.ou9
.lm 10 rm 70 ju 1 eq 1 ls 4
\ce\TWO EXAMPLES IN NON-COMMUTATIVE PROBABILITY \ce\
\ce\Dror Bar-Natan\ce\
\ce\ Department of Mathematics\ce\
\ce\Tel Aviv University, Tel Aviv, Israel\ce\
f"Current address: Department of Mathematics, Princeton University,
Princeton, N.J. 08544."
.pn
Abstract: A simple non-commutative probability theory is presented, and two
examples for the difference between that theory and the classical theory are
shown. The first example is the well-known formulation of the Heisenberg
uncertainty principle in terms of a variance inequality and the second example
is an interpretation of the Bell paradox in terms of non-commutative
probability.
.ej
1.INTRODUCTION
We shall present here a simple yet representative version of the theory of
non-commutative probability, including two examples of the difference between
that theory and the classical probability theory. The first is a precise
formulation of the Heisenberg uncertainty principle, while the second
constitutes a strong indication for the existence of random phenomena in
nature explainable only by assuming that the probability space in which we
live is non-commutative.
2. THE CLASSICAL PROBABILITY THEORY
Definition: A classical probability space is a triple (X,B,p) consisting
of:
X- a collection of points,
B- a subcollection of the collection of all functions f:X%{0,1%} satisfying
some simple closure properties,
p- a probability measure on (X,B).
A random variable (real and bounded) on (X,B) is a real bounded
measurable function f:XR (we shall omit the definition of measurability).
The distribution of the variable f on (X,B,p) is the unique probability
measure pf on R satisfying:
\ce\ X q(f(x)) dp(x) = R q(x) dpf (x)\ce\
for every real polynomial q.
The expectation E and the variance V of a random variable f are defined
by:
\ce\ E(f) = X f(x) dp(x) = R x dpf (x) \ce\
\ce\ V(f) = X f(x) dp(x) - (X f(x) dp(x) ) = R xdpf (x) -(R
xdpf (x)).\ce\
The joint distribution of a pair of random variables f,g is the unique
probability measure pf%,g on R satisfying
\ce\ X q(f(x),g(x)) dp(x) = R q(x,y) dpf%,g (x,y)\ce\
for every real polynomial in two variables q(x,y).
3. A NON-COMMUTATIVE PROBABILITY THEORY1%,2%)
Definition: A non-commutative probability space is a triple (H,P,w)
consisting of:
H- a Hilbert space,
P- a subcollection of the collection of all orthogonal projections in H
(henceforth we shall always take the full collection, and thus we shall not
list the simple closure properties that P must satisfy),
w- a unit vector in H.
A random variable (real and bounded) on (H,P) is a bounded self-adjoint
linear operator F on H having a certain measurability property relative to
P. In the case where P is the collection of all projections, there will
be no further restriction on F.
The distribution of the variable F on (H,P,w) is the unique probability
measure pF on R satisfying
\ce\ = R q(x) dpF (x) \ce\
for every real polynomial q (the existence of this measure is exactly the
spectral theorem).
The expectation E and the variance V are defined as in the classical case:
\ce\ E(F) = R x dpF (x) = \ce\
\ce\ V(F) = R xdpF (x) - (R x dpF (x) ) = - \ce\
By a theorem of von Neumann, the joint distribution pF%,G of a pair of random
variables F,G,
\ce\ = R q(x,y) dpF%,G (x,y) \ce\
exists iff [F,G] = 0.
By taking normal instead of self-adjoint operators, we can extend our
theory to complex random variables, and by taking unbounded instead of bounded
operators, we can extend our theory to unbounded random variables.
Classical probability theory is contained in the non-commutative theory:
If (X,B,p) is a probability space, take H=L(X,p), w to be the constant
function 1, and represent every random variable f:XR as a product operator on
H:
\ce\ Fu=fu \ce\
These operators commute with each other, and the equality of the corresponding
measures pF , pF%,G with the classical measures pf , pf%,g is easily seen.
