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\author{Gabriel C. Drummond-Cole}
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\title{RTG Seminar\\October 23, 2006\\
Leon Takhtajan}
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\maketitle

The set of mathematicians can e divided into two disjoint subsets.  One subset understands classical mechanics, and the second subset understands some quantum mechanics.  Many excellent mathematicians fall into the first subset.

Before we discuss the basic ideas of quantum mechanics, we'll start with some basic ideas, and then move forward to classical mechanics and then quantum mechanics.  We should have
\begin{itemize}
\item A set of states
\item observables
\item dynamics (time evolution).
\end{itemize}
In classical mechanics these are, well,
We'll start with a phase spaces, which is synonymous with a symplectic manifold, or, more generally, a Poisson manifold with invariant volume form.  Denote this by $M.$  We will have $\{,\}$  The set of states will be probability measuers on $M.$  The observables will be $C^\infty(M),$ and the algebra of classical observables will be the Poisson algebra, a commutative algebra with unit and Lie algebra structure given by the Poisson bracket.  The bracket should be a derivation of the multiplication, and this should satisfy the Liebnitz rule.  Denote the observables by $A.$

Any function on the phase space is an observable in the following sense.  Denote the set of probability measures $S.$ 

What about measurement?  Every observable in every state has a value, given by integration, $S\times A\to \rR,$ namely $(\mu,f)\mapsto E_\mu(f)=\int_M fd\mu.$  A typical example would have $\mu$ the Dirac measure at $x,$ for some $x\in M.$  Then $x=(p,q).$  These are called pure states.  The number of particles is very small.  We can have all other states, called mixed states.  These are used in statistical mechanics.  The number of particles is about $10^{23}$ or so.  This is the classical version of dealing with many particles, statistical mechanics.

Mechanics has a finite number of particles.  Just for comparison purposes, taking $N\sim 10^{69},$ this is the number of atoms in the universe.  Another number is $10^{500},$ that is the number of all possible string theories.  This is not very serious, but these theories explain so many atoms.  But we are concentrating on $N\sim 10^{23},$ or maybe perhaps $N\sim 3.$  State means the system is in some state.  The measure represents the state of the system.  The system may be one particle, and that particle may be at $(p,q),$ so that's the state.  Now instead of assigning an expectation value, we can assign an probability measure on $\rR$ so that $0 \le \mu_f((a,b))\le 1$ and this is the probability at the states $\mu$ that the value of $f$ is in $(a,b).$

So then $\mu_f((-\infty,\lambda))=\int_{\{f(x)<\lambda\}}d\mu$ so that $\int_M fdu=\int_\rR \lambda d\mu_f(\lambda).$

One more remark before dynamics.  How do you characterize pure states using this language?  There is another notion, how the values of the probability measure differ from those of the expectation value.  The dispersion variance is $E_\mu((f-E_\mu(f))^2)\ge 0.$  There is a fact which states that $\mu$ is pure if and only if its dispersion $\sigma^2_\mu(f)=0.$  This theory is dumb, it doesn't use $\{\}.$  Now let's discuss dynamics.  There are two descriptions, Hamilton's description and Liouville's description.  Hamilton's is the one we use when we have few particles.  So $\frac{d\mu}{dt}=0$ for $\mu\in S.$  This all depends on $(M,\{\},H),$ where $H$ in $A$ is a Hamiltonian function (this is abused terminology).

Then observables change in time according to $\frac{df}{dt}=\{H,f\}$ for $f\in A.$  So $f|_{t=0}=f.$  In order that this makes sense, we need that $X_H=\{D,\cdot\}$ such that this is complete, namely that there is some one-parameter family $g^t$ which hits $H$ at time zero. This is the phase flow.  This can be written $f_t(x)=f(g^t\cdot x).$

So the states are fixed, the observables change, and $E_\mu(f_t)=\int_M f(g^t x)d\mu(x).$

As an example take $M=\rR^2$ with $p$ and $q$ and $\omega=dp\wedge dq.$  

Liouiville, now.

You allow states to depend on time $\frac{df}{dt}=0$ for $f\in A.$  Let's assume that the states, for simplicity, suppose that, well I need the phase space to have an invariant volume form, which I'll call $dx.$

we can take $\mu=\rho(x)dx,$ where $p(x)$ is smooth and $\rho$ is $d\mu/dx$

Then $\frac{dp}{dt}=-\{H,\rho}$ with $\rhoSt(x)=\rho(q^{-t}x).$

Then $E(f)$ is $\int_M f(x)\rho(g^{-t}x)dx.$

This is the structure of classical mechanics.  So we can introduce the theory of classical mechanics with the followming generalization.

Let's see what, well, a Poisson algebra is the algebra of function on some manifold.  Then there is a theorem that every $C^*$ algebra, semisimple, is the spectrum of some algebras.

Let's switch from this commutative algebra to the noncommutative algebra of operators on a Hilbert space.

Let me draw the same sort of, what can I, I will delete Hamilton and Liouville and replace them with different descriptions.

I will start with Dirac-von Neumann ``axioms.''

We will define a quantum mechanical system.  \begin{enumerate}
\item The states are, well, pure states were parametized by points.  A quantum mechanical system is described as follows: First, it is a separable complex Hilbert space.
This space, in fact, could be finite dimensional.  
\item The observables are self-adjoint operators on $H.$  What is a self-adjoint operator on a Hilbert space?  Well there is a sscalar product.  But what is a self-adjoint operator?

