As we showed in the last
section, when the Hermitian
operators representing two
physical observables commute,
they share eigenfunctions.
But what happens when the
operators do not commute?
To understand this, we are
going to go back and look
again at the uncertainty
principle, but this time,
using the power of the linear
algebra we have introduced.
We're going to be able to draw
a very general conclusion
here, though it will involve a
rather mathematical proof.
This proof is straightforward,
but it itself does not offer
much physical insight.
However, the ultimate result
here has profound and broad
physical consequences.
First, we need to set up the
concepts of the mean and
variance of an expectation
value.
We're going to use more of a
statistical notation here.
A bar will denote the mean
value of a quantity A.
So imagine, statistically
speaking, we perform multiple
measurements on A. And we get
an average value A bar.
Now these are going to be
our quantum mechanical
measurements when we prepare the
system in the same state
again and again and try to
measure the quantity A. And
we'll get an average
value A bar.
And in bra-ket notation, for a
measurable quantity associated
with the Hermitian operator A,
then when the system is in the
state f that we keep preparing
it in to repeat our experiment
here, A bar will be the same
thing as what we've been
calling the expectation value
of A. And for a state f,
that's just this entity here.
f has a bra vector.
A is an operator.
And f is a ket vector.
So let's define a new operator
that we're going to call the
delta A. So this is a
mathematical operator.
It's related to the operator
we had for A.
And we're going to have this
operator associated with the
difference between the measured
value of A in any
particular experiment and
its average value.
So, that is, this new operator
we're going to define, delta
A, is just the operator A
minus the average value.
And to be strictly correct
here, we ought to have an
identity operator in here
because this is an operator.
But in a somewhat loose notation
we tend to use in
this kind of algebra, we don't
bother to put the identity
operator in.
So we should strictly
write it this way.
But in this algebra we mostly
won't bother to write that,
taking it as obvious it should
be there if we've got an
expression that's basically
an operator.
So we take this identity
operator mostly to be
understood in our algebra.
Now, note that this new
mathematical operator we've
created is also Hermitian.
A was a Hermitian operator.
A bar, which is just a number
times the identity operator,
is obviously also Hermitian.
So this entire new operator
here is also Hermitian.
Now, in statistics, variance is
the mean squared deviation
from the average.
So to examine the variance of
this quantity A that we're
going to keep on measuring--
statistically, we keep repeating
our experiment,
setting up in the same state,
measuring some value for A. We
get different answers
in general each
time we do that possibly.
And we're looking at
the variance of the
answers that we get.
So we're going to be examining
the expectation value of a new
operator delta A squared.
And let's just check what
this does for us.
So we're going to expand our
usual arbitrary function f on
the basis of the eigenfunctions
psi i of A, the
operator representing the
physical quantity.
So, as usual, we perform
our expansion here.
And we're going to formally
evaluate the expectation value
of this operator when the system
is in the state f to
check out what it does for us.
So we have the expectation value
of this operator delta
A, all squared.
By definition, if we're
in the state f, it's
just this form here.
And we can expand f in
both cases here.
Here's our expansion of
f that we wrote down.
Here's the expansion of the
bra vector version of f.
Of course, we use a different
index in here.
And here's our delta A squared
operator written out.
Well we can operate with the
first one of these, the one on
the right in this square here.
Operating on this function
just gives us A j the
eigenvalue minus A the number
here times psi j.
That's straightforward.
And then we can do this again.
We can operate with A minus
A bar on this vector here.
We can break it into parts.
So A minus A bar operating on
psi j just gives us, again, Aj
minus A bar times psi j.
So this Aj minus A bar here
becomes squared now.
And, of course, because the
psi i and the psi j are
members of this orthonormal set
here, we only have terms
surviving in this sum over i
and j when i is equal to j.
By the orthonormality of these,
that gives us 1 in each
case when i and j are the
same, and 0 otherwise.
And so we've just shown that the
expectation value of this
operator delta A, all squared,
is the sum over all the
members of the set here, with
these ci squareds, the modular
squares of them, the expansion
coefficients, times the
eigenvalue, minus the average.
Answer just a number,
all squared here.
So this will be the average
value of this squared quantity
in our statistics.
