Class calendar, week 8

Mon., 10/18:
# Notes on Stern-Gerlach:
1) B_0 must be large enough so that the force in the x-direction averages to zero.
2) Besides the splitting of the beam by the magnetic field,
there is also the quantum spreading of the wave function.
For the device to operate effectively this spreading must be less than the splitting.
This turns out to be the same condition as the requirement that the energy difference
between the spin up and spin down states at the top of the beam, minus the same difference
at the bottom of the beam, be greater than hbar divided by the time spent in the field.
3) The device won't work for free electrons, due to the Lorentz force.
The deflection per se is not a problem, since it could just shift the overall beam position.
The problem is that the Lorentz deflection in the z-direction is uncertain due to the necessary
uncertainty of the velocity in the x-direction. It turns out that this always washes out the splitting
of the beams for an electron (or proton). For an ion, however, the inertial mass of the ion
controls the response to the Lorentz force, whereas the electron mass sets the scale for the
magnetic moment interaction, so the Lorentz spreading can be negligible.
For a discussion of points 2) and 3) see G. Baym, Lectures on Quantum Mechanics,  Ch. 14, p. 324.
# Spin resonance:
Important in measuring magnetic moments, nuclear magnetic resonance (NMR), magnetic resonance
imaging (MRI), and the resonance mechanisim and mathematics is identical to that in other systems, e.g. the Raman coupling of the two hyperfine levels of the Beryllium ion in the Schrodinger cat paper.
NMR is used in chemistry to identify compounds, and in physics for all kinds of things including identifying textures such as vortices in superfluid helium-3...
The applications are endless, so it's worth taking a close look at this.
# Most of the discussion we can do with a spin-j system with arbitrary j.
A spin in a magnetic field B_1 x^  will precess about the x-axis with angular frequency  -gamma B,
where gamma is the gyromagnetic ratio.
This evolution conserves energy, as the hamiltonian -gamma B_1 J_x is time independent.
If we impose a strong magnetic field B_0 z^ in the z-direction,
then the spin states with different projection on the z-axis no longer have the same energy,
so it should not be possible to rotate them into each other with a time-independent hamiltonian.
Indeed, we know what happens: the spin precesses about the net magnetic field B_0 z^ + B_1 x^.
If B_0 >> B_1, the spin barely changes direction. It initially tries to precess about x^,
but as soon as it gets a component perpendicular to z^  that starts to precess about z^.
So we could make the spin flip over, in the presence of B_0 z^, if we were to make the direction
of the applied field B_1 co-rotate with the precession about z^, applying the time-dependent
field B_1(cos wt x^ - sin wt y^). If the frequency w is matched to the precession frequency
w_0=gamma B_0, it is plausible that the spin will flop over from up to down and back again in the z-direction, while it is rotating about z^. Evolving between these states of different energy is now possible because the hamiltonian has become time-dependent. If w is not matched perfectly to w_0, we expect the flopping amplitude will be related to w - w_0 and will diminish as this difference grows.
# The Schrodinger eqn. for the spin state:
(set hbar=1): i d/dt|v> = H(t)|v>, with H(t) = -gamma ( B_0 z J_z  +  B_1(cos wt J_x - sin wt J_y)).
If H(t) were a commuting operator at different times,
we could immediately write down the solution, |v(t)>=exp(-i\int_0^t dt' H(t') dt')|v(0)>.
But this is not valid if [H(t'),H(t'')] is nonzero, which it isn't in the present case.
In general, we must resort to perturbation theory in this situation,
but in the present case we can find the exact solution by "transforming to the rotating frame",
since in the corotating frame the spin simply precesses about the rotating x-axis.
To see this, we write the lab frame state |lab> as |lab> = exp(iwtJ_z) |rot>,
and plug into the Schrodinger eqn, finding
i d/dt |rot> = [w J_z + exp(-iwt J_z)H(t)exp(iwt J_z)] |rot> =[(w-w_0) J_z  - gamma B_1 J_x] |rot>.
Thus |rot> indeed satisfies the Schrodinger eqn with a time-independent Hamiltonian.
We can understand the appearance of  the term wt J_z  because in the rotating frame a
fictitious magnetic field appears, corresponding to the uniform rotation about z^.
The expectation that the time dependence vanishes in the rotating frame is borne  out.
To see it mathematically, note that (cos wt J_x - sin wt J_y) =(exp(iwt) J_+  +  exp(-iwt) J_-)/2.
