Quantum Field Theory

We can deal with the infinity in (2.24) in a more practical way. In physics we’reonly interested in energy differences. There’s no way to measure E 0 directly, so we cansimply redefine the Hamiltonian by subtracting off this infinity,∫H =d 3 p(2π) 3 ω ⃗p a † ⃗p a ⃗p (2.27)so that, with this new definition, H |0〉 = 0. In fact, the difference between this Hamiltonianand the previous one is merely an ordering ambiguity in moving from the classicaltheory to the quantum theory. For example, if we defined the Hamiltonian of the harmonicoscillator to be H = (1/2)(ωq −ip)(ωq +ip), which is classically the same as ouroriginal choice, then upon quantization it would naturally give H = ωa † a as in (2.27).This type of ordering ambiguity arises a lot in field theories. We’ll come across a numberof ways of dealing with it. The method that we’ve used above is called normal ordering.Definition: We write the normal ordered string of operators φ 1 (⃗x 1 ) . . .φ n (⃗x n ) as: φ 1 (⃗x 1 ) . . .φ n (⃗x n ) : (2.28)It is defined to be the usual product with all annihilation operators a ⃗pplaced to theright. So, for the Hamiltonian, we could write (2.27) as∫d 3 p: H :=(2π) ω 3 ⃗p a † ⃗p a ⃗p (2.29)In the remainder of this section, we will normal order all operators in this manner.2.3.1 The Cosmological ConstantAbove I wrote “there’s no way to measure E 0 directly”. There is a BIG caveat here:gravity is supposed to see everything! The sum of all the zero point energies shouldcontribute to the stress-energy tensor that appears on the right-hand side of Einstein’sequations. We expect them to appear as a cosmological constant Λ = E 0 /V ,R µν − 1 2 g µν = −8πGT µν + Λg µν (2.30)Current observation suggests that 70% of the energy density in the universe has theproperties of a cosmological constant with Λ ∼ (10 −3 eV ) 4 . This is much smaller thanother scales in particle physics. In particular, the Standard Model is valid at least upto 10 12 eV . Why don’t the zero point energies of these fields contribute to Λ? Or, ifthey do, what cancels them to such high accuracy? This is the cosmological constantproblem. No one knows the answer!– 26 –