Class calendar, week 5
Mon., 9/27:
# The general uncertainty relation is a consequence of the Schwarz
inequality,
which is just the statement that for any two vectors |v> and |w> in
the Hilbert space,
the norm of the part of w perpendicular to v is non-negative:
0 < = || |w> - |v><v|w>/<v|v>||^2 = <w|w> - |<v|w>|^2/<v|v>,
so <v|v><w|w> >= |<v|w>|^2,
with equality only if |w>= z |v> for
some complex number z.
# The variance Delta A is zero in an eigenstate of
A,and Delta A = ||(A-<A>)|v>|| (assuming <v|v>=1).
# Proof of uncertainty relation: define A' := A - <A>, and
B':=B - <B>, the expectation value being taken in some fixed state|v>.
Then Delta A=||A'|v>||, and Delta B = ||B'|v>||. Thus
(Delta A)^2 (Delta B)^2
= ||A'|v>||^2 ||B'|v>||^2
>= |<v|A'B'|v>|^2
(Schwarz inequality)
= |<C> + i <D>|^2, where C:= {A',B'}/2 and
i D:=[A',B']/2 = [A,B]/2
= <C>^2 + <D>^2
>= <D>^2
(since C and D are hermitian so <C> and <D> are real)
= ( (1/2) |<[A,B]>|)^2.
Taking the square root we obtain the general uncertainty relation.
# When is the inequality an equality, i.e., when is the uncertainty
as small as it can be according to the general relation?
(1) Schwarz inequality is an equality, i.e.
B'|v> = z A'|v> for some complex number z.
(2) <C>=0.
If (1) holds, then (2) implies (z+z*)<A'^2>=0, so eithe r
A'|v>=0=B'|v>, so |v> is a common eigenstate of A and B and both sides
of the uncertainty relation vanish, or z+z*=0, i.e. z=iw for some real
number w.
# Minimum uncertainty wavepacket: apply this to A=x and B=p.
The minimum incertainty condition becomes:
(p - <p>)|v> = iw (x- <x>)|v>, which is a 1st order ODE
in the x-representation (or p-representation),
whose general solution is
<x|v> = c exp(-w(x-<x>)^2/2hbar)exp(i<p>x/hbar), i.e., a Gaussian
of any width, centered on any point <x>, and shifted in momentum
to have the average value <p>. Examples: the ground state of any
harmonic oscillator, centered on any point, translating with
any velocity <p>/m. By contrast, the excited states of an oscillator
are not minimum uncertainty wavepackets.
Wed., 9/29:
# Heretofore observables were time independent (unless they had
explicit dependence on the tme parameter), and states evolved as |v(t)>
= U(t,t0)|v(t0)>.
This is called the Schrodinger picture or Schrodinger representation
Thus an observable A(t) (with possibly explicit time dependence)
would have a time-dependent
expectation value <A>(t) = <v(t)|A(t)|v(t)> = <v(t0)|U(t,t0)*A(t)U(t,t0)|v(t0)>.
We can get precisely the same result if we
instead think of the states as tim-independent and the observables
as evolving.
This is called the Heisenberg picture or Heisenberg representation.
(There is also an interaction picture, in which the observables
evolve via the free part of the Hamiltonian, and the states evolve via
the interation part.)
# Putting now a subscript S for Schrodinger picture quantities
and H for Heisenberg picture quantities,
we have
A_H = U* A_S U, and |v_H> =|v_S(t0)>.
Note a change of the fiducial time t0 just makes an overall unitary
transformation to the states and operators.
# The change from Schrodinger to Heisenberg pictures preserves the
algebra of operators:
if A_S B_S = C_S, then A_H B_H = U* A_S U U*B_S U = U* A_S B_S U =
U* C_S U = C_H.
# The Schrodinger eqn is replaced by another eqn in the Heisenberg
picture. To find it, recall from 9/10 that
i hbar dU/dt = H U. Thus we find
i hbar dA_H/dt = i hbar (d/dt) U*A_S U = U* [A_S,H_S] U + ihbar U*
dA_S/dt U so
i hbar dA_H/dt = [A_H,H_H] + ihbar U* dA_S/dt U.
The last term vanishes if A_S has no explicit time dependence. Otherwise,
you can think of it as the partial derivative of A_H with respect to the
explicit time dependence.
# Example: H = p^2/2m + V(x,t). Using [x_H,p_H]=i hbar, we find that
the Heisenberg picture operators evolve by dx_H/dt = p_H/m,
dp_H/dt = -dV_H/dx, i.e.,
the same form as the classical equations of motion!
Taking the expectation values of these equations we recover Ehrenfest's
theorem.
Fri., 10/1:
Computation tips:
# If you are finding the eigenstates and or eigenvalues for a Hamiltonian
of the form H = c I + H', where c is a number, just find them
for H' and add c to all the eigenvalues.
This can save writing and help eliminate errors. Examples of this on
previous homeworks were
(i) the B0-B0bar oscillations, where H had the form H = c I +
d sigma_x
(ii) the particle in the chain of potential wells, where H =
Sum_n [E0 |n><n| + V0 (|n+1><n| + |n><n+1|)].
(By the way, the V0 terms represent the possibility of tunneling between
nearest neighbor wells.
This is called the ``tight binding approximation" in condensed matter
physics.
Note that if one left out, say, the |n><n+1| term, the Hamiltonian
would not be hermitian,
so this term must be there if the other one is.)
# Instead of diagonalizing the 2x2 Hamiltonian matrix in the BBbar
problem, one could also just
evaluate the evolution operator U = exp(-iHt/hbar) = exp(-ict/hbar)exp(-i(dt/hbar)sigma_x).
