Fermi's interaction

W⁻
boson

(which then decays to electron and antineutrino) is not shown.

In

Fermi first introduced this coupling in his description of beta decay in 1933.[3] The Fermi interaction was the precursor to the theory for the weak interaction where the interaction between the proton–neutron and electron–antineutrino is mediated by a virtual W⁻ boson, of which the Fermi theory is the low-energy effective field theory.

According to Eugene Wigner, who together with Jordan introduced the Jordan–Wigner transformation, Fermi's paper on beta decay was his main contribution to the history of physics.^[4]

History of initial rejection and later publication

Fermi first submitted his "tentative" theory of beta decay to the prestigious science journal Nature, which rejected it "because it contained speculations too remote from reality to be of interest to the reader."^[5][6] It has been argued that Nature later admitted the rejection to be one of the great editorial blunders in its history, but Fermi's biographer David N. Schwartz has objected that this is both unproven and unlikely.^[7] Fermi then submitted revised versions of the paper to Italian and German publications, which accepted and published them in those languages in 1933 and 1934.^{[8][9][10][11]} The paper did not appear at the time in a primary publication in English.^[5] An English translation of the seminal paper was published in the American Journal of Physics in 1968.^[11]

Fermi found the initial rejection of the paper so troubling that he decided to take some time off from theoretical physics, and do only experimental physics. This would lead shortly to his famous work with activation of nuclei with slow neutrons.

The "tentativo"

Definitions

The theory deals with three types of particles presumed to be in direct interaction: initially a “heavy particle” in the “neutron state” (${\displaystyle \rho =+1}$), which then transitions into its “proton state” (${\displaystyle \rho =-1}$) with the emission of an electron and a neutrino.

Electron state

${\displaystyle \psi =\sum _{s}\psi _{s}a_{s},}$

where ${\displaystyle \psi }$ is the single-electron wavefunction, ${\displaystyle \psi _{s}}$ are its stationary states.

${\displaystyle a_{s}}$ is the operator which annihilates an electron in state ${\displaystyle s}$ which acts on the Fock space as

${\displaystyle a_{s}\Psi (N_{1},N_{2},\ldots ,N_{s},\ldots )=(-1)^{N_{1}+N_{2}+\cdots +N_{s}-1}(1-N_{s})\Psi (N_{1},N_{2},\ldots ,1-N_{s},\ldots ).}$

${\displaystyle a_{s}^{*}}$ is the creation operator for electron state ${\displaystyle s:}$

${\displaystyle a_{s}^{*}\Psi (N_{1},N_{2},\ldots ,N_{s},\ldots )=(-1)^{N_{1}+N_{2}+\cdots +N_{s}-1}N_{s}\Psi (N_{1},N_{2},\ldots ,1-N_{s},\ldots ).}$

Neutrino state

Similarly,

${\displaystyle \phi =\sum _{\sigma }\phi _{\sigma }b_{\sigma },}$

where ${\displaystyle \phi }$ is the single-neutrino wavefunction, and ${\displaystyle \phi _{\sigma }}$ are its stationary states.

${\displaystyle b_{\sigma }}$ is the operator which annihilates a neutrino in state ${\displaystyle \sigma }$ which acts on the Fock space as

${\displaystyle b_{\sigma }\Phi (M_{1},M_{2},\ldots ,M_{\sigma },\ldots )=(-1)^{M_{1}+M_{2}+\cdots +M_{\sigma }-1}(1-M_{\sigma })\Phi (M_{1},M_{2},\ldots ,1-M_{\sigma },\ldots ).}$

${\displaystyle b_{\sigma }^{*}}$ is the creation operator for neutrino state ${\displaystyle \sigma }$.

Heavy particle state

${\displaystyle \rho }$ is the operator introduced by Heisenberg (later generalized into isospin) that acts on a heavy particle state, which has eigenvalue +1 when the particle is a neutron, and −1 if the particle is a proton. Therefore, heavy particle states will be represented by two-row column vectors, where

${\displaystyle {\begin{pmatrix}1\\0\end{pmatrix}}}$

represents a neutron, and

${\displaystyle {\begin{pmatrix}0\\1\end{pmatrix}}}$

represents a proton (in the representation where ${\displaystyle \rho }$ is the usual ${\displaystyle \sigma _{z}}$ spin matrix).

