Quantum Mechanics - Some Math

Serious discussion of science, skepticism, and evolution
User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Quantum Mechanics - Some Math

Post by lpetrich » Sat Aug 25, 2012 8:37 am

This thread is intended to be a resources thread, like the threads on evolution here. If you wish to comment on this thread, there's Quantum Mechanics - Some Math - Peanut Gallery - Secular Café Any additional subjects for this thread? Anything to clarify?

This thread contains some mathematics, up to calculus, but it should not be terribly difficult for someone who majored in physics or something similar in college.


Some quantum-mechanics math made visible again.

Let's start out with Max Planck's quantization condition.

Energy = h * f
f = frequency

A wave X(t) oscillates in time as X(t) = X0*exp(-2*pi*i*f*t)
where t is the time and exp(x) = ex, e ~ 2.71828183

De Moivre's theorem: relation between exponential and trig functions.
exp(i*x) = cos(x) + i*sin(x)

It's often convenient to fold the 2*pi into the f to get the angular frequency, w = 2*pi*f.

Likewise, hbar = h/(2*pi).

Thus, X(t) = X0*exp(-i*w*t)
X(t) solves d2X/dt2 = - w2 X

The harmonic-oscillator equation, the equation that small oscillations obey, ignoring dissipation.

So how does one measure the energy of a quantized wave?

To get w, one has to take i*d/dt of it: i*dX/dt = w*X

The energy E = h*f = hbar*w, and one gets the energy in X by applying an "energy operator" on it, a function on it that gets the energy. Thus, E as an operator yields: E(X) = i*hbar*dX/dt

In similar fashion, one gets a momentum value by applying momentum operator p: p.X = -i*hbar*dX/dx
where x is the position

The space and time operators are multiply by x and t: x(X) = x*X, t(X) = t*X.

Let's now see if the order of operation makes any difference. Consider position and momentum.

p(x(X)) = p(x*X) = -i*hbar*X - i*hbar*x*dX/dx = x*p(X) - i*hbar*X = x(p(X)) - i*hbar*X

So operators x and p don't commute with each other - their order of operation makes a difference. This fact may be expressed as
[x,p] = x.p - p.x = i*hbar
Likewise,
[t,E] = t.E - E.t = -i*hbar

Notice that their noncommuting is proportional to hbar, which is about 10-34 in meter-kilogram-second units. VERY, VERY TINY.

That's why we don't see macroscopic quantum effects.

That noncommuting also leads to the Uncertainty Principle.
Last edited by lpetrich on Sun Dec 23, 2012 2:31 pm, edited 1 time in total.
Reason: Added purpose

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sat Aug 25, 2012 8:42 am

Let's now backtrack to classical mechanics. Consider Newtonian mechanics: a particle with position x and mass m moving in a potential V over time t:

m*(d2x/dt2) = - (dV/dx)

It can be derived by the Calculus of Variations from an "action principle", by minimizing the "action" I, defined by this integral:

I = integral of L over t

where L is the "Lagrangian", which is here (kinetic energy) - (potential energy) or
(1/2)m(dx/dt)2 - V

One gets the Euler-Lagrange equation of motion:
p = dL/d(dx/dt)
dp/dt - dL/dx = 0

In this problem, p = m(dx/dt), the momentum.

Lagrangians are nice because they summarize equations of motion, and just about all the more fundamental physical theories involve Lagrangians.


From the Lagrangian, one can derive another quantity, the "Hamiltonian":

H = p*(dx/dt) - L

The equations of motion become
dx/dt = dH/dp
dp/dt = - dH/dx

In our current problem, H = (kinetic energy) + (potential energy) = p2/(2m) + V

The Hamiltonian thus gives the total energy of a system. There's an interesting consequence: dH/dt = (dH/dt)explicit, so H is conserved if it has no explicit time dependence.


There's a further wrinkle called the Hamilton-Jacobi equations. It involves a function S that behaves as follows:
p = dS/dx
b = dS/da (a and b are constants of the motion)
H + dS/dt = 0

What does the potential problem look like in the Hamilton-Jacobi equations? Its solution is

S = - E*t + integral of sqrt(2*m*(E - V)) over x

The momentum p = sqrt(2*m*(E - V))

For b = dS/da, constants a = energy E, b = -t0, the reference time:

-t0 = -t + integral of sqrt(m)/sqrt(2*(E - V)) over x

Let's return to the Newtonian equations of motion:

m*d2x/dt2 = - dV/dx

Integrate over x, using the energy as an integration constant:
(1/2)*m*(dx/dt)2 = E - V

Rearrange, and one gets
t = t0 + integral of sqrt(m)/sqrt(2*(E - V)) over x

where t0 is an integration constant. That's the H-J equation b = dS/da again.


All these results are easy to generalize to multiple variables.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sat Aug 25, 2012 8:54 am

Back to quantum mechanics.

Let's consider wavefunction X, a function of t and x.

operator x on X: x*X
operator p on X: - i*hbar*dX/dx

Note that x and p are closely related, as we've found from the Hamiltonian equations of motion. In fact, we can use the Hamiltonian to get the total energy:

operator H on X: - (hbar2/(2m))*(d2X/dx2) + V(x)*X

Since it gives the total energy, we thus have (H on X) = i*hbar*(dX/dt)

This is Schroedinger's equation, published by Erwin Schroedinger in 1926.

Now for something very cute. Let X ~ exp(i*S/hbar), and let's consider the limit of S >> hbar. We find:

p = dS/dx
H = - dS/dt
The Hamilton-Jacobi equations again!


Published in 1925, Werner Heisenberg's formulation is essentially a mathematical reorganization of Schroedinger's formulation that makes operators x and p functions of time and the wavefunction fixed. Its equation of motion for an operator A is

dA/dt = (dA/dt)explicit + (i/hbar)*[H,A]
where [A,H] is the commutator A.H - H.A

The equations of motions for operators x and p one finds to be the classical Hamiltonian equations of motion.


Path integrals were published by Richard Feynman in 1948, drawing from the work of P.A.M. Dirac in 1933. They are yet another formulation of quantum mechanics, one that uses the action I, the integral of the Lagrangian over time:

An operator A has the value (<A>Z)/(<1>Z), where
<A>Z = integral of A*exp((i/hbar)*I) over x(t1), x(t2), x(t3), ... : t1, t2, t3, ... being some of the times along the way

It's an integral over all possible paths of the system variables x from one time to another.

In the limit of I >> hbar, one finds the action principle (minimize I) and the Euler-Lagrange equations again.


Looking at the S >> hbar and I >> hbar limits in the Schroedinger and the path-integral formulations, it is evident that classical mechanics is the geometrical-optics or eikonal limit of quantum mechanics.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sat Aug 25, 2012 8:57 am

Let's now do a simple problem: the harmonic-oscillator problem. Imagine a ball on a spring, and imagine pulling the ball. It will oscillate back and forth, and slowly settle down. But you notice that its oscillation has nearly constant period. Why is that?

A similar sort of experiment is a pendulum. If it does not swing too much, its swing period will be close to constant as it settles down.

If the force on the object is a smooth function of position, something that often happens, it has form
F(x) = F0 + F1*x + F2*x2 + ...

If it is zero at x = x0, then we can express it as
F(x) = F1*(x-x0) + F2*(x-x0)2 + ...

For a small oscillation, we can omit higher terms, and we get Hooke's law:
F(x) ~ F1*(x-x0)

or for convenience, setting x0 = 0 -- getting F(x) ~ F1*x

Let's now try to solve that object's equation of motion:
dx/dt = p/m
dp/dt = F = F1*x
We guess a form of the solution: x = x0*exp(q*t), p = p0*exp(q*t)
Substituting it in, we get
p0 = m*q*x0
q = +/- sqrt(F1/m)

q > 0 -- exponential growth, which we don't see in these examples. That's for F1 > 0, so F1 < 0 is necessary. That makes q imaginary, and that turns those exponential functions into trig functions. Letting q = i*w, we get

x = x(0)*cos(w*t) + p(0)/(m*w)*sin(w*t)
p = - m*w*x(0)*sin(w*t) + p(0)*cos(w*t)
F1 = - m*w2

Oscillations with constant angular frequency w, independent of the size of the oscillations.

