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Lecture 6: Symmetries I

Lecture 6: Symmetries I. Symmetry & Unifying Electricity/Magnetism Space-Time Symmetries Gauge Invariance in Electromagnetism Noether’s Theorem Isospin Parity. Useful Sections in Martin & Shaw:. Sections 5.3, 6.1, Append C.1, C.2. Symmetry. When something is changed,.

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Lecture 6: Symmetries I

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  1. Lecture 6:Symmetries I • Symmetry & Unifying Electricity/Magnetism • Space-Time Symmetries • Gauge Invariance in Electromagnetism • Noether’s Theorem • Isospin • Parity Useful Sections in Martin & Shaw: Sections 5.3, 6.1, Append C.1, C.2

  2. Symmetry When something is changed, something else is unchanged

  3. Symmetry & Electromagnetism +  Symmetry: The effect of a force looks the same when viewed from reference frames boosted in the perpendicular direction +  F (pure magnetic) +  +  In Frame of Test Charge +  + +   +q F (pure electrostatic) + UNIFICATION !! (thanks to Lorentz invariance)  + Lorentz expanded Lorentz contracted Lab Frame I v +q B  Electricity & Magnetism are identically the same force, just viewed from different reference frames

  4. Symmetry & Electromagnetism +  Symmetry of Maxwell’s Equations +  F (pure magnetic) +  +  In Frame of Test Charge +  + +   +q F (pure electrostatic) + Relativity !!  + Lorentz expanded Lorentz contracted Lab Frame I v +q B

  5. F m1 m2 x1 x2 | | ¨ force felt at x2 : F = m2 x2 ¨ recoil felt at x1 :F = m1x1 ¨ ¨ subtracting: m2 x2 + m1x1 = 0 ˙ ˙ d/dt [ m2 x2 + m1 x1 ] = 0 Translational Invariance  Conservation of Linear Momentum Space-Time Symmetries Translational Invariance If the force does not change when translated to a different point in space, then m2 v2 + m1 v1 = constant

  6. Time Invariance assume this basic description also holds at other times 1 2 E = m x2 + V dE dV dx dt dx dt = m x x + dV dx = m x x + x butdV/dx = F = mx (Newton’s 2nd law) dE dt  = m x x m x x = 0  E =constant Time Invariance  Conservation of Energy Consider a system with total energy

  7. Gauge Invariance in EM ''local" symmetry  (x,t) t    E = ∇A ∇[ x,tA + ∇(x,t) t t t  = ∇A = E t Gauge Invariance Conservation of Charge Gauge Invariance in Electromagnetism: A A + ∇(x,t) B = ∇A ∇ [A + ∇(x,t)] = ∇A= B

  8. Necessity of Charge Conservation To see this, assume charge were not conserved (Wigner, 1949) So a charge could be created here by inputing energy E And destroyed here, with the output of some energy E PoP ! PoP ! q x1 x2 E = q(x1) E = q(x2) So, to preserve energy conservation, if is allowed to vary as a function of position, charge must be conserved Thus we will have created an overall energyE E = q { (x2) (x1) }

  9. Noether’s Theorem Noether’s Theorem Continuous Symmetries  Conserved ''Currents" (Emmy Noether, 1917)

  10. Importance of Gauge Invariance  eiq Take the gauge transformation of a wavefunction to be where  is an arbitrary ''phase-shift" as a function of space and time ∂  i  ∇2 ∂t 2m here’s the problem! /t  = i ( E +q + q /t )  Gauge invarianceREQUIRESElectromagnetism !! Gauge symmetry from another angle... Say we want the Schrodinger equation to be invariant under such a transformation clearly we’re in trouble ! Consider the time-derivative for a simple plane wave:  = Aei(px-Et) /t  = i ( E + q /t )   Aei(pxEt+q) Note that if we now introduce an electric field, the energy level gets shifted by q But we can transform /t, thus cancelling the offending term! (a similar argument holds for the spatial derivative and the vector potential)

  11. Gauge Invariance and Gravity Require an interaction to make this work GRAVITY! Another example... Invariance with respect to reference frames moving at constant velocity Special Relativity:  global symmetry Generalize to allow velocity to vary arbitrarily at different points in space and time (i.e. acceleration)  local gauge symmetry

  12. Chicken or egg ? All known forces in nature are consequences of an underlying gauge symmetry !! or perhaps Gauge symmetries are found to result from all the known forces in nature !!