The operators that change a heavy particle from a proton into a neutron and vice versa are respectively represented by

${\displaystyle Q=\sigma _{x}-i\sigma _{y}={\begin{pmatrix}0&1\\0&0\end{pmatrix}}}$

and

${\displaystyle Q^{*}=\sigma _{x}+i\sigma _{y}={\begin{pmatrix}0&0\\1&0\end{pmatrix}}.}$

${\displaystyle u_{n}}$ resp. ${\displaystyle v_{n}}$ is an eigenfunction for a neutron resp. proton in the state ${\displaystyle n}$.

Hamiltonian

The Hamiltonian is composed of three parts: ${\displaystyle H_{\text{h.p.}}}$, representing the energy of the free heavy particles, ${\displaystyle H_{\text{l.p.}}}$, representing the energy of the free light particles, and a part giving the interaction ${\displaystyle H_{\text{int.}}}$.

${\displaystyle H_{\text{h.p.}}={\frac {1}{2}}(1+\rho )N+{\frac {1}{2}}(1-\rho )P,}$

where ${\displaystyle N}$ and ${\displaystyle P}$ are the energy operators of the neutron and proton respectively, so that if ${\displaystyle \rho =1}$, ${\displaystyle H_{\text{h.p.}}=N}$, and if ${\displaystyle \rho =-1}$, ${\displaystyle H_{\text{h.p.}}=P}$.

${\displaystyle H_{\text{l.p.}}=\sum _{s}H_{s}N_{s}+\sum _{\sigma }K_{\sigma }M_{\sigma },}$

where ${\displaystyle H_{s}}$ is the energy of the electron in the ${\displaystyle s^{\text{th}}}$ state in the nucleus's Coulomb field, and ${\displaystyle N_{s}}$ is the number of electrons in that state; ${\displaystyle M_{\sigma }}$ is the number of neutrinos in the ${\displaystyle \sigma ^{\text{th}}}$ state, and ${\displaystyle K_{\sigma }}$ energy of each such neutrino (assumed to be in a free, plane wave state).

The interaction part must contain a term representing the transformation of a proton into a neutron along with the emission of an electron and a neutrino (now known to be an antineutrino), as well as a term for the inverse process; the Coulomb force between the electron and proton is ignored as irrelevant to the ${\displaystyle \beta }$-decay process.

Fermi proposes two possible values for ${\displaystyle H_{\text{int.}}}$: first, a non-relativistic version which ignores spin:

${\displaystyle H_{\text{int.}}=g\left[Q\psi (x)\phi (x)+Q^{*}\psi ^{*}(x)\phi ^{*}(x)\right],}$

and subsequently a version assuming that the light particles are four-component Dirac spinors, but that speed of the heavy particles is small relative to ${\displaystyle c}$ and that the interaction terms analogous to the electromagnetic vector potential can be ignored:

${\displaystyle H_{\text{int.}}=g\left[Q{\tilde {\psi }}^{*}\delta \phi +Q^{*}{\tilde {\psi }}\delta \phi ^{*}\right],}$

where ${\displaystyle \psi }$ and ${\displaystyle \phi }$ are now four-component Dirac spinors, ${\displaystyle {\tilde {\psi }}}$ represents the Hermitian conjugate of ${\displaystyle \psi }$, and ${\displaystyle \delta }$ is a matrix

${\displaystyle {\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&1\\0&0&-1&0\end{pmatrix}}.}$

Matrix elements

The state of the system is taken to be given by the tuple ${\displaystyle \rho ,n,N_{1},N_{2},\ldots ,M_{1},M_{2},\ldots ,}$ where ${\displaystyle \rho =\pm 1}$ specifies whether the heavy particle is a neutron or proton, ${\displaystyle n}$ is the quantum state of the heavy particle, ${\displaystyle N_{s}}$ is the number of electrons in state ${\displaystyle s}$ and ${\displaystyle M_{\sigma }}$ is the number of neutrinos in state ${\displaystyle \sigma }$.

Using the relativistic version of ${\displaystyle H_{\text{int.}}}$, Fermi gives the matrix element between the state with a neutron in state ${\displaystyle n}$ and no electrons resp. neutrinos present in state ${\displaystyle s}$ resp. ${\displaystyle \sigma }$, and the state with a proton in state ${\displaystyle m}$ and an electron and a neutrino present in states ${\displaystyle s}$ and ${\displaystyle \sigma }$ as

${\displaystyle H_{\rho =-1,m,N_{s}=1,M_{\sigma }=1}^{\rho =1,n,N_{s}=0,M_{\sigma }=0}=\pm g\int v_{m}^{*}u_{n}{\tilde {\psi }}_{s}\delta \phi _{\sigma }^{*}d\tau ,}$

where the integral is taken over the entire configuration space of the heavy particles (except for ${\displaystyle \rho }$). The ${\displaystyle \pm }$ is determined by whether the total number of light particles is odd (−) or even (+).