That's equivalent to a potential-energy term V = - (1/2)F1*x2 = (1/2)m*w2*x2

The total energy one can find from the Hamiltonian, which is (kinetic energy) + (potential energy):
H = p2/(2m) + (1/2)m*w2*x2

E = (1/2)*(m*w2*x(0)2 + p(0)2/m)

Energy ~ (oscillation amplitude)2

This is all in the classical limit, and I'll get to the quantum-mechanical version in my next post.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sat Aug 25, 2012 8:57 am

Now for the quantum-mechanical harmonic-oscillator problem. First with the Schroedinger formulation. Position and momentum become operators on the wavefunction X(x,t):

x(X) = x*X
p(X) = - i*hbar*(dX/dx)
You can see that operators x and p satisfy [x,p] = i*hbar

The Hamiltonian:
H(X) = - (hbar^2/(2*m))*(d2X/dx2) + (1/2)*m*w2*x2*X
satisfying
H(X) = i*hbar*(dX/dt)

Since its Hamiltonian has no explicit time dependence, we can set X(x,t) = X(x)*exp(-i*E*t/hbar)
where E is the energy of the state. Thus,

H(X) = E*X

One can solve this differential equation for X(x), and one gets

X(x) = 1/sqrt(2n*n!*pi*xh) * exp(-x2/(2*xh2)) * Hermite(n,x/xh)

where
length scale xh = sqrt(hbar/(m*w))
energy E = (n + 1/2)*hbar*w
Hermite(n,x) is a Hermite polynomial with degree n

Graphing the wavefunctions shows that they look wavy for |x| < xh*sqrt(n+1/2), which is the object's classically-allowed area of motion for energy E. It's a standing wave inside that region. Outside that area, the wavefunction falls off, meaning that the object goes into classically-forbidden areas.

Notice also that in its ground state, the object isn't stationary, that it moves just a little, over a distance of about xh.

Try to calculate xh for some macroscopic objects some time and see what you get.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sat Aug 25, 2012 8:59 am

I'll now do an elegant ladder-operator version.

Define new operators a and a+ from x and p:

x = sqrt(hbar/(2*m*w))*(a + a+)
p = - i*sqrt((hbar*m*w)/2)*(a - a+)
or
a = sqrt((m*w)/(2*hbar))*x + i*sqrt(1/(2*hbar*m*w))*p
a+ = sqrt((m*w)/(2*hbar))*x - i*sqrt(1/(2*hbar*m*w))*p

The commutator [x,p] = i*hbar gives us [a,a+] = 1
The Hamiltonian becomes (1/2)*hbar*w*(a*a+ + a+*a) = hbar*w*(N + 1/2)

where the number operator N = a+*a
with [N,a+] = 1 and [N,a] = -1
Thus, a+ is a raising operator and a is a lowering operator.

Define a ground state X(0) by a.X(0) = 0 -- it can't be lowered further.

Get a state from another state by X(n) = 1/sqrt(n)*a+.X(n-1)
Then, X(n) = 1/sqrt(n!)*(a+)n.X(0)
and number operator N.X(n) = n*X(n)

Thus, X(n) contains n quanta, and its energy is hbar*w*(n+1/2).
The operator a+ adds a quantum, and the operator a subtracts a quantum.

Going to the Heisenberg formulation, x, p, a, and a+ are all functions of time. x and p have their classical forms, while
a = a(0) * exp(-i*w*t)
a+ = a+(0) * exp(i*w*t)

One can approximate the classical limit with a "coherent state":
a.X(c) = c*X(c)
where c can be complex. In terms of the states with well-defined energies,

X(c) = exp(-|c|2/2) * sum over n of 1/sqrt(n!)*cn*X(n)
= exp(-|c|2/2 - c*a).X(0)

This ladder-operator approach is also useful in many-body problems and in quantum field theory.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sun Aug 26, 2012 10:34 am

Now for quantum-mechanical angular momentum, which has a most elegant theory.

First, commutation of position and momentum operators:
[xi,pj] = i*hbar*dij

dij is the Kronecker delta, and it is 1 if i = j, 0 otherwise.

Position and momentum measurements will interfere with each other if their directions are not orthogonal (right-angled).


Angular momentum = cross product of position and linear momentum:
L = x x p
Li = eijk xj pk

eijk is the antisymmetric symbol: e123 = 1, e132 = -1, e122 = 0, etc.

L1 = x2p3 - x3p2
etc.

The components do not commute with each other, meaning that trying to measure one component interferes with trying to measure another component. So one makes it definite on only one direction, z or 3.

[Li,Lj] = i*hbar*eijk*Lk
[L1,L2] = i*hbar*L3, [L2,L3] = i*hbar*L1, [L3,L1] = i*hbar*L2

However, they commute with their absolute square: L2 = L12 + L22 + L32
So that quantity has a well-defined value even if its components don't.

One gets some fairly complicated expressions for the angular-momentum operators by using spherical coordinates:
x = r*(sin(theta)*cos(phi), sin(theta)*sin(phi), cos(theta))

for radius r, polar angle theta, and azimuthal angle phi.

L1 = i*hbar*(sin(phi)*d()/d(theta) + cot(theta)*cos(phi)*d()/d(phi))
L2 = i*hbar*(-cos(phi)*d()/d(theta) + cot(theta)*sin(phi)*d()/d(phi))
L3 = - i*hbar*(d()/d(phi))
L2 = - hbar2*(d2()/d(theta)2 + cot(theta)*d()/d(theta) + 1/sin(theta)2*d2()/d(phi)2)

We can now look for a wavefunction that gives definite values of L2 and L3, and we find one.

Spherical harmonics Y with total angular momentum j and projected angular momentum m:
L2Y(j,m,theta,phi) = j(j+1)*hbar2*Y(j,m,theta,phi)
L3Y(j,m,theta,phi) = m*hbar*Y(j,m,theta,phi)
Y = (complicated function of theta) * exp(i*m*phi)

j is a nonnegative integer and m is an integer between -j and j.


Now for a relationship between angular momentum and rotation. It seems obvious, but there is an interesting quantum-mechanical relation. Take a wavefunction X(phi) and rotate it by angle a in the z-axis, making X(phi+a). Then expand in powers of a, making a Taylor series:

X(phi+a) = X(phi) + (a/1!)*(dX/d(phi)) + (a2/2!)*(d2X/d(phi)2) + (a3/3!)*(d3X/d(phi)3) + ...
= exp(a*d()/d(phi)).X(phi) = exp((i*a)/hbar*L3).X(phi)

So quantum-mechanical angular-momentum operators are also rotation operators. This is true not only of orbital angular momentum, but also angular momentum built into field geometries, or spin. How can something be a spinning wave? Circularly polarized light has its field directions continually spinning, and it carries angular momentum.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sun Aug 26, 2012 2:47 pm

One can do a ladder-operator construction of angular momentum also:

L+ = L1 + i*L2, L- = L1 - i*L2
[L3,L+] = hbar*L+, [L3,L-] = - hbar*L-, [L+,L-] = 2*hbar*L3

For any angular-momentum operators, and not just orbital ones, one finds a solution:

L+X(j,m) = hbar*sqrt(j(j+1)-m(m+1))*X(j,m+1)
L-X(j,m) = hbar*sqrt(j(j+1)-m(m-1))*X(j,m-1)
L3X(j,m) = hbar*m*X(j,m)
L2X(j,m) = hbar2*j(j+1)*X(j,m)

m can equal -j, -j+1, -j+2, ..., j-1, j, and that constrains j to be half an integer. It can be either an integer or half-odd, and m is likewise either an integer or half-odd. For orbital angular momentum, j is an integer, but what do we find for spins?

The putative Higgs particle: likely spin 0
The photon (quantum of electromagnetism): spin 1
The graviton (inferred quantum of gravity): spin 2
The electron: spin 1/2
The gravitino (putative relative of the graviton): spin 3/2
Etc.


Since angular momentum is related to rotation, let's see about rotating 360 degrees. A field X(phi) can be decomposed into a sum of states exp(i*m*phi)*X(m), where the m values are all for some total angular momentum j. Now rotate by angle a round z:
X(phi+a) = sum of exp(i*m*phi + i*m*a)*X(m)

That looks like a big mess, but try rotating 360 degrees:
exp(i*m*2*pi) = (-1)2m = (-1)2j

Thus, X(phi + 360d) = (-1)2j X(phi)

If j is an integer, one gets the same sign of field again, as one would expect. But if j is half-odd, the field reverses sign, yet more quantum-mechanical weirdness.

So if one rotates an electron 360d, it reverses sign.


The ladder-operator approach works not only for orbital angular momentum and spin, but also for sums of angular momenta. These can be angular momenta of separate particles, spin and orbital angular momentum for the same particle, or any combination.

Imagine two particles with wavefunctions X1 and X2 and angular momenta j1 and j2. Spin-orbit works essentially the same, and the discussion carries over as appropriate.