Now use [J_z, J_+/-] = +/- J_+/-,
and the Campbell-Baker-Hausdorff identity exp(A)Bexp(-A) = exp([A, ])B,
to see that
exp(-iwt J_z)(J_+/-)exp(iwt J_z) =exp(-/+iwt) J_+/-, hence
exp(-iwt J_z)(exp(iwt) J_+  +  exp(-iwt) J_-)exp(iwt J_z) = J_+  +   J_  = 2J_x.

 Wed., Oct. 20:
# still hbar=1. where do the hbars go?
# i d/dt |rot> =  [(w-w_0)t J_z  - gamma B_1 J_x] |rot>. (*)
If we tune the frequency of the applied field w to the precession frequency w_0=gamma B_0
we kill off the J_z term, leaving only the J_x term. In this case, the spin precesses about the x-axis,
with the frequency gamma B_1, flopping from up to down and back again in the z-direction.
(In the lab frame, it is also precessing about the z-axis while it flops.)
If B_0 >> B_1 then we have to tune w very close indeed, so the resonance is sharper. So the
resonance is a sensitive probe of gamma B_0.
# The solution to the eqn (*) above is
|rot(t)> = exp(- it [(w-w_0)t J_z  - gamma B_1 J_x]) |rot(0)>
which is a time-dependent rotation, through an angle \Omega t, with
\Omega = Sqrt[(w-w_0)^2 + (gamma B_1)^2],
about the axis
n^ = ((w-w_0)/\Omega) z^ - gamma B_1/\Omega x^.
The exact form of the rotation matrices for any spin was worked out by Majorana,
but is a bit complicated. Let's just write the result here for spin-1/2. Then
J_z/hbar = (1/2) sigma_z and J_x/hbar = (1/2) sigma_x, so
|rot(t)> = [cos (\Omega t/2) - i sin(\Omega t/2)n^.sigma] |rot(0)>.
# If the spin starts out |+z> at t=0, then the probability of finding it down at time t is
Prob(down,t) = |<-z| lab(t)>|^2
                      =  |<-z| rot(t)>|^2   (|lab> = exp(iwt J_z) |rot> just involves a (different) phase factor for |+/-z>)
                      = sin^2(\Omega t/2) (gamma B_1)^2/((w-gamma B_0)^2 + (gamma B_1)^2).
This is Rabi's formula for the Rabi oscillations.
The probability of "down" oscillates with frequency \Omega/2. If w=w_0, it oscillates between
zero and unity. Otherwise it only partially oscillates. For example, if omega=0, it still oscillates.
This is because the spin started out up in the z-direction, and is precessing about a titled axis n^.
Classically, of course, the spin vector only dips down a bit in this case, but quantum mechanically
this means there is a nonzero probability to actuall find the spin in the down state.
The maximum probability of down is
P_max(down, w) = (gamma B_1)^2/((w-gamma B_0)^2 + (gamma B_1)^2).
The smaller the ratio B_1/B_0, the more sharply peaked this function of w is.
# Note we have solved a general time-dependent two-state problem, even if it isn't spin:
consider the hamiltonian
H = E1 |1><1| + E2 |2><2| + b e^iwt |2><1| + b* e^-iwt |1><2|.
This can be identified with the spin problem in the magnetic field, by rewriting it as
H = (1/2)(E1+E2) I + (1/2)(E1- E2) sigma_z + Re(b e^(iwt)) sigma_x - Im(b e^iwt) sigma_y.
The first term, proportional to the identity, just produces an overall phase, and the x & y terms
allow for an extra phase shift compared to what we had before. The solution is essentially
identical therefore, so in particular we have that if the stateis initially |1>, then the probability
of finding it |2> at time t is
Prob(2,t) = |<2| v(t)>|^2
               = sin^2(\Omega t/2) (|b|^2/((w-w_0)^2 + |b|^2),
with w_0 = (E1-E2)/hbar and \Omega = Sqrt[(w-w_0)^2 + |b|^2].
# This is relevant in the Schrodinger cat article, where the two states |1> and |2> are hyperfine
states of the Beryllium ion. They have different energies, and by exposing the atom to a pair
of laser fields with a beat frequency tuned to w_0, with each individual frequency tuned away from
any transitions, the atom can make "Raman transitions" between |1> and |2> with a frequency determined
by the amplitude of the laser fields. By adjusting the time that the laser beams are on, one can choose
the amount of rotation of |1> into |2>. A "pi-pulse" sends |1> to |2>, while a "pi/2 pulse" sends
|1> to an equal linear combination of |1> and |2>. Such a pi/2 pulse is the first step in preparing the Schrodinger cat state of the atom.
