Kapitza Spring 2018

Second 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 particles, fails in the relativistic regime. By using , 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 , which avoids many of the problems of the first quantization.

2 Second Quantization of the Klein-Gordon Equation

2.1 The Klein-Gordon We can write the Klein-Gordon field as

3 ( ik x) (ik x) (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 ) =0 (3) This can be split into positive and negative energy components as follows: (+) ( ) (x)= (x)+ (x)

(+) 3 ( ik x) (x)=C d ke · a(k) (4) Z ( ) 3 (ik x) (x)=C d ke · a†(k) Z The conjugate momentum for the field operator is therefore ⇧(x)=˙(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 (x)+i⇧(x) Z ✓ ◆ 3 ik x ik x = iC d x e · @ (x) @ e · (x) (6) t t Z ✓ ◆

3 ik x 0 a†(k)=C d xe · k (x) i⇧(x) Z ✓ ◆ 3 ik x ik x = iC d x e · @ (x) @ e · (x) (7) t t Z ✓ ◆ Taking the time derivative of equation 6:

3 ik x ik x @ a(k)=iC@ d x e · @ (x) @ e · (x) t t t t Z ✓ ◆ 3 ik x ik x 2 ik x 2 ik x = iC d x @ e · @ (x)+e · @ (x) @ e · @ (x) @ e · (x) t t t t t t Z ✓ ◆ 3 ik x 2 2 ik x = iC d x e · @ (x) @ e · (x) t t Z ✓ ◆ must satisfy the Klein-Gordon equation, so

3 ik x 2 2 0 2 ik x @ a(k)=iC d x e · ( m )(x)+(k ) e · (x) t r Z ✓ ◆ 3 0 2 2 2 ik x = iC d x (k ) k m e · (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) [(x),(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) ⇥ ⇤ ◆ ⇥ ⇤ (8)=0

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) ⇥ ⇤ ◆ ⇥ ⇤ (9)=0

0 3 3 i(k x+k y) [(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) ⇥ ⇤ ◆ = i3(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 [(x),(y)] ZZ ✓ + ik0[(x), ⇧(y)] + ik00[⇧(x),(y)] [⇧(x), ⇧(y)] ◆ 3 3 i(k x+k y) 0 0 = CC0 d x d ye · 0· ik [(x), ⇧(y)]+ik0 [⇧(x),(y)] ZZ ✓ ◆ 0 0 3 3 i(k x+k y) 3 = CC0(k k0 ) d x d ye · 0· (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 · (k k0) (11)=0

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) = ⇤(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 m22 +( ) ( )+⇧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 x(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) (k + k0) 2 pk0 p 00 ZZ k ✓ i(k0 k00)x0 3 i( k0+k00)x0 3 + e a(k)a†(k0) (k k0)+e a†(k)a(k0) ( k + k0) i( k0 k00)x0 3 + e a†(k)a†(k0) ( 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 is found from equation 1:

3 ( ik x) (ik x) (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 (x) (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 energy due to an additive constant. Clearly, a system consisting of an infinite number of such oscillators will have infinite energy in its ground state alone, which is nonphysical. The way around this thorny issue is to take advantage of the fact that the infinite energy comes from an additive constant, and simply shift the energy scale down to make the ground state have zero energy. We accomplish this by imposing the normal order: ordering the annihilation and creation operators a and a† such that a† is always left of a.Itisdenotedby asuperscriptN.O. or surrounding colons.

N.O. aa† =: aa† := a†a ✓ ◆ The normal-ordered Hamiltonian is

N.O. 1 3 H = d kE : a(k)a†(k):+:a†(k)a(k): 2 k Z ✓ ◆ 3 = d kEka†(k)a(k)(22) Z 3 = d kEkN(k)(23) Z

6 where the number operator N(k)=a†(k)a(k). The total number operator is N = d3kN(k)(24) Z The commutator relations follow easily:

[a(k),N(k0)] = a(k),a†(k0) a(k) 3 = (k k0)a(k)(25) ⇥ ⇤ [a(k),N]=a(k)(26)

a†(k),N(k0) = a†(k) a†(k),a(k0) 3 = (k k0)a†(k)(27) ⇥ ⇤ ⇥ ⇤ a†(k),N = a†(k)(28) ⇥ ⇤

2.4 Operations on Eigenstates Consider an energy eigenstate E .Wewouldliketoseehowtheannihila- tion and creation operators interact| i with the Hamiltonian on this eigenstate. Applying the commutator of a and H to E ,weobtain | i [a(k),H] E = E a(k) E | i k | i Ha(k) E = E a(k) E a(k)H E | i k | i | i H(a(k) E )=(E E )(a(k) E )(29) | i k | i using the fact that H E = E E .Similarly, | i | i a†(k),H E = E a†(k) E | i k | i H(a†(k) E )=(E + E )(a†(k) E )(30) ⇥ ⇤| i k | i So annihilation lowers the energy of the system by Ek and creation raises the energy by Ek.Sincethegroundstateofthesystemhaszeroenergy,the energy must be nonnegative. So, there must exist a minimum energy state Emin such that | i a(k) E =0 | mini

