Neutrino theory of light

The neutrino field satisfies the Dirac equation with the mass set to zero,

${\displaystyle \gamma ^{\mu }p_{\mu }\Psi =0.}$

The gamma matrices in the Weyl basis are:

${\displaystyle \gamma ^{0}=\left({\begin{array}{cccc}0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{array}}\right),\;\;\;\;\gamma ^{1}=\left({\begin{array}{cccc}0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{array}}\right),}$

${\displaystyle \gamma ^{2}=\left({\begin{array}{cccc}0&0&0&-i\\0&0&i&0\\0&i&0&0\\-i&0&0&0\end{array}}\right),\;\;\;\;\gamma ^{3}=\left({\begin{array}{cccc}0&0&1&0\\0&0&0&-1\\-1&0&0&0\\0&1&0&0\end{array}}\right).}$

The matrix ${\displaystyle \gamma ^{0}}$ is Hermitian while ${\displaystyle \gamma ^{k}}$ is antihermitian. They satisfy the anticommutation relation,

${\displaystyle \gamma ^{\mu }\gamma ^{\nu }+\gamma ^{\nu }\gamma ^{\mu }=2\eta ^{\mu \nu }I}$

where ${\displaystyle \eta ^{\mu \nu }}$ is the Minkowski metric with signature ${\displaystyle (+---)}$ and ${\displaystyle I}$ is the unit matrix.

The neutrino field is given by,

${\displaystyle \Psi (x)={1 \over {\sqrt {V}}}\sum _{\mathbf {k} }\left\{\left[a_{1}(\mathbf {k} )u_{+1}^{+1}(\mathbf {k} )+a_{2}(\mathbf {k} )u_{-1}^{+1}(\mathbf {k} )\right]e^{ikx}\right.}$

${\displaystyle \left.+\left[c_{1}^{\dagger }(\mathbf {k} )u_{-1}^{-1}(\mathbf {-k} )+c_{2}^{\dagger }(\mathbf {k} )u_{+1}^{-1}(\mathbf {-k} )\right]e^{-ikx}\right\},}$

where ${\displaystyle kx}$ stands for ${\displaystyle \mathbf {k} \cdot \mathbf {x} -k_{0}t}$. ${\displaystyle a_{1}}$ and ${\displaystyle c_{1}}$ are the fermion annihilation operators for ${\displaystyle \nu _{1}}$ and ${\displaystyle {\overline {\nu }}_{1}}$ respectively, while ${\displaystyle a_{2}}$ and ${\displaystyle c_{2}}$ are the annihilation operators for ${\displaystyle \nu _{2}}$ and ${\displaystyle {\overline {\nu }}_{2}}$. ${\displaystyle \nu _{1}}$ is a right-handed neutrino and ${\displaystyle \nu _{2}}$ is a left-handed neutrino. The ${\displaystyle u}$'s are spinors with the superscripts and subscripts referring to the energy and helicity states respectively. Spinor solutions for the Dirac equation are,

${\displaystyle u_{+1}^{+1}(\mathbf {p} )={\sqrt {{E+p_{3}} \over 2E}}\left({\begin{array}{c}1\\{{p_{1}+ip_{2}} \over {E+p_{3}}}\\0\\0\end{array}}\right),}$

${\displaystyle u_{-1}^{-1}(\mathbf {p} )={\sqrt {{E+p_{3}} \over 2E}}\left({\begin{array}{c}{{-p_{1}+ip_{2}} \over {E+p_{3}}}\\1\\0\\0\end{array}}\right),}$

${\displaystyle u_{+1}^{-1}(\mathbf {p} )={\sqrt {{E+p_{3}} \over 2E}}\left({\begin{array}{c}0\\0\\1\\{{p_{1}+ip_{2}} \over {E+p_{3}}}\end{array}}\right),}$

${\displaystyle u_{-1}^{+1}(\mathbf {p} )={\sqrt {{E+p_{3}} \over 2E}}\left({\begin{array}{c}0\\0\\{{-p_{1}+ip_{2}} \over {E+p_{3}}}\\1\end{array}}\right).}$