4. FIRST EXAMPLE: THE HEISENBERG UNCERTAINTY PRINCIPLE1%)
Question: Does there exist a pair of random variables f,g whose product of
variances is bounded from below independently of the probability measure?
The classical answer is trivially NO. Take the probability measure to be
the Dirac measure on any point of X, and then V(f)=V(g)=0 for every pair of
random variables f,g. For the non-commutative case, however, we have the
following theorem:
Theorem: Let F,G be any pair of random variables on H. The following then
holds for every unit vector w in H:
\ce\ V(F) V(G) 1/4 |E([F,G]) | \ce\
Proof: See e.g. ref 10.
In quantum mechanics, a vector w in L(R) represents the state of a
quantum particle, the operator (Fu)(x) = xu(x) represents the random variable
whose distribution is the position of the particle, and the operator
\ce\ Gu = -i /x u ( is the Planck constant) \ce\
represents the random variable whose distribution is the momentum distribution
of the same particle. The identity [F,G] =iI now makes our theorem the
Heisenberg uncertainty principle:
.lm 15 rm 65
"In any state of a quantum particle:
(variance of position)x(variance of momentum) 1/4 "
.lm 10 rm 70
5. SECOND EXAMPLE: THE BELL PARADOX3,4,5,9%)
Phrases like,"choose at random one of three given random variables", have
no direct meaning in classical probability, but they can easily be given a
meaning by defining a new probability space consisting of the disjoint union
of the three original spaces and some weighted sum of the three original
measures.
The non-commutative analog of this construction is simple; you just take
the direct sum of the three original spaces, some weighted sum of the original
unit vectors, and the rest is clear. In the following, we shall use such
phrases with no additional comments. We can now ask a question that separates
between classical and non-commutative probability:
Question: Does there exist a probability space with six distinguished
random variables A1%,2%,3 , B1%,2%,3 that get only the values %{1%} and
satisfy:
a) For each j the distribution of Aj - Bj is the Dirac measure on 0.
b) Choosing at random (with probability 1/3 , 1/3 , 1/3 ) one of the Aj 's,
and independently choosing (again with probability 1/3 , 1/3 , 1/3 ) one of
the Bk 's, we get
\ce\ p(Aj =Bk ) = 1/2 ? \ce\
Classical answer: NO.
Proof. Condition (a) means that Aj = Bj for every j, so that we have to
calculate p(Aj =Ak ), where j and k are independent and uniform. We shall
separate cases according to the possible values of the triple (A1 ,A2 ,A3
):
\ce\p(Aj = Ak |A1 =A2 =A3 =0 or A1 =A2 =A3 =1) =1 \ce\
\ce\ p(Aj =Ak | otherwise, e.g. A1 =A2 =0, A3 =1) = 5/9 \ce\
and hence p(Aj =Bk )=p(Aj =Ak ) 5/9 , contradicting (b).
Non-commutative answer: YES. The above argument fails because the triple
(A1 ,A2 ,A3 ) does not necessarily have a joint distribution.
Construction: Let H =R be the two dimensional Hilbert space with the
standard basis i=(10 ), j=(01 ). Define on H
\ce\ St = (cos2tsin2tsin2t%-cos2t ) \ce\
Now, take H= H O H, At = St O I, Br = - I O Sr and take the measure
\ce\ w = 1/2 (iOj-jOi). \ce\
At commutes with Br, so they have a joint distribution. We shall try to
calculate it. Take any t,r in R. At first,
\ce\ 2E(At) = \ce\
\ce\ = * + = cos 2t - cos 2t =0,\ce\
and, similarly, E(Br ) =0.
Now, let us calculate the correlation between At and Br :
\ce\ 2E(At Br ) = 2E(-St O Sr ) = =\ce\
After a straightforward calculation, one obtains
\ce\ =1/2 cos 2(t-r)\ce\
and hence
\ce\ E(At Br ) = 1/4 cos 2(t-r).
The eigenvalues of St are 1, and by that we get that the joint spectral
measure, e.g., the joint distribution of At ,Br , is supported on %{1%}
x%{1%}, and together with the correlation and expectations found we get that
the joint distribution is
\ce\1/2 (cos(t-r)sin(t-r)(sin(t-r)cos(t-r)))\ce\
In particular, p(At =Br ) = cos(t-r).