Define 0 Emin . The ground state is such that H 0 = N 0 =0. Consider a| one-particlei⌘| i state with nonzero energy: | i | i

k = a†(k) 0 (31) | i | i 7 We can construct any higher state in a similar way: with repeated applica- tions of the creation operator. With normalization, these higher states can be formed by

(a (k ))n1 (a (k ))n2 (a (k ))nl ⌅ = † 1 † 2 ... † l 0 (32) | i pn1! pn2! pnl! | i where there are ni particles with momentum ki.Itisclearthatthereshould l be i=1 ni particles in this state, so

P N ⌅ =(n + n + + n ) ⌅ (33) | i 1 2 ··· l | i n n 1 n i 1 i Moreover, using the identity [A, B ]= i=0 B [A, B]B and equation 13, we can see that there are indeed ni particles with momentum ki,aswe expect. P

nj nj N(k), (a†(kj)) = a†(k) a(k), (a†(kj)) n 1 j ⇥ ⇤ ⇥ nj i ⇤1 i = a†(k) (a†(kj)) a(k),a†(kj) (a†(kj)) i=0 nX1 ⇥ ⇤ j n 1 3 = a†(k) (a†(k )) j (k k ) j j i=0 X n 3 = n (a†(k )) j (k k )(34) j j j

(a (k ))n1 (a (k ))n2 (a (k ))nl N(k) ⌅ = N(k) † 1 † 2 ... † l 0 | i pn1! pn2! pnl! | i n1 n2 nl (a†(k1)) 3 (a†(k2)) (a†(kl)) = n1 (k k1)+N(k) ... 0 pn ! pn ! pn ! | i 1 ✓ ◆ 2 l n1 n2 nl (a†(k1)) (a†(k2)) 3 3 (a†(kl)) = n1 (k k1)+n2 (k k2)+N(k) ... 0 pn ! pn ! pn ! | i 1 2 ✓ ◆ l n1 n2 nl (a†(k1)) (a†(k2)) (a†(kl)) 3 = ... n1 (k k1) pn ! pn ! pn ! 1 2 l ✓ + n 3(k k )+ + n 3(k k )+N(k) 0 2 2 ··· l l | i ◆ n1 n2 nl (a†(k1)) (a†(k2)) (a†(kl)) 3 = ... n1 (k k1) pn ! pn ! pn ! 1 2 l ✓ (35) + n 3(k k )+ + n 3(k k )) 0 2 2 ··· l l | i ◆

8 We can also introduce a momentum operator

P = d3k kN(k)(36) Z The actions of the Hamiltonian and momentum operators on our state ⌅ can be determined easily from equations 35: | i

H ⌅ = d3kE N(k) ⌅ | i k | i Zl = n E ⌅ (37) i ki | i i=1 X

P ⌅ = d3kkN(k) ⌅ | i | i Zl = n k ⌅ (38) i i | i i=1 X The energy and momentum of the state, the eigenvalues of the Hamiltonian and momentum operators, are simply the sum of the energies and momenta of all of the particles in the state.

Consider the state in equation 14, k .Thisrepresentsaoneparticlestate µ | i with four-momentum k =(Ek, k). Clearly,

N k = k | i | i H k = E k (39) | i k | i P k = k k | i | i In a similar way, we can apply the field operator itself to the ground state:

( ) (x) = (x) 0 = (x) 0 (40) | i | i | i (+) The phi term is zero because it involves a and not a† (equation 4). The projection onto onto k is k (x) .Thisprojectionisasolutiontothe Klein-Gordon equation,| i just| i likeh | itself:i

(@ @µ + m2) k (x) = k (@ @µ + m2)(x) 0 µ h | i h | µ | i =0 k 0 h | i =0 (41)

9 So k (x) has a relation to the solutions of the Klein-Gordon equation. h | i What is this relation? We proceed by calculating (x) k = k (x) ⇤. h | i h | i ( ) (x) k = 0 a(k) (x) 0 ⇤ h | i h | | i 3 ik x ⇤ = C0 d k0 e 0· 0 a(k)a†(k0) 0 h | | i ✓ Z ◆ 3 ik x = C0 d k0 e 0· 0 a†(k0)a(k) 0 + 0 [a(k),a]†(k0) 0 h | | i h | | i Z ✓ ◆ 3 ik x 3 = C0 d k0 e 0· (k k0) Zik x = Ce · (42)

This is recognized as the plane-wave solution of the Klein-Gordon equation in the first quantization, connecting the two quantizations. Note that the probability of the system being in the state k is | i k (x) 2 = k (x) (x) k |h | i| h | ih | i = C2 0(43) since k0 = Ek > 0. The probability for any composite state can also be found to be nonnegative.

3 Conclusion

Framing the Klein-Gordon equation within quantum field theory results in the second quantization. As we have seen, the field can be described as an infinite set of uncoupled oscillators, one for each point in space-time. Moreover, the probabilities to be in any state and the energies of those states are all nonnegative. In this way, the second quantization of the Klein-Gordon avoids many of the issues that plague the first quantization.

References

[1] Das A. 2008. Lectures on . Hackensack, NJ: World Scientific Publishing.

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