polarization vector

English

polarization vector (plural polarization vectors)

(quantum field theory) One of a set of three spacelike (and possibly also one timelike) four-dimensional vectors which are orthonormal within some inertial reference frame.
- 2003, A. Zee, Quantum field theory in a nutshell, Princeton, New Jersey: Princeton University Press, §I.5, page 32:
  A massive spin 1 particle has three degrees of polarization for the obvious reason that in its rest frame its spin vector can point in three different directions. The three polarization vectors ${\\displaystyle \\epsilon _{\\mu }^{(a)}}$ are simply the three unit vectors pointing along the x, y, and z axes, respectively (a = 1, 2, 3): ${\\displaystyle \\epsilon _{\\mu }^{(1)}=(0,1,0,0)}$ , ${\\displaystyle \\epsilon _{\\mu }^{(2)}=(0,0,1,0)}$ , ${\\displaystyle \\epsilon _{\\mu }^{(3)}=(0,0,0,1)}$ . In the rest frame ${\\displaystyle k^{\\mu }=(m,0,0,0)}$ and so
  ${\\displaystyle k^{\\mu }\\epsilon _{\\mu }^{(a)}=0}$
  Since this is a Lorentz invariant equation, it holds for a moving spin 1 particle as well.

单词	polarization vector
释义	polarization vector English Noun polarization vector (plural polarization vectors) (quantum field theory) One of a set of three spacelike (and possibly also one timelike) four-dimensional vectors which are orthonormal within some inertial reference frame. 2003, A. Zee, Quantum field theory in a nutshell, Princeton, New Jersey: Princeton University Press, §I.5, page 32: A massive spin 1 particle has three degrees of polarization for the obvious reason that in its rest frame its spin vector can point in three different directions. The three polarization vectors ${\\displaystyle \\epsilon _{\\mu }^{(a)}}$ are simply the three unit vectors pointing along the x, y, and z axes, respectively (a = 1, 2, 3): ${\\displaystyle \\epsilon _{\\mu }^{(1)}=(0,1,0,0)}$ , ${\\displaystyle \\epsilon _{\\mu }^{(2)}=(0,0,1,0)}$ , ${\\displaystyle \\epsilon _{\\mu }^{(3)}=(0,0,0,1)}$ . In the rest frame ${\\displaystyle k^{\\mu }=(m,0,0,0)}$ and so ${\\displaystyle k^{\\mu }\\epsilon _{\\mu }^{(a)}=0}$ Since this is a Lorentz invariant equation, it holds for a moving spin 1 particle as well.
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