Thursday, March 27, 2014

Harmonic Oscillators, Vacuum Energy, Pauli Exclusion Principle


There are two reasons why the harmonic oscillator plays such a pivotal role in Quantum Mechanics (QM).

i) It is one of the few problems for which there exists an exact solution.

ii) The whole mathematical apparatus is transferred directly to Quantum Field Theory (QFT).

Sidney Coleman famously said: “The career of a young theoretical physicist consists of treating the harmonic oscillator in ever-increasing levels of abstraction.”

Classical Oscillator

We begin with Hooke’s Law for a body oscillating under a restoring force:

(1) F = − kx, where k is a constant for the restoring force

(2) From Newton’s law of motion,

mx'' +kx = 0, where x'' is the acceleration ( second derivative wrt time).

(3) We write this as,

mx'' +ω2x = 0, where ω = (k/m)½, is the classical frequency of oscillation.

The general solutions for this equation, aka the wave equation is,

(4) x = Acos(ωt) + Bsin(ωt), where A and B are constants whose value will be determined by initial conditions.

We can also write the general solution, but this time slightly different,

(5) x = x0cos(ωt + φ), where x0 is the amplitude and φ is the phase.

Hamiltonian Formalism

(6) H = T + V,

where H is the energy of the system, T is the kinetic energy, and V is the potential energy.

= p2/2m + ½mω2x2
Substituting (5) into (6),

(7) E= ½ m(x')2 + ½mω2x2

= ½mx02ω2sin2(ωt+φ)+ ½mω2x02cos2(ωt+φ)
= ½ mω2x02[sin2(ωt+φ) + cos2(ωt+φ)]
(8) E = ½mω2x02

We see that the energy depends on the amplitude x0 and the frequency ω, and it is independent of the phase φ. Since the amplitude is a continuous variable, so is the energy E.

Solving for the velocity, x' in equation (7),

(9) (x')2 = 2E/m − ω2x2

(10) Or, x' = (2E/m − ω2x2)½

Substituting (8),

(11) x' = ω (x02 − x2)½

We see that the particle will oscillate and x= ± x0 are the turning points, after which the particle speeds up until it reaches the origin, slows down until it reaches the next turning point, then reverses direction to repeat the motion in endless oscillations.

Quantization of the Oscillator

As we have seen in The Essential Quantum Mechanics (EQM) , we must express our system into some eigenvalue equation. We could choose position, momentum or energy as a basis. For the oscillator, we will choose the energy as our basis. We write,

(12) H│E> = E│E>, where │E> is the ket basis, and E is the eigenvalue, a real number. Our Hamiltonian operator H can be expressed in terms of the position operator X and the momentum operator P from equation (6) as,

(13) H = P2/2m + ½mω2X2

(14) P = −iℏ∂x (equation 35 in The Essential Quantum Mechanics (EQM) , )

(15) [X,P] = i ℏ (same reference as above)

We define the annihilation operator a and its adjoint, the creation operator a, as,

(16) a = fX + igP

(17) a = fX – igP

(18) where f = (mω/2ℏ)½ and g = (2mωℏ)–½. Note that f and g are not operators and fg = (2ℏ)–1.

We now calculate the commutation between these two operators,

(19) [a,a] = aa – aa

= (fX + igP)(fX – igP) – (fX – igP)(fX + igP)
= –2i(fg)[X,P]
= 1, using (15) and (18)
Next we calculate the operator aa,

(20) aa = (fX – igP)(fX + igP)

= f2X2 + g2P2 + ifg[X,P]
using (21) and (18)

=(mω/2ℏ)X2 + (2mωℏ)–1P2 + i(2ℏ)–1(iℏ) ,
= (ℏω)–1(½mω2X2 + P2/2m) – ½
= H/ ℏω – ½, using (16)

(21) H = ℏω(aa + ½)
= aa + ½, (from now on, ℏω = 1)
The commutation between a and H is,

(22) [a,H] = [a,aa + ½]
= [a,aa]
= aaa – aaa
= aaa – (aa – 1)a
= a
Similarly, the commutation between a and H is,

(23) [a,H] = [a,aa + ½]
= [a,aa]
= aaa – aaa
= a(aa – 1) – aaa
= – a
We now get an interesting unexpected result if we consider,

(24) Ha│E> = (aH – [a,H])│E>
= (aH + a)│E>, using (23)
= (E + 1)a│E>, using (12)

(25) Ha│E> = (aH – [a,H])│E>
= (aH – a)│E>, using (22)
= (E – 1)a│E>, using (12)
We see that a│E> is an eigenstate of the Hamiltonian H, with eigenvalue E+1; and a│E> is also an eigenstate of the Hamiltonian H, with eigenvalue E– 1 . By repeatingly acting the operator a or a,we can get all the eigenvalues as,

(26) E + 1, E + 2, E + 3… E + ∞ , (upward chain)

(27) E – 1, E – 2, E – 3… E – ∞, (downward chain)

IMPORTANT: The downward chain must break at some point. There must be a state │0> that cannot be lowered any more. That is,

(28) a│0> = 0.

This is how we come to the concept of a ground state or vacuum energy state, which is one of the most important concepts in QFT. And it will turn out that such a definition will cause us trouble when dealing with QFT in curved space-time (QFTCST).

Using the same scheme, we can define the concept of a particle:

(29) a│0> = N│1>,

where N is a normalizing factor, and │1> is a state with one particle with momentum k.

Usually, we would write ak│0> = N│1k>. For our purpose we can drop the index k. It is also why a is called a creation operator as it creates a particle from the vacuum state. Similarly, the operator a is the annihilation operator because it annihilates a particle when acting, and if that state is the vacuum, we get equation (28).

Applying the Hamiltonian on the vacuum state, (and restoring ℏω)

(30) H│0> = ℏω(aa + ½)│0>, using equation (24)
= ½ ℏω│0>, using equation (21)
We can see that the vacuum energy is not zero, which is another potential problem in QFT, where we must sum up over all modes. However this is to be expected since there are an infinite number of points in space. But restraining our arguments to energy density, rather than energy, we can avoid that problem.

