## Monday, July 14, 2014

### Quantum Vacuum Fluctuations - The Basics

Quantum vacuum fluctuations are facts in the real world - they were a fundamental prediction of Quantum Mechanics revealed in several processes. The spectrum of quantum fluctuations is a neat mathematical formulation that embodies this concept. It serves as a pivotal point between the very large (cosmological scale) and the very small (subatomic scale). No cosmological model that aspires to describe the real world can have the luxury of ignoring this fundamental reality.

In Flat Spacetime

Consider a scalar field in flat spacetime (equation 37 in The Essential Quantum Field Theory)

(1) Φk = ∫d3k(2π)-3(2ωk) [akeik∙x + ake-ik∙x]

We define the quantum fluctuations as,

(2) δΦk ≡ (< |Φk|2 >)½

Substitute 1 into 2,

(3) δΦk ~ (ωk)

Field values cannot be measured at a point - in a realistic experiment, only their values averaged over a finite region of space can be measured. Consider the average value of a field Φ(x) in a cubed-shape region of volume L3.

(4) ΦL ≡ L-3-L/2+L/2dx ∫-L/2+L/2dy ∫-L/2+L/2dz Φ(x)

For k = L-1

(5) δΦL ~ {(δΦk)2k3}½ (See Appendix A)

Using equation 1, we get,

(6) δΦL ~ ( kL3kL )½, kL ≡ L-1

We see that δΦL diverges for small L (large k, large mass), decays for large L (small k, small mass). Quantum vacuum fluctuations have been observed in the spontaneous emission of radiation by atoms, the Lamb shift, and the Casimir effect. These observations cannot be explained by any other known physics.

In Curved Space-Time

Recall equation 21 from Quantum Fields in Curved Space-Time , written below,

(7) Χ(x,η) = 2∫d3k(2π)-3/2[eikxv*k(η)ak-
+ e-ikxvk(η)ak+]
Given a quantum state, the amplitude of quantum fluctuations is always well-defined irrespective of whether the particle interpretation of the field is available (see discussion after equation 47 in Quantum Fields in Curved Space-Time )

Let us consider the equal-time correlation function for the vacuum state

(8) < 0|Χ(x,η)Χ(y,η)|0 > ~ k3 (sinkL)/(kL)|vk(η)|2 (See appendix B)

We can see that at k ~ L-1

(9) (sinkL)/(kL) ~ 1

(10) Therefore, < 0|Χ(x,η)Χ(y,η)|0 > ~ k3|vk|2

Fluctuations of Spatially Averaged Fields

There is another way to characterize the average fluctuations in a box of size L. This is done by using a window-averaged operator,

(11) XL(η) = ∫ X(x,η)WL(x) d3x

This function must satisfy the condition,

(12) ∫ WL(x) d3x = 1

A typical example of a window function is the Gaussian function,

(13) WL(x) = (2π)-3/2L-3 exp{ -|x|2/(2L2)}

This has the following scaling property,

(14) WL'(x) = (L3/L'3)WL(x)((L/L')x)

The Fourier image is then defined as,

(15) w(kL) = ∫WL(x)e-ikxd3x

Which satifies,

(16) w|k=0 = 1 and decays rapidly for |k|≥ L-1.

We can now calculate,

(17)δXL2(η)= < 0| [∫ X(x,η)WL(x) d3x]2 |0 >, (from equ. 11)

Using equation 7 and 15, this reduces to,

(18) δXL2(η) ~ ∫ d3k|w(kL)|2|vk|2

Applying conditions 16, we get

(19) δXL2(η) ~ k3|vk|2

Which is the same result for the two-point correlation we found in curved space-time (equation 10). Therefore we define,

(20) δ(k) ≡ (2π)-1 k3/2|vk|

as the spectrum of quantum fluctuations, which characterizes the typical amplitude of quantum fluctuations on scales L.

Complimentary Notes: In a de Sitter space, the spectrum of quantum fluctuations ~ as the Hubble constant. One can show that the quantization of gravitational waves in an expanding universe is reduced to the problem of quantizing a massless scalar field. Furthermore, the production of primordial inhomogeneities can be found in a nearly scale-invariant spectrum.

Appendix A

(A1) ΦL = L-3L3 Φ(x)d3x

~ L -3L3 d3x ∫d3k eikxΦk
The integral over d3x can be computed first. Consider that (x = x,y,z), then the x-component of that integral is,

(A2) ∫-L/2+L/2 dx eikxx = ∫-1 1 d(cosθ)eikxLcosθ

= (2/kxL) sin(kxL/2)
≡ f(kx)
We get similar expression for the y and z components. The expectation value for ΦL2 is

(A3) < ΦL2 > ~ ∫d3kd3k'< ΦkΦk' >f(kx)f(ky)f(kz)f(k'x)f(k'y)f(k'z)

(A4) But < ΦkΦk' > = (δΦk)2δ(k+k') (equ.2)

(A5) Therefore, < ΦL2 > ~ ∫d3k (δΦk)2 {f(kx)f(ky)f(kz)}2

The function f(k) is of order 1 for |kL| ≤ 1, and very smal for |kL| >>1. We may take δΦk to be constant over the integration. Then

(A6) < ΦL2 > ~ ∫d3k (δΦk)2 ~ k3(δΦk)2, k = L-1

Appendix B

From equation 7 reproduced below,

(B1) Χ(x,η) = 2∫d3k(2π)-3/2[eikxv*k(η)ak-
+ e-ikxvk(η)ak+]
(B2) Calculating

< 0|Χ(x,η)Χ(y,η)|0 > = < 0| 2 ∫d3k (2π)-3/2[eikxv*k(η)ak- + e-ikxvk(η)ak+ ]
x 2 ∫d3k' (2π)-3/2[eik'yv*k'(η)ak'- + e-ik'yvk'(η)ak'+]|0 >
= < 0|2-1(2π)-3 ∫d3k ∫d3k' [eikxv*k(η)ak-eik'yv*k'(η)ak'- + eikxv*k(η)ak-e-ik'yvk'(η)ak'+
+ e-ikxvk(η)ak+eik'yv*k'(η)ak'- + e-ikxvk(η)ak+e-ik'yvk'(η)ak'+] |0 >
Recall that,

(B3) ak-|0 > = < 0|ak+ = 0

The only surviving term in B2 is,

(B4) < 0|Χ(x,η)Χ(y,η)|0 > = < 0|2-1(2π)-3 ∫d3k ∫d3k' e i(kx - k'y) v*k(η)ak-vk'(η)ak'+ |0 >

Recall equation 25 from Quantum Fields in Curved Space-Time , written below,

(B5) [ak-,ak'+] = δ(kk'), [ak-,ak'-]= 0, [ak+,ak'+]= 0

From the first, we have

(B6) ak-ak'+ = δ(kk') + ak'+ak-

Substituting B6 into B4, taking care of B3

(B7) < 0|Χ(x,η)Χ(y,η)|0 > = < 0|2-1(2π)-3 ∫d3k ∫d3k' ei(kx - k'y)δ(kk')v*k(η)vk'(η) |0 >

= 2-1(2π)-3 ∫d3k eik∙(x - y) v*k(η)vk(η)
= 2-1(2π)-3 ∫d3k eik∙L v*k(η)vk(η), where L = |x - y|
~ k3 (sinkL)/(kL)|vk(η)|2, (Using equation A2)

## Tuesday, June 17, 2014

### Quantum Fields in Curved Space-Time

(1) ds2 = dt2 - a2(t)δijdxidxj

(See equation 10 in The Essential General Relativity )

(2) Define the conformal time dη(t) ≡ dt/a(t)

Substitute into 1, we get,

(3) ds2 = a2(η)[dη2 - δijdxidxj]

= a2(η)ημνdxμdxν
(4) where
Borrowing equations 12,13 in The Essential Quantum Field Theory

(5) ℒ = ½ ημνμϕ∂νϕ − ½ m2ϕ2

(6) ∂μμϕ + m2ϕ = 0

The corresponding action is,

(7) S = ½∫d4x ημνμϕ∂νϕ − m2ϕ2

To generalize this action to the case of a curved spacetime, we need to,

(8) replace ημν with the metric gμν. In the conformal time η, gμν = a2ημν (equation 3).

(9) Instead of the usual volume d4x (≡ d3xdt), use the covariant volume element d4x(-g)½.

The action (7) becomes,

(10) S = ½∫d4x(-g)½[ gμνμϕ∂νϕ - m2ϕ2]

(11)Using the conformal time (equation 2) and
note: det(g) = -a8, or (-g)½ = a4, the action becomes.

(12) S = ½∫d3xdηa2[ϕ'2 - (∇ϕ)2 - m2a2ϕ2]

(13) Define the auxiliary field as,

Χ ≡ a(η)ϕ

(14) Using the above, the action is,

S = ½∫d3xdη[Χ'2 - (∇Χ)2 - (m2a2 - a"/a)Χ2]
(See appendix A for derivation)

(15) Define the effective mass as,

meff(η) = m2a2 - a"/a,

(16) Equation 14 now reads as,

S = ½∫d3xdη[Χ'2 - (∇Χ)2 - meff(η)Χ2]

The basic difference between equation 6 (flat spacetime) and equation 14 (curved spacetime) is that,

(17) m → meff(η), which is time-dependent.

We find that the field Χ obeys the same equation of motion as a massive scalar field in Minkowski space, except that the effective mass becomes time-dependent. This means that the energy is not conserved, and in QFT, this leads to particle creation: the energy of the new particles is supplied by the classical gravitational field.

Quantization

We repeat the steps (14 to 22) in The Essential Quantum Field Theory .