Consider a combined state X = X1*X2. Its angular-momentum vector will be the sum of those for the two particles. Its total angular momentum j can be
|j2-j1|, |j2-j1|+1, |j2-j1|+2, ..., j1+j2-1, j1+j2

However, its angular-momentum states X(j,m) will be mixtures of the X1(m1)*X2(m2) states.

To see how this works, consider the spins of two electrons. They both have spin 1/2, and their combined spin is thus either 0 or 1. Let's now construct the states:

The topmost state is for j = 1, m = 1, and there is only one possible state:

X(j,m) = X(1,1) = X1(m1)*X2(m2) = X1(1/2)*X2(1/2)

Using the lowering ladder operator will generate the remaining states - it's the sum of the lowering operators for both states. We find

X(1,1) = X1(1/2)*X2(1/2)
X(1,0) = 1/sqrt(2) * (X1(1/2)*X2(-1/2) + X1(-1/2)*X2(1/2))
X(1,-1) = X1(-1/2)*X2(-1/2)

Note how the m's add.

There's a state that's orthogonal to the X(1,0) state, and it's the X(0,0) state:

X(0,0) = 1/sqrt(2) * (X1(1/2)*X2(-1/2) - X1(-1/2)*X2(1/2))

The X(1,0) has a + sign in the middle, the X(0,0) a - sign.

Though one can find combinations of the states for total spin 1 that can be decomposed into each of the electrons' states, one cannot do so for total spin 0. The particles' spins are thus entangled, another quantum-mechanical oddity.

Something similar can happen to photons. An electron in an atom may jump down in energy by emitting a pair of photons in opposite directions, especially if it is jumping from 0 to 0 angular momentum. The photons' polarizations are correlated:

Linear: directions 1 and 2
1/sqrt(2)*(X1(1)*X2(1) + X1(2)*X2(2))

This is equivalent to this circular case:
1/sqrt(2)*(X1(left)*X2(left) + X1(right)*X2(right))

Notice the mixed state and the photon polarizations' entanglement.

Alain Aspect*: Shedding new light on light and atoms - CNRS Web site - CNRS

Alain Aspect did some experiments on the entanglement question, and he showed that emitted photon pairs could stay in mixed states over the size of his lab equipment. Each photon's polarization by itself was completely random, but the two photons together had correlated polarizations. Furthermore, their correlation was in agreement with the large-mixed-state hypothesis and not with alternatives, like "local hidden variables".

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Mon Aug 27, 2012 3:06 am

Now to quantum field theory and spin/statistics.

To help illustrate what goes on in it, let's consider a 1D string of N balls connected by springs, with the ends looping around - periodic boundary conditions. For position offset z at location k, the equation of motion is

m*(d2z(k)/dt2) = - m*w02*(2*z(k) - z(k - 1 mod N) - z(k + 1 mod N))

for k = 0: (2*z(0) - z(N-1) - z(1))
for k = N-1: (2*z(N-1) - z(N-2) - z(0))

Take a Fourier transform over position k:
z(k) = sum over n of Z(n)*exp(2*pi*i*k*n/N)
Its equation of motion:

m*(d2Z(n)/dt2) = - m*w(n)2*Z(n)
with oscillation angular frequency
w(n) = w0*2*sin(pi*n/N) ~ w0*(2*pi*n/N) for small N

It's a wave solution: offsets z(k) ~ exp(2*pi*i*n*k/N - i*w(n)*t)
For n << N, the wave's offsets do not change much from neighbor to neighbor, and the wave has a constant speed independent of n -- a sound wave. In fact, this is what sound waves in solid materials are.


It's evident that this problem is a multiple harmonic-oscillator problem, and it can be quantized in the same way. Each mode gets its own raising and lowering operators, independent of the others. One can go to the continuum limit and still quantize this problem, though one has to be careful about offset-variable quantization.

I've thus outlined a quantum field theory of sound, and quanta or "particle" of sound are called phonons. Yes, sound also has wave-particle duality.

Though we can't resolve audible-range sound into phonons (remember energy = h*frequency), there is less-direct evidence of phonons for sound with wavelengths not much greater than condensed-material atom separations. Most of the heat energy of condensed materials resides in random phonons with such short wavelengths, judging from heat-capacity observations.

Phonons are one kind of collective-effect "quasiparticle", and there are some others known to exist in condensed materials, quantization and all.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Mon Aug 27, 2012 6:56 am

Going from collective-effect quasiparticles like phonons (quanta of sound) to "true" elementary particles, we find that we can quantize their fields also.

The spin-0 case is fairly easy; one quantizes it like one quantizes sound waves, with harmonic-oscillator raising and lowering operators for each wave mode.

Other integer spins closely parallel the spin-0 case, but with complications involving the fields' internal geometries and something called "gauge invariance".

But to get field quantization to work correctly for particles with half-odd spins, one has to make their raising and lowering operators behave differently. Instead of commutation relations, one has to define anticommutation relations for them.


Harmonic oscillator, integer-spin particles: commutation
[A,B] = A.B - B.A
Raising and lowering operators: [a,a] = 0, [a+,a+] = 0, [a,a+] = 1
Number operator N = a+a; [N,a+] = a+, [N,a] = - a

States: X(n) = (a+)n/sqrt(n!).X(0)


Half-odd-spin particles: anticommutation
{A,B} = A.B + B.A
Note the + instead of a -
Raising and lowering operators: {b,b} = 0, {b+,b+} = 0, {b,b+} = 1
Number operator N = b+b; [N,b+] = b+, [N,b] = - b

What states? Let's define states N.X(n) = n*X(n), and let's define a ground state with b.X(0) = 0

The first excited state is X(1) = b+.X(0), but is there a second one?

X(2) ~ b+.X(1) = (b+)2.X(0) = 0
There is no second excited state.

There are only 0 or 1 quanta per state, an effect called the Pauli Exclusion Principle. Only one electron at a time can occupy a state, and likewise for other half-odd-spin particles.

That's important in the structure of atoms, because in them, electrons pile up on top of each other in energy state, rather than collecting in a single state. It's also important in the structure of nuclei (protons and neutrons have spin 1/2), and in hadrons (quarks have spin 1/2).


So we have a fundamental difference between particles with different spins, something expressed as the spin-statistics theorem.
[table]Statistics | Bose-Einstein | Fermi-Dirac
Particles | bosons | fermions
Quanta/state | 0, 1, 2, ... | 0, 1 only
X(x2,x1) | +X(x1,x2) | -X(x1,x2)
Spin | 0, 1, 2, ... | 1/2, 3/2, 5/2, ...
360d rotate | + | -[/table]
The X(x1,x2) and X(x2,x1) part is about wavefunction symmetry for exchanging particles
360d rotate = wavefunction sign change after a 360-degree rotation

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Tue Aug 28, 2012 1:06 pm

Now for hydrogenlike or hydrogenic atoms. They are two-body inverse-square-force systems, and unlike many-body systems with that sort of force, they are tractable in the classical limit. This gives hope that they may also be tractable in quantum mechanics.

In 1913, Niels Bohr successfully accounted for hydrogen atoms' energy levels with quantum mechanics. However, he used "old quantum mechanics", which was based on semiclassical kludges. A decade later, the work of Louis de Broglie, Erwin Schroedinger, Werner Heisenberg, Paul Adrien Maurice Dirac, and others got quantum mechanics into its modern form, and that enabled the present solution of the hydrogenlike-atom problem. It gets not only the energy correct, but also many other details correct.

First, the classical limit.

Its Hamiltonian or total-energy function is
H = p2/(2m) - K/r

r = |x| is the distance for separation x
m is the reduced mass, m1m2/(m1 + m2) for particles 1 and 2
p is the momentum, (reduced mass) * (relative velocity)
K is the force constant, Ze2 for one electron orbiting a nucleus

It has two additional conserved quantities:

The angular momentum
L = x x p
Conserved for any central force (function-of-radius force)

The Laplace-Runge-Lenz vetor
A = (p x L)/m - K*x/r
Conserved only for inverse-square-law force

These quantities have these constraints:
L.A = 0
A2 = (2H/m)*L2 + K2

Counting conserved quantities, we have 1 (H) + 3 (L) + (3 - 1 - 1 = 1) (A) = 5.This is one less than the number of variables, 6, and the remaining quantity is a time reference.

Solving in the elliptical case gives:
x = a*(cos(u) - e)*P + a*sqrt(1-e2)*sin(u)*Q
r = a*(1 - e*cos(u))
w*(t - t0) = u - e*sin(u)
where w = sqrt(K/(m*a3))
a = semimajor (or major) axis
e = eccentricity
u = "eccentric anomaly", a convenient intermediate variable
and P, Q, and W are an orthonormal set of vectors with W = P x Q. Orthonormal = length 1, perpendicular to each other.