The neutrino spinors for negative momenta are related to those of positive momenta by,

${\displaystyle u_{+1}^{+1}(\mathbf {-p} )=u_{-1}^{-1}(\mathbf {p} ),}$

${\displaystyle u_{-1}^{-1}(\mathbf {-p} )=u_{+1}^{+1}(\mathbf {p} ),}$

${\displaystyle u_{-1}^{+1}(\mathbf {-p} )=u_{+1}^{-1}(\mathbf {p} ),}$

${\displaystyle u_{+1}^{-1}(\mathbf {-p} )=u_{-1}^{+1}(\mathbf {p} ).}$

De Broglie[1] and Kronig[14] suggested the use of a local interaction to bind the neutrino–antineutrino pair. (Rosen and Singer[16] have used a delta potential interaction in forming a composite photon.) Fermi and Yang[17] used a local interaction to bind a fermion–antiferminon pair in attempting to form a pion. A four-vector field can be created from a fermion–antifermion pair,[18]

${\displaystyle \Psi ^{\dagger }\gamma _{0}\gamma _{\mu }\Psi .}$

Forming the photon field can be done simply by,

${\displaystyle A_{\mu }(x)=\sum _{\mathbf {p} }{-1 \over 2{\sqrt {Vp_{0}}}}\left\{\left[Q_{R}(\mathbf {p} )u_{-1}^{-1}(\mathbf {p} )^{\dagger }\gamma _{0}\gamma _{\mu }u_{+1}^{+1}(\mathbf {p} )+Q_{L}(\mathbf {p} )u_{+1}^{+1}(\mathbf {p} )^{\dagger }\gamma _{0}\gamma _{\mu }u_{-1}^{-1}(\mathbf {p} )\right]e^{ipx}\right.}$

${\displaystyle \left.+\left[Q_{R}^{\dagger }(\mathbf {p} )u_{+1}^{+1}(\mathbf {p} )^{\dagger }\gamma _{0}\gamma _{\mu }u_{-1}^{-1}(\mathbf {p} )+Q_{L}^{\dagger }(\mathbf {p} )u_{-1}^{-1}(\mathbf {p} )^{\dagger }\gamma _{0}\gamma _{\mu }u_{+1}^{+1}(\mathbf {p} )\right]e^{-ipx}\right\},\quad \quad (1)}$

where ${\displaystyle px=\mathbf {p} \cdot \mathbf {x} -p_{0}t=\mathbf {p} \cdot \mathbf {x} -Et}$.

The annihilation operators for right-handed and left-handed photons formed of fermion–antifermion pairs are defined as,[19][20][21][22]

${\displaystyle Q_{R}(\mathbf {p} )=\sum _{\mathbf {k} }F^{\dagger }(\mathbf {k} )\left[c_{1}(\mathbf {p} /2-\mathbf {k} )a_{1}(\mathbf {p} /2+\mathbf {k} )+c_{2}(\mathbf {p} /2+\mathbf {k} )a_{2}(\mathbf {p} /2-\mathbf {k} )\right]}$

${\displaystyle Q_{L}(\mathbf {p} )=\sum _{\mathbf {k} }F^{\dagger }(\mathbf {k} )\left[c_{2}(\mathbf {p} /2-\mathbf {k} )a_{2}(\mathbf {p} /2+\mathbf {k} )+c_{1}(\mathbf {p} /2+\mathbf {k} )a_{1}(\mathbf {p} /2-\mathbf {k} ),\right].}$

${\displaystyle F(\mathbf {k} )}$ is a spectral function, normalized by ${\displaystyle \sum _{\mathbf {k} }\left|F(\mathbf {k} )\right|^{2}=1.}$

It is known that a photon is a boson.[25] Does the composite photon satisfy Bose–Einstein commutation relations? Fermions are defined as the particles whose creation and annihilation operators adhere to the anticommutation relations