(18) Define the canonical conjugate momentum,

π = ∂ℒ/∂Χ' = Χ'

(19)The commutator between Χ and π is,

[Χ(x,η),π(y,η)] = iδ(xy)

(20) The Hamiltonian is,

H(η) = ½∫d3x [π2 + (∇Χ)2 + meff(η)2Χ2]

The field operator Χ is expanded as a Fourier expansion (equation 37 in The Essential Quantum Field Theory ), but in this case, we need to take care that the Hamiltonian is time-dependent. So we write,

(21) Χ(x,η) = 2∫d3k(2π)-3/2[eikxv*k(η)ak-

+ e-ikxvk(η)ak+]
(22) where for the functions vk(η) and v*k(η), are time-dependent but the Wronskian, W[vk,v*k] ≠ 0, is time -independent (see appendix B)

The equation of motion, which corresponds to equation (6), is

(23) v" + ωk2(η)vk = 0,

(24) Where ωk(η)≡ (k2 + meff2(η))½

(25) Substitute 21 into 19, we get the following

[ak-,ak'+] = iδ(kk'), [ak-,ak'-]= 0, [ak+,ak'+]= 0

Making the a±k's the creation and annihilation operators (see equation 28,29 in Harmonic Oscillators, Vacuum Energy... ), provided that the functions vk(η) and v*k(η) also satisfy,

(26) Im(v'kv*k) = 1,

This is referred as the normalization condition (See appendix C).

Recap

Comparing our solution in curved spacetime to flat space:

(27) The oscillator equation has an effective mass, which is time-dependent (equation 15).

(28) The Hamiltonian is time-dependent (equation 20).

(29)The field Χ has extra functions vk(η) and v*k(η), (equation 21)

with condition 26.

Bogolyubov Transformations

The quantum states acquire an unambiguous physical interpretation only after the particular mode functions vk(η) are selected. The normalization (26) is not enough to completely satisfy the differential equation (23). In fact, one can argue that,

(30) uk(η)= αkvk(η) + βkv*k(η),

Also satisfy equation 23, where αk and βk are time-independent complex coefficients. Moreover if they obey the condition,

(31) |αk|2 - |βk|2 = 1,

then the uk(η) satisfy the normalization condion (26). See appendix D.

In terms of the mode uk(η), the field operator Χ(x,η), equation 21, is now as,

(32) Χ(x,η) = 2∫d3k(2π)-3/2[eikxu*k(η)bk-

+ e-ikxuk(η)bk+]
Where the b±k's are another set of creation and annihilation operators, satisfying equation 25. Note for the two expressions ( 21 and 32) for the same field operator Χ(x,η), then

(33) u*k(η)bk- + uk(η)bk+ = v*k(η)ak- + vk(η)ak+

Using equation 30, we get,

(34A) ak- = α*kbk- + βkbk+

(34B) ak+ = αkbk+ + β*kbk-

These are called the Bogolyubov transformations. We can reverse these as,

(35A) bk- = αkak- - βkak+

(35B) bk+ = α*kbk+ - β*kak-

The a-particles and the b-particles

Both the a±k's and the b±k's can be used to build orthonormal bases in the Hilbert space. We define the vacuum in the standard way, (see reference in 25)

(36) a-k|(a)0 > = 0, b-k|(b)0 > = 0, for all k.

Note: we have an a-vacuum and a b-vacuum, and two sets of excited states,

(37A) |(a) mk1 ,nk2... > = N(a) [(ak1+)m(ak2+)n...] |(a)0 >

(37B) |(b) mk1 ,nk2... > = N(b) [(bk1+)m(bk2+)n...] |(b)0 >

Where N(a) and N(b) are just normalized factor.

The b-vacuum can be expressed as a superposition of the excited a-particle states, (Appendix F)

(38) |(b)0 > = [ ΠkCk exp{(βk/2αk)a+ka+-k}]|(a)0 >

(39) Note: quantum states which are exponential of a quadratic combination of creation operators acting on the vacuum are called squeeze states.

It is clear that the particle interpretation of the theory depends on the choice of the mode functions. Also, the b-vacuum, a state without b-particles, nevertheless can contain a-particles! The question is, which set of mode functions is preferable to describe the real physical vacuum and particles?

The Instantaneous Lowest-Energy State

In flat space we defined the eigenstate with the lowest possible energy of a Hamiltonian that was independent of time. (See the discussion after equation 10 in The Essential Quantum Field Theory). However, from the above equation 20, we have a Hamiltonian in curved space-time that is time dependent. We could circumvent this by looking at a given moment of time η0, and define the instantaneous vacuum |η00 > as the lowest energy state of the Hamiltonian H(η0).

Problems

(i)Substitute 18 and 21 into 20 we get,

(40) H(η) = (1/4)∫d3k [ a-ka--kF*k+a+ka+-kFk+(2a+ka-k + δ(3)(0) Ek)]

(41) Ek(η) ≡ |v'k|2 + ω2k(η)|vk|2

(42) Fk(η) ≡ v'k2 + ω2k(η)vk2

When we compare this with our result in flat space-time, Equation 44 in The Essential Quantum Field Theory, reproduced below,

(43) H = ∫ d3k(2π)-3ωk(ak ak + ½(2π)3δ(3)(0))

Note: ak → a+k and ak → a-k

We see that in equation 40 we get an extra term with Fk(η). Unless Fk(η)=0, the vacuum state cannot remain an eigenstate of the Hamiltonian. See appendix G.

(ii)Starting with a vacuum at η0, the vacuum expectation value would be (from equation 43 and omitting factors of 2π),

(44) < 0) 0 |H(η0)| 0) 0 > = ∫d3k0)(a+k0)a-k0) + ½δ(3)(0) )

However at a later time η1, the Hamiltonian H(η1) in the vacuum state |0) 0 > would be,

(45) < 0) 0 |H(η1)| 0) 0 > = ∫d3k1)(a+k1)a-k1) + ½δ(3)(0) )

Now the a±k0) and the a±k1) are related by the Bogolyubov transformations {Equations 34A, 34B with a±k → a±k1) and b±k → a±k0)}

(46A) ak-1) = α*kak-0) + βkak+0)

(46B) ak+1) = αkak+0) + β*kak-0)

Substituting 46A, 46B into 45, we get (see appendix H)

(47) < 0) 0 |H(η1)| 0) 0 > = δ(3)(0)∫d3k1){½ +|βk|2}

Unless βk = 0 for all k, this energy is larger than the minimum possible value and the state
| 0) 0 > contains particles at time η1.

Ambiguity of the Vacuum State

i) The usual definition of the vacuum and particle states in Minkowski (flat) spacetime is based on a decomposition of fields in plane waves (eikx-iwkt, equation 31 in The Essential Quantum Mechanics ). In this argument, a particle is localized with momentum k, described by a wave packet with momentum spread ∆k. That is, the momentum is well-defined only if ∆k << k, which implies that (λ ~ 1/∆k) λ >> 1/k. In curved spacetime, the geometry across a region of size λ could vary significantly, and plane waves are no longer good approximations.

ii) The vacuum and particle states are not always well-defined for some modes.

ω2k(η) = k2 + m2a2 - a"/a

Certain modes can be negative for k2 + m2a2 < a"/a, in particular the excited states. The argument that there is a tower of energy states (see equation 32 in The Essential Quantum Field Theory ) and these must have a ground state (a least positive energy level) no longer holds.

iii) An accelerated detector in flat spacetime can register particles even when the field is in a true Minkowski vacuum state (see The Unruh Effect). Therefore, the definition of a particle state depends on the coordinate system. In curved spacetime, there is no preferable coordinate system - this is what GR was fundamentally based on. In the presence of gravity, energy is no longer bounded below, and the definition of a true vacuum state as the lowest energy state fails.

iv) We can still have an approximate particle state definition in a spacetime with slowly changing geometry. In this description, in the case that ωk(η) tends to a constant both in the remote past (η << η1) and in the future (η >> η2), one can unambiguously define "in" and "out" states in the past and future respectively.

On the other hand, the notion of a particle state is ambiguous in the intermediate regime, η1 < η < η2, when ωk(η) is time-dependent. The reason is that the vacuum fluctuations are not only excited but also deformed by the external field. This latter effect is called the vacuum polarization. Nevertheless, the absence of a generally valid definition of the vacuum and particle states does not impair our ability to make predictions for certain specific observable quantities in a curved spacetime, one of which is the amplitude of quantum fluctuations, which has played a pivotal role in filtering out the cosmological models on pre-bang activities. More to say on the spectrum of quantum fluctuations later on.

Appendix A

(A1) ϕ = Χ/a (equation 13)

(A2) (-g)½ = a4 (equation 11)

(A3) (-g)½m2ϕ2 = m2a2Χ2 (equations A1 and A2)

(A4) Take the derivative of equation A1,

ϕ' = Χ'/a - Χa'/a2

(A5) Square A4, and multiply throughout by a2

ϕ'2a2 = Χ'2 - 2ΧΧ'(a'/a) + Χ2(a'/a)2

(A6) ϕ'2a2 = Χ'2 + Χ2(a"/a) - [Χ2(a'/a)]'

The last term is a total derivative and can be omitted.

(A7) ϕ'2a2 = Χ'2 + Χ2(a"/a)

Appendix B

For the oscillator equation,

(B1) x" + ω2x = 0 , (see equation 3 in
Harmonic Oscillators, Vacuum Energy... )

Consider taking the derivative of x'1x2 - x1x'2,
where x1(t) and x2(t) are two solutions to B1,

(B2) = x"1x2 - x1x"2

= ω2x1x2 - x1ω2x2, using equation B1

= 0

This means that the solutions x1(t) and x2(t) are linearly dependent, since we can express,

(B3) x2(t) = λx1(t) , where λ is a constant, and this is true for ALL t.

The Wronskian, W(x1(t),x2(t))≡ x'1x2 - x1x'2

= x'1λx1(t) - x1λx'1(t), using B3

= 0

Therefore, if W(x1(t),x2(t))≠ 0, we can say that the two solutions are time-independent.

