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Computing the Spontaneous Emission Rate of a Laser's Gain Atoms

The spontaneous emission (loss) rate of our upper laser energy level can be dominated by a combination of the vacuum photon flux, `phi_0`, and the temperature-dependent black body photon flux, `phi_T`, given by Plancks Law. We need to compute the combined `phi_0+phi_T` photon flux per Hertz at our laser frequency, `f_l`. The value of `f_l` will be

`f_l=(E_2-E_1)/h`

where `E_2` and `E_1` are the upper and lower excitation energies of the atomic medium of the laser and `h` is Planck's constant. Flux here is defined as photons per second per meter squared of area that they are crossing. Both `phi_0` and `phi_T` will be omnidirectional. The directions of the photon flux is not important because all photon directions have equal average probabililty of causing a transition between the laser atomic levels of the randomly oriented atoms. This flux will be similar to Planck's black body irradiance but expressed in photon flux instead of radiant power per steradian per Hertz. The flux calculation involves both the number of orthogonal photon modes at `f_l` and the thermally dependent occupation number, `o`, of a mode at that frequency.

Electromagnetic Field Modes

The traditional way of computing the black body spectrum was to assign the electromagnetic field (emf) to a box that had totally reflecting walls. That method led to a radiation spectrum that diverged at short wavelengths so did not agree with experiments. In order to obtain agreement, Max Planck made the strange assumption that each of the various wavelengths (modes) could be represented only by an integer number of discrete energy particles which we now call photons. These photons would each have energy `E=hc/lambda` where `h` is the constant that emerged from Planck's theory, `c` is the speed of light and `lambda` is the wavelength of the mode that depends on the size of the box since the reflecting box had to have emf nodes at its inside walls. Although the size of a box containing the photons will cancel out when computing photon density and flux
(`"meter"^(-2)sec^(-1)`) we will define a box size as `L` on each side as the container for the photons and their electromagnetic field. Since the emf is defined as having nodes at each inner wall of the box then the largest wavelength,`lambda`, that can have nodes at all inner walls of the box will have `lambda_0=2L` (i.e. `L=1/2` wavelength). Since frequency `f=c/lambda_0` that means that the lowest frequency wave will be `f_0=c/(2L)` The other waves will then have, for each `(x,y,z)`, direction, an integer number of half waves, `lambda_0`. We will name these integers `(n_x,n_y,n_z)`. These integers are either zero or positive . That means that the wavelengths of waves with these indices are

`lambda_(n_x,n_y,n_z)=lambda_0/sqrt(n_x^2+n_y^2+n_z^2)`

with corresponding frequency, `f_(n_x,n_y,n_z)`:

`f_("n_x,n_y,n_z")=csqrt(n_x^2+n_y^2+n_z^2)/lambda_0=f_0sqrt(n_x^2+n_y^2+n_z^2)`

Then, to simplify notation, let's define `n=sqrt(n_x^2+n_y^2+n_z^2)` so that for any mode, `n`, the mode energy will be

`e_n=(1/2+o)hf_0n`

where `o` is the temperature-dependent occupation number of the mode and `h` is Planck's constant. Since photons are bosons, every mode has to contain at least `1/2` vacuum photon.
The Casimir effect is experimental evidence of the existence of these vacuum photons. Boson mode occupation numbers have no intrinsic limit, so `o` can be in the range `o=0-">"oo`.

Thermal Occupation Distribution of Photons

For now we will ignore the the ground state occupation of 1/2 photon per mode and concentrate on the thermally induced occupation. Using the probability distribution of Boltzmann, we may define a normalized probability, P(o,n), as

`P(o,n)=exp(-obetanhf_0)/(Z(beta,n))`

Also we define the un-normalized quantity:

`p(o,n)=exp(-obetanhf_0)`

where `beta=1/(k_BT_m)`, `k_B` is Boltzmann's constant and `T_m` is the absolute temperature of the laser gain medium so that`(k_BT_m)` is the thermal energy of a particle in the medium, and `Z(beta,n)` is just the Boltzmann normalization for `p(o,n)` such that

`sum_(o=1)^(o=oo)p(o,n)=Z(beta)`

An appropriate reference for derivation of Planck's Law

is this link. We can remove the `o` dependence of `p` by summing p over `o` from zero to infinity where the sum `sum_(o=1)^(o=oo)p(o,n)=Z(beta)` is just the Geometric Series that has a simple solution. For the entire sum of p over o which is the same as Z, we have:

`Z(beta,n)=p_("sum")(beta,n)=sum_(o=0)^(o=oo)p(o,n)=1/(1-exp(-betanhf_0))`

Then it is clear that

`(del)/(delbeta)log(Z(beta))=(del)/(delbeta)[int(dZ)/(Z(beta))]`

Boltzmann's equation for the expected energy, `"<"E_n">"`, of a mode is to compute the derivative with respect to `beta` of `log(Z(n,beta))`.

