THE DIRAC EQUATION (A REVIEW) We will try to find the relativistic wave equation for a particle.

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1 THE DIRAC EQUATION (A REVIEW) We will try to find the relativistic wave equation for a particle. First, we introduce four dimensional notation for a vector by writing x µ = (x, x 1, x 2, x 3 ) = (ct, x, y, z). The metric tensor of special relativity is η νµ = This allows us to introduce 1 x ν = η νµ x µ where x ν = (x, x 1, x 2, x 3 ) = (x, x 1, x 2, x 3 ). The summation convention has been used; when the same index appears as a superscript and a subscript, then that index is summed over. Thus, x ν = η ν x + η ν1 x 1 + η ν2 x 2 + η ν3 x 3. The scalar product x x is defined by x x = x ν x ν = (x ) 2 (x 1 ) 2 (x 2 ) 2 (x 3 ) 2 = (x ) 2 x x. For a free relativistic particle, E 2 = p 2 c 2 + m 2 c 4. Using the correspondence that E = i / t = i c / x and p = i leads to 2 2 φ t = 2 2 c 2 2 φ (x ) = 2 2 c 2 2 φ + m 2 c 4 φ. (1) Substituting the plane wave solution φ exp( ip x/ ) into Eq. (1), yields Date: April 1,

2 2 THE DIRAC EQUATION (A REVIEW) E 2 φ = (p 2 c 2 + m 2 c 4 )φ, (2) and we recover E 2 = p 2 c 2 +m 2 c 4, or E = ± p 2 c 2 + m 2 c 4 as is required. Eq. (1), an equation second order in all derivatives, was originally rejected because of the negative energy solution. Later, it was applied to mesons. Another attempt to find a relativistic wave equation was made by Dirac. Observe that the Schrödinger equation is linear in the time derivative, but of second order in the spatial derivatives. Try to find a relativistic wave equation which is linear in all derivatives. For a free particle, try ic φ/ x = ic (α 1 φ/ x 1 +α 2 φ/ x 2 +α 3 φ/ x 3 )+βmc 2 φ = Hφ. (3) where α 1, α 2, α 3, and β are to be determined, and H is the hamiltonian. Notice that if φ is replaced by the usual plane wave, then the coefficients of φ after differentiation have the dimensions of energy if α i (i = 1, 2, 3) and β are dimensionless. Apply the operator below to Eq. (3) ic / x ic (α 1 / x 1 + α 2 / x 2 + α 3 / x 3 ) + βmc 2, (4)

3 THE DIRAC EQUATION (A REVIEW) 3 and find 2 c 2 2 φ (x ) = 2 2 c 2 3 i,j=1 (α j α i + α i α j ) 2 i mc 3 3 i=1 φ x j x i (α i β + βα i ) φ x i + β2 m 2 c 4 φ. (5) Eq. (5) shall be required to reduce to Eq. (1). Notice that the first term on the right hand side of Eq. (5) will contain (α 1 ) 2 2 φ (x 1 ) + 2 (α2 ) 2 2 φ (x 2 ) + 2 (α3 ) 2 2 φ (6) (x 3 ) 2 plus cross terms such as (α 1 α 2 + α 2 α 1 2 φ ) x 1 x2. (7) In order that Eq. (5) reduces to Eq. (1), (α 1 ) 2 = (α 2 ) 2 = (α 3 ) 2 = 1, β 2 = 1, (α 1 α 2 + α 2 α 1 ) =, and the other cross terms vanish including α i β +βα i. The cross terms can not vanish if the α i and β are numbers. Matrices can anti-commute, so we assume they are matrices. This means that the 1 in the equations (α i ) 2 = β 2 = 1 is actually the unit matrix, and φ/ x in Eq. (3) is multiplied by the unit matrix. In addition, take φ to be a column matrix. This allows spin to appear in the theory naturally.