Appendix C

Definition of the Wronskian for equation 23,

(C1) W[vk,v*k] ≡ v'kv*k - vkv*'k] We will show that for equation 23,

(C2) W[vk,v*k] = 2iIm(v'kv*k)

Proof:

We will drop the subscript k as it is not relevant in this case. A solution to equation 22 ( using η → t)

(C3) v → eiω(t)t , v* → e-iω(t)t

Taking derivatives,

(C4) v' = iωeiω(t)t + ω'(it)eiω(t)t,
v*' = -iωe-iω(t)t - ω'(it)e-iω(t)t

Calculating the RHS of C1,

(C5)v'kv*k - vkv*'k

= (iωeiω(t)t+ω'(it)eiω(t)t)e-iω(t)t-eiω(t)t(-iωe-iω(t)t-ω'(it)e-iω(t)t

= 2i(ω + ω't)

Calculating the RHS of C2,

(C6) 2iIm(v'kv*k) = 2i Im((iω eiω(t)t + ω'(it)eiω(t)t)e-iω(t)t)

= 2i(ω + ω't)

(C7) Therefore,W[vk,v*k] = 2iIm(v'kv*k)

Appendix D

(D1) uk(η)= αkvk(η) + βkv*k(η) , Equation 30

(D2) Take the complex conjugate of D1,

u*k(η)= α*kv*k(η) + β*kvk(η) ,

Take the derivative of D1,

(D3) u'k(η)= αkv'k(η) + βkv*'k(η) ,

Take the derivative of D2,

(D4) u*'k(η)= α*kv*'k(η) + β*kv'k(η) ,

For the normalization condition, we need to calculate equation C1,

(D5) u'k(η)u*k(η) - uk(η)u*'k(η) =

kv'k(η) + βkv*'k(η))(α*kv*k(η) + β*kvk(η))
- [kvk(η) + βkv*k(η))(α*kv*'k(η) + β*kv'k(η)]
= αkv'k(η)α*kv*k(η)+ βkv*'k(η)α*kv*k(η)
+ αkv'k(η)β*kvk(η) + βkv*'k(η)β*kvk(η)

- [αkvk(η)α*kv*'k(η) + βkv*k(η)α*kv*'k(η)
+ αkvk(η)β*kv'k(η) + βkv*k(η)β*kv'k(η)]
Rearranging,

= |αk|2v'k(η)v*k(η)+ α*kβkv*'k(η)v*k(η)
+ αkβ*kv'k(η)vk(η) + |βk|2v*'k(η)vk(η)

- |αk|2vk(η)v*'k(η) - α*kβkv*k(η)v*'k(η)
- αkβ*kvk(η)v'k(η) - |βk|2v*k(η)v'k(η)
= |αk|2v'k(η)v*k(η) + |βk|2v*'k(η)vk(η)

- |αk|2vk(η)v*'k(η) - |βk|2v*k(η)v'k(η)
= |αk|2 (v'k(η)v*k(η)- vk(η)v*'k(η))
+ |βk|2 (v*'k(η)vk(η) - v*k(η)v'k(η))
= (|αk|2 - |βk|2) (v'k(η)v*k(η)- vk(η)v*'k(η))

If condition 31 is met, that is,

(D6) |αk|2 - |βk|2 = 1

Then equation D5 becomes,

(D7)u'k(η)u*k(η)-uk(η)u*'k(η)= v'k(η)v*k(η)-vk(η)v*'k(η)

(D8) Or Im(v'kv*k) = 1 = Im(u'ku*k)

Appendix E

Consider any two operators A, B, and the commutator between them, (See equation 13 in The Essential Quantum Mechanics EQM)

(E1) [A,BC] = ABC - BCA
= ABC - BAC + BAC - BCA
= [A,B]C + B[A,C]
Consider,

(E2) [q,p2] = [q,p]p + p[q,p] (equ. E1)
= (iℏ)p + p(iℏ) (equ. 36 in EQM)
= (iℏ)2p
We can generalize this result to,

(E3) [q,pn] = (iℏ)npn-1

Consider a generalized term in the form of qapbqc. Then,

(E4) [q,qapbqc] = (iℏ)(b)qapb-1qc, (using equ. E3)

≡ (iℏ)∂qapbqc/∂p
We can generalize this to any function of q and p as

(E5) [q,f(q,p)] = (iℏ)∂f(q,p)/∂p

The analogous relation with p is automatically obtained by interchanging, q → p and iℏ → -iℏ

(E6) [p,f(q,p)] = (-iℏ)∂f(q,p)/∂q

For any two operators that obey a similar commutation relationship as q and p,that is (ℏ =1) ,

(E7) [ak-,ak'+] = iδ(kk'), (equ. 25)

We can further generalize equation E5 as,

(E8) [ak-, f(ak-,ak+)] = i∂f(ak-, ak+)/∂ak+

Appendix F

We consider the quantum state of a single mode ϕk. The b-vacuum can be expanded as a linear combination of the a-vacuum,

(F1) |(b)0 k,-k > = Σm,n=0 Cmn |(a)mk,n-k >

(F2) where from (Equ. 37A)
|(a)mk,n-k > = N(a)(ak+)m(a-k+)n|(a)0k,-k >,

This implies that the b-vacuum is a combination of operators acting on the a-vacuum. We denote this combination as f(ak+,ak-). We can find an expression for this function from,

(F3) (αkak- - βka-k+)f(ak+,ak-)|(a)0k,-k > = 0

(F4) (αka-k- - βkak+)f(ak+,ak-)|(a)0k,-k > = 0

From E8, we can use the derivative of f(ak+,ak-) with respect to ak+ and write F3 as,

(F5) (αk∂f/∂ak+ - βka-k+f) |(a)0k,-k > = 0

We now have an equation with only creation operators. Therefore,

(F6) (αk∂f/∂ak+ - βka-k+f) = 0

A solution to this equation is,

(F7) f(ak+,ak-) = C(a-k+)exp{(βkk)a+ka+-k}

A similar equation can be written with F4 and the derivative of f(ak+,ak-) with respect to ak- to show that C is a constant, independent of a-k+. So F1 becomes,

(F8) |(b)0 k,-k > = f(ak+,ak-)|(a)0k,-k >
= [Cexp{(βkk)a+ka+-k}]|(a)0k,-k >
= [C{Σn=0kk)n(a+k)n(a+-k)n}]|(a)0k,-k >
However, the b-vacuum state is a tensor product of all the modes. Secondly, each pair is counted twice for ϕk, ϕ-k. So in addition to the product, we need to take the square root.

(F9) |(b)0 > = [πkCkn=0kk)n(a+k)n(a+-k)n}½]|(a)0 >
= [ ΠkCk exp{(βk/2αk)a+ka+-k}]|(a)0 >
Appendix G

The mode vk must satisfy the normalization condition (equation 26 reproduced below),

(G1) Im(v'kv*k) = 1,

(G2)Where W[vk,v*k] = 2iIm(v'kv*k)(equ. C2)

(G3)And W[vk,v*k] ≡ v'kv*k - vkv*'k] (equ. C1)

However the function Fk(η) must equal to zero to define the eigenstate of vacuum for the Hamiltonian in equation 40

(G4) Fk(η) ≡ v'k2 + ω2k(η)vk2 = 0 (equ. 42)

This differential equation has the exact solution,

(G5) vk(η) = C exp{±i∫ωk(η)dη}

And this does not satisfy the normalization condition if ωk(η) is dependent on time.

For the proof, we will consider just the positive in the exponent (the negative will follow logically). That is,

(G5') vk(η) = C exp{+i∫ωk(η)dη}

Take the logarithm of each side,

(G6) lnvk(η)-C = i∫ωk(η)dη

Take the derivative with respect to the conformal time η,

(G7) -Cv'k(η)/vk(η)= iωk(η)

Rearranging,

(G8) v'k(η)/vk(η)= -iC-1 ωk(η)

It follows for the complex conjugate,

(G9) v*'k(η)/v*k(η)= +iC-1 ωk(η)

The Wronskian is,

(G10)W[vk,v*k] ≡ v'kv*k - vkv*'k](equ. G3)

Substitute G8 and G9 into G10,

(G11)W[vk,v*k]=-iC-1ωk(η)vk(η)v*k - vkv*k(η)iC-1ωk(η)

= -2iC-1|vk(η)|2ωk(η)
= -2iCωk(η)
For the normalization condition to be satisfied (G1), W has to be a constant, and hence time-independent, but it is dependent on ωk(η), which is time-dependent (equation 24). We see that the exact solution to the equation, Fk(η)=0, does not satisfy the normalization condition.

Appendix H

We just need to calculate, < 0) 0 |(a+k1)a-k1)| 0) 0 >

Substitute 46A, 46B, we get

(H1) < 0) 0 |(a+k1)a-k1)| 0) 0 > = < 0) 0 |[αkak+0) + β*kak-0)] [α*kak-0) + βkak+0)]| 0) 0 >

Recall that ak-0)| 0) 0 > = 0 and < 0) 0 |ak+0) = 0. The only surviving term in H1 is,

(H2) < 0) 0 |β*kak-0kak+0)| 0) 0 > = |βk|2 δ(3)(0)

## Tuesday, June 03, 2014

### Effective Field Theory Made Simple

Math Background

A functional is a function of a function: F[x(t)] is a function of x, which is a function of t. The use of square brackets is standard practice.

Also δ F[x(t)]/δx(t) will denote the derivative of F[x(t)] with respect to x(t).

A Wick rotation is given by t → −iτ . If we substitute this into non-Euclidean geometry, more specifically, a Minkowski geometry with signature (−+++),

ds2 = -c2dt2+ dx2 + dy2 + dz2

(See equation 10 in The Essential General Relativity )

We get,

(1) ds2 = c22+ dx2 + dy2 + dz2,

And that gives Euclidean geometry.

Probability Amplitude

(2) < q2,t2| q1,t1 >Heisenberg picture

= < q2|U(t2,t1|q1 > Schroedinger picture

=< q2|e(t2−t1)H/(iℏ)|q1 > Schroedinger picture

(See equation 4 in The Path Integral Simplified)

= ∫D[q(s)] e iS[q]/ℏ

(See equation 25 The Path Integral Simplified)

(3) U(t) = e tH/iℏ = Σn |n >< n| e tEn/iℏ

Here we assume that the spectrum is discrete,

E0< E1< E2< … En

Looking at time as a complex number, the operator U(t) is bounded if Im(t) ≤0. So it is well-defined in the bottom half of the complex plane.

(4) Let t = − iτ , where τ is the Euclidean time.

(5) U(t = − iτ) = e −τH/ℏ ≡ UE(τ)

Note that UE(τ) has the same form as the density operator for a quantum state in a mixed state at temperature T > 0.

(6) ρT = (e−βH)/Z, where β = 1/kBT,

(7) where Z = Tr(e−βH), and Z is the partition function.

(8) Note that we have made the equivalence,

1/kBT ↔ τ/ℏ

We want to calculate a path integral for < q2|UE(τ)|q1 > in the Euclidean geometry.

