# Quantum Field Theory 6b Interacting Fields II

## Jul 26, 2021 12:00 · 2484 words · 12 minute read

In video 4 we developed Fermi s golden rule. If a quantum system with Hamiltonian H-hat-zero has stationary states phi-n, e-to-the minus-i omega-n-t.

00:23 - 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 M-f-i where to first-order, M-f-i 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.

01:50 - and expressed it as a sum of two terms, H-hat-i prime, and H-hat-i double-prime.

02:00 - We further separated H-hat-i prime into absorption and emission terms.

02:08 - 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 M-a-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.

02:30 - The emission term is similar, with a coefficient M-e-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.

03:14 - The size of the atom is determined by the spatial extent of the electron orbital.

03:23 - For the hydrogen wave functions, this is typically less than about 10 to the minus-9 meters.

03:33 - 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 1 plus i-0. So, to a reasonable approximation, e-to-the i-k dot-x is equal to one. This is called the dipole approximation.

04:11 - With this approximation the absorption and emission coefficients are identical.

04:21 - Now let s analyze the absorption of a photon by a hydrogen atom using quantum field theory.

04:30 - We will focus on three of the lowest-energy orbitals.

04:35 - These are the 1-s, 2-s, and 2-pz orbitals with the spatial wave functions shown. Here a is the Bohr radius, about 53 pico-meters. The s orbitals depend only on the radial distance r , so they are spherically symmetric. The 2-pz orbital has a factor of z which is responsible for its two-lobe structure. The wave function is positive for positive z and negative for negative z.

05:09 - The absorption and emission coefficients contain a factor with terms having the form of an integral over all space, of the conjugate of the m-th atomic wave function, times the derivative with respect to x , y , or z , of the n-th wave function.

05:27 - 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 zero, 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 p-z 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 1-s 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 2-s orbital, and indeed for all s orbitals. The p-z 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 of the product of one of the blue curves times one of the red curves.

07:12 - For a transition between the 1-s and 2-s 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.

07:30 - For a transition between the 1-s and 2-p-z 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 non-zero and produce a non-zero transition rate. This consideration of the even and odd symmetries of the wave functions and their derivatives are summarized in so-called selection rules.

08:11 - For the wave functions we have examined. The integrals for transitions between 1-s and 2-s 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 1-s and 2-p-z orbitals, the x-and-y integrals are zero. But the z integral is not.

08:50 - And, in general, first-order dipole transitions are possible between s and p orbitals.

09:01 - 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 a hydrogen orbital is characterized by four quantum numbers, n , l , m-sub-l, and m-sub-s. n fixes energy.

09:32 - And, l fixes orbital angular momentum magnitude, but not direction in 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.

09:52 - s orbitals have l equal to zero and therefore no orbital angular momentum. p orbitals have l equal to one, and orbital angular momentum of square-root-two h-bar, in standard units.

10:10 - Photons carry angular momentum of square-root-two h-bar in the form of spin.

10:17 - Therefore, a photon plus an s-orbital electron have total angular momentum of square-root-two h-bar. If these are destroyed and a p-orbital electron is created, the total angular momentum of square-root-two h-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.

11:03 - 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 2 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.

11:46 - 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 m-p-c. For a 1-s to 2-p transition this is about 3 meters per second. If the proton mass was infinite, then this velocity would be zero.

12:12 - 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.

12:24 - 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 k-alpha, to be absorbed by an electron in the 1-s hydrogen orbital, and transition to the 2-p-z orbital.

12:51 - 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 1-s orbital as the first electron state.

13:01 - 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 k-alpha state, which contains n-k-alpha photons. As a shorthand we write this as a one ket for the atom and a n-k-alpha ket for the radiation. In the final state the electron is in the 2-p-z orbital, which is the third electron state in our bookkeeping, and there is one less k-alpha photon.

Our shorthand notation for this is a 3 ket for the atom and a n-k-alpha minus 1 ket for the radiation.

13:52 - The term from the interaction Hamiltonian that connects these states is M-a prime of 3, 1, k, alpha, times b-hat-3 plus, b-hat-1 minus, a-hat k-alpha minus.

14:10 - We need the squared magnitude of H-hat-i, operating 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.

14:31 - The b-hat-1 minus operator destroys the electron in the first electron state. Then the b-hat-3 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.

14:49 - The a-hat k-alpha minus operator converts the n-k-alpha ket into square-root of n-k-alpha times the n-k-alpha minus 1 ket, as we showed in video 1 of this series. That is projected onto the n-k-alpha minus 1 bra of the final state.

15:11 - The two projection factors are both one, so we obtain n-k-alpha times the squared magnitude of the M-a-prime 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.

15:37 - Plugging in our expression for the M-a-prime 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 zero unless its argument is zero, 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.

16:15 - There is also a factor of the electron s charge to mass ratio squared.

16:20 - And, a geometric factor, e-k-alpha dot e-z, squared. The dot product of two unit vectors equals the cosine of the angle between them. This is one for an angle of zero, and decreases to zero 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.

16:47 - 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 1-s and 1-p-z orbitals produces a time-varying probability distribution for the electron, that oscillates at the difference of the orbital frequencies, which equals the radiation frequency.

17:14 - 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.

17:36 - We start with an electron in the 2-p-z orbital. A photon in the k-alpha state is emitted, and the electron transitions to the 1-s orbital. The initial state is an electron in the third electron state, and n-k-alpha photons in the k-alpha photon state.

17:59 - The final state is an electron in the first electron state, and n-k-alpha plus one photons in the k-alpha photon state. The relevant part of the interaction Hamiltonian is the emission term, M-e-prime of 1, 3, k, 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.

18:38 - 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 n-k-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 itself, which is one. The second operation converts the n-k-alpha photon state into square-root of n-k-alpha plus one, times the n-k-alpha plus one photon state.

This leads to the projection of the n-k-alpha plus one state onto itself, which is one, times a factor square-root n-k-alpha plus 1.

19:31 - After squaring, we end up with n-k-alpha plus 1, times the magnitude squared of M-e-prime of 1, 3, k, alpha. Plugging in our expression for the M-e-prime coefficient, we obtain the transition rate predicted by Fermi s golden rule.

19:53 - As it did for absorption, the delta function enforces conservation of energy, and we have the same charge-to-mass ratio and geometric factors.

20:06 - The transition rate is proportional to n-k-alpha plus one. The plus one term tells us that even if n-k-alpha equals zero, that is, there are initially no photons present, there is a non-zero probability of emission. This is spontaneous emission.

20:28 - The n-k-alpha term tells us that the more photons are initially present, the larger is the transition rate. This is stimulated emission.

20:39 - 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 n-k-alpha and n-k-alpha plus one factors, the absorption and emission rates are identical. .