Pauli Exclusion Principle

We can now construct states with multiple particles just by operating the creation operator multiple times on the vacuum state. This is called a Fock space. We denote the number of particles, along with their momentum.

Example: │1k1, 3k2, 7k3>, this state denotes that it contains 1 particle with momentum k1, 3 particles with momentum k2, and 7 particles with momentum k3.

This can be represented as

│1k1, 3k2, 7k3> = (ak1†)(a k2†)3(ak3†)7│0>, leaving out normalizing factor.

So generally speaking, we have a state denoted by │nk1, nk2, nk3 …nkj>. Now, for bosons, the n’s can take any value 0,1,2,…. ; but for fermions, they can only take the values 0,1 due to the Pauli Exclusion Principle.

For identical particles, how do we identify the state if we exchange the particles? Take a state with two particles, denoted by │ab>. We can interchange the particles and get the state │ba>. Note,

(31) │S> = N(│ab> + │ba>), where again N is a normalizing factor. If we interchange a ↔b, we get the same state. The state │S> is said to be symmetrical under the exchange.

(32) │A> = N(│ab> – │ba>). Now consider the same interchange a ↔b, we get a new state │A'> = – │A>. The state │A> is anti-symmetrical under the exchange.

It turns out that in nature, we can classify all particles as symmetric or anti-symmetric. Which ones are symmetric (bosons) and which ones are anti-symmetric (fermions) is determined by observation.

Note that in the case of identical particles, if a = b, then │A> = – │A>, and therefore │A> must be zero, which is the essence of the Pauli Exclusion Principle that regulates the chemical properties of atoms: Two fermions cannot occupy the same state.

Friday, February 21, 2014

The Essential Quantum Mechanics

Mathematical formulation of QM

We will focus our attention on the wavefunction, the operative part is "function", as this object is mathematical in nature, and not to be taken as a real wave. If you keep that in mind, a lot of confusion about QM will dissipate.

(1) We will use the Dirac notation, that is, the wavefunction is a vector V denoted by a "ket", | V >.

(2) A ket can be multiplied by any scalar α, β... denoted by α| V >, β| V >...

Note we can also write α| V > as | αV >, that is, a new vector | V' >.

(3)There is a dual vector, the "bra", denoted by < V |.

(4) Note that < V' | = < αV | = < V | α*, where α* is the complex conjugate of α.

(5) Vectors can also be represented by matrices. In the case of the ket vector, | V >, it will be a column matrix, while the bra < V | will be a row matrix, with its elements as the complex conjugate of every element of the vector | V >. See diagram below for a vector in 4-D.

(6) Any vector | V > in an n-dimensional space can be written as a linear combination of n linearly independent vectors | 1 >, | 2 >...| n >.

(7) A set of n linearly independent vectors in an n-dimensional space is called a basis, and so we can write,

| V > = Σ vi | i >,

where i = 1... n, vi are the components of the vector, and the | i >'s are the basis vectors.

(8) We can construct an inner product between two vectors. The analog of the dot product between two vectors V and W, VW = VW cosθ, is < V | W >. In diagram 1, we have,

< V | V > = v1v1* + v2v2* + v3v3* + v4v4* = |V|2

(9) Note that

< V | W > = < W | V >* = Σ Σ vi* wj < i | j >.

(10) If the basis vectors have unit length, then they form an orthonormal base. From (7) above, we can find the jth component of the vector | V > as such,

| V > = Σ vi | i >,

Multiply both sides by < j |

< j | V > = Σ vi< j | i >
= Σ viδij - using (6)
= vj
An important result is that,

vi = < i | V >, and (7) can now be written as,

| V > = Σ | i >< i | V > .

(11) An operator Ω is an instruction to transform a vector | V > into another vector| V' >. This can be written as,

Ω| V > = | ΩV > = | V' >.

Note that Ω acts on the right.

(12) Operators are said to be linear if they obey the following rules: for any α and β,

(12a) Ωα| V > = αΩ| V >

(12b) Ω{α| V > + β| V' >}= αΩ| V > + βΩ| V' >

The action of two successive operators Ω and Λ in general will not be commutative. We designate the commutator as such:

(13) ΩΛ - ΛΩ ≡ [Ω,Λ]

(14) Recall that | V > = Σ | i >< i | V >, we can define the projector operator as,

Pi = | i >< i |

The importance of this operator is that,

I = Σ Pi, where I is the identity operator.

(15) In regard to the dual vector, the bra < V |, we define the operator acting on it as,

< V' | = < ΩV | = < V |Ω, where Ω is called the adjoint operator.

Note that Ω acts on the left.

Also, < V' | = < αV | = < V |α*

Therefore, < V |Ω = < V |α*

In a given basis, the adjoint operation is the same as taking the transpose conjugate.

(16) An important theorem is,

(ΩΛ) = ΛΩ

(17) An operator is said to be Hermitian if,

H = H

(18) An operator is said to be unitary if,

UU = UU = I, where I is the identity operator.

Note that U = U-1, where U-1 is the inverse of U.

An important theorem is that a unitary operator preserves the inner product:

Proof: consider | V' > = U| V > and | W' > = U| W >

Then < W' | V' > = < W |UU | V > = < W | V >

(19) An important problem involves the situation when,

Ω| V > = ω| V >, where ω is a number, real or complex.

We say that the operator Ω rescales the vector | V > by a factor ω. That equation is called an eigenvalue equation: | V > is an eigenket of Ω with eigenvalue ω.

Consider < V |Ω| V > = ω < V | V >, and < V |Ω| V > = ω* < V | V >. If Ω is hermitian, then Ω = Ω. Therefore ω = ω*, and ω is real. Since what we measure are observables, and their measurements must be real numbers, Hermitian operators are perfect candidate for observables in QM.

(20) So far we have dealt with vectors. Now we want to bring in continous functions into our formalism as these will play an important role. We take a function f(x) along a certain interval between 0 and L. Divide this into equal parts, say n=20 parts. Let x=L/20, 2L/20...19L/20. See fig. 2 below.