Consider the partial argument in the exponential function of equation 2:

(9) iS[q] = i∫0tds [½m(dq/ds)2 − V(q)]

(10) Let s → − iσ ; t → − iτ

(11) iS[q] = i ∫0τ (−i)dσ [½m(i dq/dσ)2 − V(q)]

= ∫0τ dσ[−½m(dq/dσ)2 − V(q)]

= −∫0τ dσ [½m(dq/dσ)2 + V(q)]

≡ − SE[q], which is the Euclidean Action.

(12) Therefore, < q2|UE(τ)|q1 > = ∫D[q(s)] e -SE[q]/ℏ

The path integral in Euclidean time means q(0) = q1 and q(τ) = q2

Note the differences:

(13) S[q] = ∫0t ds [½m(dq/ds)2 − V(q)]
⇒ SE[q] = ∫0τ dσ[½m(dq/dσ)2 + V(q)]

(14) ∫D[q(s)]e iS[q]/ℏ ⇒ ∫D[q(σ)] e -SE[q]/ℏ

Note:

i) SE[q] is positive, and as it gets larger, e-SE[q]/ℏ becomes very small.

ii) Also, the Euclidean time τ is positive and is analogous to temperature T.

(15) Z = Tr[UE(τ)] = ∫dq< q|UE(τ)|q >

= ∫ D[q(σ)]e-SE[q]/ℏ, q(0) = q(τ).

iii) This is analogous of having periodic conditions, with τ being the period of Euclidean time.

QM at finite temperature ⇔ complex time on a cylinder with Euclidean period.

Expectation Value

We are interested in expectation values in QM. For some operator A in a thermal state β, which is a function of q, that is, A → a(q).

(16) < A >β = Tr(ρA), where β = (kbT)-1,
and density matrix ρ = UE(τ)/Z

= ∫dq< q| ρA|q > → ∫dq< q|ρ|q > a(q)

= ∫dq< q| UE(τ)/Z|q > a(q)

(17)

Two-Operator Expectation Value

(18) < A(σ1)B(σ2) >β
= (1/Z)∫ D[q]e-SE[q]/ℏ a(q(σ1)) b(q(σ2))

According to 16, if we follow the trace we get,

(19) Tr[UE(τ − σ2)BUE2− σ1)AUE1)]
= Tr[UE(τ − (σ2− σ1))BUE2− σ1)A]

i) What happens when τ goes to infinity?

Recall from (Equ. 16),
< A >β = Tr(ρβA),
ρβ = (e–βH)/Z = (1/Z) Σn|n >< n|e–βH
Z = Σn e–βEn
β = 1/kBT = τ/ℏ

Combining the above,

So when τ → ∞ , β → ∞, T → 0, we get the lowest energy, which is E0.

(21) < A >β ≈ e–βE0 <0| A|0>/ e–βE0

= < 0|A|0 >

= expectation value (ev) in the pure ground state of the system

Therefore, τ → ∞ ⇔ projecting on the ground state (vacuum).

ii) Consider 2 operator ev

What happens when Euclidean time → Real time?

Take the Wick rotation: σ1 = i t1 , σ2 = i t2

With σ1 < σ2

Also, let τ → ∞ ⇔ projecting on |0 > .

The contour is deformed: we go from U(t1), where we operate A, we go U(t2 - t1), where we operate B, then go U(-t2), we then project on the vacuum.

(22) < A(σ1)B(σ2) >β=∞
< 0|U(-t2)BU(t2-t1)AU(t1)|0 >

= < 0|U(-t2)BU(t2)U(-t1)AU(t1)|0 >

Recall: U(-t2)BU(t2) ≡ B(t2), and U(-t1)AU(t1) ≡ A(t1), in the Heisenberg picture.

(23) < A(σ1)B(σ2) >β=∞
< 0|B(t2)A(t1)|0 > Heisenberg picture if t2 > t1.

Note: this evaluation depends on the order of time. So generally,

(24) < A(σ1)B(σ2) >β=∞ = < 0| T[A(t1)B(t2)]|0 >, where T is the time-ordered operator.

Recapitulating:

QM at finite temperature ⇔ complex time on a cylinder with Euclidean period, and from vev and projecting, we recover the time-ordered operator for two operators that is, the 2-point correlation function,

(25)

We can extend this procedure to an N-point correlation function (see equation 29 below).

Effective Field Theory

In equation 15, let q ⇒ Φ, and adding a term for the source J, we get,

(26) Z[ j ] = ∫ D[Φ(x)] e −(S[Φ] − j∙Φ)/ℏ

Φ(x) can be considered as a random variable; and j(x) the source term is not a random term, but can be construed as a perturbation by an external classical field applied to the system coupled to the quantum field Φ. Also we dropped the subscript E on S[Φ] (equation 14).

(27) By definition, j∙Φ ≡ ∫ ddz j(z)Φ(z)

Expanding Z[ j ] in powers of j,

(28) Z[ j ] = ΣN((ℏ−N)/N!)∫dz1…dzNj(z1)… j(zN)Z(z1…zN)

(29)Where Z(z1…zN) = ∫D[Φ(x)] e−S[Φ]/ℏ Φ(z1)… Φ(zN)

(30) Define W[ j ] = ℏlog(Z[ j ] )

We also define φ(x) as a functional of j, called the background field. We want to know what are the properties of the quantum theory as a function of the response to j, that is, as a functional of the background field φ(x).

Legendre transformation:

(31) Γ[φ] = j∙φ − W[ j ], where Γ[φ] is the effective field of the theory.

(32) Note that j(x) = δΓ[φ]/δφ(x)

A simple application is when j(x) = 0

(33) 0 = δΓ[φ]/δφ(x) , that is, φ(x) is an extremum (minimum) of Γ[φ].

We will compute the effective field to orders of ℏ by functional integral.

Recall equation 26, rewritten below,

(34) Z[ j ] = ∫ D[Φ(x)] e −(S[Φ] − j∙Φ)/ℏ

Consider ℏ<<1. In the saddle point approximation, we want to know, what is the minimum point in the argument S[Φ] − j∙Φ,

(35) δS/δΦ(x) − j = 0, ⇒ Φc depends on j

(36) Let Φ = Φc + ℏ½ δΦ for fluctuations of O(1)

Expand to 2nd order:

(37) S[Φ] − j∙Φ = S[Φc] − j∙Φc (Saddle point)

+ ℏ½ δΦ∙ [ S'[Φc] − j] (From 35, this equals zero)

+ ℏ½ δΦ∙[ S"[Φc]∙ δΦ (fluctuations of O(1))

+… (highers order to be neglected)

Do not forget that this expression in 37 is divided by ℏ. Substituting this into the exponential functional,

(38)e−(S[Φ]−j∙Φ)/ℏ=e−(S[Φc]−j∙Φc)/ℏ−½ δΦ∙[S"[Φc]∙δΦ+ …)

Integrating to get Z[ j ] (See equation F in Appendix, ignoring factors of 2π)

(39) Z[ j ] = e–(S[Φc] − j∙Φc)/ℏ det(S"[Φc])

(40) From equation 30, W[ j ] = ℏlog[Z[ j ] ]
⇒ −(S[Φc] − j∙Φc)− ½ℏTr[log(S"[Φc])]

We want to know what is the background field.

(41)Consider W[ j ] ≡ Wc[ j∙Φ]φ=Φc

(42) φ(x) = δW[ j ]/δj(x)
= Φc(x) + ∫dy {δΦc(y)/δj(x)} {δWc[ j ]/δΦ(y)}

(43) Calculating the second bracket in the integral,

{δWc[ j ]/δΦ(y)} = j(y) – δS/δΦ(y) + O(ℏ)

(44) But j(y) − δS/δΦ(y) = 0 if Φ=Φc (equation 35)

(45) Therefore, line 42 can be written as,

φ(x) = Φc(x) + O(ℏ)

The background field is approximately equal to the saddle point, which is the minimum point of the action of the field, and a 1st order correction in ℏ.

Legendre Transform:

(46) Γ[φ] = j∙φ − W[ j ] (equation 31)

= j∙φ + (S[Φc] − j∙Φc) + ½ℏTr[log(S"[Φc])] + … (equation 40)

= j∙(φ − Φc) + S[Φc] + ½ℏTr[log(S"[Φc])] + …
(equation 44)

From φ(x) = Φc(x) + O(ℏ) (equation 45)

We write,

⇒ φ(x) = Φc(x) + δφ

(47) Tailor expansion on S[φ]
= S[φc] + (φ − φc)S'[φc] + ½ δφ S"[φc] δφ + ….

The last term contains a product of two factors of δφ, and by 45, is of order ℏ2.

S[φ] = S[φc] + (φ − φc)S'[φc] + O(ℏ2) + ….

From δS/δφ(x) − j = 0, (equation35)

or S'[φc] = j

Substituting that into 47, and ignoring O(ℏ2),

(48) S[φ] = S[φc] + (φ − φc)j + ….

Repeating equation 46,

(49) Then Γ[φ] = j∙(φ − Φc) + S[Φc]

+ ½ℏTr[log(S"[Φc])] + …
Substitute 48 into 49,

(50)Γ[φ] = S[φ] + ½ℏTr[log(S"[Φc])] + …

Quantum Effective Field = Classical Field
+ 1st order quantum correction
And since this is for a general result, it applies to any field. The prevailing belief is that all the laws of nature can be construed as:

Theory A (high energy) = Theory B (low energy)
+ O(1) + O(2) + O(3) + ...
Where we can calculate to any order of precision we wish.

Appendix

We will borrow from The Path Integral Simplified

(A) C1 = ∫–∞ e –½y2dy = (2π)½ (equ.A10)

(B) C2 = ∫–∞ e–½ay2dy = (2π/a)½ (equ.B5)

(C) C3 = ∫–∞ e–½ay2 + bydy = (2π/a)½eb2/2a (equ.C7)

We want to generalize this for an nxn matrix A, we rewrite the integral as,

(D) C2 → C4 = ∫–∞–∞…∫–∞ dx1dx2…dxn e–½x∙A∙x

(E) C3 → C5 = ∫–∞–∞…∫–∞ dx1dx2…dxn e–½x∙A∙x + J∙x

where x∙A∙x = xiAijxj and J∙x = Jixi , with i,j = 1,2…N, and repeated indices summed over.

