**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 (−+++),

ds

^{2}= -c

^{2}dt

^{2}+ dx

^{2}+ dy

^{2}+ dz

^{2}

(See equation 10 in The Essential General Relativity )

We get,

(1) ds

^{2}= c

^{2}dτ

^{2}+ dx

^{2}+ dy

^{2}+ dz

^{2},

And that gives Euclidean geometry.

**Probability Amplitude**

(2) < q

_{2},t

_{2}| q

_{1},t

_{1}>

_{Heisenberg picture}

= < q

_{2}|U(t

_{2},t

_{1}|q

_{1}>

_{Schroedinger picture}

=< q

_{2}|e

^{(t2−t1)H/(iℏ)}|q

_{1}>

_{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}= Σ

_{n}|n >< n| e

^{tEn/iℏ}

Here we assume that the spectrum is discrete,

E

_{0}< E

_{1}< E

_{2}< … E

_{n}

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}≡ U

_{E}(τ)

Note that U

_{E}(τ) has the same form as the density operator for a quantum state in a mixed state at temperature T > 0.

(6) ρ

_{T}= (e

^{−βH})/Z, where β = 1/k

_{B}T,

(7) where Z = Tr(e

^{−βH}), and Z is the

*partition function*.

(8) Note that we have made the equivalence,

1/k

_{B}T ↔ τ/ℏ

We want to calculate a path integral for < q

_{2}|U

_{E}(τ)|q

_{1}> in the Euclidean geometry.

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

(9) iS[q] = i∫

_{0}

^{t}ds [½m(dq/ds)

^{2}− V(q)]

(10) Let s → − iσ ; t → − iτ

(11) iS[q] = i ∫

_{0}

^{τ}(−i)dσ [½m(i dq/dσ)

^{2}− V(q)]

= ∫

_{0}

^{τ}dσ[−½m(dq/dσ)

^{2}− V(q)]

= −∫

_{0}

^{τ}dσ [½m(dq/dσ)

^{2}+ V(q)]

≡ − S

_{E}[q], which is the

*Euclidean Action*.

(12) Therefore, < q

_{2}|U

_{E}(τ)|q

_{1}> = ∫D[q(s)] e

^{-SE[q]/ℏ}

The path integral in Euclidean time means q(0) = q

_{1}and q(τ) = q

_{2}

Note the differences:

(13) S[q] = ∫

_{0}

^{t}ds [½m(dq/ds)

^{2}− V(q)]

⇒ S

_{E}[q] = ∫

_{0}

^{τ}dσ[½m(dq/dσ)

^{2}+ V(q)]

(14) ∫D[q(s)]e

^{iS[q]/ℏ}⇒ ∫D[q(σ)] e

^{-SE[q]/ℏ}

Note:

i) S

_{E}[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[U

_{E}(τ)] = ∫dq< q|U

_{E}(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 = (k

_{b}T)

^{-1},

and density matrix ρ = U

_{E}(τ)/Z

= ∫dq< q| ρA|q > → ∫dq< q|ρ|q > a(q)

= ∫dq< q| U

_{E}(τ)/Z|q > a(q)

(17)

**Two-Operator Expectation Value**

(18) < A(σ

_{1})B(σ

_{2}) >

_{β}

= (1/Z)∫ D[q]e

^{-SE[q]/ℏ}a(q(σ

_{1})) b(q(σ

_{2}))

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

(19) Tr[U

_{E}(τ − σ

_{2})BU

_{E}(σ

_{2}− σ

_{1})AU

_{E}(σ

_{1})]

= Tr[U

_{E}(τ − (σ

_{2}− σ

_{1}))BU

_{E}(σ

_{2}− σ

_{1})A]

Comments:

i) What happens when τ goes to infinity?

Recall from (Equ. 16),

< A >

_{β}= Tr(ρ

_{β}A),

ρ

_{β}= (e

^{–βH})/Z = (1/Z) Σ

_{n}|n >< n|e

^{–βH}

Z = Σ

_{n}e

^{–βEn}

β = 1/kBT = τ/ℏ

Combining the above,

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

_{0}.