# For nuclear spins, which are of the order of a "nuclear magneton" (more on this later in the semester),
w_0 = gamma B ~ 1.5 MHz (B/1000 gauss). The spin flopping resonance thus
happens in an RF field.
# Rather than drive with a rotating B_1 field, one can just use a linearly polarized field:
B_1 cos(wt) x^ = (1/2)B_1(cos wt x^ - sin wt y^) + (1/2)B_1(cos wt x^ + sin wt y^).
If the first term is co-rotating with the precession about z^, the second one is counter-rotating.
Near resonance, the second term is way off-resopnance and just produces high frequency wiggles.
Neglecting it completely is called the rotating wave approximation.
# In the 1938 PRL handed out in class,
(I.I. Rabi et al, Phys. Rev. Lett. 53, 318 (1938); see also the full paper Phys. Rev. 55, 526 (1939).)
Rabi and collaborators describe a method to measure nuclear gyromagnetic ratios,
called the molecular beam magnetic resonance (MBMR) method.
They fixed an oscillating RF field of frequency w and varied B_0 until they observed the resonant spin flip. (See the article to read about how they observed the flips.)
This told them they had reached w=w_0=gamma B_0, so they had found gamma = w/B_0.
They independently knew the nuclear spin (which is always hbar times an integer or half-integer),
so from gamma they got the nuclear magetic moments.
# If solutions or bulk samples of matter are placed in a magnetic field, and exposed to an RF field,
one will observe a resonant absrorption of energy from the field when w=w_0. This is called
nuclear magnetic resonance (NMR).
In a solution, the magnetic field seen by the nuclei is not just the imposed field B_0.
The induced electronic currents in the chemical enviroment of the nucleus
partly shield the nucleus from B_0, so each nucleus sees a
field B*_0 that depends on its chemical environment. This is called the chemical shift. Only the nuclei
with w=gamma B*_0 will be resonant at a given w. This can be used to distinguish different chemical
compounds, or superfluid textures in Helium-3, or types of tissue in the human body. In the last application,
magnetic resonance imaging (MRI), the power absorption is measured at many frequencies and with
the applied RF field at many angles relative to the body. All this data is "inverted" to reconstruct a three-
dimensional map of the distribution of chemical environments in the body...i.e. the tissue types.
# The energy difference between the nuclear spin states in, say, a 1 Tesla (=10^4 gauss) field is miniscule:
Delta E = Delta (-M.B) = Delta (-gamma B J) ~ hbar gamma B ~ (2/3 eV-10^-15 s)(15 MHz) = 10^-8 eV.
Compared to body temperature, ~1/40 eV, this is tiny. Even at 10 T, the ratio is of order 10^-5.
So only one in 10^5 of the nuclei are aligned with the field. This is enough, however, since there
are so many nuclei at body density.
# A quite new method of MRI is being developed for imaging lungs and other body cavities,
or tissues that are not well resolved by ordinary MRI, using hyperpolarized noble gas. The idea is
to spin up the nuclei of helium-3 (helium-4 has no spin!) or xenon-129 via optical pumping with
circularly polarized laser light. In the case of  xenon-129,  the polarization is accomplished via
exchange interactions in collsions with  Rubidium atoms that  have had their valence electron
spin-polarized by polarized laser light. This can take a long time (in the case of helium, on the order
of hours), but once polarized the noble gas nuclei interact very weakly with their enviroment.
The polarized gas can be stored in a bottle, and xenon can even be frozen, for later use.
No large magnetic field is required to align the spins once they are in the body to be imaged, since they
are not in thermal equilibrium with their enviroment (though I think a strong field still gives better
resolution for the power absorption resonance). A good website on Hyperpolarized Gas Research
at the University of Virginia has loads of info & links to related research. Their "What's New" page is
particularly impressive. It shows rotating and time-dependent images. A  review paper is available there.

Fri. Oct. 22:
Central (i.e. spherically symmetric) Potentials.
# Classically, orbits of a central potential are planar, say x-y plane.
The angular momentum L=mr^2 dphi/dt is conserved, and the energy takes the form
E = (m/2) (dr/dt)^2 + L^2/2mr^2 + V(r),
the same as for a one-dimensional problem with "effective" potential V_eff(r) = L^2/2mr^2 + V(r).