Note that for any function `f(beta)`

`d/(dbeta)(logf(beta)) =1/(f(beta))(df)/(dbeta)`;

Applying this to `Z(beta)` we have:

`(dlog(Z(beta)))/(dbeta)=1/(Z(beta))(del(Z(beta)))/(delbeta)`

`"<"E_n">"=(Z(beta))/(dbeta)=(nhf_0(exp(-betanhf_0)))/(1-exp(-betanhf_0))=(nhf_0)/(exp(betanhf_0)-1)`

Note that for frequencies `nf_0">>"1/(hbeta)`, `"<"E_n">` goes to zero which is the opposite of the classical physics result.

Computing the Number of Modes per Unit Frequency

In order to get the energy density and the photon flux at a given frequency, `f_l`, we multiply the result for `"<"E_n">"` by the available number of emf modes. The total number of modes of zero or positive indices `(n_x,n_y,n_z)` in the sphere octant of radius `n_l` is

`N_m(n_l)=1/8(4/3pin_l^3)`

Since photon modes all possess 2 orthogonal polarizations, the above expression can be converted into `N_p(n_l)`, the largest posible number of electromagnetic field modes in a sphere octant of radius, `n_l`, is obtained by multiplying by 2:

`N_p(n_l)=(2/8)(4pi)/3n_l^3`

We can also convert this into the number of modes in the outer layer of thickness `dn_l` of a sphere octant of radius `n_l`

`(dN_p(n_l))/(dn_l)=4pi(2/8)n_l^2`

Converting from Mode Number, `n_l`, to Frequency, `f_l`

Recalling that `n_l=(f_l/f_0)=f_l/(c/(2L))=(2L)/cf_l` then `n_l^2=(4L^2f_l^2)/c^2` and also `dn_l=(2L)/c(df_l)`. so we now convert our expression to a function of `f_l` which represents the number of modes in the spherical octant shell of thickness, `df_l`, of radius, `f_l`:

`(dN_p(f_l))/(df_l)=(2pi/8)(8L^2f_l^2)/c^2(((2L)/c)`
`(dN_p(f_l))/(df_l)=(2pi/8)(16L^3f_l^2)/c^3`

This will be the number of photon modes in the outer shell, `f_l`, of thickness `df_l` of a sphere octant of radius `f_l` in a box of volume `L^3`.

We need the photon mode density, `rho_p(f_l,beta)`, so we now divide by the volume of the cube, `L^3`.

`rho_p(f_l)=(dN_p(f_l))/(df_l)=(2pi)/8(16f_l^2)/c^3df_l`

The all-directional photon mode flux, `phi_p` is then their speed, c, multiplied by the density, `crho_p`

`phi_p(f_l)=4pi(f_l^2)/c^2df_l`.

`phi_p` has SI units `"photons " "meter"^(-2)"sec"^(-1)` as expected for a flux.

Planck's black body distribution involves the energy mode flux, `phi_u(f_l)`, due to this photon flux which is just `hf_l` times the photon mode flux.

`phi_u(f_l)=4pi(hf_l)f_l^2/c^2df_l`.

which of course has to be multiplied by the simple boson energy occupation distribution `1/(exp(betaf_l)-1)` to be relative to the temperature of the medium.

`phi_rho(f_l,beta)=4pif_l^2/c^2(hf_l)1/(exp(betahf_l)-1)df_l`.

Planck's result was the irradiance per unit solid angle per Hertz. We can convert our result to those units by dividing the present equation by the one directional solid angle, `2pi`, and setting `df_l=1 "Hertz"`

`phi_u(f_l,beta)=2h(f_l^3)/c^2(1/(exp(betahf_l)-1))`.

which agrees with the Planck's Law .

result.

Computing the photon frequency at maximum black body irradiance

First we re-state the irradiance equation:

`phi_u(f_l,beta)=2h(f_l^3)/c^2(1/(exp(betahf_l)-1))`.

Now we will take the derivative of `phi_u` with respect to `f_l` and use that result to obtain the `f_l` of the maximum of the black body frequency spectrum.

