## 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