${\displaystyle \{a(\mathbf {k} ),a(\mathbf {l} )\}=0,}$

${\displaystyle \{a^{\dagger }(\mathbf {k} ),a^{\dagger }(\mathbf {l} )\}=0,}$

${\displaystyle \{a(\mathbf {k} ),a^{\dagger }(\mathbf {l} )\}=\delta (\mathbf {k} -\mathbf {l} ),}$

while bosons are defined as the particles that adhere to the commutation relations

${\displaystyle \left[b(\mathbf {k} ),b(\mathbf {l} )\right]=0,}$

${\displaystyle \left[b^{\dagger }(\mathbf {k} ),b^{\dagger }(\mathbf {l} )\right]=0,}$

${\displaystyle \left[b(\mathbf {k} ),b^{\dagger }(\mathbf {l} )\right]=\delta (\mathbf {k} -\mathbf {l} ).\quad \quad (7)}$

The creation and annihilation operators of composite particles formed of fermion pairs adhere to the commutation relations of the form[19][20][21][22]

${\displaystyle \left[Q(\mathbf {k} ),Q(\mathbf {l} )\right]=0,}$

${\displaystyle \left[Q^{\dagger }(\mathbf {k} ),Q^{\dagger }(\mathbf {l} )\right]=0,}$

${\displaystyle \left[Q(\mathbf {k} ),Q^{\dagger }(\mathbf {l} )\right]=\delta (\mathbf {k} -\mathbf {l} )-\Delta (\mathbf {k} ,\mathbf {l} ).\quad \quad (8)}$

with

${\displaystyle \Delta (\mathbf {p} ^{\prime },\mathbf {p} )=\sum _{\mathbf {k} }F^{\dagger }(\mathbf {k} )\left[F(\mathbf {p} ^{\prime }/2-\mathbf {p} /2+\mathbf {k} )a^{\dagger }(\mathbf {p} -\mathbf {p} ^{\prime }/2-\mathbf {k} )a(\mathbf {p} ^{\prime }/2-\mathbf {k} )\right.}$

${\displaystyle \left.+F(\mathbf {p} /2-\mathbf {p} ^{\prime }/2+\mathbf {k} )c^{\dagger }(\mathbf {p} -\mathbf {p} ^{\prime }/2+\mathbf {k} )c(\mathbf {p} ^{\prime }/2+\mathbf {k} )\right].\quad \quad (9)}$

For Cooper electron pairs,[21] "a" and "c" represent different spin directions. For nucleon pairs (the deuteron),[19][20] "a" and "c" represent proton and neutron. For neutrino–antineutrino pairs,[22] "a" and "c" represent neutrino and antineutrino. The size of the deviations from pure Bose behavior,

${\displaystyle \Delta (\mathbf {p} ^{\prime },\mathbf {p} ),}$

depends on the degree of overlap of the fermion wave functions and the constraints of the Pauli exclusion principle.

If the state has the form

${\displaystyle |\Phi \rangle =a^{\dagger }(\mathbf {k_{1}} )a^{\dagger }(\mathbf {k_{2}} )...a^{\dagger }(\mathbf {k_{n}} )c^{\dagger }(\mathbf {q_{1}} )c^{\dagger }(\mathbf {q_{2}} )...c^{\dagger }(\mathbf {q_{m}} )|0\rangle }$

then the expectation value of Eq. (9) vanishes for ${\displaystyle \mathbf {p} ^{\prime }\neq \mathbf {p} }$, and the expression for ${\displaystyle \Delta (\mathbf {p} ^{\prime },\mathbf {p} )}$ can be approximated by

${\displaystyle \Delta (\mathbf {p} ^{\prime },\mathbf {p} )=\delta (\mathbf {p} ^{\prime }-\mathbf {p} )\sum _{\mathbf {k} }\left|F(\mathbf {k} )\right|^{2}\left[a^{\dagger }(\mathbf {p} /2-\mathbf {k} )a(\mathbf {p} /2-\mathbf {k} )\right.}$

${\displaystyle \left.+c^{\dagger }(\mathbf {p} /2+\mathbf {k} )c(\mathbf {p} /2+\mathbf {k} )\right].}$

Using the fermion number operators ${\displaystyle n_{a}(\mathbf {k} )}$ and ${\displaystyle n_{c}(\mathbf {k} )}$, this can be written,

