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The Dirac Equation: Introduction The most natural candidate for a relativistic wave equation is the Klein-Gordon equation: 2 2 t 2 2 2 m2 c 4 or equivalently H 2 P 2 m2 (1.1) With the usual operator definitions of H i 0 and Pk i k , this is equivalent to E 2 P 2 m 2 . But just as this equation has the two roots E P2 m2 , equation (1.1) yields two types of solution, one for positive E and one for negative E(i.e., Ae ip x Ae i ( Et p x ) . The negative energy solutions appear to be unphysical, but we cannot simply exclude them because we must require that the solutions of (1.1) form a complete set. The unwanted negative-energy solutions are produced because this is a second order equation in / t . To eliminate them, Dirac tried to find an equation first-order in / t like the Schrodinger equation, but first-order in space as well as in time so that space and time coordinates are on the same footing. He tried i or equivalently k mc 2 k t i x H = i P i m (1.2) (1.3) But if this is to be a relativistically invariant equation, then operating with H again on (1.2) must reproduce (1.1)b. In order for this to happen, the cross-terms produced by squaring (1.3) must vanish, so the k and must satisfy the following properties: (b) i2 2 1 j j i 0 (for i j) (c) i i 0 (a) (1.4) Equation (1.4)b tells us that i and cannot be ordinary numbers, they must be matrices in order to anticommute. (Consequently the “1” in equation (1.4)a must be interpreted as the identity matrix.) Since these matrices operate on , must be a column matrix. This tells us that has multiple components which is encouraging since we know that the electron has two spin states and we are used to representing these as column matrices. From (1.3), the i and must be Hermitian matrices, like H. From (1.4)a they must have eigenvalues of 1 . Equations (1.4)b and c tell us that these matrices must be traceless. The trace of a matrix is the sum of its diagonal elements [ Tr(A) Tr(Ai j ) Aii , with implied summation over i]. The trace of a product of matrices is invariant under cyclic permutations of the matrices, e.g. Tr(ABC)=Tr(CAB). (Proof: Tr ( ABC ) Tr ( Ai j B j k C k l ) Ai j B j k C k i C k i Ai j B j k Tr (CAB) ). So starting from (1.4)c, for example, we can right-multiply both sides by and take the trace to get (using 1 ): 1 Tr ( i ) Tr ( i ) Tr ( i ) Tr ( i ) (1.5) requiring Tr ( i ) 0 . A similar operation, but right-multiplying by i , shows Tr ( ) 0 . Since the trace of a hermitian matrix is just the sum of its eigenvalues, and since these eigenvalues must be 1 for the and matrices, the dimension N of these matrices must be even. The obvious choice is N=2, especially since we know the electron has two spin components. The Pauli matrices are the clear choice for the i ’s: 0 1 0 i 1 0 , 2 , 3 1 0 i 0 0 1 1 (1.6) These satisfy the -parts of equations (1.4), but the only linearly independent matrix satisfying (1.4)a is the unit matrix, and this is not traceless. There is no set of 2 2 matrices that will work. The next possibility is N=4, and here we find success. The canonical choice is 0 i i i 1 0 , 0 0 1 (1.7) where each matrix element is a 2 2 matrix. You should check that these do in fact satisfy (1.4). Current Conservation If the Dirac Equation is to be the equation for an electron (or fermion, in general), then we must be able to define a conserved current from it. To create this, we follow the same procedure that we used to produce a conserved current from the Schrodinger Equation. First we take the hermition conjugate of equation (1.3): † † (1.8) i k k † mc 2 t x i Note that since is a 4-component column vector, † is a 4-component row vector, † 1* , *2 , *3 , *4 . Now left-multiply (1.8) by , right-multiply (1.2) by , and subtract. This gives † † i † † k k k k t t i x x † k † k t x (1.9) Identifying † and J k † αk then equation (1.9) becomes the equation for current conservation: J 0 t (1.10) (1.11) This is a very significant result, and must have been critical in indicating to Dirac that he was on the right track. 