The Gamow factor, Sommerfeld factor or Gamow–Sommerfeld factor,^[1] named after its discoverer George Gamow or after Arnold Sommerfeld, is a probability factor for two nuclear particles' chance of overcoming the Coulomb barrier in order to undergo nuclear reactions, for example in nuclear fusion. By classical physics, there is almost no possibility for protons to fuse by crossing each other's Coulomb barrier at temperatures commonly observed to cause fusion, such as those found in the sun. When George Gamow instead applied quantum mechanics to the problem, he found that there was a significant chance for the fusion due to tunneling.

The probability of two nuclear particles overcoming their electrostatic barriers is given by the following equation:

${\displaystyle P_{g}(E)=e^{-{\sqrt {\frac {E_{g)){E))))}$^[2]

where ${\displaystyle E_{g))$ is the Gamow energy,

${\displaystyle E_{g}\equiv 2m_{r}c^{2}(\pi \alpha Z_{a}Z_{b})^{2))$

Here, ${\displaystyle m_{r}={\frac {m_{a}m_{b)){m_{a}+m_{b))))$ is the reduced mass of the two particles. The constant ${\displaystyle \alpha }$ is the fine structure constant, ${\displaystyle c}$ is the speed of light, and ${\displaystyle Z_{a))$ and ${\displaystyle Z_{b))$ are the respective atomic numbers of each particle.

While the probability of overcoming the Coulomb barrier increases rapidly with increasing particle energy, for a given temperature, the probability of a particle having such an energy falls off very fast, as described by the Maxwell–Boltzmann distribution. Gamow found that, taken together, these effects mean that for any given temperature, the particles that fuse are mostly in a temperature-dependent narrow range of energies known as the Gamow window.

Derivation

Gamow^[3] first solved the one-dimensional case of quantum tunneling using the WKB approximation. Considering a wave function of a particle of mass m, we take area 1 to be where a wave is emitted, area 2 the potential barrier which has height V and width l (at ${\displaystyle 0<x<l}$), and area 3 its other side, where the wave is arriving, partly transmitted and partly reflected. For a wave number k and energy E we get:

${\displaystyle \Psi _{1}=Ae^{i(kx+\alpha )}e^{-i{\frac {E}{\hbar ))t))$

${\displaystyle \Psi _{2}=B_{1}e^{-k'x}+B_{2}e^{k'x))$

${\displaystyle \Psi _{3}=(C_{1}e^{-i(kx+\beta )}+C_{2}e^{i(kx+\beta ')})e^{-i{\frac {E}{\hbar ))t))$

where ${\displaystyle k={\sqrt {2mE))}$ and ${\displaystyle k'={\sqrt {2m(V-E)))}$. This is solved for given A and α by taking the boundary conditions at the both barrier edges, at ${\displaystyle x=0}$ and ${\displaystyle x=l}$, where both ${\displaystyle \Psi }$ and its derivative must be equal on both sides. For ${\displaystyle k'l\gg 1}$, this is easily solved by ignoring the time exponential and considering the real part alone (the imaginary part has the same behavior). We get, up to factors depending on the phases which are typically of order 1, and up to factors of the order of ${\displaystyle {\frac {k}{k'))={\sqrt {\frac {E}{V-E))))$ (assumed not very large, since V is greater than E not marginally):

${\displaystyle B_{1},B_{2}\approx A}$

${\displaystyle C_{1},C_{2}\approx {\frac {1}{2))A\cdot {\frac {k'}{k))\cdot e^{k'l))$

Next Gamow modeled the alpha decay as a symmetric one-dimensional problem, with a standing wave between two symmetric potential barriers at ${\displaystyle q_{0}<x<q_{0}+l}$ and ${\displaystyle -(q_{0}+l)<x<-q_{0))$, and emitting waves at both outer sides of the barriers. Solving this can in principle be done by taking the solution of the first problem, translating it by ${\displaystyle q_{0))$ and gluing it to an identical solution reflected around ${\displaystyle x=0}$.

Due to the symmetry of the problem, the emitting waves on both sides must have equal amplitudes (A), but their phases (α) may be different. This gives a single extra parameter; however, gluing the two solutions at ${\displaystyle x=0}$ requires two boundary conditions (for both the wave function and its derivative), so in general there is no solution. In particular, re-writing ${\displaystyle \Psi _{3))$ (after translation by ${\displaystyle q_{0))$) as a sum of a cosine and a sine of ${\displaystyle kx}$, each having a different factor that depends on k and α, the factor of the sine must vanish, so that the solution can be glued symmetrically to its reflection. Since the factor is in general complex (hence its vanishing imposes two constraints, representing the two boundary conditions), this can in general be solved by adding an imaginary part of k, which gives the extra parameter needed. Thus E will have an imaginary part as well.

The physical meaning of this is that the standing wave in the middle decays; the emitted waves newly emitted have therefore smaller amplitudes, so that their amplitude decays in time but grows with distance. The decay constant, denoted λ, is assumed small compared to ${\displaystyle E/\hbar }$.

λ can be estimated without solving explicitly, by noting its effect on the probability current conservation law. Since the probability flows from the middle to the sides, we have:

${\displaystyle {\frac {\partial }{\partial t))\int _{-(q_{0}+l)}^{(q_{0}+l)}\Psi ^{*}\Psi dx=2\cdot {\frac {\hbar }{2mi))\left(\Psi _{1}^{*}{\frac {\partial \Psi _{1)){\partial x))-\Psi _{1}{\frac {\partial \Psi _{1}^{*)){\partial x))\right),}$

Note the factor of 2 is due to having two emitted waves.