4 4 THE DIRAC EQUATION (A REVIEW) Suppose now that λ i j is an eigenvalue of the N by N matrix α i where N is to be determined. Here j indicates which eigenvalue. Let χ i j be the eigenvector associated with the eigenvalue λ i j. χ i j will be a 1 by N column matrix. The eigenvalue equation takes the form α i χ i j = λ i jχ j j. Apply α i to the above equation, and find (α i ) 2 χ i j = χi j = αi λ i j χi j = (λi j )2 χ i j. (8) Thus (λ i j) 2 = 1, and λ i j = ±1. It will be shown next that N is even. Start with α i β = βα i, then βα i β = β 2 α i = α i. Take the trace of both sides, and find Trβα i β = -Trα i. Use the trace property that Trβα i β =Trα i ββ, and β 2 = 1 to show that Trα i = Trα i, or Trα i =. The Trace of a matrix is just the sum of its eigenvalues. Since the trace is zero, N must be even. N = 2 is ruled out because there are not four 2 by 2 anti-commuting matrices. We proceed with N = 4. Thus, the α i and β are four by four matrices, and φ is a one by four column matrix. It is convenient to multiply Eq. (3) by the matrix β. Introduce the notation γ = β, and γ i = βα i. Then Eq. (3) can be written i ( γ x + γ1 x 1 + γ2 x 2 + γ3 x 3 ) φ mcφ = (9)

5 THE DIRAC EQUATION (A REVIEW) 5 where mcφ is multiplied by the unit matrix. We assume the plane wave solution φ u exp( ip x/ ) where u is the one by four column matrix u 1 u 2 u 3 u 4 with u i to be determined. Notice that (γ i ) 2 = β α i β α i = α i β 2 α i = (α i ) 2 = 1 and (γ ) 2 = 1. Similarly, it can be shown that γ i γ j + γ j γ i =. These results can be written 1 γ µ γ ν + γ ν γ µ = 2g µν where g µν = 1 1. The product of a four by four matrix with a four by four matrix is a four 1 by four matrix, so each entry in g µν is a four by four diagonal matrix. σ A possible representation for the matrices is γ i i = σ i and 1 γ = where the σ 1 i are the Pauli matrices. Recall that 1 i 1 σ 1 =, σ 1 2 =, and σ i 3 =. As an illustration, 1 we now calculate γ 1 γ 3 + γ 3 γ 1 in the chosen representation. σ γ 1 γ 3 1 σ 3 = σ 1 σ 3 = Similarly, γ 3 γ 1 = σ 3 σ 1 σ 3 σ 1. Next, note that σ 1 σ 3 σ 1 σ 3. (1) Similarly, σ 3 σ 1 = σ 1 σ 3 = 1 1 = Finally, 1 1. (11) 1

6 6 THE DIRAC EQUATION (A REVIEW) γ 1 γ 3 + γ 3 γ 1 = σ 1 σ 3 σ 3 σ 1 σ 1 σ 3 σ 3 σ 1 = The other identities can be similarly verified. For p =, the Dirac equation takes the form. (12) ( i γ x mc) φ = (13) where φ u exp( ip x / ) and u is a 1 by 4 matrix described above. Eq. (13) can be written 1 u 1 i exp( ip x / ) 1 u 2 x 1 u 3 1 u 4 1 mcexp( ip x / ) Carrying out the matrix multiplication leads to u 1 u 2 u 3 u 4 =. (14) (p mc)u 1 =, (p mc)u 2 =, ( p mc)u 3 =, ( p mc)u 4 =. (15) One possible set of solutions is found by setting three of the u i to zero, and setting the remaining one equal to one. Thus, the four solutions to Eq. (13) are

