Class calendar, week 3 Mon. 9/13:
# To see that the entangled state |w>=a |00> + b |11> cannot assign
a definite state to the first qubit,
ask what is the probability to find the state of the first qubit is
|v>? [Note on notation: We use a single "ket" for both the
state of the two qubit system and for the state of a single qubit subsystem.
This may lead to confusion but usually does not.]
Prob(v) = ||a<v|0>|0> + b<v|1>|1>||^2 = |a|^2 |<v|0>|^2 +
|b|^2 |<v|1>|^2.
Since |a|^2 + |b|^2 | = 1 = |<v|0>|^2 + |<v|1>|^2,
prob(v) can not be 1 unless a=0 or b=0, i.e. unless the state |w> is
a product state.
Thus if both a and b are nonzero no definite state can be assigned
to the first qubit by itself.
Note that if a=b=1/SquareRoot[2], then prob(v)=1/2 for any state
of the first qubit.
In this case the state |w> contains no information whatsoever about
the state of the first qubit by itself.
All the information is in the correlations between the states of the
two qubits!
# Density matrices: The average value of an observable
A in a state |v> is <v|A|v>=Tr(A|v><v|).
If the state is not definite but is |v_i> with probability
p_i, then the average value of any observable A is Sum_i p_i
Tr (A|v_i><v_i|) = Tr A rho, where rho:= Sum_i p_i |v_i><v_i|.
rho is called the density matrix for the system. It is a generalization
of the notion of a state to allow for indeterminateness as to what the
state is.
For a maximally specified state, the density matrix is just a projection
onto a one dimensional subspace.
In this case the state is said to be a pure state. Otherwise
the state is called a mixed state, or a statistical mixture.
Note that a "mixture" and a "superposition" are very different notions.
The latter describes a linear combination of pure states which is again
a pure state.
Density matrices satisfy Tr rho = 1, rho*=rho, <v|rho|v> is
nonnegative, i.e. they are nonnegative hermitian matrices (in particular
they have nonnegative eigenvalues) with unit trace.
Pure state density matrices have one eigenvalue =1 and the rest =0.
Mixed states have all eigenvalues less than 1 and greater than or equal
to zero.
Pure state density matrices thus satisfy rho^2=rho, and mixed states
do not.
Wed., 9/15:
# A computational note: if A|a>=a|a>, then <a|A* = a*<a| (where
A* = adjoint of A).
# The eigenvectors of a hermitian operator span the Hilbert space.
Proof: Suppose the eigenvectors
span a subspace V, and let W be the orthogonal subspace. For any v
in V and w in W we have
(v,Aw)=(Av,w)=0 since Av is a linear combination of eigenvectors so
is in V. Thus A can be restricted
to W, i.e. sends W to itself. But this restriction of A must have at
least one eigenvector in W, since any linear
transformation always has at least one eigenvector (since the characteristic
polynomial always has
at least one root). But this contradicts the fact that W is orthogonal
to the span of the eigenvectors, unless
in fact W is empty.
# If A and B are two observables with a common set of eigenvectors,
then they commute. The converse is
true and is the most important mathematical theorem of quantum mechanics:
If two observables commute there exists a basis of common eigenvectors.
Proof: Suppose A|i>=a_i|i>. Then A(B|i>)=B(A|i>)=a_i B|i>, so B|i> is an
eigenvector with the same eigenvalue. If a_i is nondegenerate then B|i>=b_i|i>
for some b_i, i.e., |i> is also an eigenvector of B.
If a_i is degenerate, the corresponding eigensubspace V is more than
one-dimensional. B maps this
subspace into itself, so B possesses an ON basis of eigenvectors in
this subspace, all of which
are also eigenvectors of A with (eigenvalue a_i). Hence we have found
a common eigenbasis of A and B.
# A complete set of commuting observables is a set of commuting
observables with the property that
in their common eigenbasis, each eigenvector has a distinct set
of eigenvalues. That is, no degeneracy
remains.
An example of such a complete set for a nonrelativistic particle with
spin-1/2 in a central potential
is {H, L^2, L_z, S_z}, with corresponding eigenvalues {E, l(l+1)hbar^2,
m hbar^2, m_s}.
