From: magnus@thep.lu.se (Magnus Olsson)
Subject: Re: Vector Bosons?
Organization: Theoretical Physics, Lund University, Sweden
Date: Thu, 13 Aug 1992 17:35:23 GMT

In article <BsxHtJ.B8p@helios.physics.utoronto.ca> dooley@helios.physics.utoronto.ca (Kevin Dooley) writes:
>In article <d==y5bh@rpi.edu> caer@iear.arts.rpi.edu (Charlie Figura) writes:
>>
>>Perhaps some of you brilliant minds out there can explain a thing
>
>I hope I'll do instead....
>
>>to me...  somewhere (dont ask me, I dont know where) I heard of such
>>a thing called a vector boson.  Now, I understand what bosons are,
>>i.e, particles that obey the Bose-Einstein statistics.  I don't know,
>>however, what a *vector* boson is.  
>
>In a nutshell, the 'vector' refers to the angular momentum of the particle.
>A vector transforms under Lorentz transformations according to certain rules
>which you can find in any book on high energy physics.  In particular, a
>vector boson has spin of 1 (in units of hbar).  A scalar has spin 0, a
>tensor has spin 2.  The term also implies a certain parity.  If I reflect
>a vector in the mirror I get a vector of the same length pointing in the
>opposite direction so a vector boson such as a photon or a W or Z has
>negative parity.  Tensors and scalars have positive parity.  A particle
>with the wrong parity gets 'pseudo' tacked on in front.  So, for example
>a pion which has zero spin but negative parity is called a 'pseudoscalar'
>while a b meson (not B, but b) which has spin 1 and positive parity is
>called a 'pseudovector'.


To clarify this just a little bit more:

In Schroedinger's version of quantum mechanics, a particle is described
by a wave function. Once we combine QM with special relativity, it
turns out that we get particles with an intrinsic angular momentum (not
realted to the particle's motion), usually called spin.

The interesting thing is that the spin is directly related to the way
the wave function changes under Lorentz transformations.

In the simplest case, the wave function is a scalar, i.e. it is not
changed by a Lorentz transformation. This corresponds to a spin of 0. 

A pseudoscalar particle has a wave function that doesn't change under
"proper" LT's (those that don't involve mirror reflections) but
changes its sign if you reflect it. This also corresponds to a spin of
0. 

As said above, vectors and pseudovectors correspond to spin 1, while
tensors (of rank 2) correspond to spin 2.

For particles with non-integral spin (fermions), the wave function is
what is known as a Dirac spinor (for spin 1/2), or a combination of a
spinor and a vector (spin 3/2).


Finally: What do I mean by "transforms as a vector under Lorentz
transformations"? This is not trivial, because there are two
definitions of "vector". 

In ordinary, three-dimensional space, the most elementary definition
of a vector is as a "quantity with a magnitude and a direction", i.e.
as a triplet of real numbers.

Another definition is "a triplet of real numbers that changes in the
following way under a rotation: If the rotation matrix is a[i,j] and
the vector is v[j], then the rotated vector is  v' [i] = sum(j)
a[i,j]v[j]". 

The second definition means that not all "vectors" (def. 1) are
vectors (def. 2).

In relativity, one always uses definition 2 (modified for
four-dimensional Minkowski space). So scalars are one number, vectors
are four numbers, spinors also are four numbers, tensors are 16
numbers, and so on, and they all transform in characteristic ways
under LT's. 

Magnus Olsson                   | \e+      /_
Dept. of Theoretical Physics    |  \  Z   / q
University of Lund, Sweden      |   >----<           
Internet: magnus@thep.lu.se     |  /      \===== g
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