Now, because as we said these
modular squareds of ci are
just a probability that the
system is to be found on
measurement to be in this
state psi i, then this
quantity here for that state
simply represents the squared
deviation of the value of the
quantity A that we are
measuring this time from
its average value.
And these are the probabilities
that we get
these various answers when we
do this experiment here.
Then by definition, this
quantity here is just this
quantity on the right.
In other words, we have indeed
found that the expectation
value of this operator delta A
squared is indeed the mean
squared deviation.
So we were looking for the
average mean squared deviation
that we get.
So every time we do our
experiment, we prepare it in
the state f.
We repeat it.
We take the answers we get for
A here, the measured value.
And this operator's expectation
value, or average,
here is indeed the expectation
value of that mean square
deviation from the average.
In statistical language, of
course, this quantity is
called the variance.
And the square root of the
variance is known as the
standard deviation.
And it is written like
this, obviously.
And we can call it delta
A if we want to.
It's just a number here.
The square root of the average
value of delta A squared.
And in statistics, of course,
the standard deviation gives a
very well-defined measure of the
width of some statistical
distribution.
So it's a good measure of width
in the answers that we
are getting.
We can also consider some other
quantity B, some other
physical quantity associated
with the Hermitian operator B.
And similarly, we can write
the average value of B
that we would see.
That's just the same thing
as the expectation value.
And we use this form here,
if we're in the state f.
And with similar definitions,
we could define
a delta B all squared.
And we'd get the same answer
for what that meant.
So it would also be giving
us the variance of the
distribution.
So we have ways now of
calculating the uncertainty in
the measurements of the
quantities A and B when we do
repeated experiments, when
our starting system is
in the state f.
And we can use these results
in a general proof of the
uncertainty principle.
Suppose then, we've got two
Hermitian operators, A and B,
that do not commute
with one another.
And therefore, from our
commutator, we have a
commutation rest some operator
C. So in our above definition,
the commutator of A and B
is equal to i times C.
So now we get into a purely
mathematical proof here.
And to start that, we're
going to choose some
arbitrary real number.
This could be any real
number that we like.
We'll call it alpha.
And we're going to define
this quantity--
this is just a number--
G of alpha.
So alpha is just a coefficient
in here.
And this is what we're going to
define this quantity G of
alpha to be.
So we've got alpha times delta
A here, the operator minus i
times delta B operating on f.
And we're putting that whole
thing in bra form.
And similarly, over here,
we have it in ket form.
Now, we've written it in this
form here, in ket, to
emphasize that we mean
this vector.
All we're saying is that the
result of alpha times delta A
minus i times delta B operating
on f is just a vector.
And this expression here is just
the inner product of that
vector with itself.
So mathematically, this whole
thing here is just a vector.
It's an operator operating on
a vector, which gives us
therefore a vector.
So this whole thing here is the
inner product of a vector
with itself.
And therefore, this must be
greater than or equal to 0.
Now this is just
all mathematics
here for the moment.
But this is a mathematically
correct statement.
And we're writing it this way,
for the moment, just to
emphasize that this thing here
is just a vector, and this
thing here is just a vector.
So, as we said, it has an inner
product with itself, and
that has to be greater
than or equal to 0.
So, if we write this out here,
as we just have from a
previous view graph, we know
this has to be greater than or
equal to 0.
And now we are going to
work with this here.
The Hermitian adjoint of all of
this stuff here is just the
sum of these two Hermitian
adjoints inside here.
So the Hermitian adjoint of i
times delta B would be minus i
times delta B Hermitian
adjoint.
And with the minus sign in
front here, that ends up
flipping those signs around.
And by Hermiticity of these
operators, A and B of course,
that means the delta A
is also Hermitian.
That delta A dagger is just the
same thing as delta A. And
similarly delta B dagger is just
the same thing as delta
B. So we make that change
here in the algebra.
And now we multiply these out.
So delta A, with an alpha in
front of it, times delta A,
with an alpha in front
of it, is just alpha
squared delta A squared.
And similarly, for i delta B
times minus i delta B, that's
just also delta B squared.
And then we have two
other terms here.
We have i delta B times
alpha delta A. So
that's this term here.
We've got our minus signs here
to straighten all that out.
And we have alpha delta A times
minus i delta B. That's
this term here.