We will calculate for N=2, and then generalize to any N. We take any 2x2 matrix A' and diagonalize it to A.

Note: det A' = det A = ad – bc

We calculate xiAijxj = x1(A1jxj)+ x2(A2jxj), i=1,2

= x1(A11x1 + A12x2)+ x2(A21x1 + A22x2 ), j=1,2

But A12 = A21 = 0

Therefore, xiAijxj = x1A11x1 + x2A22x2

= (ad – bc )(x1)2 + (x2)2
= (Det[A])(x1)2+ (x2)2
So we take C4 = ∫–∞–∞ dx1dx2 e–½x∙A∙x , N=2

This becomes,

C4 = ∫–∞–∞ dx1dx2 e(Det[A])(x1)2+ (x2)2

= ∫–∞ dx2e –½x22–∞ dx1 e –½(Det[A])(x1)2
The first integral is C1, and the second is C2.

C4 = (2π)½(2π/det[A])½ = ((2π)2/det[A])½

For any nxn matrix A,

(F) C4 =((2π)N/det[A])½

(G) C5 = ∫–∞–∞…∫–∞ dx1dx2…dxn e–½x∙A∙x + J∙x

=((2π)N/det[A])½ e½ J∙A-1∙J

## Monday, May 12, 2014

### Path Integral Simplified

Transition Amplitude

Consider a normalized wavefunction ψ,

(1) ψ = A1ψ1 + A2ψ2 + A3ψ3

Where A1 is the amplitude of ψ1, A2 is the amplitude of ψ2, etc.

Given the rules of QM, the probability of measuring ψ1 is

(2) P1 = A1 A1* = │ A12

See equations 23 and 28 in The Essential Quantum Mechanics

The notion is that we started with ψ initially, then after measuring ψ1, the wavefunction underwent a transition to a new state. Equation (2) is the transition amplitude. Now, the transition amplitude in the canonical quantization approach (see The Essential Quantum Field Theory (EQFT) ) is denoted by

(3) for t± → ±∞, < f|U(t,t0)|i > ≡ < f|S|i > ≡ Sfi ( equation 64 in EQFT)

where U(t,t0) = exp(−iH(t−t0)/ℏ)is the time evolution operator (equation 57 in EQFT)

However there is an elapsed time T between the measurements of the initial state ψi and the final state ψf. So we write,

(4) K(ψif;T) = < ψf|e-iHT/ℏi >, where K is the propagator.

Wave Packets

If we take position as our eigenstates, then equation 4 is,

(5) K(xi,xf;T) = < xf|e-iHT/ℏ|xi >
= < xf| ψ >

(6) Where |ψ > = |e-iHT/ℏ|xi >, is the evolved state

As the initial state evolves into ψ, like a wave packet it spreads and its peak diminishes (area is constant). The amplitude for measuring the particle at time T can be written as,

(7) K(xi,xf;T) = ∫ δ(x - xf)ψ(x,T)dx = ψ(xf,T)

(8) and │K(xi,xf;T)│2 = ψ*(xf,T)ψ(xf,T), is the probability density at xf.

Heuristic Argument

Writing the Schrödinger Equation (with ℏ restored),

(9) iℏd|ψ>/dt = H|ψ> = E|ψ> ( equ. 19 in The Essential Quantum Mechanics )

A solution is,

(10) |ψ> = C e-i(Et – p.x)/ℏ ,

where C is a constant to be determined later on.

Now to simplify our notation, we will get rid of the Dirac notation, and simply write,

(11) ψ = Ce

(12) where ϕ = -(Et – p.x)/ℏ, is the phase angle and e is the phasor.

The wave packet peak travels at the wave group velocity v, which corresponds to the classical particle velocity (fig.1). The time rate of change of the phase angle at the peak is,

(13) dϕ/dt = -(Et – p.v)/ℏ

(14) But E = T + V, where T is the kinetic energy, and V is the potential energy.

(15) Note: T = ½mv2 and p = mv → p.v = 2T

Substituting equations (14) and (15) into (13), we get

(16) dϕ/dt = (T – V)/ ℏ
= L/ ℏ, where L is the Lagrangian.

Integrating equation (16),

(17) ϕ = ∫ Ldt/ℏ = S/ℏ where S is the action.

Therefore equation (11) can now be written as,

(18) ψ = C(T)exp(i/ℏ∫Ldt) = C(T) exp(iS/ℏ),

where we take the more generally case that C might be time-dependent.

Path Integral

The central idea of the Path Integral is that a particle/wave traveling between two events could be considered as traveling all possible paths between those two events.

Illustration

In Fig. 2, the lagrangian is simply the kinetic energy of the electron, different for each path, which do not obey the usual classical laws – least action, equal angles of incidence and reflection, and so on. But in the Path Integral, they must all be included. It turns out that if we calculate the phasor for each path we get the following:

Note that the paths that are far from the classical trajectory tend to cancel each other.

Time Slicing

We want to derive equation (18) from the basic idea of the Path Integral. To do that we need to consider finite slices of time. We also discretize space and consider a small number of paths. In our example we will consider three different paths that we label as a, b and c, each of those over two time intervals, t0 → t1 → t2.

Recall that L = T – V = ½mv2 – V(x)

For path a:

(19A) La1 = ½m(x122 – x042)/∆t – V(½(x12 + x04))
(19B) La2 = ½m(x252 – x122)/∆t – V(½(x25 + x12))

For path b:

(20A) Lb1 = ½m(x132 – x042)/∆t – V(½(x13 + x04))
(20B) Lb2 = ½m(x252 – x132)/∆t – V(½(x25 + x13))

For path c:

(21A) Lc1 = ½m(x162 – x042)/∆t – V(½(x16 + x04))
(21B) Lc2 = ½m(x252 – x162)/∆t – V(½(x25 + x16))

In terms of the phasors:

For path a:

(22) exp(i/ℏ ∫(La1 + La2)dt)
= exp(i/ℏ ∫ La1dt) exp(i/ℏ ∫ La2dt)
= exp(i/ℏ ∫ ½m(x122 – x042)/∆t – V(½(x12 + x04))dt)
x exp(i/ℏ ∫ ½m(x252 – x122)/∆t – V(½ (x25 + x12))dt)
≈ exp(iS(x04,x12/ℏ) exp(iS(x12,x25)/ℏ)
We get similar expression for paths b and c.

The sum of these three paths is:

(23) = ∑ exp(iS(x04,x1j)/ℏ) exp(iS(x1j,x25)/ ℏ), j=1,2,3

Since the transition amplitude is proportional to the above we can multiply by any constant, in this instance, C’ and ∆x1, where xL < x1 < xR.

(24) K(i,f;T)

≈ C’∑ exp(iS(x04,x1j)/ℏ) exp(iS(x1j,x25)/ℏ)∆x1
≈ C’∫xLxR exp(iS(x04,x1)/ℏ) exp(iS(x1,x25)/ℏ)dx1
≈ C’ ∫xLxR exp(i∫x04x25L/ℏ dx1
In the general case, we have intervals dx1, dx2, dx3 … dxN, where N → ∞

(25) K(i,f;T=tf–ti)= C ∫∫∫… ∫ ei∫L/ℏ dx1dx2dx3… dxn

= C(T)∫ ei∫L/ℏDx,
where the symbol D implies all paths between i and f.

Derivation of the contant

Consider equ.(5), rewritten below

(26) K(xi,xf;T) = < xf|e-iHT/ℏ|xi >

The bra and ket are Dirac delta functions (see equation 27 in The Essential Quantum Mechanics )

(27) K(xi,xf;T) = ∫ δ(x – xf)e-iHT/ℏδ(x – xi)dx

Mathematically the Dirac Delta function is:

(28) δ(x – xi) = (2π)-1∫eik(x – xi)dk, (for the ket)

= (2πℏ)-1 ∫ eip(x – xi)dp
The second line is obtained with p = ℏk

Similarly for the bra,

(29) δ(x – xf) = (2πℏ)-1 ∫ e-ip'(x – xf)dp'

= (2πℏ)-1 ∫ eip'(xf – x)dp'
The second line is obtained by taking the minus sign inside the bracket.

When we substitute equ. (28) and (29) into (27), we will get three exponential functions, and a triple integral. To simplify matters, we will examine the argument of each of the exponential function separately, from left to right, and then combine, taking care of not changing the order since we are dealing with operators.

(30) For the bra, E1 = (ip/ℏ)(x – xi)

(31) For the e-iHT/ℏ, E2 = –iHT/ℏ = – iET/ℏ

Where operating on the initial state, H = E, which is the eigenvalue. Note: we have a number(E) instead of an operator(H).

(32) For the ket, E3 = (ip'/ℏ)(xf – x)

Now we can pass E2 as it is a number. So from
exp(E1) exp(E2) exp(E3), we can now write it as
exp(E2) exp(E1) exp(E3). Let us examine the last two exponents:

(33) exp(E1) exp(E3) = e(ip/ℏ)(x – xi) e(ip'/ℏ)(xf – x)

= ei(p – p')x/ℏ eip'xf/ℏ e-ipxi/ℏ
The first exponent in equ.(33), combine with (2πℏ)-1∫dx, gives,

(34) (2πℏ)-1∫ ei(p – p')x/ℏ dx = δ(p – p')

With this result, equation (27) becomes,

(35) K(xi,xf;T)
= (2πℏ)-1∫∫e–iET/ℏδ(p – p')eip'xf/ℏe-ipxi/ℏdpdp'
= (2πℏ)-1∫e–iET/ℏeip(xf – xi)/ℏdp
= (2πℏ)-1∫e–i(p2/2m)T/ℏeip(xf – xi)/ℏdp
(36) Where the energy E is simply the kinetic energy = p2/2m .

Using appendix C,

(37) ∫-∞ e–ay2 + by dy= (π/a)½e b2/4a

Making the following correspondence,

y → p, a → iT/2mℏ, b → (i/ℏ)(xf – xi)

(38) K(xi,xf;T) = (m/(i2πℏT))½eim(xf – xi)2/(2Tℏ)

The probability density is then (see equ.(8)),

(39) │K(xi,xf;T)│2 = m/(2πℏT)

We can deduce from equation(39) that:

(i) For very large T, the probability amplitude decreases. That means that the farther away the path is from the classical path (the longer time it will take to go from xi to xf), the less it contributes to the probability amplitude (fig.3).

