 In video 4, we develop Fermi's golden rule. If a quantum system with Hamiltonian H hat 0 has stationary states, phi n, e to the minus i omega nt, then adding an interaction term, H hat i, to the Hamiltonian will cause an initial state, i, to transition to a final state, f, at a rate proportional to the squared magnitude of the matrix element, mfi. Where to first order, mfi is the projection onto the final state of H hat i operating on the initial state. To second order, we add the sum over all intermediate states of the product of first order matrix elements between the initial and intermediate states and between the intermediate and final states, divided by the energy difference between initial and intermediate states. By the conservation of energy, the energy of the final state must equal the energy of the initial state, but the intermediate states are occupied only briefly, so the uncertainty principle allows their energies to differ from the initial energy. We have worked out the details of the interaction Hamiltonian for a hydrogen atom in an electromagnetic field, and expressed it as a sum of two terms, H hat i prime and H hat i double prime. We further separated H hat i prime into absorption and emission terms. The absorption term is the sum over all pairs of electron states m and n and over all photon states k alpha of a coefficient mA prime times a creation operator for electron state m, a destruction operator for electron state n, and a destruction operator for photon state k alpha. The emission term is similar with a coefficient mE prime and a photon creation operator instead of a destruction operator. The two coefficient functions differ only by a single factor. For absorption, this factor is E to the plus i k dot x. For emission, it is E to the minus i k dot x. k dot x is zero at the origin and increases to 2 pi at a distance of lambda, one wavelength along the propagation direction E k. The size of the atom is determined by the spatial extent of the electron orbital. For the hydrogen wave functions, this is typically less than about 10 to the minus 9 meters. While the wavelengths of hydrogen transitions are larger than 10 to the minus 7 meters, much larger than the size of the atom. Therefore, k dot x is very small everywhere within the atom. And E to the i k dot x varies only slightly from one plus i zero. So to a reasonable approximation, E to the i k dot x is equal to one. This is called the dipole approximation. With this approximation, the absorption and emission coefficients are identical. Now let's analyze the absorption of a photon by a hydrogen atom using quantum field theory. We will focus on three of the lowest energy orbitals. These are the 1s, 2s, and 2pz orbitals with the spatial wave functions shown. Here A is the Bohr radius, about 53 picometers. The s orbitals depend only on the radial distance r, so they are spherically symmetric. The 2pz orbital has a factor of z, which is responsible for its two lobes structure. The wave function is positive for positive z and negative for negative z. The absorption and emission coefficients contain a factor with terms having the form of an integral over all space of the conjugate of the nth atomic wave function times the derivative with respect to x, y, or z of the nth wave function. Without computing the derivative and doing the integration over all space, we can make the following general statement. The integral over all values of x of a function f of x will be 0 if f of x is an odd function, by which we mean f of minus x equals minus f of x. The s orbitals are functions of r alone. r is an even function of the x, y, and z coordinates, meaning if we replace any of those coordinates by its negative, r is unchanged. Therefore, the s orbitals are even functions of x, y, and z. The pz orbital is an even function of x and y, since its dependence on those coordinates is only through r. However, due to its factor of z, it is an odd function of z. The 1s orbital is an even function of all three coordinates, x, y, and z. However, the derivative of an even function is an odd function, likewise for the 2s orbital and indeed for all s orbitals. The pz orbital is even in the x and y coordinates, and the corresponding derivatives are odd. But the orbital is odd in the z coordinate and the derivative of an odd function is even. For these three wave functions, our integrals will be the product of one of the blue curves times one of the red curves. For a transition between the 1s and 2s orbitals, this product will be an even times an odd function. This is an odd function, and its integral will vanish. Therefore, there can be no transitions by this mechanism between s orbitals. For a transition between the 1s and 2pz orbitals, for the x and y coordinates, we have the product of the top left and bottom left functions, even times odd, which is odd. So those terms vanish. But for the z coordinate, we have the product of the top left and bottom right functions, even times even, which is even. So that integral can be nonzero and produce a nonzero transition rate. This consideration of the even and odd symmetries of the wave functions and their derivatives are summarized in so-called selection rules. For the wave functions we have examined, the integrals for transitions between 1s and 2s orbitals are all zero. And in general, no first-order dipole transitions are possible between s orbitals. By first order, we mean the first-order form of Fermi's golden rule. For transitions between the 1s and 2pz orbitals, the x and y integrals are zero. But the z integral is not. And in general, first-order dipole transitions are possible between s and p orbitals. These selection rules have an interesting physical interpretation in terms of angular momentum. In videos 7 and 8 of the quantum mechanics series, we show that an electron in the hydrogen orbital is characterized by four quantum numbers, n, l, m sub l, and m sub s. n fixes energy, and l fixes orbital angular momentum magnitude, but not direction and space. m sub l fixes the z component of orbital angular momentum, and m sub s fixes the z component of electron spin angular momentum. s orbitals have l equal to zero, and therefore no orbital angular momentum. p orbitals have l equal to 1, an orbital angular momentum of square root 2h bar in standard units. Photons carry angular momentum of square root 2h bar in the form of spin. Therefore, a photon plus an s orbital electron have total angular momentum of square root 2h bar. If these are destroyed and a p orbital electron is created, the total angular momentum of square root 2h bar is conserved. Likewise, angular momentum is conserved if a p orbital electron is destroyed and a photon and an s orbital electron are created. On the other hand, if both electrons are in s orbitals, angular momentum will not