We denote the ket | fn(x) > as the discrete approximation of f(x).

The basis vector in this space are:

We have,

< xi | xj > = δij

Σ | xi >< xi | = I

| fn(x) > = Σ| fn(xi) > | xi > , where i = 1....n

(21) All we have to do to go from the discrete to the continuous spectrum is,

n → ∞, and Σ → ∫. For instance,

Σ | xi >< xi | = I → ∫ | x >< x | dx = I

(22) Note that < x | f > is just the projection of | f > along the basis | x >, which is just f(x).

< x | f > = f(x)

Likewise, < g | x > = g*(x).

(23) The inner product - see equation (9) - becomes,

< f | g > = ∫ < f | x >< x | g > dx - using (21)

= ∫ f*(x)g(x)dx - using (22)
Particle on a line

Consider a particle moving along a line in the x-direction. Every point on the line can be represented by a function of x, denoted by |Ψ(x)>. The observable would be the operator that locates the particle on the x-axis, say X. In this case we can write,

(24) X|Ψ(x)> = xΨ(x)

The operator X simply multiplies the function Ψ(x) by x. The next step is to find if there are eigenvalues, and what are they. To do that we would need to find the eigenvectors such that,

(25) X | λ > = λ | λ >, where | λ > is the eigenket, and λ is the eigenvalue.

Substituting and transposing,

(26)(x – λ) | λ > = 0

This tells us that whenever x ≠ λ, then the function | λ > is zero. The only place where | λ > is not zero is when x = λ. Let’s plot what | λ > looks like.

As the interval ε decreases to zero at x = λ, the function goes to infinity. This function is known as the Dirac delta function.

(27)| λ > = δ (x – λ)

Note: the area under the delta function is 1 (ε x 1/ε ).

(28) The probability of detecting the particle at position x is,

P(x) = |< x | Ψ >|2 = < x | Ψ >< Ψ | x > = Ψ(x)*Ψ(x)

The implication is that if Ψ(x) is any function of x, we can calculate the probability of finding a particle at x with the above mathematical structure we have constructed.


We are going to consider another operator, the differential ∂x, which would give another function of x, ∂xΨ(x).

(29) To be an observable, this operator needs to be Hermitian, which it isn’t in that form. However, K = – i∂x is Hermitian.


We need to show that < Ψ | K | Ψ > is real , that is,

< Ψ | K | Ψ > = < Ψ | K |Ψ >*

(i) By definition,

< Ψ | K | Ψ > = ∫ Ψ*(x)(– i∂x )Ψ(x) dx
= – i ∫ Ψ*(x) (∂Ψ(x)/∂x) dx
(ii) Integrating by parts,

< Ψ | K | Ψ > = i ∫ Ψ(x)(∂Ψ*(x)/∂x) dx

(iii) Complex conjugate the above,

< Ψ | K | Ψ >* = – i ∫ Ψ*(x)(∂Ψ(x)/∂x) dx

Which the same as (i), QED.

Our next step is to find the eigenvectors of this operator K, that is,

(30) – i∂x Ψ(x) = k Ψ(x), where k is a real number

Aside from an arbitrary constant, which we can neglect for our purposes, a solution to that equation is,

(31) Ψ(x) = e ikx

These are the eigenvectors of the operator K , which are exponential functions. Note how these functions behave differently from the eigenfunctions of the position, which were the Dirac delta functions. While the position eigenfunctions are peaks, the eigenfunctions of K extend over all spaces, oscillating everywhere with equal probability.

So what is this operator? We can see from fig 6 that after the wavelength λ, the wavefunction will repeat itself. For each cycle,

(32) kλ = 2π, or λ = 2π/k

(33) Here we appeal to history. De Broglie had hypothesized in the wave/particle duality that every wavelength was inversely proportional to the momentum, p = h/λ, where h is the Planck constant.

Substituting for λ,

(34) p = (h/2π) k = ℏ k

So we can see that our operator K differs from the momentum operator by a factor ℏ. One of the most remarkable result of quantum mechanics is that momentum along the x-axis can be represented by the operator,

(35) px = – iℏ∂x

Finally, let us calculate the commutator [x,p] ( see 13 above). Recall that we are dealing with operators, and by definition, these operate on vectors/functions.

[x,p] → [x,p]f = (xp - px)f

= {x(– iℏ∂x) - (– iℏ∂x)x}f
= x(–iℏ∂xf) + x(iℏ∂xf) +iℏf = iℏf
Therefore, [x,p] = iℏ

These two operators don't commute. We see how the Heisenberg Uncertainty Principle is revealed. If we know the position of the particle with definite precision (the Dirac delta functions), then the eigenvectors of the momentum (the exponential functions) tells us we can’t define its momentum as these wavefunctions are spread out equally all over the space (the x-axis). In this case, when the two operators don't commute, [x,p]= iℏ ≠ 0, we say that the position and momentum of a particle are incompatible observables.

On another note, the mathematical formalism of QM is quite different than classical physics. We have Hermitian operators, representing observables, operating on the wavefunction, which contains the information of the system, such as position, momentum, etc. and with this, we can calculate probabilities. A long historical debate ensued as to whether the wavefunction contains all of the information, or is QM incomplete. So far, after nearly 90 years, no one has come up with a better theory.

Wednesday, January 22, 2014

The Essential General Relativity


What are the equations needed to go from Newtonian Physics to General Relativity? In Is Newtonian Gravitational Fields Valid Within General Relativity? we showed in the case of a weak gravitational field and low velocity how the Einstein Field Equations (EFE)

(A) Gμν = 8πGc-4Tμν

yield Newton's law of gravity,

(B) ∇2φ = 4πGρ

However, historically, Einstein guessed equation 16, from the above reference reproduced below,

(C) g00 = 1+ 2φ/c2

The question is: how do you get from equation C to equation A?