(ii) As the mass m increases, so is the height, thus the width must decrease (area under envelope is constant - fig.1) That is, the wave packet approaches the classical behavior of a particle.

(iii) If ℏ were to go to zero, the peak would be infinite, giving an exact location, as it should for a classical particle.(fig.1)

Going back to the central idea of the Path Integral - which is that a particle/wave traveling between two events could be considered as traveling all possible paths between those two events - we now see that the greater deviation from the classical path, the smaller contribution we get to the probability amplitude. In addition, we also find that for large mass, or ℏ = 0, we fall into the classical regime.

Appendix A

(A1) C = ∫e–y2dy , integral is from –∞ to +∞.

Squaring both sides,

(A2) C2 = ∫e–y2dy ∫e–x2dx

In the second term, we've replaced y by x, since these are just dummy variable in the integration,

(A3) C2 = ∫∫e–(y2 + x2)dxdy

Switching to polar coordinates,

(A4) Let x = r cosθ

(A5) And y = r sinθ

(A6) Then y2+ x2 = (r cosθ)2 + (r sinθ)2

= r2 cos2θ + r2 sin2θ
= r2 (cos2θ + sin2θ)
= r2
(A7) Therefore, C2 = ∫∫e–r2dxdy

Also , the product dydx is just an element of the area of a small square. In polar coordinates, that area is rdrdθ. So,

(A8) C2 = ∫0 e–r2rdr ∫0

Make another change in variable, u = r2, so that du = 2rdr

(A9) C2 = ½ ∫0e–udu ∫0

= ½ (–) (℮–∞ – ℮–0)(2π – 0)

= ½ (–) (0 – 1)(2π)

= π

(A10) Therefore, C = ∫–∞ e –y2dy = π½

Appendix B

(B1) C = ∫–∞ e–ay2dy

In the case that a constant "a" multiplies y2, we make the following substitution,

(B2) x2 = ay2

(B3)Then x = a½y

(B4) Taking derivatives,

dx = a½dy → dy = adx

Substituting B4 into B1,

(B5) C = ∫–∞ e–ay2dy = a–∞ e–x2dx

= a π½ , from A10

= (π/a) ½

Appendix C

(C1) C = ∫–∞ e–ay2 + bydy

(C2) The exponent is –ay2 +by = – a[y2 –(b/a)y] .

(C3) We complete the square:
= –a[y2 – (b/a)y + (b/2a)2 – (b/2a)2]
= –a[(y – b/2a)2 – (b2/4a2]
= –a(y – b/2a)2 + b2/4a
(C4) Let x2 = (y – b/2a)2

(C5) x = (y – b/2a)

(C6) dx = dy

(C7) C = ∫–∞ e–ay2 + bydy , from C1

= ∫–∞ e–a(y – b/2a)2 + b2/4ady, from C3
= eb2/4a–∞ e–ax2dx, from C4, C6
= (π/a)½eb2/4a , from B5

## Monday, April 28, 2014

### The Essential Quantum Field Theory

Classically, the electron is fundamentally different from a photon. In QM, we see them not that much differently as particles/waves. How is this to be reconciled?

There are two possible ways to interpret this: one, you could envision that the particles are fundamental. The photons are particles, and if you pack enough of them, that gives rise to a classical field; two, you could imagine that it’s the field that is fundamental, and when you quantize it, it’s the ripples that give rise to particles.

So why quantum field theory?

i) Because all particles of the same type are indistinguishable. This is an amazing fact. For instance, a proton created in a far, far way galaxy, billions of years ago would be exactly the same as a proton freshly created in a collision on earth. It would mean that the universe is filled with a proton field, and when one of its ripple is converted into a particle, it is always the same particle, regardless of where and when it was created.

ii) Because QM + SR implies that the number of particles in a reaction is not conserved. Basically, this means that if you take the Schroedinger’s equation, which describes a single particle, and apply Special Relativity, you won’t go far with this scheme. You get two versions: the Klein-Gordon equation and the Dirac equation, but keeping the interpretation of the wave function to a typically single particle leads you to deep troubles. You’ll find that you get negative probabilities, of which no one can make sense, and there are infinite towers of energies, which are unbounded and not quite clear what to make of them.

What is Quantum Field Theory?

In Quantum Mechanics, you figure out the classical degrees of freedom. You then promote them to operators that act on a Hilbert space. This is also true in QFT. Here, the classical degrees of freedom are the fields, and we do the same, promote them to operators. What that means is that we have a field that is quantized, and the basic object is an operator valued function acting on the Hilbert space. Every point in space will have an operator. This immediately tells us one of the major problems in QFT. There is an infinite number of points in space. So we are dealing with an infinite number of degrees of freedom. And so infinity will come back to bite us many, many times in understanding QFT. All the richness of QFT will be in making sense of those infinities. Basically, QFT is the language in which the laws of nature are written.

Units and Scales

[ c ] = LT-1

[ ℏ ] = L2M2T-1

[ G ] = L3M-1T-2

Setting ℏ =c=1, this means we can express everything in terms of mass, or equivalently, energy (eV).

For example, λ = ℏ/mc ⇒ [ L ]= [ M-1]

For the electron, Me = .5 x 106 eV ⇔ λe = 2 x 10-12 m.

At the Planck scale, G = ℏc/Mp2 ⇒ Mp-2 , where Mp = 1018 GeV.

From Classical Physics to Quantum Mechanics

QM is formulated directly from Classical Physics according to the following recipe: construct a Lagrangian (L) from an action principle (S),

(1) S = ∫ dt L(x(t),v(t));

where x(t) is position and v(t)= ∂x/∂t is the corresponding velocity.

This yields the equation of motion, also known as the Euler-Lagrange equations,

d(∂L∂v)/dt = ∂L/∂q, (Newton's Second Law of Motion)

Define the Hamiltonian through a Legenre transformation,

(2) H = pv – L ;

where p is the canonical momentum defined as p = ∂L/∂v

The equation of motion then becomes,

(3) ∂H/∂p = v ; ∂H/∂q = -∂L/∂q = –p

The advantage is that the Hamiltonian equations are first order, while the Euler-Lagrange equations are second order.

Define the Poisson bracket,

(4) {A,B} = (∂A/∂x)(∂B/∂p) – (∂A/∂p)∂B/∂x,

where A and B are any two functions.

In particular, if B = H, then dA/dt = {A,H}

The equations of motion are further simplified as,

dx/dt = {x, H} = ∂H/∂p ; dp/dt = {p, H} = –∂H/∂p

For the position (A = x) and the momentum (B = p), the Poisson bracket becomes,

(5) {x,p} = 1

Then replace that Poisson bracket with the commutation relation

[x,p] = iℏ, (equation 35 in The Essential Quantum Mechanics (EQM) )

which yields the Heisenberg Uncertainty Principle.

And voilà, we have QM. Note: there are other ramifications in QM to keep in mind, for instance, a state "|a>" is differentiated from the observable "A", and probability is the square of the amplitude, etc. For our purposes, we should remember that we have another process of going from Classical Physics to Quantum Mechanics, which is embodied in the transformation:

(6) {x,p} = 1 → [x,p] = iℏ

Classical Field Theory

A field is a physical quantity defined at every point in space and time. In classical particle mechanics, there are a finite number of degrees of freedom. For instance, the coordinate qa(t), where a is a label that might indicate direction. In field theory, we have an infinite number of degrees of freedom. We denote it by ϕa(x,t), where a and x are considered as labels. Important Note: x is no longer a dynamical operator but just a label.

Example: Er(x,t) and Br(x,t) in E & M, where r = 1,2,3. So we have six fields. But later it was found out using Maxwell’s equations that there are 4 fields, Aμ(x,t), where μ = 0,1,2,3, and

Er = ∂Ar/∂t − ∂A0/∂xr

Br = ½ ϵrst∂At/∂xs

The dynamics of the fields are governed by the Lagrangian, which is a function of ϕ(x,t) , ∂ϕ(x,t)/∂t and ∇ϕ(x,t). We define,

(7) L(t) = ∫dx3ℒ(ϕa,∂μϕa)

⇒ S = ∫dtL(t) = ∫dx4 ℒ(ϕa,∂μϕa)

We determine the equations of motion by the principle of least action, that is,

(8) δS = 0, keeping the end points fixed.

(9) δS = ∫dx4 {(∂ℒ/∂ϕa)δϕa + (∂ℒ/∂(∂μϕa))δ(∂μϕa)}

= ∫dx4[{∂ℒ/∂ϕa − ∂μ(∂ℒ/∂(∂μϕa))}δϕa

(10) For δS = 0

(11) ∂μ(∂ℒ/∂(∂μϕa)) − ∂ℒ/∂ϕa = 0

These are called the Euler-Lagrange equations for fields.

Example: the Klein-Gordon equations

(12) ℒ = ½ ημνμϕ∂νϕ − ½ m2ϕ2

where, ημν is the Minkowski metric.

Substituting equation (12) into equation (11),

i) 1st term: ∂ℒ/∂(∂μϕa) = ∂μϕ ⇒ ∂μ(∂ℒ/∂(∂μϕa)) = ∂μμϕ

ii) 2nd term: ∂ℒ ∂ϕa = − m2ϕ

This gives,

(13) ∂μμϕ + m2ϕ = 0

Note: This is the Klein-Gordon Equation for fields. So far the term m has not been determined and is considered at this moment as a constant into the equation.

Hamiltonian Formalism For Fields

In analogy to p = ∂L/∂v and v = ∂q/∂v (see equ. 1 and 2), we define the conjugate momentum as,

(14) π(x) = ∂ℒ/(∂ϕ/∂t)

The Hamiltonian density,

(15) H(π,ϕ)= π(x)(∂ϕ(x)/∂t) − ℒ(ϕ,∂ϕ/∂t,∇ϕ)

Example:

Consider ℒ = ½ (∂ϕ/∂t)2 − ½ (∇ϕ) 2 − V(ϕ).

