Symmetires and Interactions - Lecture 6 1 The meson field The Dirac equation describes the wave function of Fermions (spin 1/2 particles). It is a relativistic equation, naturally containing spin and and also representing anti-particles. In particular, note that it describes a conserved probability density using the Dirac wave func- tions, meaning that Baryon and Lepton number are conserved. However, the first proposed relativistic wave equation was the Klein-Gordon equation (KG). It was initially rejected as it contained both positive and negative energy solutions, did not have a conserved probability current, and when applied to the hydrogen atom, did not correctly predict the atomic fine structure. This equation is now used to represent spin-zero particles (mesons), where meson number is not conserved and anti-particles have the opposity charge to par- ticles. While the equation includes negative energy solutions, these are interpreted as anti-particles simalar to what is done in the Dirac equation. The wave equation is derived by inserting the de Brolie operators for energy and momentum in the relativistic energy equa- tion for a free particle, E2 = (pc2 + (mc2)2) 2 (1/c2) ∂ ψ = 2 ψ (mc/~)2 ψ ∂t2 ∇ − The KG equation is the inhomogeneous Helmholtz (wave) equation, i(~k ~x ωt) and solutions can take the form ψ = Ae · ± . It is identical 1 to the Proca equation (PE) of electrodynamics which represents the time dependent motion of a (non existent) massive photon. The equation in natural units c = h¯ = 1is; Ψ = ψ− + ψ The solution is a superposition of inhomogeneous and homoge- neous terms chosen to satisfy the outgoing boundary conditions. Substitute the wave function above into the scattering equation; 2 2 ( + k )(ψ− + ψ) = Uψ ∇ ( 2 + k2)ψ = Uψ ∇ If a KG particle is placed in an electromagnetic field the momen- tum and energy are replaced by their conjugate values. 2 Exchange Forces In 1935 Hideki Yukawa proposed, based on the electromagnetic in- teraction, that the strong force between nucleons was due to the exchange of a heavy particle which he called a meson. Now the electromagnetic potential function decreases as the inverse distance between two charges (1/r). In this case, the exchanged particle is the zero mass photon as shown in the figure1. If the exchanged particle has mass, the range of the interaction is determined by the uncer- tainty relation, 2 photon exchange _ Q+ Q Figure 1: Virtural photon exchange produces the electomagnetic force ∆E(∆t c) = ~ c = 197 MeV-fm Let ∆E be the meson mass and ∆ tc the range of the force which is 1fm for nucleon interactions. Thus ∆E (meson mass) should be ≈ on the order of 200MeV . ∼ Then recall the equation for the electromagnetic scalar potential in the Lorentz gauge; ∂2φ 2φ (1/c2) = ρ/ǫ ∇ − ∂t2 − In the static limit the solution is; 3 ρ φ = (1/4πǫ) R d x r If the charge density is localized at a point ρ = Q δ(~r) then Q 1 φ = [4πǫ]r 3 The factor Q/(4πǫ) is gives the strength of the coupling (coupling constant) and the potential decreases as 1/r. Now suppose a nucleon field is coupled to a meson field. The Lagrangian density is composed of the meson Lagrangian, the nucleon Lagrangian, and the interac- tion Lagrangian. The nucleon is a Fermion and is a solution of the Dirac equation. The Lagrange density has the form; = + + L LDirac Lmesom Lint The variation of the action wihtout results in the free particle Lint Dirac and mesonic equations, respectively. The meson wave function is a solution of the KG equation and the fermion a solution of the Dirac equation. ∂2φ (1/c2) = 2φ (mc/~)2 φ ∂t2 ∇ − Assume that the meson field moves rapidly in a quasi-static nucleon field as the nucleon is much heavier than the meson. Thus when a variation of the action of the Lagrangian density is taken, the nu- cleon is assumed static. This leads to a time independent interaction 5 term of the scalar form, gsψφψ, or the pseudoscalar form, gpψφγ ψ, where g is the coupling constant. (Here ψ is a solution to the Dirac equation and φ is the potential obtained from the KG equation.) The coupling constant, g, is equivalent to the charge in electromagentic coupling. Averaging over the nucleon wave function, the interaction term in this case acts as a source of the mesons, ie in analogy to an effective “mesonic charge” density. The solution results in a form identical to the Proca equation, the classical equation for an electro- 4 magnetic potential with a non-zero mass term. The static limit of this equation takes the form; 2φ (mc/~)2 φ = ρ. ∇ − This equation has a solution ; µr e− φ = q r where µ = mc2/~c Therefore the nucleon-nucleon interaction is obtained by the ex- change of a virtural meson(s) in the same way that the Coulomb interaction is due to the exchange of a vitrural photon. In terms of a relativistic field theory, a nucleon creates a mesonic field which can then interact with other hadronic particles. 