Download A Primer on Quantum Mechanics and Orbitals

The Free Particle Caged: The Particle in a Box
Motivation:
We have discussed free particle wavefuctions of the form:
 k x  Ak eikx  Ak cos(kx)  i sin (kx)
where A is a normalization constant and the second equality holds by the so called Eulers relationship. This
wavefunction describes a particle free to move anywhere on the x axis. That is, the particle is not bounded
by anything. We already know that it is an eigenfunction of the momentum operator:
pˆ x  k x   i
d
d ikx
ikx
ikx
2
ikx
Ak e  i Ak
e  i Ak ik e  i k Ak e  k k x 

dx
dx
Mathematical Interlude:
The derivative of any exponential function, f(x) = e ax is always the following
d x
x
e   e ; where  is a cons tan t

dx
or putting it more succinctly
pˆ x  k x   ( k) k x  ; where k 
h 2
 momentum (DeBroglie )
2 
What is also true is that this function k is also an eigenfunction of the Hamiltonian. Remember the
Hamilton operator is composed of two parts, the kinetic energy operator and the potential energy operator.
The Hamiltonian operator can then be seen as the operator, i.e. mathematical procedure, that give the total
energy of a particle. Because classically, the kinetic energy of a particle is the square of the momentum,
p=mv, divided by two times the mass of the particle, we make the following association between the
classical result and its quantum mechanical equivalent (see the previous handout).
p2
pˆ 2
1 
d 
d  i 2  d  d   2  d 2  ˆ



i
i

  
 
  Ek
2m
2m 2m  dx  dx 
2m dx dx  2m dx 2 
So, this expression above gives the kinetic energy operator. Now because this is a free particle, there is no
potential acting on it. So the potential energy opeator V(x)=0 or all values of x. Hence the Hamiltonian
operator for the free particle is:
 2  d 2 
 2
 k 2
ikx
2
ikx
ˆ
ˆ
ˆ
ˆ


H k (x)  Ekin  V (x)  k (x)  Ekin k (x) 
(ik) Ake 
 x 
 A e  
2m dx 2  k
2m
2m k

or again more succinctly:

Hˆ  k x   Ek k x 
where the eigenvalues are Ek 
 k 2
2m
and correspond to the kinetic energy and also the total energy of
the free particle.
Mathematical Interlude:
Whenever you have two consecutive derivatives, the meaning of the operation is to take the
derivative of the derivative. This is called the second derivative So extending our previous interlude where
we took the first derivative of the function f(x)=ex , the second derivative looks like
 d 2   x  d  d  x  d  x
 d  x
2 x

 e    e    e   e
 2 
e  
dx dx 
dx 
dx 
dx 
Now we know there are some serious drawbacks to this description of our ‘free particle.’ First
among these is the total indeterminacy of the position of the particle. We know the momentum ( and
therefore the kinetic energy) of the particle exactly. This is what we mean when we say that the
wavefunction k(x) is an eigenfunction of the momentum (kinetic energy) operator. Nevertheless, because
the positon of the particle is given by the probability density
 k * k  Akeikx  Akeikx  Ak eikx Ak eikx   Ak2e ikxeikx  Ak2
*
you can see that the probability density is constant and the same everywhere in space (the constant Ak does
not depend on x because it is........well, a constant). So, we are equally likely to find the particle anywhere
in space. This is also seen if you plot the real and imaginary parts of the wavefunction and see that the
wave spreads uniformly in space
So, our solution earlier in the lecture was to combine two ‘plane waves’ with differing values of the
wavenumber k=1/ and thereby begin the process of ‘localizing’ the particle- that is, having a better idea of
where it is.
 1 
 1 
 k x     k (x)
 2 
 2 
 new  
This function is slightly more localized, we have a a better idea of where the particle is:
But we have a worse idea of its momentum (or kinetic energy) because new is not an eigenfunction of the
momentum operator.
______________________________________________________________________________________
Question 1:
Demonstrate that new is not an eigenfunction of the momentum operator.
Question 2:
Find the Expectation value for the momnentum of the more ‘localized’ particle described by new
Hint: Take advantage of the following mathematical relationships:

 1 if k  k 
ikx ik x

e
e
dx


k k 

 0 if k  k 

_________________________________________________________________________________
Background:
Wavefunctions are mathematical descriptions that, in principle, contain all information about a
particle. In order to extract this information from wavefunctions, we use mathematical procedures known
as operators. The following is a table of quantum mechanical operators and the types of wavefunction
information they correspond to. Operators are usually distinguished from the values (information)
associated with them by putting a ‘hat’ on them.
Operator
Mathematical
Procedure
xˆ
multiplication by
coordinate x
pˆ x  i
d
dx
Vˆ (x)
d
Hˆ  i
dt
Information Recieved
location about position (location) of particle along the x
coordinate
derivative of a
wavefunction at a
particular position along
x
momentum of a particle in the x direction at the particular
position
multiplication by a
potential energy at x
potential energy of a particle at the position x
derivative of a timedependant
wavefunctions (not
usually discussed in this
course)
This is another way of writing the Hamiltoninan operator in
order to get theTotal energy of a particle. Writing it in this
form is useful to use on time-dependent wavefuctions which
are usually not the topic of introductory pchem.
It turns out that getting information by just operating on a wavrfunction, say (x), at a particular
position does not really work that well. Why is that, well consider if I tried to obtain information about the
momentum of a (ground state) electron in a box at a point x=a within this box. The appropriate ground
state wavrfunction for this electron as we’ve seen in class is: (where L is the length of the box)
We’ll see where this came from later in the class. Well, then I’d simply operate on y(x) with my handydandy momentum operator and get the result below.
2  2 
x 
d 2  2 x 
  sin   i   cos 
 L 
L L  L 
dx L 
1
pˆ x x   i
1
If I then evaluate the derivative at the point x=a, I’ll (in principle) get the momentum of the particle at point
a.
2  2   a 
pˆ x a  i    cos 
L L   L 
1
Notice that all of these quantities are constants. If you know the value for L, the width of the box and the
point a at which you are evaluating the momentum operator then you can calculate the momentum at that
point, right?
So what’s the problem? Why do we need to use expectation values?
We know by the Heisenberg Uncertainty Principle that if we know the position of a particle exactly, by
saying for instance that is lies at point a, then we lose all information about the momentum. So, what is
that information we got from this big calculation, nonsense- that’s what. Using operators directly to
evaluate a momentum like this won’t work because it requires us to pin down the position of a particlewhich we know we can’t do and still conform to the HUP. This procedure simply cannot give us any
information that we can experimentally observe. So what can we do to get around this problem, Oh what
ever can we do?!?
Just for full disclosure, there are cases when you can get an ‘observable’ values by operating
directly on a function by an operator. These are situations in which the wavefunction, , are
‘eigenfunctions’ of the operator in question. That is, they satisfy equations like:
ˆ   E
H
Where E is a number not a function (in this case total energy). Note that in equations like this after the
application of the operator (in this case H, the Hamiltonian operator) to the wavefunction you get the
eigenvalue (E) and the same wavefunction back. In the case we looked at above (the momentum of the
particle in the ground state of the particle-in-a-box) we did not get the same wavefunction back when we
operated on it. So the particle in a box wavrefunction we looked at was not an eigenfunction of the
momentum operator.
The Solution:
Despite the fact that because of the uncertainty priciple we usually cannot find experimental
values by directly using operators, we can do so by using expectation values. To see why this is the case,
we have to remember what the meaning of a wavefunction is. The meaning of the wavefunction itself, ,
is difficult to discern- rather it is the square of the wavefunction * that has physical meaning as the
‘probability density’ of the particle. In short, evaluating *(a)(a) (=P(a)- P is the probability) gives us the
probability of finding a particle at a particular position a. Now, if we evaluate * for all of the values of
x that the particule can exist at (excuse poor english), then the sum of the probablies for all of the a’s has to
equal 1- because the particle has to be somewhere.
In particular
  a  a 1
*
j
j
j runs over all possible values of a
j
This has been written as a sum because I’m considering all of the values of a along the x axis. Now,
unfortunately the number of different a’s is infinite- (remember that between any two numbers is another
number)- so we really can’t use a sum and instead we have to use an integral which can be thought of as an
infinate sum that gives you the exact, as opposed to approximate, area under a curve:
L
  x   x  dx  1
*
0
This is the whole essence of normalization, the sum of the probabilities has to be 1 – the particle has to be
somewhere within the box. Within this context, the probability density, *, can be thought of as a
‘statistical weight’ for the probability of finding the electron at a particular position and therefore the
integral represents the total probability of finding the electron in the area from 0 to L. (If you feel like I’ve
said this before, just humor me).
We can take advantage of this ‘statistical weight’ property to calculate values of observables.
ˆ  , you
Knowing that when you deploy a quantum mechanical operator (say ) on a wavefunction, i.e. 
get information about the observable property corresponding to that operator, it stands to reason that when
ˆ  a, that this gives you the
you place the operator within the square of the wavefunction, i.e.  a
probability of finding a particular observable value at the position a on the x axia. So, if you evaluate
*
ˆ
 *
for all possible values of a along x and sum the results, being sure to divide by total probability of
finding the particle over all the possible values of a (hopefully =1 if the function is normalized), like:
ˆ  a 
  a 
  * a  a 
*
j
j
j
j
j
j
then you would expect that what you get is the average value of the observable corresponding to the 
operator.
Because, again, there are an infinite number of possible values of a (there is a number between any
two numbers, blah blah blah....) we cannot use a sum but have to use an integral which can be thought of as
an infinite sum. This gives us the formula for expectation values (i.e. average values). To wit for the
average value of the physical observable :
L
 
ˆ  x dx
  x * 
0
L

*
x  x dx
L

ˆ  x dx
  x * 
0
1
0
where the second equality holds if the function is normalized. If it is not normalized, then the first equality
will still be true.