2 Spin ½ If an operator is conserved, then it must commute with the Hamiltonian, since d i , H dt But using the Dirac Equation in the form of (1.3), with L R P , we find L, H R P, H k R P kˆ, i Pi kˆ k R , Pi P i kˆ i P i kˆ P i P k i i k i (1.12) (1.13) i so orbital angular momentum is not conserved! But if we define spin angular momentum in the most natural way, as 0 (1.14) 20 then , H i P (1.15) So in general, L and are not separately conserved, but the total angular momentum J L is conserved. This is exactly the behavior we expect for a spin-1/2 particle. This must have been a second powerful indication to Dirac that he was on the right track. Free Particle Solutions If we apply the Dirac Equation to a particle at rest, p=0, then from equation (1.3) I 0 (1.16) H m m 0 I so the four eigenvectors are 1 0 0 0 0, 1, 0, 0 (1.17) 0 0 1 0 0 0 0 1 with corresponding eigenvalues of m, m, m, m . So we still have negative-energy solutions! On the other hand, we have the relativistic energy-momentum relation, spin ½, and a conserved current. So let’s put aside the problem of negative energies, and investigate the behavior of the two positive-energy solutions in the presence of an electromagnetic field. Dirac Particle in an Electromagnetic Field We can extend the free-particle Hamiltonian to include the case of a particle in a general electromagnetic field using the minimal-coupling substitution p p eA / c (and for this analysis, we will explicitly include c as the speed of light). To simplify notation, we will write the 4-component wavefunction in terms of two 2-component spinors, . Then the Dirac equation becomes 2 i c e mc t (1.18) 2 c e mc 3 In the non-relativistic limit, the energy of the particle will be dominated by its rest mass, so we can write (1.19) (t ) exp i mc 2 / t where and are relatively slowly-varying functions of time. The left-hand side of equation (1.18) then becomes 2 (1.20) i i mc t t i mc2 / t Moving the second term to the right side of equation (1.18), and dividing out e gives 20 (1.21) i c e 2mc t This expression is really two coupled equations, one for the upper 2 components and one for the lower two. The second of these can be approximated, in the case of interaction energies small compared to mc 2 , by c (1.22) 2mc 2 Note that is small compared to for this positive-energy solution, as we expect from the freeparticle solution in the particle rest frame (where 0 ). Substituting this back into the upper equation gives (1.23) i e t 2m Now we can use the identities i j i ijk k and i2 to show i ai j a j aibi ai b j i ijk k a b i a b (1.24) Applying this to (1.23) gives 2 p e / c A e i B e (1.25) t 2m 2mc which is the same as the non-relativistic Pauli equation (or Schrodinger equation) for a spin-1/2 particle in an electromagnetic field. This leads immediately to correct fine structure prediction for hydrogen, including the factor of g 2 in the electrons magnetic moment. Simply expand the first term to first order in e , and use A 12 B r , giving Interpretation of the Negative-Energy Solutions (a) Dirac “Hole Theory” (b) Feynman antiparticle interpretation 4 ======================================== The Gamma Matrices In order to put E and p on the same footing, it usual to left-multiply multiply equation (1.3) by and to define the -matrices by: 1 0 0 i 0 , i 0 1 i i 0 (1.26) The gamma matrices have the following defining properties, easily derived from (1.26) and (1.4): , 2 g , I, I 0 † k † 0 0 2 I k (1.27) k 2 Note that the I in these equations is the identity matrix. Also note that equation (1.27)c is already contained in (1.27)a; it is included only for emphasis. In terms of the gamma matrices, the Dirac equation is now written as: i 0 k k m or equivalently P m 0 (1.28) x t The summation P , which is the contraction of a 4-vector with the gamma-matrices, occurs so often that Feynman introduced a shorthand, called the “Feynman slash notation” that is universally used now. We write P P , so the Dirac equation (1.28) becomes P m 0 (1.29) The Conserved Dirac Current We have already seen that current conservation follows from the Dirac equation, and is critical for the viability of the theory. The continuity equation can also be derived using the gamma-matrices, of course, and this gives a very convenient and manifestly covariant way of writing the current. This can be done as follows: first take the hermitian conjugate of (1.28), giving i † 0† † k † i k m † t x i † 0 † k i k m † t x 0 (1.30) where we have used equation (1.27) for the hermiticity of the gamma matrices. This no longer has the form of equation (1.28) because of the relative minus sign between the first two terms. We can fix this by right-multiplying by 0 and using the commutation relation (1.27) to obtain 5 † 0 0 † k 0 i i m † 0 t xk † 0 0 † 0 k i i m † 0 t xk 0 (1.31) Now we define the adjoint wavefunction by † 0 (1.32) so the adjoint Dirac equation (1.31) becomes i 0 k i m i m 0 t xk x (1.33) Right multiply (1.33) by , left-multiply (1.28) by , and subtract to get i 0 x x i 0 x (1.34) Consequently, the four quantities obtained from the matrix multiplication must transform like a 4-vector. (Note that these 4 quantities are numbers, not matrices or column vectors; . Moreover, when contracted with / x they give zero, so this four-vector is a conserved current. In order to give a quantity that represents an electric current we multiply by e : j j 0 x and this is the definition of a conserved current. This is an incredibly important step. 6 (1.35)