 We have established that the absorption and emission rates between the hydrogen 1s and 2pz orbitals are identical, except for a factor nkα for absorption and nkα plus 1 for emission. In the first video of this series, we discussed Einstein's 1917 prediction of stimulated emission in terms of A and B coefficients. Let's revisit this topic in light of our current results. We imagine an atom with two energy levels, one and two. We write the absorption rate for a single atom in level one as B12 times U of nu. Here nu is the frequency of radiation with energy equal to the difference of the atomic energies and u is the density of that radiation. The emission rate for a single atom in level two is A21 plus B21 times U of nu. The first term describes spontaneous emission. We showed that the second term, which represents stimulated emission, must be included if the absorption and emission processes are consistent with Planck's law for radiation density. And we found that the ratio of A21 to B12 is 8πH nu cubed over C cubed. And the B coefficients are equal. Using the 1s and 2pz orbitals as our two atomic energy levels, we set A21 to be the transition rate for spontaneous emission. The total emission rate is then A21 times 1 plus nkα. Comparing this to Einstein's emission rate formula, we see that A21 times nkα must equal B21 times U of nu. Therefore, the ratio of A21 over B12 must equal U of nu over nkα. To see if our results are consistent with Einstein's formula, we need to calculate the radiation density U of nu. This times d nu is the total energy per unit volume of all photons with a frequency within d nu of nu. That energy is h nu times the sum of occupation numbers nkα for all k values within that interval. In standard units, k, the magnitude of the wave vector, is ω over C, which is 2π nu over C. In k space, the set of all wave vectors with a given magnitude forms a sphere of radius k and area 4π k². We will assume the occupation number nkα is independent of the direction of propagation. We have to sum over polarizations and k vectors. There are two polarizations, so that summation results in a factor of 2. The summation over k vectors becomes 1 over 2π cubed, the integral over all k magnitudes of 4π k². For our spherical shell, this becomes nkα times 2h nu times 4π over 2π cubed times 2π nu over C² times 2π d nu over C. Canceling common factors and dropping the d nu, we arrive at U of nu equals 8π h nu cubed over C cubed times nkα. So, we see that U of nu over nkα, which is a21 over b12 in Einstein's theory, is 8π h nu cubed over C cubed. Therefore, the prediction of quantum field theory agrees with Einstein's 1917 result. Now, let's consider the interaction of a photon and a free electron that is not bound within an atom. This is the problem of photon-electron scattering. Imagine we have an electron at rest with zero momentum and kinetic energy. Suppose it spontaneously emits a photon of frequency omega. This photon has linear momentum, let's call it minus p. To conserve zero total momentum, the electron must gain momentum p. The photon has energy omega in natural units, and the electron has kinetic energy p² over 2Me. Momentum is conserved, but energy is not. If instead the electron was initially in motion, we can always use a reference frame moving with the electron, so we again have this identical scenario. Now, imagine the reverse process. A photon and electron have opposite momenta. The electron absorbs the photon and ends up with zero momentum, yet therefore has zero energy, but the initial energy of the system was not zero. Therefore, momentum is conserved, but energy is not. From these considerations, we conclude that a free electron cannot emit or absorb a single photon. In terms of Feynman diagrams, the process where a free electron is destroyed and another free electron and a photon are created is impossible. Likewise, for the process where an electron and a photon are destroyed and a new electron is created. Since these are the two types of processes supported by h-hat-i-prime, no first-order h-hat-i-prime processes are possible involving a free electron and single photon. These types of processes were possible for an electron bound in a hydrogen atom. In that case, there is no inertial reference frame in which the electronic kinetic energy is zero. Plus, there is always potential energy stored in the proton-electron interaction, and the proton is able to absorb any momentum difference between the initial and final electron and photon states. For the free electron case, what about a process involving h-hat-i-double-prime? We have said that these are typically negligible when h-hat-i-prime processes are possible, but if not, then an h-hat-i-double-prime process may represent the dominant interaction. We could have a photon and electron with opposite momenta, scattering from each other and reversing their momenta. This process would conserve both total momentum and energy. So it is possible to have a process where one free electron and photon are destroyed and another free electron and photon are created. First-order scattering processes are possible between a free electron and a photon via h-hat-i-double-prime using terms a-hat-k-alpha plus times a-hat-l-beta minus and the same operators in the reverse order. Both of these destroy a photon in state l-beta and create a photon in state k-alpha. To analyze scattering, we need to look at the possible wave functions of a free electron. In quantum mechanics, the momentum operator has components minus i times the x, y, or z derivative, and the energy operator is i times the time derivative. Consider a wave function e to the i pxx plus pyy plus pzz times e to the minus i omega t. A more compact notation is e to the i vector p dot vector x times e to the minus i omega t. Applying the x component of the momentum operator to this wave function, we obtain px times the wave function. So this is a state of definite x momentum and likewise for the y and z components. And the energy operator produces omega times the wave function, so this is a state of definite energy, omega. The Schrodinger equation is satisfied if omega equals p squared over 2 times the electron mass. For a momentum state wave function, its squared magnitude is 1 everywhere. This tells us that an electron with definite momentum is