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.

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