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\line{Physics 352 \hfill \today \hfill Handout B}
\beginsection How to sum over all states

 ``How do you sum over all configurations
in a quantum system?"  In class and in the book, the
answer is given: you should sum over a basis of quantum
states, giving a probability to each.  But another answer
appears equally plausible.  Why shouldn't we assign a
probability to each {\it linear combination} of states? 
We show below that this is un-necessary, and that it
suffices to assign probabilities to a basis of states
only.

	Suppose to the contrary that we have a basis of quantum
states $\left | j\right >$, but we have assigned a relative
probability $p_i$ to various {\it mixed} states
$\left | \psi_i\right >=\sum_j \alpha_{ji} \left | j\right >$.  We could
have many more mixed states $\left | \psi_i\right >$ than there
are basis states $\left | j\right >$.  The expectation value of
an arbitrary observable $A$ is then $$\expectation{A} =
\sum_i p_i \left <\psi_i \right | A\left | \psi_i\right
>,$$ up to normalization.  This may be rewritten in terms
of the basis states $\left | j\right >$: $$\expectation{A} =
\sum_i p_i \sum_j \alpha^*_{j,i}\sum_{j'}
\alpha_{j',i}\left <j\right | A\left | j'\right >.$$ This can be
expressed more compactly by $$\expectation{A} =
\sum_{j,j'} \rho_{j',j} \left <j\right | A\left | j'\right >,$$
where the matrix  $$\rho_{j',j}\definedas \sum_i p_i
\alpha^*_{j,i} \alpha_{j',i}.$$

In other words it is un-necessary to supply all the
$p_i$'s and $\alpha$'s individually; the combinations
$\rho_{j,j'}$ are sufficient to calculate the expectation
value of {\it any} observable $A$.  Thus they are
sufficient to calculate any measureable quantity.  Any
probability information beyond the values of the
$\rho_{j',j}$ are suspect.  This information may be
redundant, saying the same thing more than once.  It may
also be self-contradictory, violating the probabilistic
interpretation of quantum mechanics.  (Thus \eg it would
be self-contradictory to say that the probability $p_+$ for
a state $\left | +\right >$ was finite, while the probability $p_2$ for
the state $(\left | +\right > + \left | -\right >)2^{-1/2}$ was zero.)

The density matrix $\rho$ is evidently
Hermitian, in view of its defining formula above.  Thus
we may find a basis $\left | c\right >$ in which it is diagonal:
$$\tilde \rho = \sum_c \left | c\right >p_c\left <c
\right | .$$

It is certainly legitimate to assign independent and
arbitrary probabilities $p_c$ to the orthonormal states
$\left | c\right >$.  This is the prescription for quantum
configurations that we've been using all along.  The
above discussion shows that even if we try to be much more
general and assign probabilites to arbitrary mixed states,
it reduces to this same prescription in the end.

\end
	However, this discussion also reveals a shortcoming in
our class treatment of extremal and convexity properties
of the entropy $S$ in the quantum case.  We showed that
$S$ was extremal under variation of the $\{p_c\}$, but
what about variation of the basis states $\left | c\right >$?  The
answer is that it {\it is} extremal under variation of
the basis.  But the demonstration is somewhat technical,
and uses the same ideas we've discussed.  Those
interested in seeing this can look in Amnon Katz's book
on principles of statistical mechanics (Freeman 1967).