Thought Experiment

Einstein was convinced that gravity would affect light. But how? He imagined this thought experiment. Suppose that gravity does NOT affect light. Then one could construct the following scheme:

One could release a particle of mass m at rest at a distance h from the ground (fig. 1a). Its potential energy(PE) would be converted to kinetic energy (KE). Einstein knew that this particle could then be converted to a photon, and then that photon could climb against gravity, reconverted to a particle with KE ≠ 0, and would now fall, picking up more kinetic energy (fig. 1b). Einstein reasoned that the law of conservation of energy demanded that the photon must lose energy when climbing up against gravity. But how, since light always travel at a constant speed c? The only way out was to use what he had already used in his seminal paper on the photoelectric effect,

(D) E = hf, where here h is Planck's constant.

If E has to decrease, then the frequency f must also decrease, or its wavelength increase. This is known as the gravitational redshift.

Second Thought Experiment

Einstein arrived as his Equivalence Principle (EP) with the following thought experiment. In a room without windows, one would not be able to distinguish between being pulled up by an inertial force (fig.2a) from being pull down by a gravitational force near a planet (fig. 2b).

Third Thought Experiment

In fig.3a,to the observer in free falling spaceship, there is no frequency shift. However, to the observer outside the spaceship (fig.3b), the emitter is moving, and according to the Doppler effect, there should be a blueshift. Einstein reasoned that the null result would come about the cancellation between the Doppler effect and the gravitational redshift.

(1) (Δf/f)doppler = Δv/c (See equation A10 in the appendix of Relativistic Doppler Effect)

According to the EP,

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

But Δv = gΔt (definition of acceleration)
= g(h/c) (time = distance/velocity)
= Δφ/c (gh = change in gravitational field)
(3) Therefore, (Δf/f)gravity = -Δφ/c2

We need to convert this last result into a differential equation. We proceed as such,

(4) Δf/f = (frec - fem)/fem

For convenience, the emitted light happens early (em → 1), the received light happens later (rec → 2). Also the frequency is inversely proportional to time ( f ~ 1/τ). Putting this into equation (4), we get,

(5) Δf/f = (frec - fem)/fem
= (1/τ2 - 1/τ11
= τ12 - 1 = (τ1 - τ2)/τ2
= -(τ2 - τ1)/τ2 = -Δτ/τ2 = -dτ/τ2
(6) Therefore equation (3) becomes,
(dτ/τ2)gravity = +Δφ/c2
We also need to relate this result to the proper time τ and the coordinate time t. Consider two points, x at the proper time τ(x), and the second point at infinity, being the reference point where φ(∞)=0, is the coordinate time t(∞).

Therefore, (7)(dt - dτ)/dt = (φ(∞)- φ(x))/c2

Or, 1 - dτ/dt = 0 - φ(x)/c2. Rearranging, we get,

(8) dτ = (1 + φ(x)/c2)dt.

We now need one more step to link this result to the metric tensor, which we have already explored in Relativistic Doppler Effect. The metric tensor can best be expressed as,

(9) ds2 = gμνdxμdxν, (μ,ν = 0,1,2,3; or μ, ν = ct,x,y,z)

Separating the time component from the spatial components,

10) ds2 = g00dx0dx0 + gijdxidxj, (i,j=1,2,3; or i,j = x,y,z)

= g00dx0dx0 + gijdxidxj,
= η00dx0dx0 + ηijdxidxj,
= -c2dt2+ dx2 + dy2 + dz2,
Where in the last two lines we used the Minkowski metric for flat space with signature (-1,1,1,1). In the particle coordinate system, it is at rest. Therefore,

(11) dx2 + dy2 + dz2 = 0

and by definition, the time measured in that coordinate is the proper time τ,

(12) Equation 10 reduces to,

ds2 = -c22
= -c2(1 + φ(x)/c2)2dt2
Where we substitute equation (8) in the last line.

(13) For low gravity, (1 + φ(x)/c2)2 ≈ (1 + 2φ(x)/c2)

Comparing equations (10), (12), and (13), we get

(14) g00 = 1+ 2φ/c2

Which is (C), what Einstein had identified. Lucky guess or brilliant insight, most likely a combination of both spearheaded Einstein to develop the full theory of General Relativity. But how do we get the EFE,

(15) Gμν = 8πGc-4Tμν - Equation (A)

Knowing that it must yield in weak gravity field and low velocity, Newton's equation,

(16) ∇2φ = 4πGρ - Equation (B)

Without delving into differential geometry, we will outline schematically how Einstein arrived at the final destination, equation (15) - which incidentally took nearly 10 years of his life.

Matter Curves the Geometry of Spacetime

The left-hand side (LHS) of equation (15) is the geometry part of the Einstein Field Equations (EFE). It is expressed as,

(17)Gμν = Rμν - ½gμνR

Where gμν is the familiar metric tensor of equation (14), Rμν is the Ricci tensor, and R is the Ricci scalar. The last two mathematical objects are special objects obtained from the Riemann curvature tensor, which is obtained from the parallel transport of a vector.

We are familiar with the addition of two vectors. Consider two forces acting on a point. We want to know: what is the resultant net force (vector addition).

The assumption that we have made by parallel transporting F2 from the tail of F1 to the head of F1 is that space is flat. But suppose that space isn't flat.

In fig.5, we see that the parallel transport of a vector along a longitude at the equator (1), to the north pole (2), then down a different longitude to the equator (3), back to the initial position (4) brings about a 900 difference between the vectors at position (4) and (1).

In a space that is curved, the parallel transport of a vector (broken red arrow) at A to B will bring about a deviation (black arrow). Deviations will also occur at C, D and then back to A. The result of the total of these deviations ΔV going around the loop is measured by the Riemann curvature tensor Rαμβν.

The Ricci tensor is a contraction of two indices. First we raise one of the indices,

(18) gασRαμβν = Rσμβν.

Then we contract by making σ = β

(19) Rβμβν = Rμν

And the Ricci scalar is a further contraction of the Ricci tensor,

(20) R = gασRαμ = Rσμ
= Rμμ (setting σ = μ)

Historically, Einstein had chosen the Ricci tensor for the LHS of equation (15), but soon realized that the gradient of the Ricci tensor was not zero, hence the conservation of energy would not be contained in the EFE. He finally amended by using equation (17).

