Basics of quantum electrodynamics by Ioan Merches

By Ioan Merches

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On the motion of the field. We shall call it orbital angular momentum density, or current momentum density. 14) i + ˜ + ˜+ π S(r)(p)ik U(p) + U(p) S(r)(p)ik πl(r) , c l(r) where S˜(r)(p)ik = S(r)(p)ik − S(r)(p)ki ; + + + S˜(r)(p)ik = S(r)(p)ik − S(r)(p)ki . 4) of the canonical energy-momentum tensor. After a convenient rearrangement of terms, we get: (c) + + Tkl = πl(r) U(r),k + U(r),k πl(r) − Lδkl , (c) + + Til = πl(r) U(r),i + U(r),i πl(r) − Lδil . 18) is named the rotation operator. The direction of the arrow above Dik shows on which quantity the operator acts.

E. on the motion of the field. We shall call it orbital angular momentum density, or current momentum density. 14) i + ˜ + ˜+ π S(r)(p)ik U(p) + U(p) S(r)(p)ik πl(r) , c l(r) where S˜(r)(p)ik = S(r)(p)ik − S(r)(p)ki ; + + + S˜(r)(p)ik = S(r)(p)ik − S(r)(p)ki . 4) of the canonical energy-momentum tensor. After a convenient rearrangement of terms, we get: (c) + + Tkl = πl(r) U(r),k + U(r),k πl(r) − Lδkl , (c) + + Til = πl(r) U(r),i + U(r),i πl(r) − Lδil . 18) is named the rotation operator. The direction of the arrow above Dik shows on which quantity the operator acts.

These names will be justified in what follows. First, we shall derive the commutation relations satisfied by the creation and annihilation operators for bosons and fermions. a) If the particles are bosons, then ni can be any integer, including zero. This condition is fulfilled by the choice κ = 1. 20), we have: ∂ √ ci = e ∂ni µi ni ; c+ i = √ n i µ+ i e ∂ − ∂n i . 20), we may write ∂ ∂ni (ni )e ci c+ i =e that is ∂ − ∂n i = n1 + 1; [ci , c+ i ] = 1, c+ i ci = √ √ ni . 10) where 1 is the unit operator.

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