Transition probability

To calculate the lifetime of a neutron in a state ${\displaystyle n}$ according to the usual

, the above matrix elements must be summed over all unoccupied electron and neutrino states. This is simplified by assuming that the electron and neutrino eigenfunctions ${\displaystyle \psi _{s}}$ and ${\displaystyle \phi _{\sigma }}$ are constant within the nucleus (i.e., their Compton wavelength is much smaller than the size of the nucleus). This leads to

${\displaystyle H_{\rho =-1,m,N_{s}=1,M_{\sigma }=1}^{\rho =1,n,N_{s}=0,M_{\sigma }=0}=\pm g{\tilde {\psi }}_{s}\delta \phi _{\sigma }^{*}\int v_{m}^{*}u_{n}d\tau ,}$

where ${\displaystyle \psi _{s}}$ and ${\displaystyle \phi _{\sigma }}$ are now evaluated at the position of the nucleus.

According to Fermi's golden rule^{[further explanation needed]}, the probability of this transition is

${\displaystyle {\begin{aligned}\left|a_{\rho =-1,m,N_{s}=1,M_{\sigma }=1}^{\rho =1,n,N_{s}=0,M_{\sigma }=0}\right|^{2}&=\left|H_{\rho =-1,m,N_{s}=1,M_{\sigma }=1}^{\rho =1,n,N_{s}=0,M_{\sigma }=0}\times {\frac {\exp {{\frac {2\pi i}{h}}(-W+H_{s}+K_{\sigma })t}-1}{-W+H_{s}+K_{\sigma }}}\right|^{2}\\&=4\left|H_{\rho =-1,m,N_{s}=1,M_{\sigma }=1}^{\rho =1,n,N_{s}=0,M_{\sigma }=0}\right|^{2}\times {\frac {\sin ^{2}\left({\frac {\pi t}{h}}(-W+H_{s}+K_{\sigma })\right)}{(-W+H_{s}+K_{\sigma })^{2}}},\end{aligned}}}$

where ${\displaystyle W}$ is the difference in the energy of the proton and neutron states.

Averaging over all positive-energy neutrino spin / momentum directions (where ${\displaystyle \Omega ^{-1}}$ is the density of neutrino states, eventually taken to infinity), we obtain

${\displaystyle \left\langle \left|H_{\rho =-1,m,N_{s}=1,M_{\sigma }=1}^{\rho =1,n,N_{s}=0,M_{\sigma }=0}\right|^{2}\right\rangle _{\text{avg}}={\frac {g^{2}}{4\Omega }}\left|\int v_{m}^{*}u_{n}d\tau \right|^{2}\left({\tilde {\psi }}_{s}\psi _{s}-{\frac {\mu c^{2}}{K_{\sigma }}}{\tilde {\psi }}_{s}\beta \psi _{s}\right),}$

where ${\displaystyle \mu }$ is the rest mass of the neutrino and ${\displaystyle \beta }$ is the Dirac matrix.

Noting that the transition probability has a sharp maximum for values of ${\displaystyle p_{\sigma }}$ for which ${\displaystyle -W+H_{s}+K_{\sigma }=0}$, this simplifies to ^{[further explanation needed]}

${\displaystyle t{\frac {8\pi ^{3}g^{2}}{h^{4}}}\times \left|\int v_{m}^{*}u_{n}d\tau \right|^{2}{\frac {p_{\sigma }^{2}}{v_{\sigma }}}\left({\tilde {\psi }}_{s}\psi _{s}-{\frac {\mu c^{2}}{K_{\sigma }}}{\tilde {\psi }}_{s}\beta \psi _{s}\right),}$

where ${\displaystyle p_{\sigma }}$ and ${\displaystyle K_{\sigma }}$ is the values for which ${\displaystyle -W+H_{s}+K_{\sigma }=0}$.