Well this here is just the
commutator of delta A and
delta B. And we know that the
commutator of these, by
definition, we're going to
have that as iC here.
So that means that we can
rewrite this in the following
form down here.
Therefore, this whole expression
here we can rewrite
as the following, because f
delta A squared f is just the
average value of delta A
squared, or variance.
And we've got an alpha
squared here.
And similarly, if delta B
squared f is just the average
value of delta B squared, and f
alpha Cf is just alpha times
the average value we would see
for the quantity C. And that,
by definition, is the
expectation value of Cf in the
bra form, C in the operator
form, and f in
the ket form as usual.
Now, we're going to do a
rearrangement of all of these
terms here.
It's not going to be an
obvious rearrangement.
But it will work.
And it's quite easy to check
that it does work.
So we're making this non-obvious
rearrangement.
All of this set of terms here
is exactly the same as this
set of terms here.
So how does that work?
Well notice here, we've written
alpha plus C bar over
2 delta A squared bar.
And we squared all of that.
Well obviously, that gives
us an alpha squared.
And that alpha squared is this
term here, because we're
multiplying out in front.
And it also gives us a term that
looks exactly like this,
but with a plus sign
in front of it.
That's the square of
this term here.
And because we've added that
term in, we are canceling it
out by putting minus that
same term over here.
And then the rest of this square
here simply turns into
the alpha C bar.
And you can check that out.
We're going to get 2
times C bar over 2
delta A, all squared.
And because we've got the delta
A, all squared, the bar
here out in front, that
just is the equivalent
of this term here.
So we have rearranged
everything here into
this form down here.
And of course, as before, this
still must be greater than or
equal to 0.
All of this, as we said, is
greater than or equal to 0,
must be true for arbitrary
alpha.
That was one of our assumptions
up front.
Alpha could be any real
number that we liked.
And we could still see this
statement is true.
So it's true specifically if
we happen to choose alpha
equals minus C bar over
2 delta A squared bar.
And that makes this term
here just be 0.
So therefore, what we're left
with, from all of these terms
here, we can rearrange
them very slightly.
We're left with delta A squared
bar times delta B
squared bar is greater than
or equal to C bar
squared, all over 4.
Or equivalently, delta A times
delta B is greater than or
equal to mod C bar over 2.
So for two operators A and B
corresponding to measurable
quantities A and B, and for
which the commutator of A and
B is equal to i times some
operator C, in some state f
for which C bar-- that's the
expectation of C. It's written
out like this, of course--
we have the uncertainty
principle that delta A times
delta B is greater than or equal
to the modulus of C bar
all over 2.
This is our general statement of
the uncertainty principle.
And here delta A and delta B are
the standard deviations of
the values of A and B that
we would measure.
So this uncertainty principle
is quite general for any
operators A and B, where we can
evaluate their commutator.
And we see, therefore, that we
get an uncertainty principle
if the operators
do not commute.
And we've managed to prove
this without presuming
Gaussians or any other
specific form for
distributions.
This is generally true.
Only if the operators A and B
commute, that is, A and B have
a commutator 0, or strictly
speaking 0
times an identity operator.
Or possibly if they do not
commute, they have a
commutator rest.
But we're in a state f for which
this expectation value
of the commutator rest
is equal to 0.
Only in those cases is it
possible for both A and B
simultaneously to have exact
measurable values.
So we have a very
general result.
Whether or not the operators
associated with two physically
measurable quantities commute
with one another determines
whether or not there's
an uncertainty
principle between them.
This is a core concept
in quantum mechanics.
Indeed, one can take the view
that it's this absence of
commutation between such
operators that is the real
mathematical distinction between
the classical and
quantum worlds.
In the mathematics of
classical mechanics,
attributes like position and
momentum and energy and
angular momentum are represented
by numbers, or
ordinary vectors.
The multiplication of numbers
commutes, as does the dot
product of vectors.
But for the quantum mechanical
view, we have introduced
operators for these attributes,
and the
multiplication of these does
not in general commute.
Whenever that computation is
absent, we have an uncertainty
principle for those
attributes, a very
non-classical idea, at least
for those particular
attributes.
Some people even take the view
that it's the postulation of
these so-called commutation
relations that's the real core
set of postulates of
quantum mechanics.
That this is the real difference
between the
classical and quantum
descriptions.