  13. Symmetry and pragmatism True even for ''approximate" symmetries ! Pragmatism: Symmetries (and asymmetries) in nature are often clear and can thus be useful in leading to dynamical descriptions of fundamental processes

  14. )  n p )  n p proton & neutron appear to be swapping identities  ''exchange force" Isospin Note that mp = 938.3 MeVmn = 939.6 MeV Isospin Both are found in the nucleus, apparently held to each other by pions This means, to conserve charge, pions must come in 3 types: q = -1, 0, +1 So there appears to an ''approximate" symmetry here

  15. Assignment of Isospin pn I3 : 1/2 -1/2 I=1/2 System (just 2 states) Impose isospin conservation I3(+) = +1 So we can think of these particle ''states" as the result of a (continuous) ''rotation" in isospin-space Noether’s theorem says something must be conserved... Call this ''Isospin" in analogy with normal spin, so the neutron is just a ''flipped" version of the proton Some way pions can be produced p + p  p + n + +  p + p +  I3() = 0  p + p +  +  I3() = -1 I=1 for the pions(similar arguments for other particle systems)

  16. Isospin Example + p n   Example: What are the possible values of the isotopic spin and it’s z-component for the following systems of particles: a) + + p b)  + p p : I = 1/2, I3 = +1/2 + :I = 1, I3 = +1 a) total I3 = 1 + 1/2 = 3/2 thus, the only value of total Isospin we can have is also I = 3/2  : I = 1, I3 = 1 p :I = 1/2, I3 = +1/2 b) total I3 = 1 + 1/2 = 1/2 thus, possible values of total Isospin are: I = 1  1/2 = 1/2 or I = 1  1/2 = 3/2

  17. y consider the scattering probability of the following: m1 m2 x y m1 m2 x y m1 m2 x So, even though x and dx/dt are each odd under parity, the scattering probability, PS, is even (i.e.PPS = PS) Parity P F(x) = F(-x) Parity discreet symmetry(no conserved ''currents") P x = x P dx/dt =  dx/dt parity is multiplicative, not additive

  18. More on Parity Also note that parity does not reverse the direction of spin! +z z +z z flip z flip ''velocity" direction z +z z +z (stand on your head) parity But, for orbital angular momentum in a system of particles, it depends on the symmetry of the spatial wave function!! +z z z z flip x & y positions flip z flip ''velocity" direction z +z +z +z parity +z not the same ! (stand on your head) z

  19. Intrinsic Parity of the Photon thus, we must havePE(x,t) = E(-x,t) for Poisson’s equation to remain invariant But alsoE = ∇A/t = A/t (in absence of free charges) and since /t doesn’t change the parity PA(x,t) = A(x,t) ButAbasically corresponds to the photon wave function: A(x,t) = N (k) exp[i(kxt)] Thus, the intrinsic parity of the photon is 1(or  = 1 ) Intrinsic parity of the photon from ''first-principles": ∇ E(x,t) = (x,t)/0 P(x,t) = (x,t) P∇ = ∇

  20. Effective Parity of the Photon l P =  (1) (i.e. radiation could be s-wave, p-wave etc.) However,the effective parity depends on the angular momentum carried away by the photon from the system which produced it: but for an isolated photon, this cannot be disentangled!!

  21. Parity Example The (547) meson has spin 0 and is observed to decay via the electromagnetic interaction through the channels:  +  + 0 0 + 0 + 0 and From this, deduce the intrinsic parity of the  and explain why the decays: are never seen  +  and 0 + 0  1 L12 L3 L3 = ()3(1) (1) L12  2 { } L3 =()3 (1) Example: P = P P P but final state must have zero total angular momentum since the initial state has spin 0  L12 = L3 Ltot = L12 + L3 = 0 = ()3 = 1 = ()3 P = ()2 (1)L However, for 2-pion final states we would have: P = ()2 = 1 and is thus forbidden but we must have L=0, so

  22. l pn = pn (1)  so l must be even Byconvention,p = n +1 & alsoe- +1 Evidence for Parity Conservation Parity Conservation & Assignment (in strong/electromagnetic interactions) 1)Polarized protons scattering off a nucleus show no obvious asymmetry towards spin-up vs spin-down directions 2)Ground state of deuteron (np) has total angular momentum J=1 and spin S=1. Thus, the orbital angular momentum could take on values of l= 0 (m=1), l= 1 (m=0) or l= 2 (m = 1) But the observed magnetic moment is consistent with a superposition of only S and D waves (l=0, 2). This can be reconciled if P(p+n) = P(d) and the relative parities of the other particles then follow (anti-fermions have the opposite parity, anti-bosons have the same parity)

  23. Basic Approach to Parity Weak interaction violates parity !! Parity is a different animal from other symmetries in many respects... It is often impossible to determine the absolute parity (assigned +1 or -1) of many particles or classes of particles. So we essentially just assume that basic physical processes are invariant with respect to parity and construct theories accordingly, making arbitrary assignments of parity when necessary until we run into trouble. (so, ''left" and ''right" really matter... weird!)

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