(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: σ

_{1}= i t

_{1}, σ

_{2}= i t

_{2}

With σ

_{1}< σ

_{2}

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

The contour is deformed: we go from U(t

_{1}), where we operate A, we go U(t

_{2}- t

_{1}), where we operate B, then go U(-t

_{2}), we then project on the vacuum.

(22) < A(σ

_{1})B(σ

_{2}) >

_{β=∞}→

< 0|U(-t

_{2})BU(t

_{2}-t

_{1})AU(t

_{1})|0 >

= < 0|U(-t

_{2})BU(t

_{2})U(-t

_{1})AU(t

_{1})|0 >

Recall: U(-t

_{2})BU(t

_{2}) ≡ B(t

_{2}), and U(-t

_{1})AU(t

_{1}) ≡ A(t

_{1}), in the Heisenberg picture.

(23) < A(σ

_{1})B(σ

_{2}) >

_{β=∞}→

< 0|B(t

_{2})A(t

_{1})|0 >

_{Heisenberg picture}if t

_{2}> t

_{1}.

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

(24) < A(σ

_{1})B(σ

_{2}) >

_{β=∞}= < 0| T[A(t

_{1})B(t

_{2})]|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 ≡ ∫ d

^{d}z j(z)Φ(z)

Expanding Z[ j ] in powers of j,

(28) Z[ j ] = Σ

_{N}((ℏ

^{−N})/N!)∫dz

_{1}…dz

_{N}j(z

_{1})… j(z

_{N})Z(z

_{1}…z

_{N})

(29)Where Z(z

_{1}…z

_{N}) = ∫D[Φ(x)] e

^{−S[Φ]/ℏ}Φ(z

_{1})… Φ(z

_{N})

(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, ⇒ Φ

_{c}depends on j

(36) Let Φ = Φ

_{c}+ ℏ

^{½ }δΦ for fluctuations of O(1)

Expand to 2nd order:

(37) S[Φ] − j∙Φ = S[Φ

_{c}] − j∙Φ

_{c}(Saddle point)

+ ℏ

^{½ }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"[Φ

_{c}])

^{-½}

(40) From equation 30, W[ j ] = ℏlog[Z[ j ] ]

⇒ −(S[Φ

_{c}] − j∙Φ

_{c})− ½ℏTr[log(S"[Φ

_{c}])]

We want to know what is the background field.

(41)Consider W[ j ] ≡ W

_{c}[ j∙Φ]

_{φ=Φc}

(42) φ(x) = δW[ j ]/δj(x)

= Φ

_{c}(x) + ∫dy {δΦ

_{c}(y)/δj(x)} {δW

_{c}[ j ]/δΦ(y)}

(43) Calculating the second bracket in the integral,

{δW

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

(44) But j(y) − δS/δΦ(y) = 0 if Φ=Φ

_{c}(equation 35)

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

φ(x) = Φ

_{c}(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[Φ

_{c}] − j∙Φ

_{c}) + ½ℏTr[log(S"[Φ

_{c}])] + … (equation 40)

= j∙(φ − Φ

_{c}) + S[Φ

_{c}] + ½ℏTr[log(S"[Φ

_{c}])] + …

(equation 44)

From φ(x) = Φ

_{c}(x) + O(ℏ) (equation 45)

We write,

⇒ φ(x) = Φ

_{c}(x) + δφ

(47) Tailor expansion on S[φ]

= S[φ

_{c}] + (φ − φ

_{c})S'[φc] + ½ δφ S"[φ

_{c}] δφ + ….

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

^{2}.

S[φ] = S[φ

_{c}] + (φ − φ

_{c})S'[φ

_{c}] + O(ℏ

^{2}) + ….

From δS/δφ(x) − j = 0, (equation35)

or S'[φ

_{c}] = j

Substituting that into 47, and ignoring O(ℏ

^{2}),

(48) S[φ] = S[φ

_{c}] + (φ − φ

_{c})j + ….