⇒ π(x) = ∂ℒ/(∂ϕ/∂t) = ∂ϕ/∂t

(16) H = ½ π2 + ½ (∇ϕ)2 + V(ϕ)

And the Hamiltonian, H = ∫ d3x H .

Canonical Quantization of the Fields

Recall in QM, canonical quantization tells us to take the coordinates qa and the momenta pa and promote them to operators (ℏ=1).

(17) [qa,pb] = iδab ⇒ [q,p] = (2π)3δ(q-p) (in 3-D)

In field theory, we do the same for ϕ(x) and π(x).

(18) [ϕa(x),πb(y)] = iδab δ(3)(x−y)

In the Schroedinger picture, ϕ depends on space but not on time. The time-dependence resides in the states, which evolve according to the Schroedinger equation.

(19) id|ψ>/dt = H|ψ>

The question arises: what are the eigenvalues of the Hamiltonian. So, we want to know the spectrum of H. Recall that in QM, this can be solved for the harmonic oscillator and the hydrogen atom. When we try it on the helium atom, we run into trouble as we are dealing with six degrees of freedom. In QFT, we are dealing with an infinite number of degrees of freedom – at least one for every point in space. So the task becomes horrible. Thankfully, there is one class of fields in which we can do something about, and these are called free field theories. If you write the theory with "good" coordinates, the fields all decouple from each other, and each of these degrees of freedom evolves independently. In these free field theories, the Lagrangian are quadratic in the fields, and the equations of motion are linear. The simplest free field theory is the Klein-Gordon equation (K-G) for a real scalar, where m has dimension of mass.

μμϕ + m2ϕ = 0 , equation (13)

Solving for ϕ, we do a Fourier transformation,

(20) Φ(x,t) = ∫ d3p(2π)-3 e ip∙x Φ(p,t)

Substituting (20) into the K-G equation,

(21)⇒ [∂2/∂t2 + (p2 +m2)]Φ(p,t) = 0

This is the equation for a harmonic oscillator with frequency,

(22) ωp = (p2+ m2)½

The solutions to the classical K-G equations are superpositions of simple harmonic oscillators (SHO). To quantize Φ(x,t), we need to quantize SHO. (see Harmonic Oscillators, Vacuum Energy, Pauli Exclusion Principle)

In the Schroedinger picture,

(23) H = ½ p2 + ½ ω2q2 , [q,p] = i

To find the spectrum of H, we define

(24) a = i(2ω)½p + (ω/2)½q

(25) a = −i(2ω)½p + (ω/2)½q

(26) ⇒ [a, a] = 1

(27) ⇒ H = ½ ω(aa + aa)

(28) or H = ω(aa + ½ )

(29) ⇒ [H,a] = ωa

(30) and, [H,a] = −ωa

The importance of the last result means that if we have an eigenstate of E, we can construct the other states.

(31) That is,

if H|E> = E|E>, ⇒ Ha|E> = (E + ω)a|E> .

(32) Similarly, Ha|E> = (E − ω)a|E> .

We get these towers of energies, E, E + ω, E + 2ω… from above, and E − ω, E − 2ω going down below. If we want a spectrum that is bounded below, then this process must stop. This implies the existence of a ground state, |0> , such that a|0> = 0. (This is also true for <0|a = 0)

(33) ⇒ H|0> = ω(aa + ½ )|0> = ½ ω|0>, which is the ground state energy.

For all the excited states,

(34) |n> = (a)n|0>, and H|n> = (n + ½ )ω|0>

Free Field Theory

From the Klein-Gordon equation, consider the following,

(35) H = ½∫d3x[π2 + (∇ϕ)2 + m2ϕ2], (see equ. 16)

(36) ωp = + (p2+ m2)½ and [ϕ(x),π(y)] = iδ(3)(xy)

We define the creation operators ap and annihilation operators ap from the following,

(37) Φ(x)=∫d3p(2π)-3(2ωp) [apeip.x + ape-ip.x]

(38) π(x)= ∫d3p(2π)-3(-i)(ωp/2)½[apeip.x - ape-ip.x]

Note that both Φ(x) and π(x) are hermitian operators, with Φ = Φ and π = π.

( See Equ 19 in The Essential Quantum Mechanics (EQM) for the importance of Hermitian operators)

(39) ⇒ [ap,aq] = (2π)3δ(3)(pq),

Note: for p = q, [ap,ap] = (2π)3δ(3)(0)

(40) From H = ½∫d3x[π2 + (∇ϕ)2 + m2ϕ2], (equ. 35)

= ½∫d3x d3p d3q(2π)-6
{−(ωqωp/2) [apeip.x - ape-ip.x][aqeiq.x - aqe-iq.x]
+i2-1pωq)[papeip.x - pape-ip.x][qaqeiq.x - qaqe-iq.x]
+m22-1pωq)[apeip.x + ape-ip.x][aqeiq.x + aqe-iq.x]}
Using the definition of the delta function,

(41) δ(xy) = (2π)-3∫d3keik(x-y),

we integrate over x and q.

(42)H = 1/4∫d3p(2π)-3p)-1
[(−ωp2 + p2 + m2)(apa-p + apa-p)
+(ωp2 + p2 + m2)(apap + apap)]
But ωp2 = p2 + m2, therefore the first term in the square bracket vanishes, and we are left with,

(43) H = ½∫ d3p(2π)-3ωp(apap + apap)

This is the energy of an infinite number of uncoupled SHO, as expected. Also from the commutation relationship, we get,

(44) H = ∫ d3p(2π)-3ωp(apap + ½(2π)3δ(3)(0))

The last term is problematic as it is infinite.

The Vacuum

Define the vacuum as in the case of the SHO,

ap|0> = 0, for all p.

Its energy is then,

(45) H|0 > = E0 |0 > = [∫d3p ωpδ(3)(0)]|0 >

Note that we have infinity coming from two sources:

(i) from the delta function, because space extends to infinity in all directions - these are called infra-red (IF) divergences.

(ii) Consider a box of size L. In the limit that L → ∞,

(2π)3δ(3)(0) = lim ∫-L/2+L/2 d3xeip.x = V

Where V is the volume of the box.

Therefore, E0 = ½V ∫d3p(2π)-3ωp

This indicates that we should work with energy density,

(46) E = E0/V = ½ ∫d3p(2π)-3ωp

This energy density is still infinite due to space being infinitesimally small (as p → ∞, x gets smaller and smaller, that is, we have bigger and bigger oscillations over smaller and smaller distances). This is the second source of infinity, which is called an ultra-violet (UV) divergence. Here we have assumed that our theory is valid over arbitrarily short distance scales.

Since we are only interested in energy differences, we remove the infinities and redefine,

(47) H = ∫d3p(2π)-3ωpapap

(48) ⇒ H|0> = 0

Interacting Picture

i) Starting with the Schroedinger picture:

(49) id|ψ(t) >s/dt = Hs|ψ(t) >s, (equ. 19)

States are time-dependent and operators Os are time-independent.

ii) In the Heisenberg picture,

(50) OH(t) = eiHt Os e-iHt and |ψ>H = eiHt|ψ>s

States are time-independent and operators Os are time-dependent.

The interaction picture is a hybrid of the two. We write the Hamiltonian as,

(51) H = H0 + Hint

Time-dependence of operators is governed by H0. And time-dependence of states is governed by Hint.

(52) |ψ>I = eiH0t |ψ>s

(53) OI(t) = eiH0tOse-iH0t

Note: HI ≡ (Hint)I = eiH0t(Hint)se-iH0t

From the Schroedinger equation,

id|ψ>s/dt = Hs|ψ>s (equ. 49)

Substituting 52 and 53,

⇒ id(e-iH0t|ψ>I)/dt = (H0 + Hint)se-iH0t|ψ>I

⇒ ie-iH0td|ψ>I/dt = (Hint)se-iH0t|ψ>I

⇒ id|ψ>I/dt = eiH0t(Hint)se-iH0t|ψ>I

(54) id|ψ>I/dt = HI(t)|ψ>I

In solving this equation we write,

(55) |ψ(t) >I = U(t,t0)|ψ(t0) >I, where U(t,t0) is the time evolution operator.

(56) ⇒ idU/dt = HI(t)U

This leaves us with an equation of operators, instead of an equation of states.

If U and HI were ordinary functions, we could solve this equation as,

(57) U(t,t0) = exp(−i∫t0t HI(t')dt')

However, this is not correct because,

(58) [HI(t), HI(t')] ≠ 0, when t ≠ t'

Claim: the correct solution is given by Dyson’s formula,

(59) U(t,t0) = T exp(−i∫t0t HI(t')dt') , where T is the time-ordering operator defined as follow:

(60) T[O(t1)O(t2)] = O(t1)O(t2), if t1 > t2
= O(t2)O(t1), if t2 > t1
The solution is then,

(61) U(t,t0) = 1 − iexp(∫t0t HI(t')dt'
+ (−i)2/2{ ∫t0tdt'∫t0t'dt"HI(t")HI(t')
+ ∫t0tdt'∫t0t'dt"HI(t')HI(t")} + …)
Scattering

We’ll work with scalar Yukawa theory,

(62) ℒ = ½∂μϕ∂μϕ + ∂μΨ*∂μΨ − ½m2ϕ − M2Ψ*Ψ − gΨ*Ψϕ ,

where ϕ is a real field with its particle of mass m, Ψ is a complex field with its particle of mass M, and the last term is for the interaction between the two fields with coupling constant g. We’ll take g << M,m to ensure weak coupling.

(63) Hint = gΨ*Ψϕ

In this theory, particles number is not conserved. In particular,

Φ ~ a + a can create/destroy Φ-particles (mesons).

Ψ ~ b + c can destroy Ψ+particles and create Ψ- (nucleons).

Ψ ~ b + c can create Ψ+particles and destroy Ψ- (nucleons).

At first order perturbation theory, we will have terms as Ψ*Ψϕ ~ cba which will destroy a meson and create a pair of nucleon and an anti-nucleon.

ϕ → Ψ+ + Ψ (decay)

This can be represented by the following diagram:

At second order perturbation theory, this will include terms as, (cba)(cba) which will create a meson, destroy a pair; then destroy the meson and create a pair.