Define A=A,B=B,A=A,B=B,A=A,B=B, and now
(a) is immediate: p(Aj =Bj ) =cos0 =1. (Alternatively, one can calculate
the distribution of Aj - Bj directly by noticing that (At -Bt )w=0.)
Part (b) is easy too:
\ce\ p(Aj =Bj ) = 1/3 cos0 + 2 1/3 cos 120 = 1/3 + 1/6 = 1/2.\ce\
6. CONCLUDING REMARKS
A FACT:6%,7%,8%) Physicists have constructed some complicated machinery,
and made the following experiment (fig. 1; see also ref. 9): A single source
sends two signals to two receptors. At the time when the signals are
received, the two observers immediately, independently,and randomly choose and
press one of the keys 1,2,3 in front of them. One of the lights + or - now
goes on immediately on each machine. The time period between the observers'
decision on which key to press and the lighting of one of the lights is so
short that no information can be transferred between the two machines due to
the light speed limitation.
THE OBSERVED RESULTS:
.lm 15 rm 65
a) In 1/3 of the experiments both observers pressed the same key (on different
machines) and the same light went on (again, on the different machines).
b) Disregarding the keys and looking only upon the lights, we find each of the
possibilities ++; +-; -+; -- with equal probabilities.
.lm 10 rm 70
The observers are known and respectable, and we have no reason to mistrust
them.
POSSIBLE EXPLANATIONS:
1) The probability space in which we live is non-commutative. (with spacelike
separated spacetime points observed only by using commuting observables -
notice that the A's and B's of our example commute).
2) Interaction quicker than the transmission time for light is possible.
.lm 15 rm 65
Notice: It is not possible to transfer information using the machinery
described above, and therefore there is no direct contradiction with the
theory of relativity. It seems that the only way to correlate the lights is
by transferring information, and hence it seems that there is an indirect
contradiction with relativity.
.lm 10 rm 70
If we want to choose the second explanation, and still keep the theory of
relativity, we must specify what kind of interaction quicker than the
transmission time for light is allowed. A natural candidate for an answer,
which lies on physical grounds and does not contradict relativity is:
.lm 15 rm 65
"Interaction quicker than than the transmission time for light is allowed only
if the same situation can be described without interaction using non
commutative probability, with commuting observables attached to spacelike
separated points."
.lm 10 rm 70
It seems simpler to choose the first explanation.
The above experiment has many other explanations, and they all require us
to change something fundamental in our perception of the universe (the
direction of time, the uniqueness of time, the order between cause and effect,
the uniqueness of consciousness,...).
ACKNOWLEDGEMENTS
I am grateful to Ehud Ben-Reuven who introduced me to the Bell paradox,
and to Professor L.P. Horwitz for more than what I can write.
REFERENCES
1. S.P. Gudder, Stochastic Methods In Quantum Mechanics, North Holland,
Amsterdam (1979).
2. R. Jajte, Strong Limit Theorems In Non-Commutative Probability, Lecture
Notes in Mathematics 1110, Springer, Berlin (1984).
3. A. Einstein, B. Podolsky and N. Rosen, Phys. Rev. 47, 777 (1935).
4. J.S. Bell, Physics 1, 195 (1964).
5. J.S. Bell, Rev. Mod. Phys. 38, 447 (1966).
6. A. Aspect, J. Dalibard and G. Roger, Phys. Rev. Lett. 47, 460 (1981).
7. A. Aspect, J. Dalibard and G. Roger, Phys. Rev. Lett. 49, 91 (1982).
8. A. Aspect, J. Dalibard and G. Roger, Phys. Rev. Lett. 49, 1804 (1982).
9. N.D. Mermin, Phys. Today 39, No. 4, 38 (1985).
10. G.W. Mackey, The Mathematical Foundations Of Quantum Mechanics,
Benjamin, New-York (1963), pp 78.
.ej
*