${\displaystyle \Delta (\mathbf {p} ^{\prime },\mathbf {p} )=\delta (\mathbf {p} ^{\prime }-\mathbf {p} )\sum _{\mathbf {k} }\left|F(\mathbf {k} )\right|^{2}\left[n_{a}(\mathbf {p} /2-\mathbf {k} )+n_{c}(\mathbf {p} /2+\mathbf {k} )\right]}$

${\displaystyle =\delta (\mathbf {p} ^{\prime }-\mathbf {p} )\sum _{\mathbf {k} }\left[\left|F(\mathbf {p} /2-\mathbf {k} )\right|^{2}n_{a}(\mathbf {k} )+\left|F(\mathbf {k} -\mathbf {p} /2)\right|^{2}n_{c}(\mathbf {k} )\right]}$

${\displaystyle =\delta (\mathbf {p} ^{\prime }-\mathbf {p} ){\overline {\Delta }}(\mathbf {p} ,\mathbf {p} )}$

showing that it is the average number of fermions in a particular state ${\displaystyle \mathbf {k} }$ averaged over all states with weighting factors ${\displaystyle F(\mathbf {p} /2-\mathbf {k} )}$ and ${\displaystyle F(\mathbf {k} -\mathbf {p} /2)}$.

De Broglie did not address the problem of statistics for the composite photon. However, "Jordan considered the essential part of the problem was to construct Bose–Einstein amplitudes from Fermi–Dirac amplitudes", as Pryce[5] noted. Jordan[4] "suggested that it is not the interaction between neutrinos and antineutrinos that binds them together into photons, but rather the manner in which they interact with charged particles that leads to the simplified description of light in terms of photons."

Jordan's hypothesis eliminated the need for theorizing an unknown interaction, but his hypothesis that the neutrino and antineutrino are emitted in exactly the same direction seems rather artificial as noted by Fock.[26] His strong desire to obtain exact Bose–Einstein commutation relations for the composite photon led him to work with a scalar or longitudinally polarized photon. Greenberg and Wightman[27] have pointed out why the one-dimensional case works, but the three-dimensional case does not.

In 1928, Jordan noticed that commutation relations for pairs of fermions were similar to those for bosons.[28] Compare Eq. (7) with Eq. (8). From 1935 till 1937, Jordan, Kronig, and others[29] tried to obtain exact Bose–Einstein commutation relations for the composite photon. Terms were added to the commutation relations to cancel out the delta term in Eq. (8). These terms corresponded to "simulated photons". For example, the absorption of a photon of momentum ${\displaystyle \mathbf {p} }$ could be simulated by a Raman effect in which a neutrino with momentum ${\displaystyle \mathbf {p+k} }$ is absorbed while another of another with opposite spin and momentum ${\displaystyle \mathbf {k} }$ is emitted. (It is now known that single neutrinos or antineutrinos interact so weakly that they cannot simulate photons.)

In 1938, Pryce[5] showed that one cannot obtain both Bose–Einstein statistics and transversely polarized photons from neutrino–antineutrino pairs. Construction of transversely polarized photons is not the problem.[30] As Berezinski[6] noted, "The only actual difficulty is that the construction of a transverse four-vector is incompatible with the requirement of statistics." In some ways Berezinski gives a clearer picture of the problem. A simple version of the proof is as follows:

The expectation values of the commutation relations for composite right and left-handed photons are:

${\displaystyle \left[Q_{R}(\mathbf {p} ^{\prime }),Q_{R}(\mathbf {p} )\right]=0,\;\left[Q_{L}(\mathbf {p} ^{\prime }),Q_{L}(\mathbf {p} )\right]=0,}$

${\displaystyle \left[Q_{R}(\mathbf {p} ^{\prime }),Q_{R}^{\dagger }(\mathbf {p} )\right]=\delta (\mathbf {p} ^{\prime }-\mathbf {p} )(1-{{\overline {\Delta }}_{12}}(\mathbf {p} ,\mathbf {p} )),}$

