Quantum Field Theory

But is this Lorentz invariant? It’s not obvious because we only have 3-vectors. Whatcould go wrong? Suppose we have a Lorentz transformationp µ → (p ′ ) µ = Λ µ ν p ν (2.55)such that the 3-vector transforms as ⃗p → ⃗p ′ . In the quantum theory, it would bepreferable if the two states are related by a unitary transformation,|⃗p〉 → |⃗p ′ 〉 = U(Λ) |⃗p〉 (2.56)This would mean that the normalizations of |⃗p〉 and |⃗p ′ 〉 are the same whenever ⃗p and⃗p ′ are related by a Lorentz transformation. But we haven’t been at all careful withnormalizations. In general, we could get|⃗p〉 → λ(⃗p, ⃗p ′ ) |⃗p ′ 〉 (2.57)for some unknown function λ(⃗p, ⃗p ′ ). How do we figure this out? The trick is to look atan object which we know is Lorentz invariant. One such object is the identity operatoron one-particle states (which is really the projection operator onto one-particle states).With the normalization (2.54) we know this is given by∫1 =d 3 p|⃗p〉 〈⃗p| (2.58)(2π)3This operator is Lorentz invariant, but it consists of two terms: the measure ∫ d 3 p andthe projector |⃗p〉〈⃗p|. Are these individually Lorentz invariant? In fact the answer is no.Claim The Lorentz invariant measure is,∫ d 3 p2E ⃗p(2.59)Proof: ∫ d 4 p is obviously Lorentz invariant. And the relativistic dispersion relationfor a massive particle,p µ p µ = m 2 ⇒ p 2 0 = E2 ⃗p = ⃗p 2 + m 2 (2.60)is also Lorentz invariant. Solving for p 0 , there are two branches of solutions: p 0 = ±E ⃗p .But the choice of branch is another Lorentz invariant concept. So piecing everythingtogether, the following combination must be Lorentz invariant,∫∫ ∣ d 4 p δ(p 2 0 − ⃗p 2 − m 2 )d 3 p ∣∣∣p0∣ =(2.61)p0 >02p 0 =E ⃗pwhich completes the proof.□– 32 –