Normalization of the momentum eigenstates:

The appendix in the book states

momentum eigenstate, dual or BRA-momentum eigenstates

position eigenstate, dual or BRA-position eigenstates

=the position space wavefunction for a momentum eigenstate

=the momentum space wavefunction for a position eigenstate

== adjoint or dual rep. space wavefunction for a BRA-momentum eigenstate.

== adjoint or dual rep. space wavefunction for a BRA-position eigenstate.

So the interpretation depends on whether x or p is interpreted as the fixed valued or the index in the space. [e.g. fixed value of p in position space or fixed value of x in momentum space] The relationships show that they are, except for sign changes in the exponent, the same. For example

= .

The above shows what we mean by the various inner products and shows that the form of the wavefunction can be the same but there can be two interpretations.

The normalization given above for the state ensures unitarity. The required condition is then

[appendix unitarity condition].

Let us look at the units. Since delta functions are integrated in this case over or to get a 1 that is they must have inverse units of distance and inverse momentum respectively.

uncertainty relation:

The exponential is thus dimensionless and above integral has the correct units of inverse meters.

This integral has the correct units or inverse momentum.

If one chooses to use then the is removed from the normalization and needs unit of to cancel the and the integral over obtains the correct units for .

Thus the normalization factor has the correct dimensions to normalize the free particle states of definite momentum in position space and the free particle states of definite position in momentum space.

Where did the unitarity condition come from ?

If and are complete set of functions that span the space then the transformation from one to the other must be unitary.

unitarity condition i.e. a unitary transformation has an adjoint equal to its inverse.

Typically one examines the impact of the transformation in general

and then requires that as above no probability is lost or gained. Notice that this is not a valid constraint for most of the operators that we have dealt with so far If the transformation is to preserve the probability or amount of the state then if the state is normalized to one then the state must also be normalized to one and the condition holds.

Examine

which is the matrix element that takes one from a to .

which is the adjoint that takes to

which takes to and then back to

in momentum space

The condition is important and perhaps obvious. Once this condition is accepted as necessary the normalization is set.

In the book there are two approaches that lead to two normalization conditions. One is the periodic boundary condition (box normalization). The other is to normalize so that over an infinite range you find a delta function.

First let us consider the periodic boundary conditions. The goal is to have the appropriate boundary conditions for a particle in a box with rigid walls (V=∞) and have states of definite momentum. Schiff explains the problems associated with confining a particle in terms of the reflections off of the rigid walls of a finite box. Basically this problem does not conserve momentum. Twice the balls momentum is absorbed by the wall. Thus the way you require the ball to act at the boundary can lead to complications when trying to find momentum eigenstates. However, by assuming periodic boundary conditions the ball passes through the wall and its entry into a new box in some sense puts the ball at the opposite side of the original box with momentum unchanged. If the wave extends to infinity in both directions then whatever leaves on the left enters on the right. Within the box one can use these states to build energy eigenstates or any other physical states. These states are truly momentum eigenstates. Of course, outside the box the solution is no longer zero. Whatever solution you require in the box is then periodic and the same solution appears in each periodic box. One just uses these functions to predict the state in the box and ignores the solution outside. The confined particle problem localizes the quantum states while in free space the momentum wave functions are smoothly spread out over an infinite range. Momentum eigenstates must extend over all space and these periodic states are standard momentum states but only those that match the BC are used. It is important to realize that in any confined space momentum cannot be determined to arbitrary precision. We know this from the uncertainty principle. Putting the particle in a box gives you some knowledge of its location thereby making it impossible to specify the momentum exactly. So we impose boundary conditions that are periodic. This allows us to force the wavefunction to be zero at the walls but still be defined outside the box and therefore be a legitimate momentum eigenstates. The limit can be taken as to approach the free particle solution.

Problem 4.6 examines these states over a finite interval a [-a/2èa/2] with a focus on the normalization as the range goes to infinity. The detail associated with whether you choose the function to be zero outside the box or periodic is not discussed.

