4 The key fact is the Dirac equation mu p 6 pu p and m u q u q 6 q when you

# 4 the key fact is the dirac equation mu p 6 pu p and

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4. The key fact is the Dirac equation mu ( p ) = 6 pu ( p ) and m u ( q ) = u ( q ) 6 q (when you take the adjoint, remember that γ 0 γ μ γ 0 = γ μ ). Thus, we find u ( q ) γ μ u ( p ) = 1 2 m u ( q )( 6 μ + γ μ 6 p ) u ( p ) = 1 2 m u ( q ) 1 2 { 6 q, γ μ } + 1 2 [ 6 q, γ μ ] + 1 2 { γ μ , 6 p } + 1 2 [ γ μ , 6 p ] u ( p ) = 1 2 m u ( q ) q μ + p μ - 1 2 [ γ μ , γ ν ]( q ν - p ν ) u ( p ) as desired. 5. (a) The Dirac equation is i 6 ∂ψ = ( - eA · γ + m ) ψ i∂ t ψ = ( - eA 0 + γ 0 γ i ( i∂ i + eA i ) + 0 ) ψ = ( - eA 0 + γ 0 γ i ( - p i + eA i ) + 0 ) ψ
Phys 253a 5 (b) We have H + eA 0 = σ i ( p i - eA i ) m m - σ i ( p i - eA i ) ! ( H + eA 0 ) 2 = m 2 + σ i σ j ( p i - eA i )( p j - eA j ) 0 0 m 2 + σ i σ j ( p i - eA i )( p j - eA j ) ! We can simplify the block diagonals as follows σ i σ j ( p i - eA i )( p j - eA j ) = 1 2 { σ i , σ j } + 1 2 [ σ i , σ j ] ( p i - eA i )( p j - eA j ) = δ ij + i ijk σ k ( p i - eA i )( p j - eA j ) = ( p - e A ) 2 + i 2 σ k ijk [ p i - eA i , p j - eA j ] = ( p - e A ) 2 + i 2 σ k ijk ieF ij ( x ) = ( p - e A ) 2 - e σ · B Thus we have ( H + eA 0 ) 2 = m 2 + ( p - e A ) 2 - e Σ · B . Restoring factors of c and ~ so that every term has units of energy squared, we have ( H + ecA 0 ) 2 = m 2 c 4 1 + ( p - e A ) 2 m 2 c 2 - e ~ Σ · B m 2 c 2 . (c) Now we can take a square root of above expression, assuming c is large, to find H = - ecA 0 + ( p - e A ) 2 2 m - e ~ 2 m Σ · B + O 1 c 2 Note that we’ve dropped the rest energy term. We see that there’s a magnetic dipole moment with g = 2, but no electric dipole term. (d) We already know the commutation relation [ σ i , σ j ] = 2 i ijk σ k , which implies σ i 2
Phys 253a 6 satisfy the properly normalized commutation relation. Meanwhile, [ L i , L j ] = ikl jmn [ x k p l , x m p n ] = ikl jmn ( x k [ p l , x m ] p n + x m [ x k , p n ] p l ) = ikl jmn ( - ix k p n δ lm + ix m p l δ kn ) = i ( δ ij δ kn - δ in δ kj ) x k p n - i ( δ ij δ lm - δ im δ lj ) x m p l = i ( x i p j - x j p i ) = i ijk L k Where in the fourth line we’ve used abc dec = δ ad δ be - δ ae δ bd . This proof may be more long-winded than necessary. If you’re not so comfortable with the index notation, it might be better to just compute [ L x , L y ] and then argue for the general result using symmetry. (e) Let’s pick Coulomb gauge, where the vector potential corresponding to a constant field B z is A = B z 2 ( x ˆ y - y ˆ x ). Then ( p - e A ) 2 = p 2 - eB z 2 ( p y x + xp y - p x y - yp x ) + e 2 A 2 = p 2 + e 2 A 2 - eB z L z Thus, the total coupling to the magnetic field is - eB z 2 m ( L z + 2 S z ) where S z = ~ 2 Σ . We find g e = 2, as expected. One way to measure this is to begin with an electron with spin up in the z direction (say by using a Stern-Gerlach apparatus). Now apply a magnetic field in the

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