The conserved quantities:
H = - K/(2a)
L = sqrt(m*K*a*(1-e2))*W
A = K*e*P

The other cases are circular, parabolic, hyperbolic, and radial; these can be found in appropriate limits and by analytic continuation as appropriate.


Now Bohr's solution. He imagined the electrons as having circular orbits, and he found the action over the entire orbit:
integral of p over x = 2*pi*m*w*a2 = 2*pi*sqrt(m*K*a)

He equated that with a quantization condition: 2*pi*n*hbar, where n is an integer >= 1, and he found

a = n2*a0 - the Bohr radius a0 = hbar/(m*K)
E = - Ry/n2 - the Rydberg energy Ry = (m*K2)/(2*hbar2)

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Thu Aug 30, 2012 8:00 am

Now for fully-developed quantum mechanics. Solving the Schroedinger equation yields this wavefunction:

X(n,lj,lm,r,theta,phi) = (complicated function of n, lj, r) * Y(lj,lm,theta,phi)

n = principal quantum number: 1, 2, 3, ...
lj = total angular momentum: integer >= 0, < n
lm = projected angular momentum: integer |lm| <= lj
The radial function has the Bohr radius as its length scale.
Energy = - Ry/n2 (Bohr's result)

Central forces in general produce the same form of wavefunction, though their energy values are functions of both n and lj.

A 2D version is
X(n,m,r,phi) = (complicated function of n, lm, r) * exp(i*lm*phi)

This energy degeneracy for angular momentum, this equality of energy values for different angular-momentum values, exists in general only for two cases, the inverse-square force case, and the harmonic-oscillator case. Both of them have another thing in common: those cases are the only ones that have closed orbits in the classical limit (Bertrand's theorem). Those orbits are also elliptical, but with the inverse-square case having the center at one focus of the ellipse, and with the harmonic-oscillator case having the center at the center of the ellipse.


Though the solutions are very horrible-looking expressions, they have this feature: they oscillate in the classically-allowed regions and fade off outside of those regions.

Here's how some of the inverse-square orbit elements relate to the quantum numbers:

a ~ n2*a0 (Bohr radius)
e ~ sqrt(1 - (lj/n)2)
inclination ~ arccos(lm/lj)

Classically allowed:
a(1-e) <= r <= a(1+e)
(90d - incl) <= theta <= (90d + incl) (incl <= 90d)
For incl > 90d, take incl = 180d - incl

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Thu Aug 30, 2012 9:26 am

Now to operator-algebra solutions of the 2D and 3D harmonic-oscillator and inverse-square problems.

The 2D and 3D harmonic-oscillator solutions with definite angular momentum can also be expressed in operator-algebra form:

X(n,lm) ~ (a1++i*a2+)(n+m)/2*(a1+-i*a2+)(n-m)/2 X(0,0)
raising operators a1+, a2+
Quantum numbers:
lm = -n, -n+2, ..., n

X(n,lj,lm) ~ rn*Y(lj,lm,theta,phi) X(0,0,0)
raising operators a1+, a2+, a3+:
a1+ = r*sin(theta)*cos(phi)
a2+ = r*sin(theta)*sin(phi)
a3+ = r*cos(theta)
The expression is a polynomial in the raising operators * X(0,0,0)
Quantum numbers:
lj = n, n-2, n-4, ...
|lm| <= lj


For the inverse-square problem, we put the Laplace-Runge-Lenz vector A to work:

[Li,Lj] = i*hbar*eijkLk
[Li,Aj] = [Ai,Lj] = i*hbar*eijkAk
[Ai,Aj] = i*hbar*(-2H/m)*eijkLk

For 1D, there is no angular momentum and the LRL = - K*(sign of r)
The wavefunction is the radial solution for 3D for lj = 0 multiplied by r, and the energy is - Ry/n2.

For 2D, the angular momentum is a scalar, and for 3D, a vector, so we can use operator algebra there. But there is a difference.

A2 = (2H/m)*(L2 + (D-1)2/4*hbar2) + K2
where D is the number of space dimensions.

Let H = - (1/2)m*v2 and A = B*v.
Then, [Li,Lj] = i*hbar*eijkLk, [Li,Bj] = [Bi,Lj] = i*hbar*eijkAk, [Bi,Bj] = i*hbar*eijkLk

and the LRL-magnitude equation becomes
-(2H/m)*(B2 + L2 + (D-1)2/4*hbar2) = K2


In 2D, L and B act like components of an angular momentum C, with B2 + L2 = C2 = hbar2*j(j+1)

giving energy - Ry/(j+1/2)2

Since L's values are integer multiples of hbar, j must also be an integer.


In 3D, L and B act like they are combinations of two angular momenta, C1 and C2:
L= C1 + C2, B = C1 - C2
L.B = 0 -> C12 = C22 = hbar2j(j+1)

The LRL-magnitude equation becomes
-(2H/m)*(2C12 + 2C22 + hbar2) = K2

giving energy - Ry/(2j+1)2
Here, j can be half-odd as well as an integer, and we get the Bohr value again.

Ry is the Rydberg energy (m*K2)/(2*hbar2)

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Thu Sep 06, 2012 9:34 pm

Let's now consider atoms.

First, spectroscopic notation for angular-momentum states:
[table]AM value | Letter | What
0 | s | sharp
1 | p | principal
2 | d | diffuse
3 | f | faint
4 | g
5 | h[/table]

(n,L)2*S+1J
n = principal quantum number
L = angular-momentum letter
S = total spin (usually on the left side)
J = total angular momentum (spin + orbit)

Back to atoms.

The first one is hydrogen. The nucleus has charge Z = 1, and to cancel it out, it has 1 electron. Easy. It's in the lowest energy level, with n = 1 and lj = 0, or the 1s state. Thus, it has configuration 1s1

The next one is helium. Z = # electrons = 2. What is their combined wavefunction?

For their spins, let the up states be u1 and u2 for each electron, and the down states d1 and s2. Their positions will be x1 and x2.

If their total spin is 1, their wavefunction is
X(x1,x2) * {u1*u2, 1/sqrt(2)*(u1*d2+d1*u2), d1*d2} (projected AM: +1, 0, -1)
From FD stats, X(x2,x1) = - X(x1,x2)

If their total spin is 0, their wavefunction is
X(x1,x2) * 1/sqrt(2)*(u1*d2 - d1*u2) (projected AM: 0)
From FD stats, X(x2,x1) = + X(x1,x2)

To be in their lowest energy level, both electrons must be in the 1s state. This means that their combined wavefunction must be X(x1,x2) = X1s(x1)*X1s(x2), which is, of course, symmetric. This forces their spin to 0.

Thus, helium has 2 electrons in the 1s state, making its configuration 1s2.

The next one is lithium, with Z = # e's = 3. The additional electron must be in either the 2s or 2p state, because from the Pauli Exclusion Principle, it cannot be in the 1s state.

Is the third electron 2s or 2p? The higher the angular momentum, the more it avoids the nucleus, and the more the nucleus's charge gets canceled out by the inner electrons, meaning that the electron "sees" less effective charge. Thus, higher-angular-momentum states for the same principal quantum number get more energy when other electrons are present. This makes the additional electron 2s instead of 2p, making lithium 1s2*2s1 or (He)*2s1.

Beryllium is like helium, with (He)*2s2.

Boron gets an electron in its 2p state, since the 2s states are all filled. It is thus (He)*2s2*2p1

Carbon gets another 2p electron, making it (He)*2s2*2p2 However, the two 2p electrons are unpaired, with parallel spins, for a total spin of 1. This is because total spin 0 (antisymmetric) would make their combined orbital wavefunction symmetric, making the electrons tend to be near each other. An antisymmetric orbital wavefunction makes them far, and this gives them spin 1 (symmetric spins).

Nitrogen gets another 2p electron, with all three 2p electrons unpaired: (He)*2s2*2p3

Oxygen gets another 2p electron, and its spin has to pair up with another 2p electron's spin: (He)*2s2*2p4

Fluorine gets another 2p electron, and its spin pairs up with another 2p electron, giving 1 unpaired electron and 2 pairs of paired electrons: (He)*2s2*2p5

Neon gets another 2p electron, making all its 2p electrons paired: (He)*2s2*2p6


Sodium, (Ne)*3s1, to argon, (Ne)*3s2*3p6, add electrons in the 3s and 3p states in similar fashion, because the lower ones are all occupied.

