Kapitza Spring 2018 Second Quantization of the Klein-Gordon Equation Ben Saltzman Q.F.T. TERM PAPER Professor Sarada Rajeev April 20th 2017 1 Introduction The Schr¨odinger equation, successful as it is for describing non-relativistic quantum particles, fails in the relativistic regime. By using special relativity, one can derive a relativistic version of the Schr¨odingerequation, known as the first quantization of the Klein-Gordon equation. Unfortunately, this has its own problems: negative energy and probability solutions, nonphysical entities that could spell the end for a theory. Moving from quantum theory to quantum field theory, fields are introduced as fundamental. Quantizing the Klein-Gordon equation in quantum field theory leads to a model known as the second quantization, which avoids many of the problems of the first quantization. 2 Second Quantization of the Klein-Gordon Equation 2.1 The Klein-Gordon Field We can write the Klein-Gordon field operator as 3 ( ik x) (ik x) φ(x)=C d k e − · a(k)+e · a†(k) (1) Z ✓ ◆ where 1 C = (2) 2k0(2⇡)3 0 and k = Ek. The field operator satisfiesp the Klein-Gordon equation, i.e. µ 2 (@µ@ + m )φ =0 (3) This can be split into positive and negative energy components as follows: (+) ( ) φ(x)=φ (x)+φ − (x) (+) 3 ( ik x) φ (x)=C d ke− · a(k) (4) Z ( ) 3 (ik x) φ − (x)=C d ke · a†(k) Z The conjugate momentum for the field operator is therefore ⇧(x)=φ˙(x) 0 3 = ik C d k exp( ik x)a(k) exp(ik x)a†(k) (5) − − · − · Z ✓ ◆ 1 Inverting equations 1 and 5 and solving for the annihilation and creation operators a and a†: 3 ik x 0 a(k)=C d xe · k φ(x)+i⇧(x) Z ✓ ◆ 3 ik x ik x = iC d x e · @ φ(x) @ e · φ(x) (6) t − t Z ✓ ◆ 3 ik x 0 a†(k)=C d xe− · k φ(x) i⇧(x) − Z ✓ ◆ 3 ik x ik x = iC d x e− · @ φ(x) @ e− · φ(x) (7) − t − t Z ✓ ◆ Taking the time derivative of equation 6: 3 ik x ik x @ a(k)=iC@ d x e · @ φ(x) @ e · φ(x) t t t − t Z ✓ ◆ 3 ik x ik x 2 ik x 2 ik x = iC d x @ e · @ φ(x)+e · @ φ(x) @ e · @ φ(x) @ e · φ(x) t t t − t t − t Z ✓ ◆ 3 ik x 2 2 ik x = iC d x e · @ φ(x) @ e · φ(x) t − t Z ✓ ◆ φ must satisfy the Klein-Gordon equation, so 3 ik x 2 2 0 2 ik x @ a(k)=iC d x e · ( m )φ(x)+(k ) e · φ(x) t r − Z ✓ ◆ 3 0 2 2 2 ik x = iC d x (k ) k m e · φ(x) − − Z ✓ ◆ =0 0 2 2 2 since (k ) k = m .Thesameprocedurefollowsfora†,meaningthatthe annihilation− and creation operators are time independent. The commutation relations between the annihilation and creation op- erators can be deduced from the already-known relations between the field operator and the conjugate momentum when imposing the contraint x0 = y0. To reduce clutter, we define C0 C . ⌘ 0 0 k k0 ! 2 3 3 i(k x+k y) [φ(x),φ(y)] = CC0 d k d k0 e− · 0· [a(k),a(k0)] ZZi(k x k y) ✓ i( k x+k y) + e− · − 0· a(k),a†(k0) + e− − · 0· a†(k),a(k0) i( k x k y) + e− − · − 0· ⇥ a†(k),a†(k⇤0) ⇥ ⇤ ◆ = 0⇥ ⇤ (8) 0 0 3 3 i(k x+k y) [⇧(x), ⇧(y)] = k k0 CC0 d k d k0 e− · 0· [a(k),a(k0)] − i(k x k ZZy) ✓ i( k x+k y) e− · − 0· a(k),a†(k0) e− − · 0· a†(k),a(k0) − − i( k x k y) + e− − · − 0· ⇥ a†(k),a†(k⇤0) ⇥ ⇤ ◆ = 0⇥ ⇤ (9) 0 3 3 i(k x+k y) [φ(x), ⇧(y)] = ik0 CC0 d k d k0 e− · 0· [a(k),a(k0)] − i(k x kZZy) ✓ i( k x+k y) e− · − 0· a(k),a†(k0) + e− − · 0· a†(k),a(k0) − (10) i( k x k y) e− − · − 0· ⇥ a†(k),a†(k⇤0) ⇥ ⇤ − ◆ = iδ3(x y) ⇥ ⇤ − The commutation relations between the annihilation and creation operators can be calculated from equations 6 and 7, knowing equations 2.1, 