${\displaystyle \left[Q_{L}(\mathbf {p} ^{\prime }),Q_{L}^{\dagger }(\mathbf {p} )\right]=\delta (\mathbf {p} ^{\prime }-\mathbf {p} )(1-{{\overline {\Delta }}_{21}}(\mathbf {p} ,\mathbf {p} )),}$

${\displaystyle \left[Q_{R}(\mathbf {p} ^{\prime }),Q_{L}(\mathbf {p} )\right]=0,\;\left[Q_{R}(\mathbf {p} ^{\prime }),Q_{L}^{\dagger }(\mathbf {p} )\right]=0,\quad \quad \quad \quad (10)}$

where

${\displaystyle {{\overline {\Delta }}_{12}}(\mathbf {p} ,\mathbf {p} )=\sum _{\mathbf {k} }\left[\left|F(\mathbf {k} -\mathbf {p} /2)\right|^{2}(n_{a1}(\mathbf {k} )+n_{c2}(\mathbf {k} ))\right.}$

${\displaystyle \left.+\left|F(\mathbf {p} /2-\mathbf {k} )\right|^{2}(n_{c1}(\mathbf {k} )+n_{a2}(\mathbf {k} ))\right].\quad \quad \quad \quad (11)}$

The deviation from Bose–Einstein statistics is caused by ${\displaystyle {\overline {\Delta }}_{12}(\mathbf {p} ,\mathbf {p} )}$ and ${\displaystyle {\overline {\Delta }}_{21}(\mathbf {p} ,\mathbf {p} )}$, which are functions of the neutrino numbers operators.

Linear polarization photon operators are defined by

${\displaystyle \xi (\mathbf {p} )={1 \over {\sqrt {2}}}\left[Q_{L}(\mathbf {p} )+Q_{R}(\mathbf {p} )\right],}$

${\displaystyle \eta (\mathbf {p} )={i \over {\sqrt {2}}}\left[Q_{L}(\mathbf {p} )-Q_{R}(\mathbf {p} )\right].\quad \quad \quad \quad (12)}$

A particularly interesting commutation relation is,

${\displaystyle [\xi (\mathbf {p} ^{\prime }),\eta ^{\dagger }(\mathbf {p} )]={i \over 2}\delta (\mathbf {p} ^{\prime }-\mathbf {p} )[{\overline {\Delta }}_{21}(\mathbf {p} ,\mathbf {p} )-{\overline {\Delta }}_{12}(\mathbf {p} ,\mathbf {p} )],\quad \quad (13)}$

which follows from (10) and (12).

For the composite photon to obey Bose–Einstein commutation relations, at the very least,

${\displaystyle [\xi (\mathbf {p} ^{\prime }),\eta ^{\dagger }(\mathbf {p} )]=0\quad \quad \quad \quad (14)}$

Pryce noted.[5] From Eq. (11) and Eq. (13) the requirement is that

${\displaystyle \sum _{\mathbf {k} }\left[\left|F(\mathbf {k} -\mathbf {p} /2)\right|^{2}(n_{a1}(\mathbf {k} )+n_{c2}(\mathbf {k} )-n_{a2}(\mathbf {k} )-n_{c1}(\mathbf {k} ))\right.}$

${\displaystyle \left.+\left|F(\mathbf {p} /2-\mathbf {k} )\right|^{2}(n_{c1}(\mathbf {k} )+n_{a2}(\mathbf {k} )-n_{c2}(\mathbf {k} )-n_{a1}(\mathbf {k} ))\right]}$

gives zero when applied to any state vector. Thus, all the coefficients of ${\displaystyle n_{a1}(\mathbf {k} )}$ and ${\displaystyle n_{c1}(\mathbf {k} )}$, etc. must vanish separately. This means ${\displaystyle F(\mathbf {k} )=0}$, and the composite photon does not exist,[5][6] completing the proof.

