 So far in this series, we have developed quantum field theories for the electron in a hydrogen atom and for photons in the radiation field. To combine these, we need a description of how these fields interact. This should take the form of an interaction Hamiltonian, H-hat interaction. To find this, we will consider the interaction between a charged particle and the electromagnetic field, as described by classical electrodynamics. To do this requires us to apply the methods of Lagrangian and Hamiltonian mechanics. In this slide, we will look at a very brief overview of this formalism. For motivation, we will consider the simple problem of a mass falling in a gravitational field. Let q be the height of the mass above the ground. q dot, the time rate of change of q, is the upward velocity, and gravity pulls downward with a constant force F. Newton's law, mass times acceleration equals force, becomes mq double dot equals minus mg. Where m is the mass, q double dot is the upward acceleration, and g is the gravitational acceleration. In Lagrangian mechanics, we generate expressions for the kinetic energy k and the potential energy u. k equals one-half mass times velocity squared, one-half q dot squared. u equals mgq, force times height above the ground. Then, we form the function l equals k minus u, called the Lagrangian. The equation of motion is then given by the Euler-Lagrange equation. The time derivative of the derivative of l with respect to velocity equals the derivative of l with respect to position. l depends on q through the term minus mgq. So the derivative of l with respect to q is minus mg, the right side of Newton's equation. l depends on q dot through the term one-half mq dot squared. So the derivative of l with respect to q dot is mq dot, and the time derivative of that is mq double dot, the left side of Newton's law. Lagrangian mechanics does not introduce any new physics. It is simply an alternate mathematical formulation of Newtonian mechanics. Our example problem is so simple that Lagrangian mechanics is redundant. We can simply form and solve the Newtonian equations directly. However, in more complicated problems, especially those with many variables in complex geometries, it can be very difficult to develop and solve the Newtonian equations. But, very often it is much simpler to write the expressions for the kinetic and potential energies, and form the Lagrangian and evaluate and solve the Euler Lagrange equations. In Hamiltonian mechanics, we denote by p, the derivative of the Lagrangian with respect to velocity, and call this the conjugate momentum. For our example problem, this is simply mq dot, the mechanical momentum of the particle. We solve for q dot equals p over m, and form the Hamiltonian, h equals pq dot minus the Lagrangian. We substitute for q dot in terms of p to arrive in an expression for h in terms of p and q, p squared over 2m plus mgq. This is the kinetic energy plus the potential energy, which is the total energy. The Hamiltonian equations of motion are q dot equals the derivative of h with respect to p. In our example, this is q dot equals p over m, and p dot equals minus the derivative of h with respect to q. p dot is mq double dot, and minus the derivative of h with respect to q is minus mg. So again, we arrive at the Newtonian equation mq double dot equals minus mg. The Hamiltonian plays a central role in quantum mechanics. In our application so far, throughout the quantum mechanics series, and up to this point in the quantum field series, we could write down the Hamiltonian by inspection. But for a charged particle interacting with an electromagnetic field, it is not obvious how to do this. We need to work through the Lagrangian and Hamiltonian formalism to discover the form the Hamiltonian takes for this problem. So let's do that. We have a point particle with charge q at position x, with velocity v equals x dot, and electric and magnetic fields e and b. The electric force, Fe, is parallel to the electric field. The magnetic force, Fm, is perpendicular to both the magnetic field and the velocity. As we discussed in video three of this series, the total force on the charge is q times the quantity e plus v cross b. Writing mass times acceleration, mx double dot, equals this force, gives us the equation of motion. Here, x boldface is the vector position with components x, y, and z. Let's look at the first component of this vector equation. mx double dot equals q times quantity ex plus y dot bz minus z dot by. In video three, we show that the electric and magnetic fields can be derived from a single vector potential a. e equals minus time derivative of a, and b equals curl of a. From these relations, ex equals minus ax comma t, where the comma t