The right-hand side of equation (15) is a little more of a guessing game. From equation (C), Einstein reasoned that the density of matter ρ must be related to the energy-momentum tensor (See equation 18 in Is Newtonian Gravitational Fields Valid Within General Relativity? ). So Einstein proposed,

(21) Gμν = kTμν

With the only thing to fix is the constant k, which is done by considering again the case of a weak field and low velocity (the Newtonian limit). See Appendix below for the derivation of k = 8πG/c4

Comments The path from Newton's law of gravity to Einstein's field equations is anything but straight and forward. There were a number of brilliant insights (the three thought experiments) and fortunate guesses ( equations 14 and 17). Nevertheless that path was arduous and far from being obvious. Why is it that the EFE are accepted by the physics community? Empirical evidence gives sufficient support in the anomalous precession of the perihelion of Mercury's orbit, light from a distant star deflected by the sun, and the gravitational redshift - all those confirming that GR is the right stuff. It is also present in every day of our lives with GPS, which without GR the positional satellites would be less than useful. Could GR be supplanted? Of course, if some day a brilliant mind comes along with more brilliant insights, and perhaps a few lucky guesses, and puts forward a new theory of gravity, hopefully one that is compatible with QM. So far, that day doesn't seem to be close at hand.

Appendix From now on, we will borrow heavily from Is Newtonian Gravitational Fields Valid Within General Relativity? . In particular, equations 15 and 24 combine to give,

(A1) R00 = -½∇2g00

And from the same reference, equation 5, which now reads as,

(A2) R00 = k(T00 - ½Tg00)

Equating A1 and A2,

(A3) -½∇2g00 = k(T00 - ½Tg00)

LHS of A3:

-½∇2g00 = -½∇2 (1+ 2φ/c2) = -∇2φ

RHS of A3:

k(T00 - ½Tg00) = ½kT00 - using that T = g00T00

Using that T00 = -ρc2, equation 18 in same reference above.

k(T00 - ½Tg00) = -½kρc2

Equating LHS with RHS of A3:

2φ = ½kρc2

Comparing this with Newton's law of gravity:

2φ = 4πGρ

Then k = 8πG/c4

Wednesday, October 23, 2013

Description of Reality - The EPR Paper Revisited

In a complete theory there is an element corresponding to each element of reality. A sufficient condition for the reality of a physical quantity is the possibility of predicting it with certainty, without disturbing the system. In quantum mechanics in the case of two physical quantities described by non-commuting operators, the knowledge of one precludes the knowledge of the other. Then either (1) the description of reality given by the wave function in quantum mechanics is not complete or (2) these two quantities cannot have simultaneous reality. Consideration of the problem of making predictions concerning a system on the basis of measurements made on another system that had previously interacted with it leads to the result that if (1) is false then (2) is also false. One is thus led to conclude that the description of reality as given by a wave function is not complete.

If you are not familiar with this quotation, it is taken directly from the original EPR paper (See: Can Quantum-Mechanical Description of Physical Reality Be Considered Complete' ? A.EINSTEIN, B.PODOLSKY AND N.ROSEN, Institute for Advanced Study, Princeton, New Jersey (Received March 25, 1935))

There are two distinct errors in the logic to the pre-amble of one of the most famous scientific papers ever written. In the statement, If P then Q, one generally accepts that the premise P implies the conclusion Q. We are making the allusion to the following sentences:

In quantum mechanics in the case of two physical quantities described by non-commuting operators, the knowledge of one precludes the knowledge of the other. Then either (1) the description of reality given by the wave function in quantum mechanics is not complete or (2) these two quantities cannot have simultaneous reality.

(1) Knowledge is one thing, reality is something else. Even Einstein famously quoted, “Do you really believe that the moon isn't there when nobody looks?” Not knowing the position ( or momentum) of a particle does not deny its existence.

(2) Not being able to measure two things simultaneously does not deny the existence of either thing.

Yet it is this erroneous thinking that leads the EPR to conclude that QM is “incomplete”.

The beauty of the Heisenberg Uncertainty Principle (HUP) is that it conveys quite accurately the wave/particle duality. In classical physics, all things were placed into two different bins: (1) particles - those things that bounce off each other; (2) waves - those things that went through each other, amplitudes interfering constructively/destructively, but then went on their way as if nothing had happened. In the subatomic, reluctantly we found objects that don't fit into this framework - hence the need of a different framework that went under the name of Quantum Mechanics (QM).

But the HUP is at the core of this alternative framework.

Δx Δp ≈ ℏ

If we measure a particle, Δx → 0 then Δp →∞, we don’t know where the particle is going.

If we measure with certainty its momentum Δp → 0 then Δx →∞, which is what a wave is.

The best picture we get from QM of these objects is a wave packet for which there is some uncertainty in both the position and momentum.

Quantum Field Theory (QFT)

This is borne out by QFT, in which the theory gives predominance to fields, and particles are regarded as excited states of the fields. To make the theory work, particles are conceived as things being surrounded by a cloud of particles/anti-particles, sometimes interacting with themselves, sometimes with other fields or particles, and sometimes even with the quantum fluctuations of the vacuum. Indeed, these are complicated objects.


Historically speaking, the word “incomplete” as used in the EPR paper was technically correct, but it was interpreted incorrectly. If QFT is the house, then QM is the basement. It is the foundation on which QFT was built. However, there were those persuaded by the EPR paper who were drawn to conclude that QM gave the wrong description of reality and an alternative, new theory had to be developed. These attempts may have been brave but led to nowhere. In the meantime, QFT was being built (1930-1970’s), which would complete the house. The irony is that this was accomplished by incorporating into QFT, Special Relativity – Einstein’s own theory that had stunned the world in 1905.

Sunday, July 07, 2013

The Unruh Effect

In this blog, ℏ=c=kb = 1 where ℏ is the reduced Planck constant, c is the speed of light and kb is Boltzmann’s constant.


