Bosons

The set of the bosons consists of all particles that carry any integer multiple of Dirac’s constant (aitch-bar) as their inherent spin. In the fundamental particles the set includes the force mediators (such as the photon) and the mesons. If a large number of the same kind of bosons comes together in a system (such as a gas), then the overall state function describing that system is symmetric with respect to any interchange of particles; that is, if particle A occupies state-1 and particle B occupies state-2 (ψ=ψ(A,B)), then ψ(B,A)=ψ(A,B). That means that any number of the particles can occupy a single given state (in contrast with fermions, which the Pauli exclusion principle bars from such coziness) and, thus, that large collections of bosons conform to the requirements of Bose-Einstein statistics.

The state function of any particle in the set conforms to the Klein-Gordon equation for particles of spin-0 and to the Proca equation for particles of spin-1. Those equations appear respectively as

(Eq’n 1)

and

(Eq’n 2)

In the Klein-Gordon equation psi takes the form of a scalar, just as it does
in Schrödinger’s equation. In the Proca equation **A**^{ν}
takes the form of a four-vector. When m=0, that four-vector describes an
electromagnetic field in empty space as an array of virtual photons, including,
of course, any actual photons passing through that empty space. If we solve one
of those equations for a description (the state function) of the associated
field, then we can use Born’s theorem to calculate expectation values for the
properties of that field. But the Proca equation, as written, doesn’t seem to
acknowledge the existence of the particle’s spin.

Nonetheless, the complete state function must include a description of the direction in which the particle’s spin axis points, even if only implicitly. The Proca equation constrains the form that the state function can take, in accordance with the principle of least action density by way of the Euler-Lagrange equations. In the absence of an interaction with other particles, the direction in which a particle’s spin axis points does not affect the action that the particle plays out. Once we have solved Proca’s equation we can include the spin orientation by multiplying the state function by some factor that behaves as a constant with respect to the differentiation operators in the equations.

Just as there is no absolute velocity, so there is no absolute orientation. The Universe has no state of absolute rest and it has no North Pole. Thus we are free to orient our coordinate frame as we wish: we can, for example, simply draw a straight line from the center of the sun to the center of another star and say that it defines the z-direction of our Cartesian grid. By convention, physicists use the z-axis of the grid as the primary reference in the quantum theory of angular momentum.

A spin-1 particle has three eigenstates, which we identify by their eigenvalues; +1 (spin pointing in the positive z-direction), 0 (spin pointing somewhere in the x-y plane), and -1 (spin pointing in the negative z-direction). Those eigenvalues comprise a set that looks a little like a vector, {+1,0,-1}.

Three spin operators exist to extract the relevant measurements of spin from the state function. Those three exist, one for each axis of the coordinate system, because we cannot measure all three components of a spin vector at once. In mathematical terms, the spin operators don’t commute with each other; that is, we have the (non-zero) commutators ([A,B]=(AB-BA)) as

(Eq’ns 3)

The operators take the form

(Eq’n 4)

(Eq’n 5)

and

(Eq’n 6)

In order to calculate the expectation value for one of the components of a boson’s spin vector, we must start by noting that we have to include something resembling a column vector as a factor in the state function; that is, we must have the state function as

(Eq’n 7)

in which alpha, beta, and gamma represent what look like the direction
cosines, but are actually the partial probabilities, of the spin vector. We
interpret the partial probabilities, which are complex numbers, by noting that
α*α
yields the probability of finding the particle in the state s_{x}=+S
(the asterisk on the first alpha refers to the complex conjugate of that alpha).
In setting up that spinor we have tacitly assumed that we can somehow determine
the values of the partial probabilities, perhaps, for example, by putting the
boson into a magnetic field and seeing how it interacts with radio waves. We
than calculate the expectation value by the usual method, exploiting Born’s
theorem;

(Eq’n 8)

In that equation I’ve written the spinor as the Hermitian conjugate of a matrix (the complex conjugate of the transposed matrix) in which the asterisks signify the complex conjugates of alpha, beta, and gamma. Because the spinors and the spin matrix do not depend explicitly upon the spatial coordinates, they and their matrix multiplication commute with the psi-functions and the integration, so I pulled them out to the left. Then I used the fact that the remaining integral equals one by definition and carried out the matrix multiplication.

The magnitude of the spin vector itself does not equal one unit of aitch-bar, oddly enough. Angular momentum does weird things in the quantum theory. We calculate the magnitude of the spin vector from

(Eq’n 9)

in which s=1 in this case. Thus we calculate the magnitude as

(Eq’n 10)

That statement tells us that in order to have one Dirac unit of spin parallel to the z-axis the particle’s spin axis must tilt into the x-y plane in an indeterminate way.

When we discuss large numbers of bosons, we don’t refer explicitly to their spin states; rather, we describe the collection of particles with Bose-Einstein statistics, which describe the distribution of the bosons among the various energy states available to them. Bose-Einstein statistics, like Fermi-Dirac statistics, differ from the classical Maxwell-Boltzmann statistics (reflected in the Maxwell distribution) in the way in which the spins alter the count of microstates that constitute a given macrostate.

In the classical case we treat the particles as distinguishable objects (in essence each particle carries a unique license-plate number), so interchanging two particles between the two states that they occupy (moving particle A from state-1 to state-2 and moving particle B from state-2 to state-1) creates a different microstate, which must be counted toward the total in the given macrostate. Quantum particles of a given species are perfectly indistinguishable from one another, so interchanging two such particles in a collection does not create a new microstate. Thus bosons have fewer microstates in a given macrostate than classical particles do.

Because bosons are indistinguishable, we need only concern
ourselves with the number of particles occupying each single-particle state, n_{r}.
In a gas consisting of N particles we must then have

(Eq’n 11)

and

(Eq’n 12)

in which ε_{r}
represents the per-particle energy of the r-th state, R labels the possible
quantum states of the gas, and E_{R} represents the total energy
contained by the R-th macrostate. The Boltzmann factor,

(Eq’n 13)

describes the relative probability of finding the gas in a particular R-state at the absolute temperature T. With that factor we can calculate the average number of particles in state-s as

(Eq’n 14)

in which Z represents the partition function associated with the gas.

The nature of the exponential allows us to draw the factor describing the state-s away from the rest of the exponential and write the above equation as

(Eq’n 15)

in which Z_{s}(N-n_{s}) represents the partition function
associated with the second sum in both the numerator and the denominator. We
thus have

(Eq’n 16)

Because N represents an extremely large number, the natural logarithm of the partition function will change by a minuscule amount when we subtract relatively small numbers from it, so we can write

(Eq’n 17)

To a good approximation the value of
α_{s}
is independent of s, so α_{s}=α
and we have

(Eq’n 18)

That fact transforms Equation 16 into

(Eq’n 19)

Because of the way in which we define it, α=-μ/kT, represents the chemical potential of the gas. Also, the two infinite series in that equation have closed-formula manifestations, so we can write

(Eq’n 20)

Making the appropriate substitutions, we transform Equation 19 into

(Eq’n 21)

Thus we have the fundamental formula for the Bose-Einstein distribution. If we let the chemical potential go to zero, we get the Planck distribution, the fundamental formula for photon statistics, which describes the distribution of energy in blackbody radiation.

Now we have the basic information that applies to bosons.

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