One often uses these eiugenvalues to uniquely label the states as |E,l,m,m_s>.
Another example is the three commuting operators A_1,2,3 in the supplement
on nonlocality and the GHZ state. The eigenvalues of these operators are
1,-1, and the states are labeled by the eight choices of these
three eigenvalues. The Hilbert space is eight dimensional, so there
is clearly no remaining degeneracy.
Fri., 9/17:
# locations of a non-relativisitic particle are distinguishable, labeled
by real numbers, so there is a
hermitian position operator whose spectrum is the possible values
of the position,
X |x> = x |x>. Clearly <x|x'> = 0 if x is not equal to x', but what
about <x|x>? Define it so that
I = int dx |x><x|. Thus I |x'> = |x'> implies <x|x'> =
dirac delta function (x-x').
This is called continuum normalization.
The norm of |x> is <x|x> = delta(0) = infinity.
# The Dirac delta function delta(x) is zero when x is not zero
and infinity when x = 0 such that
int dx delta(x) = 1.
It can be defined as a limit of regular functions, the limit being
taken after the integral is performed.
# For any ket |v> = int dx |x><x|v>. <x|v> are the components
of |v> in the |x> basis. They are called the
position space wavefunction or just wavefunction.
# Momentum: historically, de Broglie reasoned that since E=hbar omega
(which began with Planck's oscillators and was then applied to photons
by Einstein),
according to relativity it should also be true that p=hbar k, and maybe
this should
apply to massive particles. That is, a wavefunction exp(ikx) should
have momentum hbar k.
This motivates the definition of a momentum operator p^ = -i hbar d/dx,
for which such a wave is an eigenfunction with eigenvalue p.
# A more fundamental point of view begins with the translation operator
T_a |x> = |x+a>.
T_a is unitary, and T*_a = T_(-a).
We can also express the action of T_a on the position operator: T*_a
X T_a = X + a.
As a function of a, T_a is simple: T_0 = I and T_a T_b = T_a+b,
which implies that T_a is of the exponential form T_a = exp(-iaK) for
some hermitian operator K,
the generator of translations. Expanding in a, T_a = 1
- iaK + O(a^2),
the linear term in a of the action of T_a on on X yields [X,K] = i.
The momentum operator is defined by P = hbar K,
hence the canonical commutation relation [X,P] = i hbar.
# Action of P in the X-representation: expand <x|T_a|v> = <x-a|v>
to fisrt order in a, find
<x|P|v> = -i hbar (d/dx)<x|v>.
# Conservation of momentum: If the system is invariant
under translations,
then translation commmutes with evolution, U T_a = T_a U.
Expand to linear order in a to find U P = P U.
For a momentum eigenstate |p> this implies P(U|p>) = p (U|p>),
i.e. the evolved state U|p> is still an eigenstate with momentum p.
More generally all matrix elements of P are conserved.
# Eigenstates of the translation and momentum operators are
the same
(with different eigenvalues of course):
P |p> = p |p>, and T_a |p> = exp(-ipa/hbar) |p>.
# <x|T_a|p> = <x-a|p> = exp(-ipa/hbar) <x|p>, so
<x|p> = c exp(-ipx/hbar).
This is the position space wavefunction of a momentum eigenstate
(or the conjugate of the momentum space wavefunction of a position
eigenstate).
# Normalization of momentum eigenstates: <p|p'> = c*c
int dx exp(i(p'-p)x/hbar).
Continuum normalization: <p|p'>
= delta(p-p') if c = (2pi hbar)^(-1/2).
Box normalization: If the particle
is confined to a periodic box of length L
the momentum eigenstates can be normalized to one , <p|p>
= 1 if c = L^(-1/2).
This is possible because the spectrum is not continuous in a box:
only a discrete set of momenta satisfying the boundary
conditions are allowed:
<x+L|p>=<x|p> implies exp(ipL/hbar)=1 so p = n 2pi hbar/L
for integer n.
# Fourier transformrelation between position and momentum
space wavefunctions:
<p|v> = int dx <p|x> <x|v> = int
dx exp(-ipx/hbar) <x|v>,
<x|v> = int dp <x|p> <p|v> = int
dp exp(+ipx/hbar) <p|v>.