# 2.2. Relation with classical mechanics¶

In order to see whether we can guess the structure of the Hamiltonian
for systems which have a classical analogue, we consider the time
evolution of a physical quantity . We assume that does not depend
on time *explicitly*. However, the expectation value of may vary in
time due to the change of the wave function in the course of time. For
normalised wave functions,

Using the time-dependent Schrödinger equation and its Hermitian conjugate, (note the minus sign on the left-hand side which results from the Hermitian conjugate) we obtain where is the commutator. We see that the time derivative of is related to the commutator between and . This should wake you up or ring a bell. Perhaps you have seen a derivation of the time derivative of any physical quantity in classical mechanics, based on the Hamilton equations of classical mechanics. These equations read Note that is expressed as a function of the generalised coordinates and .

As a side note, we provide a different form of this equation. Introducing the two-dimensional coordinates , we may write the Hamilton equations in the form where and

The time derivative of is given by
We see that this
equation is very similar to that obtained above for the time derivative
of the expectation value of the operator ! The differences
consist of replacing the Poisson bracket by the commutator and adding a
factor . Note that the commutator and the Poisson
bracket both are anti-symmetric under exchange of the two quantities
involved: It
seems that classical and quantum mechanics are not that different after
all. Could this perhaps be a guide to formulate quantum mechanics for
systems for which we have already a classical version? This turns out to
be the case. At this stage, we summarise the result by writing down the
time-evolution for and its classical version:
which shows the striking similarity between the two formulations. These
equations are both called *Liouville equation* -- the first is however
usually denoted as *quantum Liouville equation*.

As an example, we start by considering a one-dimensional system for
which the relevant classical observables are the position and the
momentum . Classically, we have
The second term in the expression vanished because
and are to be considered as *independent* coordinates. From this,
and from the factor occurring in the quantum evolution
equation above, we may guess the quantum version of this Poisson
relation: which should look
familiar (if it does not, please review the second year quantum
mechanics course). It seems that our recipe for making quantum mechanics
out of classical mechanics makes sense! Therefore we can now state the
following rule:

If the Hamiltonian of some classical system is known, we can use the same form in quantum mechanics, taking into account the fact that the coordinates and become Hermitian

operatorsand that their commutator relations are:

You can verify these extended commutation relations easily by working out the corresponding classical Poisson brackets.

In the second year, you have learned that
What about this
relation? It was not mentioned here so far. The striking message is that
this relation *can be derived from the commutation relation*. In order
to show this, we must discuss another object you might have missed too
in the foregoing discussion: the wave function written in the form
(for a particle in 3D). It is important to study the
relation between this and the state .
Consider a vector in two dimensions. This vector can be
represented by two numbers and , which are the components of
the vector . However, the actual values of the components depend
on how we have chosen our basis vectors. The vector is an arrow
in a two dimensional space. In that space, has a particular
length and a particular orientation. By changing the basis vectors, we
do not change the *object* , but we *do* change the numbers
and .

In the case of the Hilbert space of a one-dimensional particle, we can
use as basis vectors the states in which the particle is localised at a
particular position . We call these states .
They are eigenvectors of the position operator with eigenvalue
: It is
convenient to normalize the states as follows:
where
is the Dirac delta function. We now can define
:
that
is, are the 'components' of the 'vector'
*with respect to the basis*
. For three dimensions, we have a wave function
which is expressed with respect to the basis
.

In order to derive the representation of the momentum operator, , we first calculate the matrix element of the commutator: The last expression is obtained by having in the first term act on the bra-vector on its left, and on the ket on the right in the second term.

On the other hand, using the commutation relation, we know that This is an even function of , as interchanging and does not change the matrix element on the right-hand side. Since this function is equal to , we know that must be an odd function of .

Now we evaluate the matrix element . We recall from linear algebra that, since are the eigenstates of an Hermitian operator, they form a complete set, that is: where is the unit operator. Then we can write Now we perform a Taylor expansion around in order to rewrite : We then obtain The first term in the square brackets gives a zero after integration, as it is multiplied by , which was an odd function of . The second term gives where we have used the relation

We use the same relation for the second term. But then we obtain a term of the form in the integral over . This obviously yields a zero. The same holds for all higher order terms, so we are left with which is the required result.

Having obtained this, we can analyse the form of the eigenstates of the momentum operator: The states can be represented in the basis ; the components then are . We can find the form of these functions by using the eigenvalue equation and the representation of the momentum operator as a derivative: The first of these equation expresses the fact that is an eigenstate of the operator , and the second one follows directly from the fact that the momentum operator acts as a derivative in the -representation. Combining these two we obtain a simple differential equation with a normalised solution:

This allows us to find any state in the momentum representation, that is, the representation in which we use the states as basis states: This is nothing but the Fourier transform, the specific form of which we already encountered at the end of the previous chapter. The analysis presented here for a one-dimensional particle can be generalised to three or more dimensions in a natural way.