Potassium, (Ar)*4s1 adds an electron in the 4s instead of the 3d state, because the electron gets closer to the nucleus in the 4s state than in the 3d one. After calcium, (Ar)*4s2, the electrons go into the 3d shell: scandium, (Ar)*4s2*3d1, to zinc, (Ar)*4s2*3d10. Then the 4p electrons get added, from gallium, (Ar)*4s2*3d10*4p1, to krypton, (Ar)*4s2*3d10*4p6. There are a few complications along the way with the 4s and 3d electrons; in some of the elements, an electron gets moved from the 4s to the 3d state.

Rubidium, (Kr)*5s1, to xenon, (Kr)*5s2*4d10*5p6, work in the same way, complete with similar 5s-4d complications.

After cesium, (Xe)*6s1, and barium, (Xe)*6s2, we get the filling of the 4f electron states, from lanthanum, (Xe)*6s2*5d1 instead of (Xe)*6s2*4f1, to ytterbium, (Xe)*6s2*4f14. Lutetium, (Xe)*6s2*4f14*5d1, to radon, (Xe)*6s2*4f14*5d10*6p6, work like before.

Francium, (Rn)*7s1 to ununoctium, (Rn)*7s2*5f14*6d10*7p6, work in the same way.

(Wikipedia)Periodic table (electron configurations) has a nice table with which states have how many electrons for each element.

Notice that the number of electron "slots" in each state is 2*(2*lj+1), from the 2 spin states and the (2*lj+1) orbital angular-momentum states (lj = OAM). The 2 for spin states is from (2*(1/2)+1), since electrons have spin 1/2.

Also notice the "fill order" of the state "shells":
1s
2s, 2p
3s, 3p
4s, 3d, 4p
5s, 4d, 5p
6s, 4f, 5d, 6p
7s, 5f, 6d, 7p

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Thu Sep 06, 2012 10:30 pm

Why the odd name "ununoctium"? That is a temporary name composed from its atomic number, according to the International Union of Pure and Applied Chemistry's rules for systematic naming of elements. If the evidence of discovery is strong enough, then the IUPAC will decide on an individual name and a 2-letter abbreviation. There have been some controversies over names in recent decades, with different labs having different proposed names.

Each digit gets a name root, and they are concatenated with -ium added at the end. The abbreviation is the symbols concatenated.
[table]Digit | Root | Symbol
0 | nil | n
1 | un | u
2 | b(i) | b
3 | tr(i) | t
4 | quad | q
5 | pent | p
6 | hex | h
7 | sept | s
8 | oct | o
9 | en(n) | e[/table]
The letters in ()'s may be removed to avoid awkward combinations. The roots are from both Latin and Greek to make them more distinct.

In this system, hydrogen is unium (U), helium bium (B), neon unnilium (Un), sodium ununium (Uu), calcium binilium (Bn), fermium unnilnilium (Unn), etc.

The individually-named elements are continuous up to copernicium, Cn, #112, officially named late last year. There are two additional named ones, flerovium, Fl, #114, and livermorium, Lv, #116, officially named last May. There is some evidence for the #113, #115, #117, and #118, the latter one being the highest atomic-number one that anyone has claimed to have produced.


There was an earlier temporary-name scheme, one thought up by Dmitri Mendeleev, discoverer of the periodic table of elements. The next elements in sequence below an element would get that element's name, prefixed with eka- (1), dvi- (2), and tri- (3), from Sanskrit.

Thus, he proposed eka-boron (scandium), eka-aluminum (gallium), and eka-silicon (germanium), and when those elements were discovered, they had close to his predicted properties.

Thus, those yet-unnamed elements are
[table]#113 | ununtrium | Uut | eka-thallium
#115 | ununpentium | Uup | eka-bismuth
#117 | ununseptium | Uus | eka-astatine
#118 | ununoctium | Uuo | eka-radon[/table]

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Fri Sep 07, 2012 8:19 am

Now for molecules.

Consider a hydrogen molecule, two hydrogen atoms. They share their two electrons, so let's see what happens. From Fermi-Dirac statistics and the Pauli exclusion principle, combined electron wavefunction is either

(Symmetric orbital) * (antisymmetric spin: 0)
(Antisymmetric orbital) * (symmetric spin: 1)

Symmetric: electrons can be close to each other
Antisymmetric: electrons prefer to be far away from each other

For best results in holding a hydrogen molecule together, the electrons must be between the nuclei, and that implies symmetric, and thus total spin 0.

This happens with other "covalent" bonds; the electrons pair up and go between the nuclei.

-

Another kind of chemical bond is the "ionic" bond. Let's consider ordinary table salt, NaCl. Sodium loses electrons to chlorine:

Na: (Ne)*3s1 -> (Ne)
Cl: (Ne)*3s2*3p5 -> (Ne)*3s2*3p6 or (Ar)

and the positive sodium ions attract the negative chloride ions. Notice the filled shells here; that's what makes that configuration stable.

Strictly speaking, covalent and ionic bonding are two extremes on a continuum. Elements with greater "electronegativity", like chlorine, attract electrons from elements with smaller values of it, like sodium.

This also happens in cases with largely-covalent bonding, like water. The oxygen atoms attract electrons from the hydrogen atoms, making the hydrogen end positive and the oxygen end negative. This makes water have much higher melting and boiling points that one might expect, because the hydrogen ends tend to stick to the oxygen ends. This also makes microwave ovens work, because the microwaves interact with the charges, making the water molecules wiggle.

Hydrocarbons have much less polarity, if any at all, and that's why oil and water don't mix -- the water molecules prefer sticking to other water molecules, and not to hydrocarbon ones.

-

The third kind of chemical bond is "metallic". In that one, the atoms' outermost electrons form a continuous sea, where electrons can continuously move. This makes metals conduct electricity. However, materials with ionic or covalent bonds have outermost electrons that stay put, making them electrical insulators. This also makes metals shiny, because their electrical conduction means that incoming light can make the electrons move back and forth over sizable distances, and those electrons then reradiate that light energy as a reflection.

Most elements are metals, and nonmetals will become metals at high enough pressures. That has been observed for some of them, and the way it happens indicates that will always happen with enough pressure. One needs at least a few million times Earth atmospheric pressure, however.

Electrons in metallic bonds still obey Fermi-Dirac statistics, and they also pile up in energy. Imagine a 1D box with length L. The first electron in it goes into the lowest orbital mode, a standing wave with wavelength pi/L, and the second one does also, because that's the lowest-energy mode. As with electrons orbiting atoms, their total spin becomes zero. The third and fourth electrons go into the mode with wavelength pi/(2*L), and pair up with spin 0, and likewise with wavelengths pi/(3*L), pi/(4*L), etc.

The result is a "Fermi liquid" or a "Fermi sea". Electrons can be excited from its surface, and they can leave behind vacant states or bubble-like "holes" below the surface. Both excited electrons and holes can carry electric currents.

In semiconductor technology, one starts with a substrate material like silicon or germanium, almost always silicon because germanium is relatively rare and expensive. One then makes it super super super pure, so that its original impurities will not affect its electrical properties. To control its electrical properties, one adds "dopants", materials that either give or take electrons, the takers making holes in the Fermi sea of electrons. One adds them in suitable amounts and in suitable places to form the sort of electronic components that one wants. Computer chips nowadays have millions or even billions of components, all produced with the help of silkscreening-like masking.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Fri Sep 07, 2012 6:01 pm

The outermost electrons of an atom are called its "valence electrons", and their number and configuration determines how the atoms react with other atoms. Elements in a column in the Periodic Table of Elements tend to have the same number of valence electrons, and thus tend to be chemically similar. However, there are lots of complications along the way.

The halogens -- fluorine, chlorine, bromine, iodine, and astatine -- have valence electrons (...)*np7 for row n. They tend to accept electrons and become like the next noble gas: (...)*np8.

The alkali metals -- lithium, sodium, potassium, rubidium, cesium, and francium -- have 1 valence electron: ns1, which they tend to give. Likewise, the alkali earths -- beryllium, magnesium, calcium, strontium, barium, and radium -- have 2 valence electrons: ns2, which they also tend to give. The d-filling elements, the transition metals, are a more complicated case, as are the f-filling elements, the rare earths, the lanthanides and actinides.

When row-2 elements from boron to fluorine make compounds, their valence electrons go into mixtures of 2s and 2p states, making their valence electrons act pretty much alike. Thus, boron has 3 electrons available for making compounds, carbon has 4, nitrogen has 3, oxygen has 2, and fluorine has 1.

