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]

Comments:

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])] + …

This reads as,

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