`(dphi_p)/(df_l)=(6h(f_l^2/c^2))/(exp(betahf_l)-1)-(2hf_l^3)/c^2(betahexp(betahf_l))/(exp(betahf_l)-1)^2=0`

`(3hf_l^2/c^2)-hf_l^3/c^2(betahexp(betahf_l))/(exp(betahf_l)-1)=0`

Dividing this by `hf_l^2/c^2` and multiplying by `exp(betahf_l)-1` we obtain:

`3-((betahf_l)exp(betahf_l))/(exp(betahf_l)-1)=0`

Multiplying both sides by `exp(betahf_l)-1` we obtain:

`3(exp(betahf_l)-1)-(betahf_l)exp(betahf_l))=0`

Let `x=betahf_l`. Then we have the transcendental equation:

`3(exp(x)-1)-xexp(x)=0`

`(3-x)exp(x)=3`

A solution is `x=2.822` and this agrees with Wien's displacement law . As an example, for a temperature of 6000K, we have

`U_T=k_BT_M=1.380649 x 10^(-23)6000~8.28 x 10^(-20)`

and photon energy is `2.822U_T`

`hf_l=2.337x10^(-19) "Joules" =1.46 eV`

and

`f_l=(2.337*10^(-19))/h Hz=3.54*10^(14) Hz`

As an example of using this result, I will compute Planck's radiant power per steradian per Hz.

`phi_u(f_l,beta)=2h(f_l^3)/c^2(1/(exp(betahf_l)-1))`
`=2*6.6*10^(-34)(3.54*10^(14))^3/((3*10^8)^2)(1/(exp(2.882)-1))`
`=4.116*10^(-8) "watts per steradian per Hz"`

Radiant Power Calculation
Blackbody Radiation Experiment

Using the Photon Flux to Compute the Rate of Spontaneous Emission

What we now want is loss rate of the upper laser level due to the photon flux. The all directional photon flux per Hz of band width is:

`dphi_rho(f_l,beta)/(df_l)=4pif_l^2/c^2(hf_l)(1/2+1/(exp(betahf_l)-1))`.

where I have included the vacuum uccupancy of the photon mode since that will be dominant for high photon frequencies. The simplest way to get some frequency width, `df_l`, is that for a gas at STP. Then the width will be doppler broadening:

`df_l=v/cf_l`

where `v` is the RMS velocity width of the laser gas gain medium. For typical gas atoms this results in about `df_l=10^9 Hz` or a little lower for heavy gain atoms. However, for a gain medium like that of a dye, the frequency width is much, much broader. A favorite laser dye, rhodamine 6G, has a fluoresence wavelength width of 550 to 590 nanometers which results a frequency width of about `3.7x10^13` Hz. If we have a total density of `n_a` atoms in the upper laser state that the photon flux is impinging upon and the effective cross sectional area for transition of each photon is `sigma_a` then the rate of loss of the population `n_a` is

`(dn_a)/dt=-n_asigma_aphi_p`

The term `sigma_a` (units `m^2`) will usually be the radial cross section of the dipole or quadrupole moment of the laser transition.

Under Einstein Coefficient, wikipedia gives the following equation for the spontaneous transition rate:

`1/tau_s=(2omega_l^3e^2)/(3epsilon_0hc^3)|"<"f|bbvecr|i">"||^2`

where `e` is the electron charge and `bbvecr` is the atom's spatial charge asymmetry if any. For the `B` coefficients a similar expression is multiplied by the energy density, `u(omega_l)`, of thermal black body radiation or other artificial input at `f_l`.

`1/tau_b=u(omega_l)(4pi^3e^2)/(3epsilon_0h^2)|"<"f|vecr|i">"||^2`

Then, for the spontaneous rate expression, the photon vacuum energy density is already included. So the value of `sigma` would then be just:

`sigma_a=pi|"<"f|bbvecr|i">"||^2`

Generally `sigma_a` will be on the order of a few `(nm)^2` or about `sigma_a=pi*10^(-18)` `"meters"^2`. Using value for `sigma_a` and the flux, `(dphi_p)/(df)deltaf`, we obtain `tau_s=85*10^(-6)` or 85 microseconds.
Therefore our rate of pumping, `1/tau_p`, must be considerably larger than `1/tau_s` to see any laser action. The rate ratio `(1/tau_s)/(1/tau_b)=tau_b/tau_s` is interesting since it shows the ratio of spontaneous emission to stimulated emission.

`tau_b/tau_s=(2omega_l^3e^2)/(3epsilon_0hc^3)/ [u(omega_l)(4pi^3e^2)/(3epsilon_0h^2)]`
=`(homega_l^3/c^2 )/(4pi^2u(omega_l))`