completely unlocalized. This is required by the uncertainty principle. The product of the momentum and space uncertainties can be no smaller than Planck's constant. If the momentum uncertainty is zero, then the space uncertainty must be infinite. The same is true for a photon with a definite wave vector k. So even though we draw electrons and photons on a Feynman diagram, as if they follow precise spacetime trajectories, these considerations tell us that we cannot take Feynman diagrams literally. They are symbolic representations of terms in the interaction Hamiltonian, not accurate representations of the spacetime trajectories of electrons and photons. Now, let's look at the a squared or h hat i double prime term in the interaction Hamiltonian. Here is the expression for this. We develop for the hydrogen atom. The psi m conjugate and psi n functions are the hydrogen orbital wave functions. We need to change these to the free electron wave functions. We write the free electron wave function as a sum over all momentum states with coefficients BP. And the conjugate wave function has the coefficients and exponential factors conjugated. As mentioned previously, we assume that time dependence is absorbed into the BP coefficient. And to simplify the math, we will assume we limit consideration to a unit volume of space. Applying the second quantization concept described in video four, the wave function becomes an operator, psi hat minus, by replacing the BP coefficients with destruction operators, B hat P minus. The conjugate form has creation operators, B hat P plus. We substitute these into our h hat i double prime expression. Here's the a hat operator. We need the square of this, which will have a double sum. We use variables k alpha for the first summation and l beta for the second. Expanding the product, we will get four types of products of creation and destruction operators. One will have two creation and one two destruction operators. We are not interested in those. Instead, we focus on the two products that have one destruction and one creation operator. We will further break this into two terms. The first term has a creation operator times a destruction operator. The second term has a destruction operator times a creation operator. Assuming these operators apply to different photons, that is, k alpha is not equal to l beta, the operators commute and we flip their order. Comparing the result to the first term, we see that they are identical except for swapped summation variables which has no physical significance. So the terms are equal. Combining them will give us a factor of two which cancels the factor of two in the denominator and we arrive at what we will call the h hat i a e double prime operator. This is the part of h hat i double prime that describes the absorption of one photon and the emission of another. There are three factors with spatial dependence. We want to combine these and integrate over our unit volume. Here is that expression. We rewrite this as the integral of e to the i q plus l minus p plus k dot x. If p plus k equals q plus l, this is the integral of e to the zero which is one over a unit volume which is one. If these are not equal, the integral is zero. This assumes a certain spacing of discrete momentum and wave vector values which we will. So the integral is the so-called chronic or delta function of p plus k and q plus l. q is the initial and p the final electron momentum. l is the initial and k the final photon momentum. So q plus l is the initial and p plus k the final total momentum. This tells us that our chronic or delta function enforces conservation of momentum. For reference in standard units, the momentum of a photon is its energy divided by the speed of light which is h bar omega over c which is h bar times k in the e k direction, the direction of propagation. In natural units with h bar and c equal to one, we have p equals e equals omega equals k. Now let's look at scattering calculations. Here is the applicable part of h hat i double prime which describes the destruction of an electron of momentum q and a photon of momentum l and polarization beta and the creation of an electron of momentum p and a photon of momentum k and polarization alpha. Using Fermi's golden rule, the transition rate is proportional to the square of the matrix element MFI which is h hat i double prime operating on the initial state projected onto the final state. The chronic or delta conserves momentum and the golden rule delta function preserves energy. We take for the initial state an electron of momentum q nl beta photons in the l beta state and nk alpha photons in the k alpha state. The final state is an electron of momentum p, one less photon in the l beta state and one more photon in the k alpha state. We can choose any initial and final electron and photon states provided they satisfy the conditions for conservation of momentum and energy. Then the electron operators destroy the momentum state q and create the momentum state p. The photon operators generate a factor square root of nl beta, destroy one of the l beta photons, generate a factor square root of nk alpha plus one and create another k alpha photon. And the matrix element is e squared over 2 Me e k alpha dot el beta over square root omega k omega l square root nl beta square root nk alpha plus one. So the transition rate is 2 pi times the square of this. Here's that transition rate expression again. The dot product is the cosine of the angle between the initial and final photon polarization vectors. This tells us that the more aligned these are, the stronger is the scattering. The nl beta factor tells us that the more photons there are in the initial state, the greater is the probability of scattering. And if there are no photons, there is no possibility of scattering. The one term in the last factor tells us we do not need any photon initially in the final state for scattering to occur. This is added to nk alpha. That term predicts stimulated scattering. The more photons there are already in the final state, the more likely it is that a photon in the initial state will scatter. This is added to nk alpha. That term predicts stimulated scattering. The more photons there are already in the final state, the more likely it is that a photon in the initial state will scatter.