7 THE DIRAC EQUATION (A REVIEW) 7 1 φ 1 = exp( imcx / ), φ2 = exp( imcx / ) 1 φ 3 = exp(+imcx / ) 1, φ4 = exp(+imcx / ). (16) 1 Since p = mc for the first two solutions, we shall associate a positive energy spin up electron with φ 1, and we shall associate a positive energy spin down electron with φ 2. The two negative energy solutions will be discussed in more detail when positrons are treated. For p, assume that a solution to Eq. (9) will take the form φ u exp( ip x/ ). Here u = u 1 u 2 u 3 u 4 and the four vector p is given by p µ = (p = E/c, p 1, p 2, p 3 ). Substitution into Eq. (9) gives the following result: (p mc)u 1 (p 1 ip 2 )u 4 p 3 u 3 = (17) (p mc)u 2 (p 1 + ip 2 )u 3 + p 3 u 4 = (18) (p + mc)u 3 + (p 1 ip 2 )u 2 + p 3 u 1 = (19) (p + mc)u 4 + (p 1 + ip 2 )u 1 p 3 u 2 =. (2) For the positive energy spin up solution, set u 2 = and find u 3 = p 3 u 1 /(p + mc) and u 4 = (p 1 + ip 2 )u 1 /(p + mc). So

8 8 THE DIRAC EQUATION (A REVIEW) u 1 φ 1 exp( ip x/ ) p 3 cu 1 /E + mc 2 ) (21) (p 1 + ip 2 )cu 1 /(E + mc 2 ) In order to agree with reference (6) in the Mott Rutherford paper, u 1 is set equal to (E + mc 2 )/(2mc 2 ). Write φ 1 (x) w 1 (p)exp( ip x/ ) where w 1 (p) = 1 (E + mc 2 )/(2mc 2 ) p 3 c/(e + mc 2 ). (22) (p 1 + ip 2 )c/(e + mc 2 ) Set w = w γ and notice that w 1 w 1 = (w 1 ) γ w 1 = (E+mc 2 )/(2mc 2 ) ( 1 p 3 c/(e + mc 2 ) (p 1 ip 2 )c/(e + mc 2 ) ) p 3 c/(e + mc 2 ) = 1. (23) 1 (p 1 + ip 2 )c/(e + mc 2 ) A similar calculation yields (w 1 ) w 1 = E/mc 2. Introduce the four vector k = p/. Then φ 1 (x) w 1 (p)exp ( ik x). The normalization that (φ 1 2 ) φ 1 1 d 3 x = δ 3 (k 2 k 1 ) requires that φ 1 (x) = mc 2 /(2π) 3 E w 1 (p)exp( ik x). Alternatively, if box normalization is used, (φ 1 2 ) φ 1 1 d3 x = δ k2,k 1 and φ 1 (x) = mc 2 /(EV ) w 1 (p)exp( ik x). For the positive energy spin down solution, set u 1 = and find

9 THE DIRAC EQUATION (A REVIEW) 9 φ 2 exp( ik x) u 2 (p 1 + ip 2 )cu 2 /(E + mc 2 ) p 3 cu 2 /(E + mc 2 ) (24) Set u 2 equal to (E + mc 2 )/(2mc 2 ). Write φ 2 (x) w 2 (p)exp( ik x) where w 2 (p) = (E + mc 2 )/(2mc 2 ) 1 (p 1 ip 2 )c/(e + mc 2 ). (25) p 3 c/(e + mc 2 ) Notice that w 2 w 2 = 1, and w 1 w 2 = w 2 w 1 =. Also, (w 2 ) w 2 = E/mc 2, and (w 2 ) w 1 = (w 1 ) w 2 =. Box normalization will require φ 2 (x) = mc 2 /(EV )w 2 (p)exp ( ik x). Introduce the notation u(p, s z = + /2) = w 1 (p) and u(p, s z = /2) = w 2 (p). For an arbitrary spin s, write φ(x) = mc 2 /(EV )u(p, s)exp ( ik x) where the column matrix u(p, s) is a linear combination of w 1 (p) and w 2 (p). This paper justifies the wave functions used in the paper on Mott- Rutherford scattering. In special relativity, the three-dimensional vector potential and the potential, A = Φ, are combined to form a four-vector A = (A, A 1, A 2, A 3 ). The replacement p i γ µ / x µ eγ µ A µ yields the Dirac equation i γ µ ψ x µ eγµ A µ ψ mcψ =. (26)

10 1 THE DIRAC EQUATION (A REVIEW) for a particle in an electro-magnetic field. This is the equation that we will use to solve scattering problems in QED. The next two review papers will show how this equation is used to calculate the S-matrix.