In the early 1970’s there were several applications of QFT to GR, the most famous one was done by Hawking in which he derived that a Black Hole has entropy proportional to its area. A less famous, but equally important, is the Unruh Effect. What’s interesting is that this has certain parallel with Einstein derivation of E = mc2. In both cases, there is the use of two observers; in Einstein case, one is in motion relative to the other; in the Unruh case, one is accelerating with respect to the other. Secondly, Einstein calculated the total and kinetic energy for each of the observer, and concluded the only thing that made sense was that a decaying particle must lose mass to the equivalent of E = mc2. (See: Einstein's Derivation of the Famous Equation, E=mc2) On the other hand, Unruh calculated the vacuum energy of each observer, and concluded that the accelerating vacuum energy would radiate as a black body, and its temperature would be proportional to its acceleration.

The Unruh Effect

The standard metric in Special Relativity is (see equation (6) in Relativistic Doppler Effect ),

(1) ds2 = -dt2 + dx2

The light-cone coordinates are defined as,

(2) u = – t + x, v = t + x

Taking the differential,

(3) du = – dt + dx, dv = dt + dx

The metric becomes,

(4) ds2 = –dt2 + dx2 = dudv

Now consider two observers: Alice, who is in an accelerating frame; and Bob, who is at rest or in uniform motion.

We will write the velocity, v = dxμ/dτ = x'μ, and the acceleration, aμ = x''μ

Consider the square of the acceleration,

(5) a2 = aμ aμ = – (t'')2 + (x'')2

A solution to that equation is,

(6) t = (1/a ) sinh(aτ), x = (1/a ) cosh(aτ).

We see that in fig.1 both Alice and Bob are space-like, and Alice lies on a hyperbola.

The light-cone coordinates, equation (2), becomes,

(7) u = (1/a)e-aτ, v =(1/a)e

We want to find a coordinate system in which Alice is going nowhere, that is,

(8) her time coordinate is ξ0= τ, and her spatial one is ξ1= 0.

In this new coordinate system, the light-cone coordinates would be,

(9) U = ξ1 – ξ0 = – τ, and V = ξ1 + ξ0 = τ

Substituting equation (9), (8) into (7), we get,

(10) u = (1/a)eaU, v =(1/a)eaV

Note that,

(11) U = a-1ln(ua) and V = a-1ln(va)

Taking the derivative of equations (10),

(12) du = eaUdU, dv = eaVdV

Now the metric, equation (4), takes on the form,

(13) ds2 = dudv = ea(U+V)dUdV

Both observers will write down their action according to the coordinates they live in.

(14) For Bob, sBob = ∫ ∂uφ∂vφdudv;
for Alice, sAlice = ∫ ∂Uφ∂VφdUdV

The equation of motion is the well-known wave equation,

(15) For Bob, ∂uvφ = 0; for Alice, ∂UVφ = 0

Since both are similar equations, we will write the solution for one, keeping in mind that the other is just a matter of changing, u → U, and v →V. So the solution can be written as,

(16) φ = A(u) + B(v), where A(u) are the left-moving waves and B(v) are the right-moving waves.

In terms of QFT, these solutions are written as,

(17) For Bob, φ Bob = ∫ dω(2ω) (a−Lωe−iωu + a+Lωeiωu) + ∫ dω(2ω)(a−Rωe−iωv + a+Rωeiωv)

(18) For Alice, φ Alice = ∫ dΩ(2Ω)(b−LΩe−iΩU + b+LΩeiΩU) + ∫dΩ(2Ω)(b−RΩe−iΩV + b+RΩeiΩV)

• Where (2ω) and (2Ω) are normalizing factors.

• a−Lω and a−Rω are Bob's annihilation operators; b−LΩ and b−RΩ are Alice's annihilation operators.

• a+Lω and a+Rω are Bob's creation operators; b+LΩ and b+RΩ are Alice's creation operators.

Again, because the solution is symmetric between left-moving waves and right-moving waves, we need to focus on one of them. Thus, we will use for both observers, the left-moving waves and drop the L and R index. That is,

(19) a−Lω → aω , b−LΩ → bΩ.

Both observers will define their vacuum state as such,

(20) aω |0>Bob = 0; bΩ |0>Alice = 0

The question is: are these two vacua the same?

Let’s assume we can write,

(21)bΩ = ∫ dω (αΩωaω + βΩω a+ω), and
b+Ω = ∫ dω (α*Ωω a+ω − β*Ωω aω)

Substitute equation (21) into equation (18), again keeping only the left-moving waves, and the annihilation operator terms, that is, we want the aω coefficient which is,

(22) ∫dΩ(2Ω)∫dω(αΩω e−iΩU − β*ΩωeiΩU)

For Bob, from equation (17), the coefficient for aω is

(23) ∫dω (2ω)e−iωu

For these two to match, we need,

(24) (2ω)e−iωu = ∫dΩ' (2Ω')Ω'ωe−iΩ'U − β*Ω'ωeiΩ'U)

Multiply both sides by eiΩU and then integrate over all U,

(25) ∫(2ω)e−iωu eiΩUdU = ∫dΩ'(2Ω')∫dU(αΩ'ωe−iΩ'UeiΩU − β*Ω'ωeiΩ'UeiΩU)

On the right-hand side, the first term inside the bracket is just the definition of the Dirac delta function,

(26) ∫dU e−iΩ'U eiΩU = ∫dU ei(Ω − Ω')U = 2π δ(Ω −Ω')

And the second term vanishes at infinity. The net result is,

(27) αΩω = (2π)-1(Ω)½(ω) ∫dUe-iωueiΩU

Similarly, we get,

(28) βΩω =(−1)(2π)-1(Ω)½(ω) ∫dUe+iωueiΩU

Using equation (11), which is repeated below,

(29) U = a-1ln(ua)

(30) αΩω = (2π)-1(Ω)½(ω) ∫(du/u)e-iωuei(Ω/a)lnua

(31) βΩω =(−1)(2π)-1(Ω)½(ω) ∫(du/u)e+iωuei(Ω/a)lnua

Now we don’t need to evaluate these integrals to find the relationship between αΩω and βΩω. We note that there are very similar, and we get our answer by using the transformation u → -u in equation (31),