The next rows down are usually more complicated, however. Take phosphorus, an analogue of nitrogen with 3 valence elements. It can combine with 3 oxygens making a phosphite ion: PO3, but it usually combines with a fourth one also, making a phosphate one: PO4. That's because the two paired 3s/3p electrons get shared with an oxygen atom, filling its valence shell. Thus, phosphorus sort of has valence 5 as well as 3. Such multiple valences are common among metals; iron can give 2 or 3 electrons.

Etc.

-

One can calculate the structure of atomic nuclei much like how one does for atoms, but there are some differences. The nucleons, protons and neutrons, have spin 1/2, and thus follow Fermi-Dirac statistics, including the Pauli Exclusion Principle. They form intermixed Fermi liquids, and a good approximation to the structure of a nucleus is the liquid-drop model.

However, protons and neutrons also fill shells of orbital energy states, and their filling shells can produce additional stability, just as with electrons in atoms. The numbers of them that produce such improvements are called "magic numbers" or "shell numbers", and they are much like the numbers of electrons in the noble gases: helium, neon, argon, krypton, xenon, and radon. Protons and neutrons work separately here, and some nuclei are "doubly magic", with magic numbers of both flavors of nucleons. The numbers:

Atoms: 2, 10, 18, 36, 54, 86
Nuclei: 2, 8, 20, 28, 50, 82, 126

-

Going from nuclei to hadrons, we find that they are made of quarks, spin-1/2 electronlike particles. There are two types:
Mesons: quark-antiquark
Baryons: 3 quarks
Baryons' antiparticles have 3 antiquarks, of course.

Quarks come in 6 flavors, with these electric charges: up 2/3, down -1/3, strange -1/3, charm 2/3, bottom -1/3, top 2/3.
Proton: up-up-down
Neutron: up-down-down

But the quark model had a big problem at first. For it to describe the baryons, the quarks' combined flavor and spin wavefunctions would have to be symmetric. This is especially evident in the delta baryon with charge +2. It has spin 3/2 = 3*1/2, and the quark model makes it up-up-up, with charge 3*(2/3), so it's clearly symmetric in both spin and flavor, and a violation of the Pauli Exclusion Principle.

A spin-up proton's wavefunction would thus be
(1/sqrt(18))*(X(u+,u-,d+) + X(u-,u+,d+) + X(u+,d+,u-) + X(u-,d+,u+) + X(d+,u+,u-) + X(d+,u-,u+) - 2X(u+,u+,d-) - 2X(u+,d-,u+) - 2X(d-,u+,u+))

and a delta baryon's one would be X(u+,u+,u+).

But in the 1970's, a solution was found. Quarks have an additional degree of freedom, "color", named from there being three values of it, and hadrons are all color singlets or colorless states. For colors red, green, and blue, a colorless baryon state has wavefunction

(1/sqrt(6))*(X(r,g,b) - X(r,b,g) + X(g,b,r) - X(g,r,b) + X(b,r,g) - X(b,g,r))

Notice the reversing sign when interchanging colors -- it's antisymmetric.

Since quarks have color, flavor, and spin degrees of freedom, their combined wavefunctions in baryons are thus antisymmetric, thus rescuing Fermi-Dirac statistics.

-

Can we go further in compositeness? There isn't any evidence that Standard-Model elementary particles are composite, at least not the easier-to-study ones like electrons and up and down quarks. That's from pounding them together at energies much greater than their rest-mass energies, meaning that if they are composite, then their binding energies must almost exactly cancel out their unbound masses. For electrons, that's about 1 in 200,000, judging from the results of the previous inhabitant of the LHC's tunnels, the LEP.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Fri Sep 07, 2012 8:44 pm

Now for some issues that I'd hand-waved over earlier, issues that can easily be illustrated with 1D waves.

For a constant potential, the Schroedinger equation has a simple wave solution:

X ~ exp(i*(k*x - w*t))

where energy E = hbar*w - angular frequency
and momentum p = hbar*k - angular wavenumber

k = +- (1/hbar)*exp(2*m*(E - V))
V = potential.

If the potential is greater than the energy, we get exponential falloff:
X ~ exp((q*x - i*w*t))
q = +- (1/hbar)*exp(2*m*(V - E))
Note: q = i*k

If the potential is slowly varying, its variation length scale being much greater than the wavelength, then one gets the Wentzel-Kramers-Brillouin semiclassical approximation:

X ~ 1/sqrt(|k|)*exp(i*(a(x) - w*t))
where da(x)/dx = k(x)
k(x) = +- (1/hbar)*exp(2*m*(E - V(x)))

This is like old quantum mechanics, which is no coincidence.

-

On the opposite extreme, a potential may change very fast relative to the wavelength, and in the extreme case, it jumps. To solve that case, one must make X and dX/dx continuous across it.

However, that can still be useful in illustrating various wave effects. A potential that is constant between jumps allows for relatively simple solutions: sine waves between the jumps, and continuity conditions at the jumps.

-

Let's consider our first problem: a step potential. That is, transmission and reflection of a wave with wavenumber k, with wavenumber k' on the other side. One gets:
Reflection amplitude r = (k - k')/(k + k')
Reflection intensity R = |r|2
Transmission amplitude t = 2k/(k + k')
Transmission intensity T = |t|2

(kx/k)*T + R = 1 -- conservation

The (kx/k) is a particle-flux factor.

Now make it fading off on the other side: X ~ exp(-q*x)
r = (k-i*q)/(k+i*q)
R = 1
t = 2k/(k+i*q)
T = 4k2/(k2 + q2)

Note that R = 1 here -- no particles get through. T is made irrelevant by the exponential decay.


Let's now consider the case of a rectangular potential, one that has a different constant value between x = 0 and x = L, and with a jump at each end. The wavenumber is k outside the region, k' inside the region.

{r,t} = {(k2 - k'2)*sin(k'*L), 2i*k*k'} / ((k2 + k'2)*sin(k'*L) + 2i*k*k'*cos(k'*L))

For k' real inside the potential region (sinusoidal wave),
{R,T} = {((k2 - k'2)*sin(k'*L))2, (2*k*k')2} / (((k2 + k'2)*sin(k'*L))2 + (2*k*k'*cos(k'*L))2)
and R + T = 1

Notice that the reflection oscillates as a function of k'*L, the wave phase across the region. That's from the waves bouncing back and forth inside the region, and either being in phase and reinforcing, or else being out of phase and canceling.

Turning to the case of a potential that's greater than the energy, the wave becomes exponential in the potential region. With k' = i*q, we get

{r,t} = {(k2 + q2)*sinh(q*L), 2i*k*q} / ((k2 - q2)*sinh(q*L) + 2i*k*q*cosh(q*L))

{R,T} = {((k2 + q2)*sinh(q*L))2, (2*k*q)2} / (((k2 - q2)*sinh(q*L))2 + (2*k*q*cosh(q*L))2)

If the barrier is wide or high enough, then q*L >> 1. In that limit,
T = (4*k*q/(k2+q2))2 * exp(-2*q*L)

But since T is nonzero, we get barrier penetration or quantum-mechanical tunneling.

QM tunneling is involved in alpha radioactive decay. Nuclei shoot out alpha particles (helium-4 nuclei), and let's see what happens when we run the decay backward. In the classical limit, the alpha particle stops a little outside of the result nucleus because of electrostatic repulsion. But this is quantum mechanics, and the alpha particle is also a wave. It can leak through the electrostatic potential barrier and reach the nucleus. If it can get in by that route, it can go out by that route.

One can calculate how long alpha decay takes by that mechanism, and one finds reasonably good agreement with observations. Thus, uranium-235 and uranium-238 have survived for the age of the Earth because fragments of their nuclei have to leak across the electrostatic-potential barrier in order to escape.
Last edited by lpetrich on Fri Sep 07, 2012 10:48 pm, edited 1 time in total.
Reason: Fixed some typos

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Fri Sep 07, 2012 9:27 pm

Let's now consider a rectangular potential again, but with a particle with too low energy to travel outside of it. I'll take k -> i*q, k' -> k.

The wave function fades off as exp(-q*x) outside of the potential region, where x is the distance from its nearby boundary. But what's inside can be found from standing-wave self-consistency:
tan(k*len) = - (2*k*q)/(q2 - k2)

For a deep potential, q >> k, and tan(k*len) ~ - 2*k/q or
k*len = n*pi - 2*k/q
or
k*(len + 2/q) = n*pi

For infinite depth, k*len = n*pi.

The energy E = V + (hbar*k)2/(2m)
giving for infinite depth E(n) = E(0) + (n*pi*hbar/len)2/(2m)

-

Let's look at the behavior. In general, the wavefunction will oscillate inside the potential region and fade off outside of it.