Taking ${\displaystyle \Psi \sim e^{-\lambda t))$, this gives:

${\displaystyle \lambda \cdot {\frac {1}{4))\cdot 2(q_{0}+l)A^{2}{\frac {k'^{2)){k^{2))}\cdot e^{2k'l}\approx 2{\frac {\hbar }{m))A^{2}k,}$

Since the quadratic dependence in ${\displaystyle k'l}$ is negligible relative to its exponential dependence, we may write:

${\displaystyle \lambda \approx {\frac {\hbar k}{m(q_{0}+l))){\frac {k^{2)){k'^{2))}\cdot e^{-2k'l))$

Remembering the imaginary part added to k is much smaller than the real part, we may now neglect it and get:

${\displaystyle \lambda \approx {\frac {\hbar k}{m2(q_{0}+l)))\cdot 8{\frac {E}{V-E))\cdot e^{-2{\sqrt {2m(V-E)))l/\hbar ))$

Note that ${\displaystyle {\frac {\hbar k}{m))}$ is the particle velocity, so the first factor is the classical rate by which the particle trapped between the barriers hits them.

Finally, moving to the three-dimensional problem, the spherically symmetric Schrödinger equation reads (expanding the wave function ${\displaystyle \psi (r,\theta ,\phi )=\chi (r)u(\theta ,\phi )}$ in spherical harmonics and looking at the n-th term):

${\displaystyle {\frac {\hbar ^{2)){2m))\left({\frac {d^{2}\chi }{dr^{2))}+{\frac {2}{r)){\frac {d\chi }{dr))\right)=\left(V(r)+{\frac {\hbar ^{2)){2m)){\frac {n(n+1)}{r^{2))}-E\right)\chi }$

Since ${\displaystyle n>0}$ amounts to enlarging the potential, and therefore substantially reducing the decay rate (given its exponential dependence on ${\displaystyle {\sqrt {V-E))}$), we focus on ${\displaystyle n=0}$, and get a very similar problem to the previous one with ${\displaystyle \chi (r)=\Psi (r)/r}$, except that now the potential as a function of r is not a step function.

The main effect of this on the amplitudes is that we must replace the argument in the exponent, taking an integral of ${\displaystyle 2{\sqrt {2m(V-E)))/\hbar }$ over the distance where ${\displaystyle V(r)>E}$ rather than multiplying by l. We take the Coulomb potential:

${\displaystyle V(r)={\frac {z(Z-z)k_{e}e^{2)){r))}$

where ${\displaystyle k_{e))$ is the Coulomb constant, e the electron charge, z = 2 is the charge number of the alpha particle and Z the charge number of the nucleus (Z-z after emitting the particle). The integration limits are then ${\displaystyle r_{2}={\frac {z(Z-z)k_{e}e^{2)){E))}$, where we assume the nuclear potential energy is still relatively small, and ${\displaystyle r_{1))$, which is where the nuclear negative potential energy is large enough so that the overall potential is smaller than E. Thus, the argument of the exponent in λ is:

${\displaystyle 2{\frac {\sqrt {2mE)){\hbar ))\int _{r_{1))^{r_{2)){\sqrt ((\frac {V(r)}{E))-1))\,dr=2{\frac {\sqrt {2mE)){\hbar ))\int _{r_{1))^{r_{2)){\sqrt ((\frac {r_{2)){r))-1))\,dr}$

This can be solved by substituting ${\displaystyle t={\sqrt {r/r_{2))))$ and then ${\displaystyle t=cos(\theta )}$ and solving for θ, giving:

${\displaystyle 2\cdot r_{2}{\frac {\sqrt {2mE)){\hbar ))\cdot (\cos ^{-1}({\sqrt {x)))-{\sqrt {x)){\sqrt {1-x)))=2{\frac ((\sqrt {2m))z(Z-z)k_{e}e^{2)){\hbar {\sqrt {E))))\cdot (\cos ^{-1}({\sqrt {x)))-{\sqrt {x)){\sqrt {1-x)))}$

where ${\displaystyle x=r_{1}/r_{2))$. Since x is small, the x-dependent factor is of order 1.

Gamow assumed ${\displaystyle x\ll 1}$, thus replacing the x-dependent factor by ${\displaystyle \pi /2}$, giving: ${\displaystyle \lambda \sim e^{-{\sqrt {\frac {E_{g)){E))))}$ with:

${\displaystyle E_{g}={\frac {2\pi ^{2}m\left[z(Z-z)k_{e}e^{2}\right]^{2)){\hbar ^{2))))$

which is the same as the formula given in the beginning of the article with ${\displaystyle Z_{a}=z}$, ${\displaystyle Z_{b}=Z-z}$ and the fine structure constant ${\displaystyle \alpha ={\frac {k_{e}e^{2)){\hbar c))}$.

For a radium alpha decay, Z = 88, z = 2 and m = 4m_p, E_G is approximately 50 GeV. Gamow calculated the slope of ${\displaystyle \log(\lambda )}$ with respect to E at an energy of 5 MeV to be ~10¹⁴ joule⁻¹, compared to the experimental value of ${\displaystyle 0.7\cdot 10^{14))$ joule⁻¹.