The term L^2/2mr^2 is called the centrifugal barrier
# Quantum mechanically, we can do much the same thing.  Since V(r) is invariant under rotations,
L^i and L^2 commute with H, so we can somulataneously diagonalize H, L^2, L_z. In fact, these form
a complete commute set of observables, since it turns out no two states have the same eigenvalue
for all three of these operators. Label these states |nlm>, so
H |nlm> = E_nl |nlm>,  L^2 |nlm> = l(l+1)hbar^2 |nlm>,   L_z |nlm> = m hbar |nlm>.
The energy is degenerate in m, since the m-raising operator L_+ commutes with H.
# The position space wavefunction is: <r theta ph i| nlm> = R_nl(r) Y_lm(theta,phi).
We also use u(r), defined by R(r) = u(r)/r.
Since the spherical harmonics are orthonormal wrt the angular integration,
the norm is \int dr r^2 R*R = \int dr u*u.
The Schrodinger equation for each l reduces to the radial equation, H_l u = E u,
where the radial Hamiltonian H_l is
 
H_l   =  -(hbar^2/2m)(d^2/dr^2) + (hbar^2/2m)(l(l+1)/r^2) +V(r) 

So the problem reduces to a one-dimensional problem.
The boundary conditions on u(r) are that it vanishes faster than 1/Sqrt[r] at infinity for  bound states,
and it vanishes at r=0. The latter condition is simple to see if the centrifugal barrier dominates,
since then the solutions for u near r=0 go like r^(l+1) and r^(-l). The latter is not normalizable at r=0,
and the former vanishes at r=0. For  zero angular momentum, there is no centrifugal barrier,
so this argument does not apply. If the potential V(r) is finite however, then the wave function
should be finite at the origin, and R(r)=u(r)/r is finite only if u(r) goes to zero. If V(r) diverges,
as in the Coulomb potential -1/r, we can see by rounding off the potential to a finite one, and taking
the limit in which the round-off goes away, that the physical solution we are interested in is the one that vanishes at the origin.  (For an interesting mathematical discussion of the ambiguity in the s-wave
Schrodinger equation for the Coulomb potential, and the possibility of other boundary conditions that yield different spectra for the Hamiltonian, see On the Energy Levels of the Hydrogen Atom, by C.J. Fewster.
# The radial problem is thus a 1-d quantum problem, with an infinite potential barrier at the origin. Equivalently, we can extend the potential to negative values of r by V(-r) = V(r).
In this doubled potential, the eigenstates alternate in parity, and only the odd parity states are relevant
to the 3d problem, since it is they which vanish at the origin.
This shows that a 3d potential may not have a bound state, since it would be the first excited state of the doubled potential, but not every 1d potential has more than one bound state.
# For each l, we get a non-degenerate spectrum of eigenstates,
which we can label by the radial quantum number n.
Perhaps this integer would be better called n_l, or n_r.
Thus the three quantum numbers n,l,m fully specify the bound states.
The states with angular momentum l=0,1,2,3,4,5,... are called s,p,d,f,g,h,... waves.
# For the same n, states with higher l always have higher energy, since the l-dependent term in the Hamiltonian---the centrifugal barrier---is positive definite.
Formally, we can prove this by using the identity on derivatives of energy
eigenvalues wrt a parameter (in this case l) in the Hamiltonian, treating l as a continuous parameter:
dE_nl/dl = <dH/dl> = <(2l+1)hbar^2/2mr^2> > 0.
Thus, E_nl > E_nl'  if  l > l'.
# Free particle solutions for R(r): spherical Bessel functions, a.k.a.
spherical Bessel function j_l(kr),
Neumann function n_l(kr),
Hankel functions h_l = j_l + i n_l,   h*_l = j_l - i n_l.
Their properties were discussed in class. See any book on QM.
# Spherical square well: free particle solutions inside and outside, must be matched at the well wall.
Inside we must take only j_l, since n_l blows up at the origin. Outside, for scattering states both h_l, h*_l
appear, they are outgoing and ingoing wave solutions. For bound states, we must take the decaying Hankel
function, h_l(ikr),  k = i Sqrt[-2mE}/hbar. See any book on QM.
For an infinitely deep well, the order of the levels is 1s, 1p, 1d, 2s, 1f, 2p, 1g, 2d, 1h, 3s, 2f, ....
The letter in this notation indicates the angular momentum, while the number indicates the radial quantum number, starting with 1 as the ground state. (One also sees the convention where the radial quantum number starts at zero.)