be conserved. In the first case, angular momentum decreases, and in the second, it increases. Fermi's golden rule ensures conservation of energy, and we have seen that the selection rules ensure conservation of angular momentum. What about linear momentum? Suppose our atom is at rest, and it absorbs a photon, causing it to transition to an excited state. In video two of the quantum mechanics series, we showed that a photon has linear momentum equal to its energy divided by the speed of light. Since the atom initially has no linear momentum, after the photon is absorbed, its linear momentum must equal the linear momentum of the photon. Since momentum is mass times velocity, and the mass of a hydrogen atom is nearly the mass of a proton, the atom should end up moving with a velocity of nearly e over mpc. For a 1s to 2p transition, this is about 3 meters per second. If the proton mass was infinite, then this velocity would be zero. Essentially, the nucleus absorbs the photon's linear momentum. However, there is nothing in our quantum mechanical model of the hydrogen atom that accounts for motion of the nucleus. So it should not be surprising that nothing in our quantum field calculation enforces conservation of linear momentum. Now, let's go through the steps to calculate the transition rate for a photon in the state kα to be absorbed by an electron in the 1s hydrogen orbital and transition to the 2pz orbital. The initial quantum state of the system is simply a list of the occupation numbers of the electron and photon states. We've labeled the 1s orbital as the first electron state. So the initial state of the atom is one electron in the first electron state and zero in all the others. For the radiation field, we assume there are no photons present, except in the kα state, which contains nkα photons. As a shorthand, we write this as a 1 ket for the atom and a nkα ket for the radiation. In the final state, the electron is in the 2pz orbital, which is the third electron state in our bookkeeping, and there is one less kα photon. The shorthand notation for this is a 3 ket for the atom and a nkα-1 ket for the radiation. The term from the interaction Hamiltonian that connects these states is ma' of 3,1kα times bhatt3 plus bhatt1 minus ahattkα minus. We need the squared magnitude of hhatti on the initial state projected onto the final state. We substitute the expressions for the initial and final states and the operative part of the interaction Hamiltonian. The bhatt1 minus operator destroys the electron in the first electron state. Then the bhatt3 plus operator creates an electron in the third electron state. This converts the 1 ket into a 3 ket, which is projected onto the 3 bra of the final state. The ahattkα minus operator converts the nkα ket into square root of nkα times the nkα-1 ket, as we showed in video 1 of this series. This is projected onto the nkα-1 bra of the final state. The two projection factors are both 1, so we obtain nkα times the squared magnitude of the ma' coefficient. We see that the transition rate is proportional to the number of photons present in the radiation field, that is, to the intensity of radiation. Plugging in our expression for the ma' coefficient, we obtain the transition rate predicted by Fermi's golden rule. In addition to the integral factor, which produces the selection rules, there is a delta function. This is 0 unless its argument is 0. And this ensures that the photon frequency equals the difference of the atomic orbital frequencies. Since energy is proportional to frequency in quantum theory, this simply enforces the conservation of energy. There is also a factor of the electrons charged to mass ratio squared, and a geometric factor, eKα dot ez squared. The dot product of two unit vectors equals the cosine of the angle between them. This is 1 for an angle of 0 and decreases to 0 for an angle of 90 degrees. This factor tells us that the less the polarization of the photon is aligned with the z-axis, the smaller will be the transition rate. An analogous factor is present in the expression for the power received by a wire antenna illuminated by a plane wave. In video 6 of the quantum mechanics series, we saw how the superposition of 1s and 1pz orbitals produces a time-varying probability distribution for the electron that oscillates at the difference of the orbital frequencies, which equals the radiation frequency. This geometric factor tells us that the absorption process is most efficient when the photon polarization is aligned with the direction of this oscillation, which we have taken to be the z-axis. Now, let's look at the emission process. We start with an electron in the 2pz orbital. A photon in the k-alpha state is emitted, and the electron transitions to the 1s orbital. The initial state is an electron in the third electron state and nk-alpha photons in the k-alpha photon state. The final state is an electron in the first electron state and nk-alpha plus 1 photons in the k-alpha photon state. The relevant part of the interaction Hamiltonian is the emission term Me' of 1,3k-alpha times B-hat-1 plus B-hat-3 minus A-hat-k-alpha plus. And we need the squared magnitude of h-hat-i acting on the initial state projected onto the final state. Substituting the appropriate terms, we have an expression in which B-hat-1 plus B-hat-3 minus operate on the third electron state, and A-hat-k-alpha plus operates on the nk-alpha radiation state. The first operation destroys the third electron state and creates the first electron state. This leads to the projection of the first electron state onto a cell, which is 1. The second operation converts the nk-alpha photon state into square root of nk-alpha plus 1 times the nk-alpha plus 1 photon state. This leads to the projection of the nk-alpha plus 1 state onto itself, which is 1, times the factor square root nk-alpha plus 1. After squaring, we end up with nk-alpha plus 1 times the magnitude squared of Me' of 1,3k-alpha Plugging in our expression for the Me' coefficient, we obtain the transition rate predicted by Fermi's golden rule. As it did for absorption, the delta function enforces conservation of energy, and we have the same charge to mass ratio and geometric factors. The transition rate is proportional to nk-alpha plus 1. The plus 1 term tells us that even if nk-alpha equals 0, that is, there are initially no photons present, there is a non-zero probability of emission. This is spontaneous emission. The nk-alpha term tells us that the more photons are initially present, the larger is the transition rate. This is stimulated emission. A general result from integration theory tells us that our integral factor is unchanged if we swap the orbital functions. This means that except for the nk-alpha and nk-alpha plus 1 factors, the absorption and emission rates are identical.