Wednesday, November 19, 2014

Another argument against the BBT

In The Equivalence Principle and the Big Bang Theory, we explored the idea that the galaxies are exhibiting a redshift that can be interpreted either as a Doppler effect, that is, they are moving away, or as a gravitational shift, in which case we can safely say they are at rest with each other. An argument for the BBT that every observer in the universe will see every other galaxy moving away is often used to validate the BBT. This argument is misleading. It is only partially true if one ignores that velocity has not only magnitude but it has also a direction.



According to an observer in our galaxy, he sees every other galaxy moving away. (Black Arrows). However, an observer from another galaxy, say the blue one, sees the other galaxies moving away but in different directions than what we see (Blue Arrows).



We will agree with her concerning both of our galaxies - She sees us move away from her, we see her moving away from us along the same line, and since motion is relative, our observation agrees. However, we will disagree about the direction for galaxies Red, Green and any other galaxy not situated along the line joining our two galaxies.

On the other hand, if we agree that instead of a Doppler shift, the galaxies are exhibiting a gravitational shift and are at rest with each other, then we all agree that the net force on any galaxy is zero.



Cosmic Microwave Background

The last remaining argument in favor of the BBT is the Cosmic Microwave Background (CMB). It has been said that only a Hot Big Bang scenario can explain the CMB, and any other explanation would have to be contrived. In future blogs, I will demonstrate that the explanation of the CMB from a Hot Big Bang scenario is itself contrived. Stay tuned.

Saturday, November 01, 2014

The Equivalence Principle and the Big Bang Theory

In this blog I explore the idea that Hubble failed to notice an important aspect of Einstein’s Equivalence Principle.

Consider this thought experiment. Different emitters are placed at different heights from the ground. The earth plays the role of the source of gravity.



They emit light, which from an earth observer, would be blueshifted. According to Einstein’s Equivalence Principle: we can say that the Doppler effect is equal to the gravitational shift ( see equation 2 in The Essential General Relativity , reproduced below)

(1) (Δf/f)gravity = -(Δf/f)doppler = -Δv/c

For emitter 1, we can say,

(2) Δv1 = g(d1) Δt1 (definition of acceleration, where g(d1) is the gravitational potential field at d1.

(3) Define g(d1) = g1

(4) Δv1 = g1 (d1/c) (time = distance/velocity)

(5) However, g1 = (GMsource)/ R21
= (GMsource)/ (Rsource + d1)2
= (GMsource) (Rsource + d1)−2
= (GMsource/ R2source) ( 1 + d1 / Rsource)−2
≈ (GMsource/ R2source) ( 1 − 2d1 / Rsource)
For 2d1 << Rsource

(6) g1 = (GMsource)/ R2source

Substitute (6) into (4),

(7) Δv1 = (GMsource)/ cR2source )(d1)

We can get the same result for emitters 2 and 3,

(8) Δv2 = (GMsource)/ cR2source )(d2)

(9) Δv3 = (GMsource)/ cR2source )(d3)

We can generalize equations 7,8 and 9 as,

(10) Δv = Hd ,

where H = (GMsource)/ (cR2source)

In case you haven’t recognized this, it is Hubble’s equation. When he discovered that all galaxies have a redshifted spectrum, Hubble concluded that all the galaxies were moving away. That is the Doppler Effect. However using Einstein’s Equivalence Principle, we can say that galaxies are at rest, and photons are redshifted( they are moving against gravity). Note that Hubble discovered not a change in velocity but just a velocity. In his days, he did not have the technology to observe such a small change in the galaxies' velocities, and it took more than 80 years before it was discovered that galaxies are actually accelerating.

In our thought experiment, the emitters are at rest, so one can easily say that such emitters would start to accelerate as they cannot be “nailed” in outer space. What about the galaxies, can they be “nailed” so that we can claim they are at rest with respect to each other? Consider one galaxy against all others.



According to Gauss’ theorem (see fig.3 in Newton's Law of Gravity), a galaxy would be attracted as if all the matter inside the sphere were concentrated at the center of that sphere. One can ignore all the other galaxies outside that sphere. At the same time, one can draw an infinite number of spheres, in which the galaxy would be attracted to the center of each sphere. Here’s a diagram with just three spheres drawn.



If the universe is infinite, we can safely say that the total force on a galaxy is zero, and therefore, the galaxy is at rest.



Does it mean that the Big Bang theory is wrong? No. The Big Bang theory says that for every galaxy, all other galaxies are moving away. But Einstein’s Equivalence Principle also says that we can look at every galaxy at rest with their light being redshifted. Both pictures are equivalent.



APPENDIX

Equation 10 is true for every galaxy, since each one is emitting light from a source of gravity that has an infinite radius. This argument is only valid if the universe is infinite.

Tuesday, October 28, 2014

Riemannian Geometry and the Big Bang Theory



Actually it was Gauss who proposed how to describe the inhabitants on a sphere as if they were unaware of the third dimension. Of course, they would need only two coordinates, why it’s called a 2-sphere. Here’s an example with spherical coordinates φ and θ:



But it was Riemann, Gauss’ student, who extended this idea to higher dimensions, hence why it’s generally known as Riemannian geometry. Notice that we have a 2-sphere embedded into 3D. So if we take our 3-dimensional world then Riemannian geometry tells us that it is embedded into a 4 spatial dimensional world. In General Relativity (GR), time is considered as another coordinate. So for the Big Bang Theory, which is based on GR, we must assume that it's a theory of a universe with 4 coordinates,(3+1), but embedded into 4 spatial dimensions + one temporal dimension.

Notice that if our inhabitants on a 2-D sphere were to observe that their New York City is moving away from their Paris, they would conclude that their universe is expanding. We, the 3-D creatures observing them, would know that someone/something is blowing air or matter into the inside of their sphere, causing their world to act in that way. But what about our 3-D spatial world, which is really treated as a Riemannian 4-D spatial world, what's causing it to being blown away? The question is: is the BBT really the final word on this? Does the fourth spatial Riemannian dimension really exist? Or is it just a fabrication that makes the calculations in the BBT simpler, but as a consequence, it is in fact misleading us? Stay tuned.

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]

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


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