[I went to the restroom]

So $A$ is self-adjoint if $A=A*.$  Now $A$ is symmetric if $D(A)\subset D(A^*)$ and $(Ax,y)=(x,Ay)$ for $x,y\in D(A).$  Here $D(A^*)$ is $\{y\in H$ such that there exists $\eta\in H$ such that $(Ax,y)=(x,\eta)$ for all $x\in D(A)\}.$
\item 
The set of states consists of all bounded self-adjoint (this is automatic) positive definite trace class operators $M$ on $H$ with $Tr\ M=1.$
\end{enumerate}
So $L^2(\rR)\cong L^2(\rR^n)$ as spaces.  But when you add dynamics these won't be the same.

[Am I supposed to throw away my old idea about the waves being the states?]

Wait a second, wait a second.  So $\Psi\in H,$ with $||\Psi||=1,$ and $M=P_\Psi,$ which is the orthogonal projection on the one-dimensional subspace spanned by $\Psi,$ is then a pure state.  So then what is trace one?

Bounded $M$ on $H$ is of trace class if for every orthonormal basis $\{e_n\}$ the series $\sum |(Ae_n,e_n)|<\infty.$  Then $Tr\ M=\sum (Me_n,e_n).$  This is the operator trace.  If $M$ si self adjoint, then its eigenvalues will sum to the trace.

Now the most difficult part is, you have a state and an observable, what does it mean to measure the observable with the state?

Given an observable $A$ and a state $M,$ we want to define the expectation value $\langle A|M\rangle$ which might be in $\rR.$  We want to define $\mu_A,$ the probability measure on $\rR$ such that $\mu_A((a,b))$ is in the interval $[0,1],$ and have this be the probability that $A$ takes the value in $(a,b).$

Here we have two operators.  So how do we get a number?  We could take $Tr\ AM$ if this trace exists.  This doesn't explain how to construct the probability measure out of nothing.  We have no measure but if the operator is self-adjoint there is a measure from von Neumann's spectral theory.  The spectral theory, let me say it in this way, 
\begin{thm}(von Neumann)\\Let $A$ be a self-adjoint operator on $H.$ Then there is a projection valued measure on $\rR, \ \pPSA$ such that the following holds: (this is $E\in \Bb(\rR)$ means that $\pP(E)$ is a projection operator, $\pP^2=\pP,\ \pP^*=\pP.$  All of these projection operators commute, and in fact define $P$ on the intersection, $P(E_1)P(E_2)=P(E_1\cap E_2).$  So the domain of $A$ is the collection of vectors such that the $\int_\rR \lambda^2 d\pP_\A(\lambda)\varphi,\varphi)<\infty.$

\item Define $P(\lambda=P((-\infty,\lambda)).$ Then for $\varphi\in H,$ consider  the real valued distribution $(P(\lambda)\varphi,\varphi)$, which is left semicontinuous and has $\lim_{\lambda\to\infty\}p_\varphi(\lambda)=1.$

\item We can write $A_\varphi=\int_{-\infty}^\infty \lambda dP_A(\lambda)\varphi,$ where this means in the strong operator topology, $\varphi\in D(A).$

\item You can write $f(A)\varphi=\int_{-\infty}^\infty f(\lambda)(dP_A(\lambda)\varphi,\varphi).$

So every self-adjoint operator has a projection valued measure, and conversely given a projection valued measure you can get a self-adjoint operator. 
\end{thm}
Why is this useful?  This is extremely difficult to compute.  Let me just write one more thing.  Let me do the finite dimensional case.  Let $A$ be $n\times n$ with $\lambda_i$ eigenvalues.  Then $P_A(\lambda)=\sum_{\lambda_i<\lambda} P_{e_i}.$  Then $$A=\int \lambda dP_A(\lambda)=\sum \lambda_i P_{e_i}.$$

Everything is included, all possible spectra, in von Neumann.  To finish the discussion, finally we would like to define the measure $\mu_A.$  So this is given by the Born-von Neumann formula, $\mu_A(E)$ for $E$ Borel is defined as $Tr(P_A(E)M).$  This is trace class because it is the product of a bounded operator with a trace class operator.

This is a logical stop, in a couple of minutes.  Let me just say, $\mu_A((-\infty. \lambda))=Tr(P_A((-\infty,\lambda)M)$ in classical mechanics.  So $\mu_f((-\infty,\lambda)),$ if you recall, was $\int_{\{f(x)<\lambda\}}d\mu.$

There are subtle things about measurement and quantum mechanics.  The measurement is something, there is much more philosophy.  We have the same problem in classical mechanics.  We should change the behavior of the ball by looking at it.  In quantum mechanics when you observe, you change the parameters.  These are virtual potential observations that have not actually been made.  This takes us to an application, once you make an observation.  I avoid this paradox, collapse of the wave function.

Maybe Martin can explain this better.

[Most physicists come on two sides.  There are a few who spend a lot of time thinking about this and do not much else, and the vast majority who take your attitude, use quantum mechanics to get numbers, and as a prescription to doing that seem well-defined.  I think that the other attitude has not led to very much\ldots productive things.  This means that one merely needs to be a little careful and precise in applying the mathematical formalism correctly.  We have a way to extract the numbers from it, and for most physicists that's enough.

Planck's constant is small so we don't grow up with an intuitive grasp of quantum mechanics.  That's okay, if we can exract numbers, we should be satisfied.  That is an extreme perspective.]

[How about a break until five after?]

[I did not return.]













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