notation is a shorthand for the time derivative. By equals ax comma z minus az comma x, and bz equals ay comma x minus ax comma y. Therefore, the first component of our equation of motion can be written d by dt of mx dot equals minus q ax comma t. Plus q times the quantity y dot ay comma x minus y dot ax comma y minus z dot ax comma z plus z dot az comma x. Here's that expression again. We want to get this into the Lagrangian equation of motion form. d by dt of derivative of l with respect to x dot equals derivative of l with respect to x. The left side has the correct form, but on the right there are three terms that are derivatives of coordinates other than x. To fix this, let's consider d by dt of ax. This equals ax comma t plus ax comma xx dot plus ax comma y, y dot, plus ax comma z, z dot. A subtle point here is that this expression has two types of time derivatives, d by dt ax and ax comma t. The first is the total derivative of ax with respect to time, and the second is the partial derivative of ax with respect to time. The left side represents how the ax value seen by the particle changes with time. The four terms on the right represent different ways this can happen. First, at a fixed point in space, ax can be changing with time. That's the ax comma t term. The partial derivative of ax with respect to t, assuming no change in the spatial coordinates. Another way the particle can see ax changing with time is if ax changes in space and the particle moves through space. Those are the last three terms. ax comma x represents how fast ax changes with the x-coordinate, and x dot represents how fast the particle's x-coordinate is changing with time, and similarly for the next two terms. We solve this equation for ax comma t and substitute into the top equation. This gives us d by dt of mx dot equals q times, in parentheses, all for expression for ax comma t, plus the remaining terms from the right side of the top equation. We will combine the d by dt terms and put them on the left side. We also see that four of the terms on the right sum to zero. This leaves us with d by dt of quantity mx dot plus qax equals q times quantity x dot ax comma x plus y dot ay comma x plus z dot az comma x. Here's that expression again. This has the form of the Euler Lagrange equation of motion. On the left we have d by dt of an expression we can interpret as the derivative of l with respect to x dot, and the right side can be interpreted as the derivative of l with respect to x. Identifying the derivative of l with respect to x dot as mx dot plus qax, and denoting this as the conjugate momentum px, we can solve for x dot equals px minus qax over m. And we can identify the Lagrangian as l equals one-half m times quantity x dot squared plus y dot squared plus z dot squared plus q times quantity x dot ax plus y dot ay plus z dot az. For the derivative of l with respect to x dot, the one-half mx dot squared term gives us mx dot and the qx dot ax term gives us qax. And the derivative of l with respect to x gives us the required right-hand side of the equation of motion. Given l, we can form the Hamiltonian. h equals px x dot plus py y dot plus pz z dot minus l. Let's look at the x dot terms. For px x dot, we substitute for x dot or above expression in terms of px and ax. Minus l gives us a minus one-half mx dot squared term into which we substitute for x dot and a minus qax x dot term into which we also substitute for x dot. After a bit of algebra, we end up with quantity px minus qax squared over 2m. In addition to this, h will have corresponding terms in py ay and pz az. For a free particle, the Hamiltonian is h par equals p squared over 2m. Our derivation shows that if the particle has charged q and an electromagnetic field a as present, then the Hamiltonian becomes quantity p minus qa squared over 2m. This represents the sum of the particle and interaction Hamiltonians. If the particle moves in a potential u, the resulting Hamiltonian due to the charged field interaction is the same with the addition of a u term. Let's expand this expression. We have 1 over 2m times a sum of three terms, quantity px minus qax squared, etc. plus the potential u. Multiplying out the binomials, we get 1 over 2m times quantity px squared plus py squared plus pz squared plus u minus q over m times quantity ax px plus ay py plus az pz plus q squared over 2m times quantity ax squared plus ay squared plus az squared. We identify the first two terms as the original particle Hamiltonian. The remaining terms are therefore the interaction Hamiltonian. This interaction Hamiltonian is the expression we have been seeking that when added to our separate electron and photon quantum field theories will give us our desired single interacting theory.