Particle Physics Dirac Equations (cont), Lecture Notes - Physics, Study notes of Particle Physics

Download Particle Physics Dirac Equations (cont), Lecture Notes - Physics and more Study notes Particle Physics in PDF only on Docsity! 129 Lecture Notes More on Dirac Equation 1 Ultra-relativistic Limit We have solved the Diraction in the Lecture Notes on Relativistic Quan- tum Mechanics, and saw that the upper (lower) two components are large (small) in the non-relativistic limit for positive energy solutions. They are switched for negative energy solutions which represent (the absence of) the anti-particle. It is also useful to consider the ultra-relativistic limit E  m (or m→ 0) in order to discuss high-energy processes and massless neutrinos. We start with the Dirac equation ih̄ ∂ ∂t ψ = Hψ = [c~α · ~p+mc2β]ψ. (1) We solved it using the two-component eigenvectors of ~σ · ~p ~σ · ~pχ+(~p) = ~σ · ~p ( cos θ 2 sin θ 2 eiφ ) = +|~p|χ+(~p), (2) ~σ · ~pχ−(~p) = ~σ · ~p ( − sin θ 2 e−iφ cos θ 2 ) = −|~p|χ−(~p). (3) The positive energy solutions are ψ(~x, t) = u±(p)e −ipµxµ/h̄, u+(p) =  √E+mc22mc2 χ+(~p)√ E−mc2 2mc2 χ+(~p)  , u−(p) =  √E+mc22mc2 χ−(~p) − √ E−mc2 2mc2 χ−(~p)  , (4) while the negative energy solutions are ψ(~x, t) = v±(p)e +ipµxµ/h̄, v+(p) =  √E−mc22mc2 χ+(~p)√ E+mc2 2mc2 χ+(~p)  , v−(p) =  −√E−mc22mc2 χ−(~p)√ E+mc2 2mc2 χ−(~p)  . (5) It is easy to check that they are helicity eigenstates. We have seen before that the total angular momentum is given by ~J = ~x × ~p + h̄ 2 ~Σ, which is conserved, while the orbital and spin angular momenta are individually not 1 conserved. In order to seperate the spin degree of freedom, it is useful to project it along the direction of the momentum, ~p · ~J = ~p · h̄ 2 ~Σ. (6) The orbital angular momentum drops out and we single out the spin. The helicity is defined by h = ~p · ~J |~p| = h̄ 2 ~p · ~Σ |~p| . (7) The positive helicity +h̄/2 is called “right-handed,” while the negative he- licity −h̄/2 “left-handed.” It is important to note that the helicity is frame- dependent if the particle has a (not matter how small) a finite mass. If the mass is finite, in principle you can go faster than it an look back. The mo- mentum appears to go the opposite direction in your rest frame, while the spin remains the same. Therefore, you would observe the opposite helicity. On the other hand, if the particle is massless, you can never pass it, and hence the helicity is frame-independent. Now we take the ultra-relativistic limit E  m. Up to the normalization factor of √ E/mc2, we find u+(p) ∝ ( χ+(~p) χ+(~p) ) , u−(p) ∝ ( χ−(~p) −χ−(~p) ) , (8) v+(p) ∝ ( χ+(~p) χ+(~p) ) , v−(p) ∝ ( −χ−(~p) χ−(~p) ) . (9) They have simplified dramatically. In particular, by introducing a matrix γ5 = ( 0 I I 0 ) , (10) we find they are all eigenstates of this matrix, γ5u±(p) = ±u±(p), γ5v±(p) = ±v±(p). (11) The eigenvalue of γ5 is called “chirality,” originating from a Greek word that means “hand.” The name suggests that it is closely related to the handedness, namely the helicity of the particle. Indeed, what we see here is that, in the limit of E  m or massless limit, the chirality and the helicity are in one-to-one correspondence. 2 Here I used the fact that the derivative also changes the sign under this substitution. The question is to bring this back to the form as close as possible to the original equation. Thanks to the anti-commutation property of gamma matrices, it is easy to see that γ0γiγ0 = −γi. Therefore,( iγ0(∂0 + i e c A0(−~x, t)) + iγ0γiγ0(∂i − i e c Ai(−~x, t))−mc ) ψ(−~x, t) = 0. (26) By multiplying the equation by γ0 from the left,( iγ0(∂0 + i e c A0(−~x, t)) + iγi(∂i − i e c Ai(−~x, t))−mc ) γ0ψ(−~x, t) = 0. (27) Therefore, γ0ψ(−~x, t) almost satisfies the same equation as before, except the sign of the vector potential. Indeed, we have forgotten to change the sign of the vector potential! Under parity, the electric field should change its sign, while ~E = −~∇φ− ~̇A. Therefore the vector potential should change the sign under the parity. It is also consistent with the axial-vector nature of the magnetic field, ~B = ~∇ × ~A which changes the sign twice. Now that we remember this, the correct parity-transformed Dirac equation is( iγ0(∂0 + i e c A0(−~x, t)) + iγi(∂i + i e c Ai(−~x, t))−mc ) γ0ψ(−~x, t) = 0, (28) which is of exactly the same form as the original equation. In summary, the parity transformation is given by ψ(~x, t) → γ0ψ(−~x, t), (29) A0(~x, t) → A0(−~x, t), (30) Ai(~x, t) → −Ai(−~x, t). (31) The immediate consequence of this is the parity eigenvalues of the particle and anti-particle at rest. For the particle at rest ~p = 0, u±(p) are non- vanishing only in the first two components, while for the anti-particle at rest we use the negative-energy solutions