For n = 1, the wavefunction does not oscillate. It reaches its maximum in the middle of the potential and fades off to the edges.
/\

For n = 2, the wavefunction goes up, then down past zero, then up again.
/\/

For n = 3, the wavefunction goes up, then down past zero, then up past zero, then down again.
/\/\

For n = 4, it goes up, down, up, down, up:
/\/\/

In general, it has (n-1) zero crossings.


One can also see this behavior in the wavefunctions that I'd discussed earlier, like the harmonic-oscillator and the hydrogenlike-atom ones. Those wavefunctions are oscillatory in the classically-allowed regions, and fading-off outside those regions.

Also note that the ground state has nonzero energy relative to the classical limit, something we've also seen in the harmonic-oscillator case. For hydrogenlike atoms, the ground state has a finite energy, as opposed to classical-limit -infinity -- also nonzero.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Sat Sep 08, 2012 10:46 am

I'll now derive the wavefunctions for hydrogenlike atoms and the harmonic oscillator in an arbitrary number of space dimensions. They solve the time-independent Schroedinger equation

- (hbar2/(2m))*D2X + V*X = E*X

D2X = (d2X/dr2) + ((dim-1)/r)*(dX/dr) + (1/r2)*D'2X

for (dim) space dimensions, where D'2X is the angular second derivative. We can derive it by induction. With x = cos(a) for angle a, and D'' being for the lower-dimension angles,

D'2X = (1-x2)(d2X/dx2) - (dim-1)*x*(dX/dx) + (1/(1-x2))*D''2X = - lj*(lj+dim-2)*X
where D''2X = - lk*(lk+dim-3)*X and lj and lk are nonnegative integers.

The solution:
X = (1-x2)lk/2*Gegenbauer(lj-lk, lk+dim/2-1, x) * X(other angles)

where "Gegenbauer" is the Gegenbauer or ultraspherical polynomials, and lk <= lj. It gives the right results for dim = 2 and dim = 3 also.

The degeneracy is 2*upfac(lj,dim-2) + upfac(lj,dim-3)
where upfac(x,k) = x(x+1)(x+2)...(x+k-1)/k!

Thus,
D2X = (d2X/dr2) + ((dim-1)/r)*(dX/dr) - (lj(lj+dim-2)/r2)*X

-

Now for the radial wavefunctions.

For hydrogenlike atoms,

X = exp(-r/a) * rlj * Laguerre(n, 2lj+dim-2, 2r/a) * X(angles)
nx = n + lj + (dim-1)/2
a0 = hbar2/(m*kf)
a = nx*a0
Ry = (m*kf2)/(2*hbar2)
energy = - Ry/nx2

For the harmonic oscillator,

X = exp(-r2/(2x02)) * rlj * Laguerre(n, lj+dim/2-1, r2/x02) * X(angles)
x0 = sqrt(hbar/(m*w))
energy = hbar*w*(2n + lj + dim/2)

In both of them, "Laguerre" is the generalized Laguerre polynomials, and n is a nonnegative integer.

So I confirm the results that I'd obtained with operator algebra.

However, I've omitted normalization. That's a numerical factor that one multiplies the wavefunction by so that its absolute square's integral over space will equal 1.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Mon Sep 10, 2012 11:33 am

I wish to note a spherical-coordinates version of the constant-potential case:

X = 1/(k*r)dim/2-1 * Bessel(lj+dim/2-1, k*r) * X(angular)

where Bessel is a "Bessel functon". For the first arg equaling (integer) + 1/2, one finds the "spherical Bessel functions", which can be expressed as sums of powers of the second arg and sine and cosine of it.

For the infinite-depth case, one does the same sort of solution as before, but one imposes finiteness and zero outside of radius R. The Bessel function here is the first kind with no mixture of the second kind -- it's well-behaved at zero, and k = x/R:

X = (R/(x*r))dim/2-1 * J(lj+dim/2-1, x*(r/R)) * X(angular)
where J(lj+dim/2-1,x) = 0

Here also, it oscillates where classically allowed, and fades off where not, which is near the center for nonzero angular momentum.

Not surprisingly, one can express plane waves as sums of these spherical waves, and get a "partial wave" expansion for scattering. The inward spherical waves are as in the plane-wave case, while the outward ones get multiplied by scattering factors.

For the spherical case and zero angular momentum, the solutions are rather simple:

X = X0 * sin(k*r)/(k*r) + X1 * cos(k*r)/(k*r)
regular at the origin, singular at the origin

or
X = Xi * exp(-i*k*r)/(k*r) + Xo * exp(i*k*r)/(k*r)
inward, outward

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Mon Sep 10, 2012 7:54 pm

Let's now look at the Uncertainty Principle.

(DA)2 * (DB)2 >= (1/4)<i*[A,B]>2
or
(DA) * (DB) >= (1/2)*|<[A,B]>|

where (DA)2 = <(A - <A>)2>, the variance of the observations of A, and likewise for B. Note that
<A2> = <A>2 + (DA)2

For position and momentum, one gets this famous result:
(Dx)*(Dp) >= (1/2)*hbar

For time and energy, one gets
(Dt)*(DE) >= (1/2)*hbar

For angles and angular momentum, one gets
(Da)*(DL) >= (1/2)*hbar

and for different angular-momentum components,
(DL1)*(DL2) >= (1/2)*hbar*|<L3>|

The Uncertainty Principle more-or-less explains why quantum-mechanical ground-state energies are greater than their classical-limit counterparts. If a system had a position exactly at a classical ground-state spot, its momentum would be infinitely spread, giving it an infinite kinetic energy. So it must be spread out from such a spot, and the extra potential energy it gets from doing so is in general about equal to the kinetic energy from the corresponding momentum uncertainty.

That is why atoms don't collapse, something which was a big physical paradox when atomic nuclei were discovered in 1911 by Ernest Rutherford. If an electron tried to spiral in, there would be a point where its uncertainty-principle momentum would increase enough to counterbalance its loss of energy from spiraling in, and that point is where its ground state is. Not a definite point, of course, but a ground-state orbital wavefunction that would be spread over that point's distance.

-

Here's a hand-waving derivation for position-momentum uncertainty and similar cases. Let's start with a spatially-extended wave and let's try to measure its wavenumber k. We must measure it over a distance L, so Dx ~ L. We measure it by counting wave cycles, and that number is k*L. Since in general, there will be an uncertainty of around 1 cycle, (Dk)*L >~ 1. Thus,
(Dx)*(Dk) >~ 1

For a more precise calculation, and with k' the angular wavenumber,
(Dx)*(Dk') >= 1/2

Since momentum p = hbar*k', we get (Dx)*(Dp) >= (1/2)*hbar

-

Here is a more precise one, and a more general one. Consider two operators A and B, and subtract out their average values: A' = A - <A> and B' = B - <B>.

A wavefunction's integrated absolute square, |X|2 = X+.X >= 0, with it being zero for X = 0 everywhere.

Now consider Y = (A' + i*z*B').X where z is real and A and B Hermitian, like good operators for observables in quantum mechanics (A+ = A).

Take its absolute square:
|Y|2 = X+.(A' - i*z*B').(A' + i*z*B').X
= <A'2> + i*z*<[A',B']> + z2*<B'2>
= (DA)2 + i*z*<[A,B]> + z2*(DB)2
[A',B'] = [A,B] and is anti-Hermitian.

We now find the value of z that minimizes it: z = -(1/2)*i*<[A,B]>/(DB)2
Thus we find the uncertainty inequality
(DA)2 * (DB)2 >= (1/4)*(i*<[A,B]>)2

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Wed Sep 12, 2012 11:23 pm

It ought to be evident that quantum mechanics is a grossly counterintuitive theory. However, quantum-mechanics-based theories have had numerous successful predictions, sometimes successful to extreme accuracies.

This has provoked a lot of effort to find (Wikipedia)interpretations of quantum mechanics. That Wikipedia article lists 14 of them, with variations of several of them, complete with a table of how they answer these questions about them. Deterministic? Wavefunction real? Unique history? Hidden variables? Collapsing wavefunctions? Observer role? Local? Counterfactual definiteness? (definiteness of observations not yet made)

One of those contentious issues is the collapse of the wavefunction.

Consider light going through two slits. It will get diffracted by each slit as it passes through them; it will spread out over a range of directions. When a wave travels through the two slits, the part of it that went through one slit will meet the part that went through the other one, and they will alternately add and subtract, making an interference pattern of alternating light and dark bands. When the light falls on a detector, it will activate tiny bits of that detector, and that detector will record that interference pattern.