The second term is given by a simple formula: for any vector b
= |b| b^, where b^ is a unit vector, we have
| exp(i b.sigma) = cos |b| +
i sin |b| (b^.sigma) |
which is obtained simply by expanding the exponential in a power series,
and noting that
(b^.sigma)^2 = 1, i.e., the 2x2 unit matrix.
Applied to the BBbar problem, this yields
prob(Bbar)(t) = |<Bbar| v(t)>|^2 = |<Bbar| U(t)|B>|^2 =
sin^2(dt/hbar).
The period of oscillation is given by t = pi hbar/d ~ 10^-15 s/d(eV).
In fact, d is of order 10^-3 eV,
so the period is of order 10^-12s, which is of the same order as the
lifetime. The B and Bbar
can be distinguished by their decay products. To measure the oscillation
one would want to
see how the decay products vary as a function of time since the B was
created.
The B-mesons are created travelling near the speed of light in the
frame of the detector,
so they travel a distance of order 0.1 mm in one oscillation period.
The detectors have good enough
spatial resolution to detect this oscillation. Prof. Hassan Jawahery
of our department is involved with
experiments to measure this and can tell you much more about it.
# (Almost) everything in physics is (at least approximately) a harmonic
oscillator. For example,
small vibrations about an equilibrium point x0: V(x) = V(x0) + 1/2
V''(x0) (x-x0)^2 + ...
is an oscillator potential centered on x0, with "spring constant" k
= V''(x0), shifted by the
constant V(x0). Examples of this is are an atom in a trap potential,
or the normal modes of a molecule,
or of a crystal. In the last case, the excitations of the oscillator
modes are called phonons.
Even the electromagnetic field is a collection of oscillators, one
for each wavevector and polarization, as is easily seen by Fourier transforming
the wave equation. And the same goes for the W and Z
and gluon fields, when interactions are neglected.
# Hamiltonian: H = p^2/2m + mw^2 x^2/2.
Classical equations are dx/dt = p/m, and dp/dt = - m w^2 x, hence d^2
x/dt^2 = -w^2 x.
# Estimate the size of the ground state wave function using the uncertainty
relation:
if the size is L, then the minimum momentum uncertainty is of order
hbar/L, so the minimum
energy is of order E = (hbar/L)^2/2m + mw^2 L^2/2. Minimize wrt L to
find L = x_0 := sqrt(hbar/mw).
At this value, the energy is of order hbar w.
The quantum oscillator has a length scale, while the classical
one does not.
Note L_0 goes to zero with hbar, and inversely with m or w.
# H = 1/2 hbar w [ (x_0 p/hbar)^2 + (x/x_0)^2]. Note the location of
the x_0's and hbar's can
be determined easily by dimensional analysis.
# H looks like A^2 + B^2, which can be factorized as [(A+iB)(A-iB)
+ (A-iB)(A+iB)]/2. This turns
out to be a great idea, so define
| a = ((x/x_0) + i (x_0 p/hbar))/sqrt[2] |
Then [x,p]=ihbar implies [a,a*]=1, so we get H = hbar w (a*a + 1/2).
The spectrum of the number operator N=a*a is the nonnegative
integers (see below), so the energy spectrum
is E = (n + 1/2) hbar w, and the ground state has energy 1/2 hbar w,
the zero-point energy.
# Note N is hermitian, so the spectrum is real. Also if a were
a complex number,
a*a would be non-negative. In fact this is true for the operator as
well: <v|a*a|v> =||a|v>||^2 >=0,
so in particular the eigenvalues of N are non-negative.
Moreover, if |v> is an eigenvector with eigenvalue v,
N|v> = v|v>, it follows from [N,a*] = a* and [N,a] = -a
that
N a*|v> = (v+1)|v> and Na|v> = (v-1)|v>.
Thus a* and a are raising and lowering operators, or ladder
operators,
and the eigenstates come in infinite chains,
...|v-2>,|v-1>,|v>,|v+1>,|v+2>,...
The only thing that can prevent us from getting negative eigenvalues
(which we showed previously cannot occur) is if somewhere on
the chain we have a|v>=0. This implies
that N|v>=a*a|v>=0, hence the eigenvalue v must be zero, and all the
other eigenvalues in the chain
must be positive integers. Thus the spectrum of N is 0,1,2,3,...,
with corresponding number eigenstates
|0>, |1>, |2>, |3>, ...
# We found all this just from the algebra of the operators. It could
be that there are multiple copies of this
semi-infinite chain of eigenvectors---nothing in our calculation so
far rules this out. However, appealing to
the x-representation, and the nondegeneracy of the spectrum of the
x operator, we have the conclusion that
the bound states are non-degenerate, so there is only one chain.
# Using the notation |n> for a normalized eigenvector of N with
eigenvalue n, we have
| |n> = (1/sqrt[n!]) (a*)^n |0>,
a*|n> =sqrt[n+1] |n+1>,
a|n> = sqrt[n] |n-1> |
I have found it useful to remember that the argument of the square root
in the first two formulae is always the larger of the two n's.
# The parity operator P satisfies, Px = -xP, Pp = -pP,
hence Pa = -aP and Pa* = -a*P.
Since P commutes with the hamiltonian H, we can always simulataneously
diagonalize P and H.
And since in one dimension the eigenstates of H are nondegenerate,
they must be parity eigenstates,
P|n> = P(n)|n>, where P(n) = +- 1.
The parity alternates: P|n+1> ~ Pa*|n> ~ -a*P|n> ~ - P(n) |n+1>.
The ground state has even parity (if we choose the overall phase of
the parity operator appropriately),
so the state |n> has parity (-1)^n.