2.1, and 10: 3 3 i(k x+k y) 0 0 [a(k),a(k0)] = CC0 d x d ye · 0· k k0 [φ(x),φ(y)] ZZ ✓ + ik0[φ(x), ⇧(y)] + ik00[⇧(x),φ(y)] [⇧(x), ⇧(y)] − ◆ 3 3 i(k x+k y) 0 0 = CC0 d x d ye · 0· ik [φ(x), ⇧(y)]+ik0 [⇧(x),φ(y)] ZZ ✓ ◆ 0 0 3 3 i(k x+k y) 3 = CC0(k k0 ) d x d ye · 0· δ (x y) − − − ZZ 0 0 3 i(k+k ) x = CC0(k k0 ) d xe 0 · − − Z 0 0 i(k0+k00) x 3 = CC0(k k0 )e · δ (k k0) − − − = 0 (11) 3 In the last step, we use the fact that k0 = k00.Similarly,wefindthat a†(k),a†(k0) =0 (12) 3 a(k⇥),a†(k0) = δ⇤(k k0)(13) − So the creation and annihilation⇥ operators⇤ commute with themselves for any two k and k0,butdonotcommutewitheachother. 2.2 Relationship with the Harmonic Oscillator The Hamiltonian is the integral of the Hamiltonian density over all space: H H = d3x H Z 1 = d3x m2φ2 +( φ) ( φ)+⇧2 (14) 2 r · r Z ✓ ◆ Recall that k0 = k00. Using equation 1: 3 3 3 2 1 3 d k d k0 i(k+k ) x − 0 · 0 d xφ(x)= 3 d x e a(k)a(k ) 2(2⇡) pk0 pk00 Z i(k Zk ) x ZZ i(✓k+k ) x + e− − 0 · a(k)a†(k0)+e− − 0 · a†(k)a(k0) i( k k ) x + e− − − 0 · a†(k)a†(k0) 3 3 ◆ 1 d k d k0 i(k0+k00)x0 3 = e− a(k)a(k0)δ (k + k0) 2 pk0 p 00 ZZ k ✓ i(k0 k00)x0 3 i( k0+k00)x0 3 + e− − a(k)a†(k0)δ (k k0)+e− − a†(k)a(k0)δ ( k + k0) − − i( k0 k00)x0 3 + e− − − a†(k)a†(k0)δ ( k k0) − − ◆ 1 3 2ik0x0 0 = d k e− a(k)a( k)+e a(k)a†(k) 2k0 − Z ✓ 0 2ik0x0 + e a†(k)a(k)+e a†(k)a†( k) − ◆ 1 3 2ik0x0 = d k e− a(k)a( k)+a(k)a†(k) 2k0 − Z ✓ (15) 2ik0x0 + a†(k)a(k)+e a†(k)a†( k) − ◆ Similarly, using equation 5: 4 1 3 2 3 3 3 0 0 i(k+k0) x d x ⇧ (x)= d x d k d k0 pk k0 e− · a(k)a(k0) −2(2⇡)3 Z i(k kZ) x ZZ i( k+k ) x ✓ e− − 0 · a(k)a†(k0) e− − 0 · a†(k)a(k0) − − i( k k ) x + e− − − 0 · a†(k)a†(k0) ◆ 0 2 (k ) 3 2ik0x0 = d k e− a(k)a( k) a(k)a†(k) − 2k0 − − Z ✓ (16) 2ik0x0 a†(k)a(k)+e a†(k)a†( k) − − ◆ The gradient of φ is found from equation 1: 3 ( ik x) (ik x) φ(x)= iCk d k e − · a(k) e · a†(k) (17) r − − Z ✓ ◆ So, we see that 2 3 3 3 (i) 3 d k d k0 d x φ(x) φ(x)= 3 d x k r ·r 2(2⇡) pk0 p 00 Z Z ZZ k i(k+k ) x i(k k ) x k0 e− 0 · a(k)a(k0) e− − 0 · a(k)a†(k0) · − ✓ i( k+k ) x i( k k ) x e− − 0 · a†(k)a(k0)+e− − − 0 · a†(k)a†(k0) − ◆ 2 k 3 2ik0x0 = d k e− a(k)a( k)+a(k)a†(k) 2k0 − Z ✓ (18) 2ik0x0 + a†(k)a(k)+e a†(k)a†( k) − ◆ Substituting equations 2.2, 2.2, and 2.2 into equation 14, we can derive an equation for the Hamiltonian: 1 3 2 2 0 2 2ik0x0 H = d k (m + k (k ) ) e− a(k)a( k) 4k0 − − Z ✓ 2ik0x0 2 2 0 2 + e a†(k)a†( k) +(m + k +(k ) ) a(k)a†(k)+a†(k)a(k) − ◆ ✓ ◆ 1 3 2ik0x0 = d k 0 e− a(k)a( k) 4k0 ⇥ − Z ✓ 2ik0x0 0 2 + e a†(k)a†( k) +2(k ) a(k)a†(k)+a†(k)a(k) − ◆ ✓ ◆ 5 1 3 0 = d kk a(k)a†(k)+a†(k)a(k) 2 Z ✓ ◆ 1 3 d kE a(k)a†(k)+a†(k)a(k) (19)= 2 k Z ✓ ◆ Recalling the commutation relations from equations 2.1, 12, and 13, the commutation relations of the annihilation and creation operators with the Hamiltonian are easily calculated to be [a(k),H]=Eka(k)(20) a†(k),H = E a†(k)(21) − k So the annihilation and creation⇥ operators⇤ annihilate and create a quantum of energy Ek,respectively.Thisisidenticalinformtotheharmonicoscillator. Equations 20 and 21 therefore naturally lead to the description of the Klein- Gordon field as an infinite set of uncoupled harmonic oscillators, one for each point in space-time x. 2.3 Normal Ordering The problem with the harmonic oscillator description of the Klein-Gordon field is that harmonic oscillators have nonzero ground state energy due to an additive constant.