Perkins[15][22] reasoned that the photon does not have to obey Bose–Einstein commutation relations, because the non-Bose terms are small and they may not cause any detectable effects. Perkins[12] noted, "As presented in many quantum mechanics texts it may appear that Bose statistics follow from basic principles, but it is really from the classical canonical formalism. This is not a reliable procedure as evidenced by the fact that it gives the completely wrong result for spin-1/2 particles." Furthermore, "most integral spin particles (light mesons, strange mesons, etc.) are composite particles formed of quarks. Because of their underlying fermion structure, these integral spin particles are not fundamental bosons, but composite quasibosons. However, in the asymptotic limit, which generally applies, they are essentially bosons. For these particles, Bose commutation relations are just an approximation, albeit a very good one. There are some differences; bringing two of these composite particles close together will force their identical fermions to jump to excited states because of the Pauli exclusion principle."

Brzezinski in reaffirming Pryce's theorem argues that commutation relation (14) is necessary for the photon to be truly neutral. However, Perkins[22] has shown that a neutral photon in the usual sense can be obtained without Bose–Einstein commutation relations.

The number operator for a composite photon is defined as

${\displaystyle N(\mathbf {p} )=Q^{\dagger }(\mathbf {p} )Q(\mathbf {p} ).}$

Lipkin[19] suggested for a rough estimate to assume that ${\displaystyle F(\mathbf {k} )=1/\Omega }$ where ${\displaystyle \Omega }$ is a constant equal to the number of states used to construct the wave packet.

Perkins[12] showed that the effect of the composite photon’s number operator acting on a state of ${\displaystyle m}$ composite photons is,

${\displaystyle N(\mathbf {p} )(Q^{\dagger }(\mathbf {p} ))^{m}|0\rangle \;=\left(m-{m(m-1) \over \Omega }\right)(Q^{\dagger }(\mathbf {p} ))^{m}|0\rangle ,}$

using ${\displaystyle N(\mathbf {p} )|0\rangle =0}$. This result differs from the usual one because of the second term which is small for large ${\displaystyle \Omega }$. Normalizing in the usual manner,[31]

${\displaystyle Q^{\dagger }(\mathbf {p} )|n_{\mathbf {p} }\rangle \;={\sqrt {(n_{\mathbf {p} }+1)\left(1-{n_{\mathbf {p} } \over \Omega }\right)}}|n_{\mathbf {p} }+1\rangle ,}$

${\displaystyle Q(\mathbf {p} )|n_{\mathbf {p} }\rangle \;={\sqrt {n_{\mathbf {p} }\left(1-{(n_{\mathbf {p} }-1) \over \Omega }\right)}}|n_{\mathbf {p} }-1\rangle ,\quad \quad \quad \quad (15)}$

where ${\displaystyle |n_{\mathbf {p} }\rangle }$ is the state of ${\displaystyle n_{\mathbf {p} }}$ composite photons having momentum ${\displaystyle \mathbf {p} }$ which is created by applying ${\displaystyle Q^{\dagger }(\mathbf {p} )}$ on the vacuum ${\displaystyle n_{\mathbf {p} }}$ times. Note that,

${\displaystyle Q^{\dagger }(\mathbf {p} )|0\rangle =|1_{\mathbf {p} }\rangle ,}$

${\displaystyle Q(\mathbf {p} )|1_{\mathbf {p} }\rangle =|0\rangle ,}$

which is the same result as obtained with boson operators. The formulas in Eq. (15) are similar to the usual ones with correction factors that approach zero for large ${\displaystyle \Omega }$.

The main evidence indicating that photons are bosons comes from the Blackbody radiation experiments which are in agreement with Planck's distribution. Perkins[12] calculated the photon distribution for Blackbody radiation using the second quantization method,[31] but with a composite photon.