Going from classical to quantum, the detector works by a photon activating a tiny bit of it. Though the photons activate the detector bits at random, their probability function is what one expects from a classical-limit interference pattern. This means that each photon must have gone through both slits, but then got shrunk down by the detector. This is the collapse of the wavefunction.


But what makes it happen? According to one popular interpretation, the Copenhagen one, an act of observation causes it. But what counts as an observation?

According to another interpretation, the objective-collapse one, interaction with a macroscopic system with all its complexity can count as an observation. There is actually some theoretical support for this interpretation in the form of "quantum decoherence". This interpretation thus means that it's the observation apparatus that causes the collapse, thus supporting Copenhagen-like behavior.

But a variant of the Copenhagen interpretation states that it is mind / consciousness that causes the collapse. Many quantum woomeisters have liked this interpretation, because it seems to imply a form of psychokinesis, a form of "wishing will make it so". However, there is no evidence of such quantum psychokinesis, and the original form of this interpretation has consciousness causing a collapse without otherwise influencing it.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Thu Sep 13, 2012 10:06 pm

Where are we now?

Quantum mechanics has been a very successful paradigm so far, a paradigm in the sense of Thomas Kuhn's The Structure of Scientific Revolutions.

The Standard Model of particle physics is based on relativistic quantum field theory, quantum mechanics applied to elementary-particle fields and made consistent with special relativity. However, the SM is one of an infinite number of possible RQFT-based theories, and RQFT by itself gives no indication of why the Universe might follow the SM instead of some other RQFT-based theory. Furthermore, the SM is rather complicated, which suggests that it is not the final theory. It also has some order in it, something that gives hints of a theory that includes it as a special case.

How did we get here?

The first successful RQFT-based theory was quantum electrodynamics, the quantum-mechanical theory of electromagnetism. It had some awkward divergences at first, a result of integrating over high interaction energies. But physicists like Richard Feynman worked out how to handle them with a technique called "renormalization", and QED proved to be a great success. Renormalization works by recognizing that the divergences are always present, and folding them into various observed quantities, like masses and charges.

It was also possible to come up with a simple RQFT theory of the weak interactions, but it would produce nonsensical results at high interaction energies, nonsensical results that could not be subtracted out by renormalization.

The hadrons were even worse, and in the 1960's, some physicists tried to avoid RQFT outright for a while. They tried developing a theory of the scattering matrix or S-matrix:

X(particles coming out) = S . X(particles going in)

using self-consistency and various plausible hypotheses. Geoffrey Chew proposed a "bootstrap model" in which all the hadrons are co-equal elementary particles, with each one of them being formed from the others like pulling oneself up with one's bootstraps. However, that and other S-matrix-related efforts did not get very far, though Gabriele Veneziano and others developed an early version of string theory in S-matrix form.

Other physicists continued to work with RQFT-based theories, and by the 1970's, they succeeded with electromagnetic-weak unification and quantum chromodynamics, and the likes of the bootstrap model were left behind. In honor of this success, someone decided to call those two theories and their particles the Standard Model, some time in the early 1980's.

What next?

The Standard Model has a lot of complexity, and that suggests that it's incomplete, a special case of some larger theory. But what? There's been a lot of work on Grand Unified Theories, though it's been difficult to test them.

Closer to home, there are extensions of the Standard Model, like supersymmetry, making the likes of the Minimal Supersymmetric Standard Model. Some of those are more easily testable, and some of the particles that the Large Hadron Collider is searching for are supersymmetric partners of known particles.

All of this theorizing is done with RQFT-based theories, but RQFT has been a successful-enough paradigm in the Standard Model, and alternatives would face the problem of coexistence with a RQFT SM.

-

There's also the problem of a quantum theory of gravity. The classical one is well-understood: general relativity or some superset like generalized Brans-Dicke. However, it's been VERY difficult to come up with a self-consistent quantum theory that contains GR or GBD in its classical limit.

The most prominent approach so far has been string theory, and that offers the possibility of also getting the Standard Model. However, string theory has plenty of problems of its own, like an enormous number of possible ground states and low-energy theories.

There are various others, like loop quantum gravity, but they have not gotten very far either.

Here also, much of this theorizing uses RQFT or something that yields RQFT in some limit.

User avatar
lpetrich
Posts: 14453
Joined: Mon Feb 23, 2009 6:53 pm
Location: Lebanon, OR, USA

Post by lpetrich » Thu Sep 13, 2012 11:59 pm

An important part of the Standard Model is "symmetry breaking". Let's see how it works. It works by the Higgs mechanism, and I'll work with a simplified version of it called the Stueckelberg mechanism. Imagine the electromagnetic field and a spin-0 particle with charge q. Its Lagrangian looks like this:

L = - (1/4)*F.F + (1/2)*Dq(X).Dq(X)* + V(|X|2)

The terms are the electromagnetic field's kinetic energy, the particle's kinetic energy and electromagnetic interactions, and the particle's potential energy. The second term is a sort of absolute square of this gradient/interaction term:

Dq(X)= D(X) + i*q*A*X

where D is the ordinary gradient, q is the charge, and A is the electromagnetic field's potential. Note that this is a 4-vector quantity, a vector in space and time, and that the ordinary gradient and electromagnetic potential are also. That's typical of relativity, where space and time are sort-of-coequal. Likewise, momentum and energy form a 4-vector, and various other quantities similarly combine, though usually with a more complicated structure.

It's easiest to work with the Lagrangian, certainly easier than with the explicit equations of motion.

Let's try decomposing X into magnitude and argument: X = f*exp(i*a). Then,

L = - (1/4)*F.F + (1/2)*D(f).D(f) + (1/2)*f2*(D(a)+q*A).(D(a)+q*A) + V(f2)

Electromagnetism has what is called gauge symmetry. Let us do a gauge transformation: A -> A + D(P) and a -> a + q*P for some function P. Then P will cancel out.

That will also happen for *any* charged field: X -> X*exp(i*q*P). In fact, there is something interesting that one finds for it when several particles interact. Let's consider particles going in and coming out of a reaction:
X(in) = X1(in) * X2(in) * ...
Do a gauge transformation: X(in) -> X(in) * exp(i*P*sum of in q1, q2, ...)

Do the same for the particles coming out, X(out). One gets:

sum of in q1, q2, ... = sum of out q1, q2, ...

Conservation of electric charge. Thus related to electromagnetic gauge symmetry.

-

Returning to the Stueckelberg / Higgs mechanism, we get an equation of motion for f:
D2(f) = dV/df

The field will be static over space and time if dV/df is zero. At first sight, that means that f must be zero. But can f be nonzero? Yes it can, for certain shapes of V. To visualize this, imagine the field value as a ball and the potential as a surface that the ball moves on. The surface is radially symmetric, with f the distance from the center, and a the angle around the center.

For a massless particle, V = 0, and the potential is flat. For a massive particle without self-interaction, the potential is shaped like a bowl:
V = (1/2)*m2*f2

The particle moves back and forth in that bowl. We get the equation of motion for a noninteracting spin-0 particle, the Klein-Gordon equation:
D2(f) = m2*f

If the potential has a different shape, its motion will be different, and often more complicated. If the potential is shaped like a bowl with a central hump, sort of like a juice squeezer, then the ground states will have some nonzero value of f, and also any value of a. For
V = (1/2)*V2*f2 + (1/4)*V4*f4
we get equation of motion
D2(f) = V2*f + V4*f3

If V2 < 0, then dV/df = 0 if f = sqrt(-V2/V4) everywhere in space-time away from excitations. Let f0 be that value, and f = f0 + f'. Then

D2(f') = 2*V4*f02*f' + 3*V4*f0*f'2 + V4*f'3

So f' has a mass equal to sqrt(-2*V2) and interaction with itself.

-

Let's see what effect this has on the electromagnetic field. Leaving out f', we get Lagrangian

L = - (1/4)*F.F + (1/2)*m2*(A + (1/q)*D(a)).(A + (1/q)*D(a))

where the photon has acquired a mass m = q*f0.

Note also that the particle phase a can be absorbed into the photon by a gauge transformation. Without the photon, it would be a massless particle, a "Goldstone particle" that is produced by the symmetry breaking here.

The Higgs mechanism works in the same way, though with a more complicated field structure, and with some photonlike particles getting masses rather than the photon itself. These particles are the W+, the W-, and the Z.
Last edited by lpetrich on Fri Sep 14, 2012 1:41 pm, edited 1 time in total.
Reason: Fixed typos, added some details

Locked