Marjan said:
As i understand spin of a photon can be -1 or +1.
We talk about a photon as having "spin 1". Similarly, we talk about an electron as having "spin 1/2". In general, for a particle with "spin s" (note: according to Quantum Mechanics, s must be restricted to an
integer or
half-integer), the associated
angular momentum is given by
h
barsqrt{s(s+1)}. Thus, the total
spin angular momentum for a photon
(s = 1) is h
bar√2.
Now, when you say that the spin of a photon can be ±1 in connection with "spinning"
right (+1) or
left (-1), what you are talking about are two
specific quantum-mechanical photon spin-states. These states are called "helicity" sates.
Say, for example, that we have a photon which is traveling in the +z direction. If the photon is in the
+1 helicity state, then this means that the
z-component of the spin vector is
positive (and equal to +h
bar). On the other hand, if the photon is in the
-1 helicity state, then this means that the
z-component of spin vector is
negative (and equal to -h
bar). Observe that the z-component of spin is
quantized, which is to say that, in any measurement of that component, it can be found to have a value of ±h
bar, and nothing else.
Notice that, in the above, I have mentioned only a single
component of spin and not the
total spin vector itself. No doubt, you have heard of an "uncertainty principle" regarding the
position and the
momentum of a particle. Well, as it turns out, such a principle also applies to the
spin of a particle. More precisely, a pair of
spin components along two
nonparallel axes are subject to an "uncertainty principle". Even more precisely, and in particular, if the particle is in a spin state such that, say, the
z-component of spin is
completely definite (as is the case for our two helicity states), then the
direction of the spin component in the
xy-plane will be
completely uncertain (only the
magnitude will be certain).
From these remarks, you will understand that helicity states refer to the direction (+ or -) of the spin
component (parallel to the direction of propagation), and
not to the
total spin vector itself. This is what
ZapperZ was alluding to in post #2. Of course, from the description given there and the fact that s = 1 for a photon, we would expect that the allowable eigenvalues for S
z are m = -1, 0, and 1. If so, we would then also have a
zero helicity state (S
z eigenstate with m = 0) for the photon ... but as pointed out by
zefram_c (post #3) and
vanesch (post #4), for a
physical photon, this m = 0 case is
excluded.
______________
In a classical view we can say that one is spinning right and other to the left ?
Yes, viewing the photon in a
'classical' way (note: this is a 'simplification' to be used only with
caution!), we can say that the +1 helicity state corresponds to the rotation of a "right-handed" screw (again, this is said with regard to the
z-component of spin
only, and not the
total spin vector), and -1 to a "left-handed" screw.
______________
What would be the difference between light with all photons -1 and light with all photons 1 ?
Let us denote the +1 helicity state as |R>, and the -1 helicity state as |L>. Then, as
vanesch (post #4) has pointed out, |R> (|L>) corresponds to right- (left-) circularly polarized light.
______________
Can be spin of a photon also 0 ?
As already indicated, the m = 0 eigenstate of S
z is
excluded. However, this does
not mean that the
only possible photon states are eigenstates of S
z with m = ±1. In general, any (normalized) linear combination of |R> (eigenstate of S
z with m = +1) and |L> (eigenstate of S
z with m = -1) is
also a possible state. Each such (nontrivial) linear combination will correspond to either
elliptical or
linear polarization. For example,
|x> ≡ (1/√2)(|R> + |L>)
and
|y> ≡ (-i/√2)(|R> - |L>)
are suitable choices for
linearly polarized photons along the x- and y- axes, respectively. In both cases <S
z> = 0.
Note that |x> is an eigenvector of S
y with eigenvalue m = 0, and |y> is an eigenvector of S
x with eigenvalue m = 0.