Ψ+ + Ψ → ϕ → Ψ+ + Ψ (scattering)

To calculate amplitudes for these processes, we need an important assumption: the initial and final states are non-interacting particles, meaning that the initial state |i > at t → −∞ and the final state |f > at t → +∞ are eigenstates of H0.

Some caveats:

i) For instance, an electron and a proton interact and form a hydrogen atom. The final state is an interacting state.

ii) Particles are never alone. They can have a cloud of pairs of particles/anti-particles surrounding them.

(64) Definition: t± → ±∞ < f|U(t,t0)|i > ≡ < f|S|i >, where S is a unitary operator, called the S(cattering)-Matrix. And U(t,t0) was found in equation 61.

Meson Decay

(65) |i > = 2Epap|0 >, ( a meson with momentum p)

(66) |f > = 4Eq1Eq2bq1cq2 |0 > (two nucleons with momenta q1 and q2)

⇒ < f|S|i > = − ig< f|∫dx4 Ψ(x)Ψ(x)ϕ(x) |i > (using equ. 63 and 64)

Using the definition of the field operator(from equ. 37),

Φ(x) = ∫d3k(2π)-3(2Ek)[ake-ik.x + akeik.x]

And using equation (65), we get

(67)< f|S|i > = −ig< f|∫dx4 Ψ(x)Ψ(x)∫d3k(2π)-3(2Ek)[ake-ik.x + akeik.x] 2Epap|0 >,

The second term from the square bracket will give a term that contains < f|akap|0 >. When equation (66) is substituted, the ak term will hit the < 0| on the left and give zero. What's left will be,

< f|S|i > = − ig < f| ∫dx4Ψ(x)Ψ(x)∫d3k(2π)-3
x (2Ep/2Ek)akape ik.x|0>
Then we then pass the ak through ap to kill the vacuum |0 > but pick up a delta function, which when integrated is non-zero for k = p, making the term in the square-root equal to 1.

(68) < f|S|i > = − ig < f|∫dx4Ψ(x)Ψ(x)eip.x|0 >

Using equation 37 for the field operators Ψ(x)Ψ(x), taking into consideration that Ψ produces "b" and "c" particles instead of "a" particles

(69) < f|S|i > = −ig< f|∫dx4∫d3k1d3k2(2π)-6(4Ek1Ek2)
x bk1ck2ei(k1+k2-p).x|0>
Substituting for < f| = (4Eq1Eq2)½ < 0|cq1bq2 from equation (69), we do as previously, passing these two annihilator operators through bk1ck2 to kill the vacuum |0 > and pick up a delta function which when integrated is non-zero for k1 = q1, and k2 = q2 making the term in the square-root equal to 1.

(70) < f|S|i > = − ig < 0|∫dx4ei(q1+q2- p).x|0 >

= − ig(2π)4δ(4)(q1+ q2− p)
The delta function expresses the conservation of momentum p = q1+ q2

Normal Ordering

In free field theory we define a normal ordered string of operators ϕ(x1)…. ϕ(xn), we write it as :ϕ(x1)…. ϕ(xn): to be the usual product of operators with all the annihilation operators moved to the right. For the Hamiltonian, this would be :H:

Wick’s Theorem

(71) Consider the operator; O = T exp(−i∫HI(x)dx4),

We’ll need to compute terms like < f|T{ HI(x1) HI(x2)… HI(xn)} |i > . When we do these calculations, we need to move the annihilator operators to the right, each time we get a delta function through the commutation relations, in order to make many terms vanish, that is, the annihilators will destroy the vacuum state. In other words we want to rewrite the time-ordered operators as normal-ordered operators, where all the annihilation operators are to the right.

Example: For a real scalar field, from equ. 37, we write,

(72) Φ(x) = Φ+ (x) + Φ(x), where

(73) Φ+(x)) = ∫d3p(2π)-3 (2Ep)-1ape-ip.x, (1st term)

(74) Φ(x) = ∫d3p(2π)-3(2Ep)-1ap+eip.x, (2nd term)

(75) Recall (equ. 60) that

TΦ(x)Φ(y) = Φ(x)Φ(y) , when x0 > y0

= {Φ+(x) + Φ-(x)} {Φ+(y) + Φ-(y)}

= Φ+(x)Φ+(y) + Φ+(x)Φ-(y) + Φ-(x)Φ+(y) + Φ-(x)Φ-(y)

= Φ+(x)Φ+(y) + Φ+(x)Φ-(y) + Φ-(x)Φ+(y) + Φ-(x)Φ-(y)

Note that Φ+(x)Φ-(y) = Φ-(y)Φ+(x) + [Φ+(x),Φ-(y)]. Substituting in the above,

(76) TΦ(x)Φ(y) = Φ+(x)Φ+(y) + Φ-(y)Φ+(x)

+ [Φ+(x),Φ-(y)] + Φ-(x)Φ+(y) + Φ-(x)Φ-(y)

= :Φ(x)Φ(y): + [Φ+(x),Φ-(y)]

But we already know what the commutation will do when acting on the vacuum.

< 0|Φ(x)Φ(y)|0 >

= ∫d3pd3p'(2π)-6(4Ep Ep')< 0|apap' |0 > e-i(p.x- p'.y)

(77) < 0|Φ(x)Φ(y)|0 > = ∫d3p(2π)-3(2Ep)-1e-ip.(x-y) (see appendix)

(78)< 0|Φ(x)Φ(y)|0 > ≡ D(x – y), called the propagator

Combining equations (76) and (78),

For x0 > y0, TΦ(x)Φ(y) = :Φ(x)Φ(y): + D(x – y)

And for y0 > x0, TΦ(x)Φ(y) = :Φ(x)Φ(y): + D(y – x)

In general,

(79) TΦ(x)Φ(y) = :Φ(x)Φ(y): + ΔF(x – y), where

(80) ΔF(x – y) = ∫d4k(2π)-4 (ie-ip.(x-y))/(k2 – m2 +ie), called the Feynman Propagator

We define a contraction of a pair of field in a string of n operators as,

|-----------|
Φ(x1)...Φ(xi)... Φ(xj)... Φ(xn)= {Φ(xi)Φ(xj)}= ΔF(xi – xj)

For a complex field:

{Ψ(x)Ψ(y)} = ΔF(y – x)

{Ψ(x)Ψ(y)} = 0

{Ψ†(x)Ψ(y)} = 0

(81) Define: Φ(x1) = Φ1 and Φ(x2) = Φ2, etc.

Theorem: For any collection of fields,

(82) T(Φ1… Φn) = :Φ1… Φn: + :all possible contractions:

Example: Take a string of four fields

T(Φ1Φ2Φ3Φ4) = :Φ1Φ2Φ3Φ4:

+{Φ1Φ2}:Φ3Φ4:+ {Φ1Φ3}:Φ2Φ4: + 4 similar terms

+ {Φ1Φ2} {Φ3Φ4} + {Φ1Φ3} {Φ2Φ4} + {Φ1Φ4} {Φ2Φ3}

The second line, we contract one field with another → 6 terms

The third line, we contract one pair of fields with another pair → 3 terms

Nucleon Scattering

Ψ + Ψ → Ψ + Ψ

|i> = (2Ep1)(2Ep2)bp1bp2|0> ≡ |p1,p2 >,

|f> = (2Ep'1)(2Ep'2)bp'1bp'2|0> ≡ |p'1,p'2 >,

We want to compute to orders of g2.

⇒ Recall Hint = gΨ*Ψϕ (equ. 63)

Second order is,

⇒ (− ig)2/2 ∫d4x1d4x2T(Ψ(x1)Ψ(x1)ϕ(x1(x2)Ψ(x2)ϕ(x2))

There is a term from Wick’s theorem which looks like,

(x1)Ψ(x1(x2)Ψ(x2):{ϕ(x1)ϕ(x2)}

Leaving the contraction aside for the moment,

⇒ < p'1,p'2|:Ψ†(x1)Ψ(x1) Ψ†(x2)Ψ(x2):| p1,p2 >

⇒ e-ix1.(p1'- p1) + ix2.(p2'- p2) + e -ix1.(p2'- p1) + ix2.(p2'- p1) + term (x1 ↔ x2)

Inserting the contraction term

< f|S|i > = (− ig)2/2 ∫d4x1d4x2 [ all exp terms ] ∫d4p(2π)-4 (i e-ip.(x1- x2)) /(k2 – m2 +ie)

⇒ (−ig)2[ ( (p1 – p'1)2 – M2 ) -1 + ( (p1 – p'2)2 – M2 ) -1 ] (2π)4 δ(4) ( p1+ p2 – p'2 – p'2)

Notice again that the delta function expresses the conservation of momentum

Feynman Diagram

This is an alternative to Wick’s theorem. Each diagram corresponds to one of these terms that arise from Wick’s theorem.

General Rules for scalar field (spin zero):

- Draw an external line for each particle in the |i> and |f>

- Draw an arrow on the line to Ψ+, Ψ particles to denote its charge. Choose incoming (outgoing) arrow for Ψ+) in the |i>, and opposite for |f>.

- Join the lines together with vertices.

- Add a momentum k to each internal line

- To each vertex, there is a factor of (−ig(2π)4δ(4)iki)

- For each internal line with momentum k, there is a factor of ∫d4k(2π)-4 D(k2),

where D = i(k2 – m2 +ie)-1 for ϕ

And D = i(k2 – M2 +ie)-1 for Ψ

Appendix

(A)[ap,ap'] = (2π)3δ(3)(pp'), (equ. 39)

(B)< 0|apap'|0 > = < 0|ap'ap + [ap,ap']|0 >

=< 0|ap'ap +(2π)3δ(3)(pp')|0 >
Where we used equ.(A) in the second line.

Now with ap|0 > = 0 and < 0|0 > = 1, we get,

(C)< 0|apap'|0 > = (2π)3δ(3)(pp')

Rewriting the line above equation (77),

< 0|Φ(x)Φ(y)|0 >

= ∫d3pd3p'(2π)-6(4Ep Ep')< 0|apap' |0 > e-i(p.x- p'.y)

Substituting equ.(C),

= ∫d3pd3p'(2π)-6(4Ep Ep')(2π)3δ(3)(pp')e-i(p.x- p'.y)

Integrating over p', we get

= ∫d3p(2π)-3(2Ep)-1e-ip.(x-y),

This is equation (77).