The atoms in the walls of the cavity are taken to be a two-level system with photons emitted from the upper level β and absorbed at the lower level α. The transition probability for emission of a photon is enhanced when n_p photons are present,

${\displaystyle w_{\alpha \beta }(n_{\mathbf {p} }+1\leftarrow n_{\mathbf {p} })=(n_{\mathbf {p} }+1)\left(1-{n_{\mathbf {p} } \over \Omega }\right)w_{\alpha \beta }(1_{\mathbf {p} }\leftarrow 0),\quad (16)}$

where the first of (15) has been used. The absorption is enhanced less since the second of (15) is used,

${\displaystyle w_{\beta \alpha }(n_{\mathbf {p} }-1\leftarrow n_{\mathbf {p} })=n_{\mathbf {p} }\left(1-{n_{\mathbf {p} }-1 \over \Omega }\right)w_{\beta \alpha }(0\leftarrow 1_{\mathbf {p} }).\quad (17)}$

Using the equality,

${\displaystyle w_{\beta \alpha }(0\leftarrow 1_{\mathbf {p} })=w_{\alpha \beta }(1_{\mathbf {p} }\leftarrow 0),}$

of the transition rates, Eqs. (16) and (17) are combined to give,

${\displaystyle {w_{\alpha \beta }(n_{\mathbf {p} }+1\leftarrow n_{\mathbf {p} }) \over w_{\beta \alpha }(n_{\mathbf {p} }-1\leftarrow n_{\mathbf {p} })}={(n_{\mathbf {p} }+1)\left(1-{n_{\mathbf {p} } \over \Omega }\right) \over n_{\mathbf {p} }\left(1-{(n_{\mathbf {p} }-1) \over \Omega }\right)}.}$

The probability of finding the system with energy E is proportional to e^−E/kT according to Boltzmann's distribution law. Thus, the equilibrium between emission and absorption requires that,

${\displaystyle w_{\alpha \beta }(n_{\mathbf {p} }+1\leftarrow n_{\mathbf {p} })e^{-E_{\beta }/kT}=w_{\beta \alpha }(n_{\mathbf {p} }-1\leftarrow n_{\mathbf {p} })e^{-E_{\alpha }/kT},}$

with the photon energy ${\displaystyle \omega _{p}=E_{\beta }-E_{\alpha }}$. Combining the last two equations results in,

${\displaystyle n_{\mathbf {p} }={2 \over u+(u+2)/\Omega +{\sqrt {u^{2}(1+2/\Omega )+(u+2)^{2}/\Omega ^{2}}}},}$

with ${\displaystyle u=e^{\omega _{p}/kT}-1}$. For ${\displaystyle \Omega (\omega _{p}/kT)>>1}$, this reduces to

${\displaystyle n_{\mathbf {p} }={1 \over e^{\omega _{p}/kT}\left(1+{1 \over \Omega }\right)-1}.}$

This equation differs from Planck’s law because of the ${\displaystyle 1/\Omega }$ term. For the conditions used in the Blackbody radiation experiments of Coblentz,[32] Perkins estimates that 1 / Ω < 10⁻⁹, and the maximum deviation from Planck's law is less than one part in 10⁻⁸, which is too small to be detected.

Experimental results show that only left-handed neutrinos and right-handed antineutrinos exist. Three sets of neutrinos have been observed,[33][34] one that is connected with electrons, one with muons, and one with tau leptons.[35]

In the standard model the pion and muon decay modes are:

π+	→	μ+	+	ν μ
μ+	→	e+	+	ν e	+	ν μ

To form a photon, which satisfies parity and charge conjugation, two sets of two-component neutrinos (i.e., right-handed and left-handed neutrinos) are needed. Perkins (see Sec. VI of Ref.[15]) attempted to solve this problem by noting that the needed two sets of two-component neutrinos would exist if the positive muon is identified as the particle and the negative muon as the antiparticle. The reasoning is as follows: let
ν
₁ be the right-handed neutrino and
ν
₂ the left-handed neutrino with their corresponding antineutrinos (with opposite helicity). The neutrinos involved in beta decay are
ν
₂ and
ν
₂, while those for π–μ decay are
ν
₁ and
ν
₁. With this scheme the pion and muon decay modes are:

π+	→	μ+	+	ν 1
μ+	→	e+	+	ν 2	+	ν 1

There is convincing evidence that neutrinos have mass. In experiments at the SuperKamiokande researchers[13] have discovered neutrino oscillations in which one flavor of neutrino changed into another. This means that neutrinos have non-zero mass. Since massless neutrinos are needed to form a massless photon, a composite photon is not possible.