$. The matrices are $(2s+1)\times(2s+1)$ hermitian matrices ($S_{1}$ and $S_{3}$ are real symmetric) with trace equal to 0 satisfying the commutation relations

[TABLE]

The eigenvalues of $S_{1}$, $S_{2}$, $S_{3}$ are $s,s-1,\dots,-s$ for a given $s$. We have

[TABLE]

where $I_{2s+1}$ is the $(2s+1)\times(2s+1)$ identity matrix. Furthermore we have

[TABLE]

and

[TABLE]

In general we have that $\mbox{tr}(S_{j}^{n})=0$ if $n$ is odd and $\mbox{tr}(S_{j}^{n})=\frac{1}{3}s(s+1)(2s+1)$ if $n$ is even. It follows that $\mbox{tr}((S_{j}S_{k})\otimes(S_{\ell}S_{m}))=0$ if $j\neq k$ and $\ell\neq m$.

The Pauli-Lubanski vector operator (see [2, 3, 4] and references therein) is given by

[TABLE]

where $\mu=0,1,2,3$ and the underlying metric tensor field is

[TABLE]

The $W^{\mu}$ ($\mu=0,1,2,3$) are $(2s+1)\times(2s+1)$ matrices with

[TABLE]

and $S=(S_{\rho\lambda})$ denotes the $(4\cdot(2s+1))\times(4\cdot(2s+1))$ matrix

[TABLE]

Here $0_{2s+1}$ is the $(2s+1)\times(2s+1)$ zero matrix. The inverse matrix of $S$ is given by

[TABLE]

We also write $S=T_{1}+T_{2}$ with

[TABLE]

where $T_{1}$ is a skew-hermitian matrix and $T_{2}$ is a hermitian matrix. We study the eigenvalue problem for the matrix $S$ and the commutators generated by $T_{1}$ and $T_{2}$. First we note that the matrix $S$ is nonnormal. We obtain

[TABLE]

The trace $\mbox{tr}(S^{N})$ can also be found in closed form. We obtain

[TABLE]

Based on the properties of matrices $T_{1}$ and $T_{2}$, it is possible to show that the trace of $S^{N}$ is

[TABLE]

The formula above shows the eigenvalues along with their multiplicity.

For spin-$\frac{1}{2}$ we find the eight eigenvalues

[TABLE]

For spin-1 we find the twelve eigenvalues

[TABLE]

For spin-$\frac{3}{2}$ we find sixteen eigenvalues

[TABLE]

For spin-2 we find the twenty eigenvalues

[TABLE]

For spin-$\frac{5}{2}$ we find the twenty four eigenvalues

[TABLE]

For the general case with spin-$s$ the eigenvalue $s$ is $(2s+1)+2=2s+3$ degenerate. For spin $s\geq 1$ we find the eigenvalue $(-s-1)$ which is $(2s+1)-2)=2s-1$ degenerate. The complex eigenvalues

[TABLE]

are $(2s+1)$ times degenrate.

Consider now the commutators and anti-commutators for $T_{1}$ and $T_{2}$. We obtain

[TABLE]

Furthermore we find

[TABLE]

and

[TABLE]

The anticommutator of $T_{1}$ and $T_{2}$ is $[T_{1},T_{2}]_{+}=-T_{1}$. We note that $T_{1}$ and $T_{2}$ are both separately normal with $T^{*}_{1}=-T_{1}$ $T^{*}_{2}=T_{2}$ and $[S,S^{*}]=2T_{3}$.

For spin-$\frac{1}{2}$ and eigenvalue $\frac{1}{2}$ (four times degenerate) the set of normalized linearly independent eigenvectors is given by

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Applying the $n$-tangle introduced by Wong and Christensen ([5, 6]) we find that the four eigenvectors are non-entangled. For the second and fourth this also seen from the fact that the eigenvectors can be written as Kronecker products of normalized vectors in ${\mathbb{C}}^{2}$ and ${\mathbb{C}}^{4}$. Now forming combinations of these vectors (which are also then eigenvectors) we find the following, where we normalize the linear combinations. ${\bf v}_{1}+{\bf v}_{3}$ and ${\bf v}_{1}-{\bf v}_{3}$ are non-entangled as well as ${\bf v}_{2}+{\bf v}_{4}$ and ${\bf v}_{2}-{\bf v}_{4}$ are non-entangled. On the other hand ${\bf v}_{1}+{\bf v}_{4}$, ${\bf v}_{1}-{\bf v}_{4}$, ${\bf v}_{2}+{\bf v}_{3}$, ${\bf v}_{2}-{\bf v}_{3}$ are entangled.