Master's Thesis in Theoretical Physics - Universiteit Utrecht

Note that the matrices Ω and ˜Ξ are hermitean. Suppose now that we are in the inflationaryregime where φ 1 ,φ 2 ≫ v 1 , v 2 as in the ordinary Higgs model of the previous section. It iseasy to see that we can then write our potential (4.67) asV (φ 1 ,φ 2 ) = λ 1 (φ ∗ 1 φ 1) 2 + λ 2 (φ ∗ 2 φ 2) 2 + λ 3 (φ ∗ 1 φ 1)(φ ∗ 2 φ 2), (4.72)where I have actually redefined λ 3 + λ 5 cos 2 θ + λ 6 sin 2 θ → λ 3 . Thus we have obtained theLagrangian (4.70) with a relatively simple potential.As a consequence of the off-diagonal kinetic terms the equations of motions for φ 1 and φ 2will be coupled through the derivative terms. In order to solve the equations of motion, wewant each equation of motion to contain only derivatives of one field. Therefore we wantto diagonalize the kinetic term. This was also done by Garbrecht and Prokopec[28] for asimilar Lagrangian. In that case they were able to solve the field equations analytically. Wewill try to do the same here. First we want to diagonalize the kinetic term by solving theeigenvalue equationΩU = UD, (4.73)where D is a 2 × 2 diagonal matrix with the eigenvalues of Ω on the diagonal, and U is thecorresponding matrix with the two eigenvectors as columns. Solving the eigenvalue matrixfrom Eq. (4.73) gives( 1 + |ω| 0D =0 1 − |ω|), (4.74)with |ω| = ω ∗ ω. Multiplying both sides of Eq. (4.73) from the right with U −1 , we see thatwe can writewhere the matrices are given byU =Ω = UDU −1 , (4.75)( √ √ω− ωω ∗ ω ∗1 1⎛√ω ∗U −1 = 1 ⎝ √2 −ω1ω ∗ω1⎞)(4.76)⎠. (4.77)Now we plug Eq. (4.75) into the derivative term of the kinetic term of the Lagrangian (4.70)and we writewhere I have defined new fieldsΦΦ †≡≡(φ+L kin = −gg µν ∂ µ ˜Φ † UDU −1 ∂ ν ˜Φ= −gg µν ∂ µ Φ † ∂ ν Φ, (4.78)⎛)= D 2U −1 ˜Φ = 1 ⎝φ − 2( φ∗+φ ∗ −) T= 1 2˜Φ † Û D = 1 2⎛⎝ [√ ]1 + |ω|ω ∗ω φ 1 + φ 2 [ √ ]1 − |ω|ω−∗ω φ 1 + φ 2 [√1 + |ω|ωφ ∗ ω ∗ 1 2]+ φ∗ []1 − |ω| −√ωφ ∗ ω ∗ 1 + φ∗ 2⎞⎠⎞⎠T. (4.79)