(32)βΩω =(−1)(2π)-1(Ω)½(ω) ∫(du/u)e-iωuei(Ω/a)ln(-ua)

We need to look at the last term in the integral,

(33) ei(Ω/a)ln(-ua) = ei(Ω/a)ln(ua)ei(Ω/a)ln(-1)

Recall Euler’s famous equation:

(34) e = -1 → iπ = ln(-1). Therefore,

(35) ei(Ω/a)ln(-ua) = ei(Ω/a)ln(ua)ei(Ω/a)ln(-1) = ei(Ω/a)ln(ua)e-(πΩ/a)

Putting this back into equation (32),

(36)βΩω =(−1)(2π)-1(Ω)½(ω) ∫{(du/u)e-iωu ei(Ω/a)ln(ua)

x e-(πΩ/a)}

Now comparing (33) and (30), we get

(37) βΩω = − αΩωe-(πΩ/a)

Squaring both sides and dropping the indices,

(38) |β|2 = |α|2e-2π(Ω/a)

This is where Unruh noticed that this result is the same as for a black body radiating at a temperature T.

More specifically, the density of states between any two state m and n is,

(39) ρmn = <ψm|e-H/T| ψn> = |α|2 e-Em/T, where H is the Hamiltonian, and E is the energy eigenvalue of H.


(40) ρnm = <ψn|e-H/T| ψm> = |β|2 e-En/T

The density state is symmetric under the indices. That is,

(41) ρmn = ρnm

We get,

(42) |β|2 = |α|2e-(En - Em)/T = |α|2e-E/T

But from the Einstein equation,

(43) E = ℏΩ = Ω (ℏ = 1)


(44) |β|2 = |α|2e-Ω/T

Comparing the above with equation (38), we get,

(45) T = a/(2π)

Now putting back the constants ℏ,c, and kb to make the equation dimensionally consistent,

(46) T = aℏ/(2πckb)

An accelerating detector would see particles in a thermal bath. This is the Unruh effect.

We know that the action of accelerating electrons gives off electromagnetic radiation, what we need to transmit to TV’s, satellites, phones, etc. As for the Unruh effect, a typical acceleration of 10m/s2, (earth’s acceleration due to gravity), would yield a temperature of the order, T ~ 10-20 K, or to boil water at 373K, we would need to accelerate it at a ~ 1022 m/s2, both cases being beyond our present technology.

Monday, June 10, 2013

Relativistic Doppler Effect

One of the assumption of Special Relativity is that the velocity of light is constant in every inertial frame of reference. Consider two frames: one at rest (t,x,y,z); and a second one (t',x',y',z') is in motion with velocity V with respect to the first one. We've omitted the z- and z'-axis as representation of 4 dimensions on a 2-D surface is not possible.

Now suppose a light was turn on, and our two observers would be looking at this beam of light spreading throughout space as a sphere. (See fig above)

In the rest frame we have:

(1) x2 + y2 + z2 = (ct)2

Similarly in the moving frame:

(2) x'2 + y'2 + z'2 = (ct')2

Rewriting both equations (1) and (2) as:

(3) -(ct)2 + x2 + y2 + z2 = 0

(4) -(ct')2 + x'2 + y'2 + z'2 = 0

Since both expressions are equal to zero, they are equal to each other. But more fundamental, these two expressions are suggesting that they represent the same fundamental quantity in two different frames. That there are equal means something doesn't change from one frame to the other. Let's represent this quantity as:

(5) s2 = -(ct)2 + x2 + y2 + z2

= -(ct')2 + x'2 + y'2 + z'2

To keep in line with convention, we will express these quantities as small differences. We use the symbol d, which is common use in calculus. We also set c =1, as it is a constant and we can always put it back if we need to do a calculation.

(6) ds2 = -dt2 + dx2 + dy2 + dz2

We define the metric as the coefficient of each of the terms in the above:

(7) η00 = -1, η11 = 1,η22 = 1,η33 = 1,and ηij = 0 for i≠j

We also define the proper time τ as,

(8) dτ2 = - ds2

= dt2 - (dx2 + dy2 + dz2)

= dt2 ( 1 - (dx2/dt2 + dy2/dt2 + dz2/dt2))

= dt2 ( 1 - v2)

Where we have use the definition of velocity,
v = (dx/dt,dy/dt,dz/dt).

Taking the square root on both sides,

(9) dτ = dt/γ

(10) where γ = (1 - v2)

Now, in Newtonian physics, it was assumed that time was different from space. But in Relativity, this separation cannot be upheld any longer. We can see in the definition of the proper time, that both time and space form one manifold. So we need to redefine our quantities with this new perspective.

Convention: we use latin indices for 3-dimensional objects. For instance, the velocity, vi, where i = 1,2,3. So a velocity's components which were written as v = (vx,vy,vz), now will be written as v = (v1,v2,v3).

We use Greek letters (α,β,γ,δ...) to denote objects with 4 components. They will take values 0,1,2,3 for t,x,y,z. So, for example, uβ =(u0,v1,v2,v3). Sometimes, we want to break up the components as temporal and spatial. So we write,uβ =(u0,vi), where it is understood, i = 1,2,3, or uβ =(u0,v)

We will measure the velocity with respect to the proper time τ, not the ordinary time t.

(11) uβ =dxβ/dτ

Using the chain rule and equation (10),

(12) uβ = (dt/dτ)dxβ/dt = γ(dt/dt,dx1/dt,dx2/dt,dx3/dt) = γ(1,v1,v2,v3) = γ(1,v)= (γ,γv) .

The dot product between two vectors is now defined with the metric (see equation (7)),

(13) u2 = u•u = ηαβuαuβ

Note that in u2, the 2 means squaring, u2 = u squared, not the 2nd component of u. When there's confusion, we will point out what is meant by an upper index.