Fermi makes three remarks about this function:

• Since the neutrino states are considered to be free, ${\displaystyle K_{\sigma }>\mu c^{2}}$ and thus the upper limit on the continuous ${\displaystyle \beta }$-spectrum is ${\displaystyle H_{s}\leq W-\mu c^{2}}$.
• Since for the electrons ${\displaystyle H_{s}>mc^{2}}$, in order for ${\displaystyle \beta }$-decay to occur, the proton–neutron energy difference must be ${\displaystyle W\geq (m+\mu )c^{2}}$
• The factor

${\displaystyle Q_{mn}^{*}=\int v_{m}^{*}u_{n}d\tau }$

in the transition probability is normally of magnitude 1, but in special circumstances it vanishes; this leads to (approximate)

selection rules

for ${\displaystyle \beta }$-decay.

Forbidden transitions

As noted above, when the inner product ${\displaystyle Q_{mn}^{*}}$ between the heavy particle states ${\displaystyle u_{n}}$ and ${\displaystyle v_{m}}$ vanishes, the associated transition is "forbidden" (or, rather, much less likely than in cases where it is closer to 1).

If the description of the nucleus in terms of the individual quantum states of the protons and neutrons is accurate to a good approximation, ${\displaystyle Q_{mn}^{*}}$ vanishes unless the neutron state ${\displaystyle u_{n}}$ and the proton state ${\displaystyle v_{m}}$ have the same angular momentum; otherwise, the total angular momentum of the entire nucleus before and after the decay must be used.

Influence

Shortly after Fermi's paper appeared,

leads to the electromagnetic force. He found that the force would be of the form ${\displaystyle {\frac {\text{Const.}}{r^{5}}}}$, but that contemporary experimental data led to a value that was too small by a factor of a million.^[13]

The following year, Hideki Yukawa picked up on this idea,^[14] but in his theory the neutrinos and electrons were replaced by a new hypothetical particle with a rest mass approximately 200 times heavier than the electron.^[15]

Later developments

Fermi's four-fermion theory describes the weak interaction remarkably well. Unfortunately, the calculated cross-section, or probability of interaction, grows as the square of the energy ${\displaystyle \sigma \approx G_{\rm {F}}^{2}E^{2}}$. Since this cross section grows without bound, the theory is not valid at energies much higher than about 100 GeV. Here G_F is the Fermi constant, which denotes the strength of the interaction. This eventually led to the replacement of the four-fermion contact interaction by a more complete theory (

.

The interaction could also explain

In the original theory, Fermi assumed that the form of interaction is a contact coupling of two vector currents. Subsequently, it was pointed out by

The inclusion of parity violation in Fermi's interaction was done by

Fermi constant

The most precise experimental determination of the Fermi constant comes from measurements of the muon lifetime, which is inversely proportional to the square of G_F (when neglecting the muon mass against the mass of the W boson).^[21] In modern terms, the "reduced Fermi constant", that is, the constant in natural units is^[3][22]

${\displaystyle G_{\rm {F}}^{0}={\frac {G_{\rm {F}}}{(\hbar c)^{3}}}={\frac {\sqrt {2}}{8}}{\frac {g^{2}}{M_{\rm {W}}^{2}c^{4}}}=1.1663787(6)\times 10^{-5}\;{\textrm {GeV}}^{-2}\approx 4.5437957\times 10^{14}\;{\textrm {J}}^{-2}\ .}$

Here, g is the

, which mediates the decay in question.

In the Standard Model, the Fermi constant is related to the Higgs vacuum expectation value

${\displaystyle v=\left({\sqrt {2}}\,G_{\rm {F}}^{0}\right)^{-1/2}\simeq 246.22\;{\textrm {GeV}}}$.^[23]

More directly, approximately (tree level for the standard model),

${\displaystyle G_{\rm {F}}^{0}\simeq {\frac {\pi \alpha }{{\sqrt {2}}~M_{\rm {W}}^{2}(1-M_{\rm {W}}^{2}/M_{\rm {Z}}^{2})}}.}$

This can be further simplified in terms of the Weinberg angle using the relation between the W and Z bosons with ${\displaystyle M_{\text{Z}}={\frac {M_{\text{W}}}{\cos \theta _{\text{W}}}}}$, so that

${\displaystyle G_{\rm {F}}^{0}\simeq {\frac {\pi \alpha }{{\sqrt {2}}~M_{\rm {Z}}^{2}\cos ^{2}\theta _{\rm {W}}\sin ^{2}\theta _{\rm {W}}}}.}$

References