v±(p) which are non-vanishing only in the lower two components. Because γ0 = diag(1, 1,−1,−1), it follows that the particle at rest has parity eigenvalue +1, while the anti-particle at rest has parity eigenvalue −1. Of course, the distinction between particle and anti-particle is a convention. If you change your convention that you call positrons as particles and electrons anti-particles, even though it is awfully 5 inconvenient (we live in the world of anti-matter in that case!), the positron has the even parity while the electron odd. It may sound troublesome that the parity eigenvalue is convention dependent, but it turns out the physical quantities won’t. To do the reality check of the parity eigenvalue we just found, let us consider the charged pion π+ = ud̄. There is no orbital angular momentum L = 0, and the spins are anti-aligned S = 0 to form spin-zero meson. Under parity, u, being a particle, returns a positive sign, while d̄, being an anti- particle, returns a negative sign. Therefore the pion has the odd parity 0−, as we talked about already several times. Note that even if you switch to the other convention, you still get the overall minus sign. The spin parallel counter part ρ+ = ud̄ also has the same odd parity: 1−. In general, if the meson has the orbital angular momentum L between the quark and the anti-quark, it gives the parity (−1)L from the spherical harmonics. In addition, either quark or anti-quark, depending on your con- vention, gives the minus sign, and hence the parity eigenvalue of the meson is (−1)L+1. Using the parity and the spin of the meson, you can figure out the orbital angular momentum. For example, when L = 1, S = 0, the spin must be J = 1, while when S = 1 for the same L, the spin can be J = 0, 1, 2 (addition of the angular momentum 1 and 1). All of them must share the same positive parity. Indeed, the I = 1 mesons b1(1235) are the L = 1, S = 0 case, and a0(1450), a1(1260), and a2(1320) are the L = 1, S = 1 cases. They are all parity even. Look at http://pdg.lbl.gov/2002/quarkmodrpp.pdf for a table of mesons, their quantum numbers, and their spectroscopic clas- sifications. It is also important to note that γ5 is parity-odd, and hence pseudo-scalar. It follows simply because γ5 anti-commutes with γ 0. 4 Charge Conjugation Another important symmetry is the interchange of particles and anti-particles. The world of anti-matter would look just the same as ours, made up of anti-nucleons and positrons forming anti-atoms. Indeed, Athena experiment at CERN (http://athena.web.cern.ch/athena) has produced many anti- hydrogen atoms. The overall switch of particles and anti-particles is called “charged conjugation” C, because that would flip the charges such as electric charge, strangeness, baryon number and so on. 6 What we want to do then is to find a way to change the Dirac equation such that the positive- and negative-energy solutions are interchanged while the form of the equation remains the same. The good guess is that it involves a complex conjugation so that e−iEt/h̄ becomes e+iEt/h̄ and vice versa. Given this guess, let us take the complex conjugate of the whole equation,( −iγµ∗(∂µ − i e c Aµ(~x, t))−mc ) ψ∗(~x, t) = 0. (32) The obvious problem is that −iγµ∗. We have to bring them back to iγµ. Recall that γ0,1,3 are all real, while γ2 is imaginary. The only one that changes its sign under the complex conjugation is therefore γ2. Using the anti-commutation of gamma matrices, we can write the general equation (iγ2)γµ∗(iγ2) = −γµ. (33) Note that iγ2 is hermitean and unitarity. By multiplying the equation by (iγ2) from the left, we find( iγµ(∂µ − i e c Aµ(~x, t))−mc ) iγ2ψ∗(~x, t) = 0. (34) Therefore the equation is the same as before, except that the sign of the electromagnetic potential is the opposite. Well, under the charge conjugation, we should have also flipped the sign of the electromagnetic potential! Doing so, the Dirac equation maintains exactly the same form as before. In other words, the electromagnetic potential is odd under the charged conjugation, and the photon has the odd eigenvalue under C. You can also check that the positive- and negative-energy solutions are interchanged by ψ(~x, t) → iγ2ψ∗(~x, t). It is interesting to apply the charge conjugation to mesons. Some particles are eigenstates of the charge conjugation, namely that the particle and its anti-particle are the same. The photon was indeed a good example, C|γ〉 = −|γ〉. In order for this to be possible the particle must be neutral under all charges. Another such example is π0. Its wave function consistent with I = 1 and S = 0 is |π0〉 = | 1√ 2 (uū− dd̄) 1√ 2 (↑↓ − ↓↑)〉 = 1 2 |u↑ū↓ − u↓ū↑ − d↑d̄↓ + d↓d̄↑〉. (35) Under the charge conjugation, it becomes C|π0〉 = 1 2 |ū↑u↓ − ū↓u↑ − d̄↑d↓ + d̄↓d↑〉. (36) 7