λ� 2 π 3 δ 3 vector p vector q λ� 1 2 3 646 But for the timelike annihilation

# Λ? 2 π 3 δ 3 vector p vector q λ? 1 2 3 646 but

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λλ (2 π ) 3 δ (3) ( vector p vector q ) λ,λ = 1 , 2 , 3 (6.46) But for the timelike annihilation and creation operators, we have [ a 0 vector p ,a 0 vector q ] = (2 π ) 3 δ (3) ( vector p vector q ) (6.47) This is very odd! To see just how strange this is, we take the Lorentz invariant vacuum | 0 ) defined by a λ vector p | 0 ) = 0 (6.48) Then we can create one-particle states in the usual way, | vector p,λ ) = a λ vector p | 0 ) (6.49) For spacelike polarization states, λ = 1 , 2 , 3, all seems well. But for the timelike polarization λ = 0, the state | vector p, 0 ) has negative norm, ( vectorp, 0 | vector q, 0 ) = ( 0 | a 0 vector p a 0 vector q | 0 ) = (2 π ) 3 δ (3) ( vector p vector q ) (6.50) Wtf? That’s very very strange. A Hilbert space with negative norm means negative probabilities which makes no sense at all. We can trace this negative norm back to the wrong sign of the kinetic term for A 0 in our original Lagrangian: L = + 1 2 ˙ vector A 2 1 2 ˙ A 2 0 + ... . At this point we should remember our constraint equation, μ A μ = 0, which, until now, we’ve not imposed on our theory. This is going to come to our rescue. We will see that it will remove the timelike, negative norm states, and cut the physical polarizations down to two. We work in the Heisenberg picture, so that μ A μ = 0 (6.51) makes sense as an operator equation. Then we could try implementing the constraint in the quantum theory in a number of different ways. Let’s look at a number of increasingly weak ways to do this We could ask that μ A μ = 0 is imposed as an equation on operators. But this can’t possibly work because the commutation relations (6.39) won’t be obeyed for π 0 = μ A μ . We need some weaker condition. – 133 –
We could try to impose the condition on the Hilbert space instead of directly on the operators. After all, that’s where the trouble lies! We could imagine that there’s some way to split the Hilbert space up into good states | Ψ ) and bad states that somehow decouple from the system. With luck, our bad states will include the weird negative norm states that we’re so disgusted by. But how can we define the good states? One idea is to impose μ A μ | Ψ ) = 0 (6.52) on all good, physical states | Ψ ) . But this can’t work either! Again, the condition is too strong. For example, suppose we decompose A μ ( x ) = A + μ ( x ) + A μ ( x ) with A + μ ( x ) = integraldisplay d 3 p (2 π ) 3 1 radicalbig 2 | vector p | 3 summationdisplay λ =0 ǫ λ μ a λ vector p e ip · x A μ ( x ) = integraldisplay d 3 p (2 π ) 3 1 radicalbig 2 | vector p | 3 summationdisplay λ =0 ǫ λ μ a λ vector p e + ip · x (6.53) Then, on the vacuum A + μ | 0 ) = 0 automatically, but μ A μ | 0 ) negationslash = 0. So not even the vacuum is a physical state if we use (6.52) as our constraint Our final attempt will be the correct one. In order to keep the vacuum as a good physical state, we can ask that physical states | Ψ ) are defined by μ A + μ | Ψ ) = 0 (6.54) This ensures that ( Ψ | μ A μ | Ψ ) = 0 (6.55) so that the operator μ A μ has vanishing matrix elements between physical states.