Master's Thesis in Theoretical Physics - Universiteit Utrecht

of motion (5.4). If we now substitute our solution for v k from Eq. (5.27) with u k from Eq.(5.26) into (5.29), we find the condition|α k | 2 − |β k | 2 = 1. (5.30)Note that we could have written our expansion (5.28) in terms of the solutions u k asˆφ k (t) = ˆb − k u k(t) + ˆb + −k u∗ k(t). (5.31)A simple exercise shows that the new creation and annihilation operators ˆb + k and ˆb − can bekexpressed in terms of the "old" creation and annihilation operators asˆb + k= α ∗ kâ+ k + β kâ − −kˆb − k= α k â − k + β∗ kâ+ −k . (5.32)This is a Bogoliubov transformation. If the commutation relations (5.15) are satisfied forâ − k and â+ k , then the same commutators are satisfied for ˆb − k and ˆb + if the relation (5.30) iskfulfilled. If we now act with the total number operator, which in this case is ∫ d 3 k ˆb + ˆb − k k , onthe vacuum defined in Eq. (5.15), we find∫∫∫〈0| d 3 k ˆb + ˆb − k k |0〉 = 〈0| d 3 k|β k | 2 |0〉 = d 3 k|β k | 2 . (5.33)So instead of finding zero particles in the vacuum, we now find ∫ d 3 k|β k | 2 particles. Sincewe only have the condition (5.30) on these coefficients, |β k | 2 can run from zero to infinity.Our particle number apparently depends on the choice of coefficients in the mode expansionin Eq. (5.28)! The question now remains: can we now, based on physical arguments, makesome choice for α k and β k ? The answer is: yes, we can. We calculate again the energy byacting with the Hamiltonian on the vacuum of Eq. (5.15). This time the Hamiltonian isexpressed in terms of the fields defined in Eq. (5.28) (or Eq. (5.31) together with Eq. (5.32)).We find after a small calculation that the energy of the vacuum is∫E vac = 〈0|Ĥ|0〉 = 〈0|d 3 k(1 + 2|β k | 2 ) ω k2 |0〉 = ∫d 3 k(1 + 2|β k | 2 ) ω k2 . (5.34)Now we can use our physical argument: if we want our vacuum to be the state of lowestenergy, then the only choice we have is to take β = 0, which reduce our lowest energy toE vac = E 0 from Eq. (5.24). In this case, also our particle number does not depend on thechoice of the vacuum.Another important observation is that if the vacuum is the state of lowest energy at onepoint in time, then it will also be the state of lowest energy at some later time. The reasonis that the Hamiltonian is constant in time, which can easily be seen when concerning theexpression for H from Eq. (5.21). The creation and annihilation are constants in time, andso is our frequency ω k . Another way to see that the Hamiltonian is time independent is byconsidering the (classical) expression for H and taking the time derivativeddt H = d ∫ ( 1d 3 kdt 2 |π k| 2 + 1 )2 (k2 + m 2 )|φ k | 2∫ ( 1= d 3 k2 ˙φ k ¨φ−k + 1 2 (k2 + m 2 ) ˙φk φ −k + 1 2 ˙φ −k ¨φk + 1 )2 (k2 + m 2 ) ˙φ−k φ k= 0,where we have used the equation of motion Eq. (5.5) in the final step. So we conclude thatthe state of lowest energy is given by the vacuum (5.15) when we expand our field as in Eq.