3 Lagrangian The Lagrange density, which is a Lorentz scalar, is composed of bi- linear forms composed of solutions to the particle equations so that a scalar quantity proportional to energy is obtained. Thus for example a vector solution must be contracted (dot product) with a another vector, or in the case of mesons, a pseudo-scalar must be multipled by a pseudoscalar, as observed in the last section, and detailed below. Recall that a Dirac wave function has 4-components, and that ~γ, ~α 5 Bilinear Form Transformation Property ψ ψ Scalar ψγn ψ Vector ψγ5 ψ Pseudoscalar ψγ5γn ψ Pseudovector and β are the 4 4 matricies used in the Dirac equation. As an × aside note that the current ~j = cψ~αψ leads to an expectation value of the velocity. Write the following bilinear forms which have the transformation properites listed below. These forms are then coupled with another field with similar trans- formation properties in order to form a Lorentz scalar. This is then inserted in the Lagrangian. For example, in the section above it was assumed that the interacting Dirac field was either a scalar or a pseu- doscalar coupled to a meson which had either a scalar or pseudoscalar transformational properties. 4 Noether’s Theorem Noether’s theorem states that for every continuous variation of the action which leaves it invariant, There is a conserved current (quan- tity). Einstein remarked that Emmy Noether was the “most signifi- cant mathematician of all time”. Her work, particularly this theorem, has been the foundation of all modern physics theory. 6 For the Lagrange density (φ (x ), ∂ φ (x )), where φ (x ) are L k u u k µ k µ fields, The variation of action is; b δs = 0 = δ R dtd3x a L 3 ∂ ∂ 0 = R dtd x[ δφk + δ(∂µφk)] ∂φLk ∂(∂Luφ) Integrate by parts and note the integral over time is between fixed end points. The Euler-Lagrange equation results. ∂ ∂ = ∂µ[ ] ∂φLk ∂(∂µLφk) Then as an example suppose a displacement, x x + ǫ . The µ → µ µ xµ and ǫµ are 4 vectors. This is substituted into the Euler-Lagrange equation. The result becomes; ∂ ǫµ∂µ = ǫnu∂µ[ L ∂nuφk] L ∂(∂µφP k) ∂ Define the tensor, Tµnu = gµν + L ∂nuφk so that; − L ∂(∂µφk) ptµTµν = 0 The tensor defines a momentum density = T and a 4-momentum, Pµ 0µ P = R d3x which has a conderved current. µ Pµ 7 5 Symmetry breaking There are many examples of broken symmetry. Perhaps the most familiar is symmetry breaking due to the application of a magnetic field in a Hamiltonian. To begin, assume that the Hamiltonian, H0, is rotationally symmetric. In this case the particle angular momen- tum, L~ , commutes with H0 and the particle energy is independent of spatial rotations. However, suppose that a magnetic field is ap- plied. This field interacts with a magnetic moment ~µ = (q/2mc)L~ so that, E = ~µ B~ . L~ does not commute with the full Hamilto- int − · nian. Thus the energy depends on the value of L~ . In this case, the symmetry breaking term could be small compared to the symmetric term, so that the symmetry breaking term adds only a small change in energy to include the asymmetry. 6 Spontaneous symmetry breaking Spontaneous symmetry breaking is different than the symmetry break- ing described in the last section. In spontaneous symmetry breaking the dynamics of the interaction break the symmetry in the system ground state. The classical example is a ferromagnet. All spatial directions of an isolated ferromagnet are equilivent, but the minium potential energy of the isolated, local ferromagnetic can be displaced from zero but it remains rotationally invarient, even though in its ground state the ferromagnetic field is polarized in some arbitrary direction. This is illustrated in figure 2. In a ferromagnetic sys- tem the global interaction of all the ferromagnetics selects a state of 8 minimum energy which is no longer locally invarient. However, as the energy of the system is increased, rotational symmetry can be restored as the individually aligned atoms break their ferromagnetic bonds. Consider a Lagrangian density for a complex scalar field, φ = (1/√2)(φ1 + φ2); 2 = ∂ φ†∂ φ m φ†φ L µ µ − Now for the static case the potential energy term has the form; 2 ~ φ† ~ φ + m φ†φ ∇ · ∇ This has a minimum when φ1 = φ2 = 0. Then consider the fol- lowing functional form for the mass term which is inserted in the Lagrange density 2 2 2 = ∂ φ†∂ φ (m /2φ )[φ†φ φ ] L µ µ − 0 − 0 At φ, the energy is not zero but decreases as φ φ where it van- → 0 ishes.