Repeating equation 46,

(49) Then Γ[φ] = j∙(φ − Φ

_{c}) + S[Φ

_{c}]

+ ½ℏTr[log(S"[Φ

Substitute 48 into 49,
_{c}])] + …(50)Γ[φ] = S[φ] + ½ℏTr[log(S"[Φ

_{c}])] + …

This reads as,

*Quantum Effective Field = Classical Field*

*+ 1*^{st}order quantum correction

*Theory A (high energy) = Theory B (low energy)*

**+ O(1) + O(2) + O(3) + ...****Appendix**

We will borrow from The Path Integral Simplified

(A) C

_{1}= ∫

_{–∞}

^{∞}e

^{–½y2}dy = (2π)

^{½}(equ.A10)

(B) C

_{2}= ∫

_{–∞}

^{∞}e

^{–½ay2}dy = (2π/a)

^{½}(equ.B5)

(C) C

_{3}= ∫

_{–∞}

^{∞}e

^{–½ay2 + by}dy = (2π/a)

^{½}e

^{b2/2a}(equ.C7)

We want to generalize this for an n

*x*n matrix A, we rewrite the integral as,

(D) C

_{2}→ C

_{4}= ∫

_{–∞}

^{∞}∫

_{–∞}

^{∞}…∫

_{–∞}

^{∞}dx

_{1}dx

_{2}…dx

_{n}e

^{–½x∙A∙x}

(E) C

_{3}→ C

_{5}= ∫

_{–∞}

^{∞}∫

_{–∞}

^{∞}…∫

_{–∞}

^{∞}dx

_{1}dx

_{2}…dx

_{n}e

^{–½x∙A∙x + J∙x}

where x∙A∙x = x

_{i}A

_{ij}x

_{j}and J∙x = J

_{i}x

_{i}, 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 2

*x*2 matrix A' and diagonalize it to A.

Note: det A' = det A = ad – bc

We calculate x

_{i}A

_{ij}x

_{j}= x

_{1}(A

_{1}

_{j}x

_{j})+ x

_{2}(A

_{2}

_{j}x

_{j}), i=1,2

= x

_{1}(A

_{11}x

_{1}+ A

_{12}x

_{2})+ x

_{2}(A

_{21}x

_{1}+ A

_{22}x

_{2}), j=1,2

But A

_{12}= A

_{21}= 0

Therefore, x

_{i}A

_{ij}x

_{j}= x

_{1}A

_{11}x

_{1}+ x

_{2}A

_{22}x

_{2}

= (ad – bc )(x

_{1})^{2}+ (x_{2})^{2} = (Det[A])(x

So we take C_{1})^{2}+ (x_{2})^{2}_{4}= ∫

_{–∞}

^{∞}∫

_{–∞}

^{∞}dx

_{1}dx

_{2}e

^{–½x∙A∙x}, N=2

This becomes,

C

_{4}= ∫

_{–∞}

^{∞}∫

_{–∞}

^{∞}dx

_{1}dx

_{2}e

^{(Det[A])(x1)2+ (x2)2}

= ∫

The first integral is C_{–∞}^{∞}dx_{2}e^{–½x22}∫_{–∞}^{∞}dx_{1}e^{–½(Det[A])(x1)2}_{1}, and the second is C

_{2}.

C

_{4}= (2π)

^{½}(2π/det[A])

^{½}= ((2π)

^{2}/det[A])

^{½}

For any n

*x*n matrix A,

(F) C

_{4}=((2π)

^{N}/det[A])

^{½}

(G) C

_{5}= ∫

_{–∞}

^{∞}∫

_{–∞}

^{∞}…∫

_{–∞}

^{∞}dx

_{1}dx

_{2}…dx

_{n}e

^{–½x∙A∙x + J∙x}

=((2π)

^{N}/det[A])^{½}e^{½ J∙A-1∙J}
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