Expanding the above into the temporal and spatial components, and using the metric in (7),

(14) u•u = η00u0u0 + ηijuiuj
= (-1)γ2 + γ2v2 = (-1)γ2 ( 1 - v2)

But from (10),

(15) γ2 = (1 - v2)-1


(16) u2 = u•u = -1

Energy and Momentum

Momentum is defined as mass x velocity,

(17) pβ = muβ

Similarly, we define a 4-vector momentum as,

(18) pβ =(p0,pi) = (p0,p)

An important result is to calculate p2, where the 2 means squaring, not the component 2. First using equation (17)

(19)p2 = muβmuβ = m2u2 = - m2

Then using equation (18) and the metric in (7),

(20) p2 = p•p = η00p0p0 + ηijpipj
= (-1)(p0)2 + (p)2

We define p0 = E/c = E , (using the convention, c=1). Putting this altogether, we get,

(21) p2 = - m2 = (-1)E2 + (p)2


(22) E2 = m2 + (p)2.

Putting c into the equation,

(23) E2 = m2c4 + p2c2.

Notice when the particle is at rest, p = 0, and we get, E = mc2.

Relativistic Doppler Effect

In the last blog, Einstein's Derivation of the Famous Equation, E=mc2 , we were given the energy of the photon as,

(24) γB+ = ½ E(1 + (V/c)cosΦ)(1 – V2/c2) for the incoming photon

γB- = ½ E(1 - (V/c)cosΦ)(1 – V2/c2) for the outgoing photon

The energy is,

(25) Etotal = - pβ uβ = - (η00p0u0 + ηijpiuj)

= - (-1)Eγ - (pcosθ, psinθ, 0)(-γV,0,0)

= Eγ + (pcosθ)γV = Eγ(1 + (V/c)cosθ),

Where the last step , we use p = E/c.

Each photon released will carry half the energy in opposite direction. For the incoming photon,

(26) γB+ = ½ Etotal = ½ E γ(1 + (V/c)cosΦ)
= ½ E(1 + (V/c)cosΦ)(1 – V2/c2)

Where we use equation (10) in the last step.

For the outgoing photon, we get an extra minus sign in the momentum,

(27) pi = (-pcosθ, -psinθ, 0)


(28)γB- = ½ E(1 - (V/c)cosΦ)(1 – V2/c2)

Notice that for the energy of the incoming photon is greater than the energy of the outgoing photon. This is the Doppler Effect, as a light coming towards you will appear blueshifted, while a photon moving away from you will be redshifted.


From here on, the speed of light is c.

We will use the following conventions:

(A1) β = v/c

(A2) γ = (1 - v2/c2) = (1 - β2)

For a wave, it is proportional to ei(k∙x - ωt),where,

(A3) k = |k| = 2π/λ, ω = 2πf, and ω = kc

Just like we can define a 4-vector position, xμ = (ct,xi), we can also define a 4-vector wavenumber, kμ = (ω/c,ki). The phase factor, (k∙x - ωt), can now be written as,

(A4) k∙x - ωt = ημνxμkν, where ημν is the Minkowski metric tensor with signature (-1,1,1,1).

The importance of the phase factor, which basically counts the number of peaks and troughs of the wave, must be frame-independent, that is, a Lorentz scalar.

(A5) kμ → k'μ = Lμνkν

Where Lμν is the Lorentz transformation (See diagram below).

In particular, under a Lorentz boost in the +x direction, (A6) k'x = γ(kx - βω/c)

(A7) ω'x = γ(ω - βckx)
= γ(ω - βckcosθ)

Where θ is the angle between the boost direction in the +x direction and the direction of the wave propagation k.

Using ω = kc (equ. A3), and γ = (1 - β2) (equ. A2) then equation A7 becomes,

(A8) ω' = γ(ω - βckcosθ) = ω(1 - βcosθ)(1 - β2)

Note: if θ = π/2, then ω' = 0. There is no Doppler shift in the transverse direction.
if θ = 0, then ω'/ω = [(1- β)/(1+β)]½
For low velocity, v < c, then

(A9) ω'/ω ≈ (1- β)


(A10) Δω'/ω ≈ -v/c

Saturday, June 01, 2013

Einstein's Derivation of the Famous Equation, E=mc2

Here’s a quick rundown on one of the most famous equation in physics, E = mc2. Einstein knew from experiments previously done that a particle could decay and release gamma rays. He reasoned that when this happened, the particle would lose kinetic energy, and this could only be accounted by a loss of mass. So how did he come to that conclusion? He analysed the situation both in a rest frame and in a moving frame.

The value of γ was already known, but we can derive it from Relativistic Doppler Effect (to be the topic for another blog).

By the law of conservation of energy:

The energy before decay = the energy after decay

(1) In the rest frame:

A0 + KA0 = A1 + KA1 + ½ E + ½E = A1 + KA1 + E

(2) In the moving frame:
B0 + KB0 = B1 + KB1 + ½ E(1 – (v/c)cosΦ)(1 – v2/c2)
+ ½ E(1 + (v/c)cosΦ)(1 – v2/c2)

= B1 + KB1 + E(1 – v2 /c2)

Now taking a look at the energy difference of the particle in each of the frame:

(3) In the rest frame:

A1 – A0 = KA0 – E – KA1 = – ΔKA – E

(4) In the moving frame:

B1 – B0 = KB0 – KB1 – E(1 – v2 /c2)
= – ΔKB – E(1 – v2 /c2)

Whether the observer is at rest or moving with respect to the particle, the energy difference should be the same.

(5) – ΔKA – E = – ΔKB – E(1 – v2 /c2)

Calculating the difference in kinetic energy:

((6) ΔK = ΔKA – ΔKB = E(1 – v2 /c2) – E

= E ((1 – v2/c2) – 1)

≈ E ((1 + ½v2/c2) – 1)

= ½ E (v2/c2)

Einstein had reasoned that if the kinetic energy of the particle is smaller by ½ E (v2/c2), the only way this can happen is that the particle must lose mass when emitting radiation.

By definition the kinetic energy is,

(7) K = ½ mv2

If the particle was initially at rest,

(8) ΔK = K = ½ mv2 = ½ E (v2/c2)

(9) Therefore E = mc2.
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