130_notes.dvi

Note that the variational calculation still uses first order perturbation theory. It just adds a variable parameter to the wavef ...

shift up in energy. We see that the atomic shells fill up in the order 1s,2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p ...

Molecules canrotatelike classical rigid bodies subject to the constraint that angular momentum is quantized in units of ̄h. Erot ...

In a gauge in which∇·~ A~= 0, the perturbation is V= e mc A~·~p+ e 2 2 mc^2 A^2. For most atomic decays, theA^2 term can be negl ...

considered. Applying periodic boundary conditions in a cubic volumeV, the integral over final states can be done as indicated be ...

This can be used to easily compute decay rates for Hydrogen, for example the 2p decay rate. Γ 2 p→ 1 s= 4 αωin^3 9 c^2 ∣ ∣ ∣ ∣ ∣ ...

This is a field that falls off much faster than^1 r.A massive scalar field falls off exponentially and the larger the mass, the ...

H = (Fμ 4 ) ∂Aμ ∂x 4 −L = 1 2 (E^2 +B^2 ) if there are no source terms in the region. Gauge symmetry may be used to put a condit ...

Pk,α=− iω c (ck,α−c∗k,α) for the harmonic oscillator at each frequency. We assume that a coordinate and its conjugate momentum h ...

States of the field are given by the occupation number of each possible photon state. |nk 1 ,α 1 ,nk 2 ,α 2 ,...,nki,αi,...〉= ∏ ...

To scatter a photon the field must act twice, once to annihilate the initial state photon and once to create the final state pho ...

infinite although it can be rendered finite if we accept the fact that out theories are not valid up to infinite energies. The q ...

self energy correction, at least when relativistic quantum mechanics is used. Our non-relativistic calculation gives a qualitati ...

Define the4 by 4 matricesγμare by. γi = ( 0 −iσi iσi 0 ) γ 4 = ( 1 0 0 − 1 ) With this definition, the relativistic equation can ...

This“Schr ̈odinger equation”, derived from the Dirac equation, agrees well with the one we used to understand the fine structure ...

Eis positive for solutions 1 and 2 and negative for solutions 3 and 4. Thespinors are orthogonal u(~pr)†u(r ′) ~p = |E| mc^2 δrr ...

backward in time) into the conventional positive exponent solution ifwe change the charge to +e. We can interpret solutions 3 an ...

Classification Covariant Form no. of Components Scalar ψψ ̄ 1 Pseudoscalar ψγ ̄ 5 ψ 1 Vector ψγ ̄μψ 4 Axial Vector ψγ ̄ 5 γμψ 4 ...

and the current we have found for the Dirac equation, the interaction Hamiltonian is. Hint=ieγ 4 γkAk This is simpler than the n ...

The Dirac equation is invariant under charge conjugation, defined as changing electron states into the opposite charged positron ...