This shows that the integral of the proposed wave function is properly normalized as the range goes to infinity.

Notice, however, for a finite range that for any arbitrary values of k and k’ with and for a finite interval a, the integral is not zero. If one chooses the k values such that then these k’s have an integral equal to 0 and are orthogonal.

It is also interesting to note that for a particle in a box the energy can have eigenvalues and it appears to be made up of eigenstates of right and left going momentum but the state

Choose range a=10 and plot an example. The plot shows a cosine function or the real part of the state above. The plot emphasizes that wavefunctions go to zero outside the box.

Are these momentum eigenstates? From the above discussion the answer is clearly no. Also note that at the boundary the derivative of the function is discontinuous. Thus the momentum operator in position space is not defined at this point.

A second approach is to examine the states normalized over an infinite range

the normalization leads to a delta function.

Using this state as a representation of a momentum eigenstate wrt the position basis we find:

I verify that the limit is a delta function via some handwaving arguments in deltafunction.doc which is included in the box below.

In the homework examining particle in a box solutions as .

There are two approaches:

1. Examine the problem with the normalization factor included an encounter

which shows that the functions in the limit are orthonormal in the traditional sense. As or kèk’ for finite {case a)}the integral is 1 but when {case b)} the integral approaches 0 for any as long as . So considering the PIB states they seem to converge to the correct behavior as you make the box larger and larger i.e. normalized and orthogonal.

2. Or we encounter the following integral without the normalization factor and this leads to the delta function.

which based on the mathematics of delta functions is a proper definition for this function. The properties of this function point a) is straightforward L’hopitale’s rule allows one to find the limit as goes to zero and then the limit as a goes to infinity is straightforward but point b) does not appear as obvious. It does become a sinusoidal oscillation with a wavelength of zero. If you integrate a function against an oscillating function with 0 wavelength the contribution will be zero. The integral oscillates so rapidly because two waves that are almost the same frequency and take a very long distance a to become out of phase. Thus as the integrals are performed over a large range the results of the integrals become very different the larger a becomes:

if a is small. The small difference is based of the fact that the sine functions are almost identical and so their integral will be almost identical. If the range is greater than the wavelength then any complete oscillation integrates to zero. Thus the integral can be viewed as due to the fraction of a wavelength left in the last stretch of the range of x. Since two waves that are close in frequency remain almost identical for a long range the difference will be small. However, if then even the smallest difference results in a significant phase difference and thus the integral has a very different value. Thus the dependence of the integral on leads to a rapidly changing result. As this rapid oscillation has an frequency or zero wavelength. If you integrate a convolution of a function of with this rapidly changing sinusoidal function then it must result in a zero contribution if the function is smoothly varying. The rapid oscillation causes additions and subtractions at such a rapid rate that the function hasn’t changed. Thus the impact of the function away from zero b) has the proper behavior.

So the question becomes how are the two methods of normalization equivalent?

To clarify this relationship I have added a subscript to the momentum label in the first set of states. This is because the limit as will require a transition of the momentum eigenvalues from discrete to continuous.

The integral is over a finite range and again only values of p allowed are .

There should be no problem representing a continuous function of x over a finite region by using a Fourier series. If the region becomes infinite the sum become an integral or Fourier transform. So we have not made the transition to continuous momentum but the original problem encompasses continuous x.

It may be necessary to consider in more detail the relationship between a sum and an integral.

You sum a series of numbers. For example, you can imagine these are individual water molecule volumes. You sum up the volumes of all of the water molecules to reach the volume of a glass of water. However the size of each molecule is so small that you make the approximation that the water is a continuous medium and you represent the volume V as

One defines the density and sums all the contributions from all molecules which you treat as continuous and replace it as an integral.

In order to consider how a sum becomes an integral we imagine a volume V broken into J subvolumes

and the total volume is the sum

Now as

What we are considering is twofold.

1. Real physical situations that are discrete have been treated with continuous mathematics. [Water molecules] In this case as J gets large we start to sum volumes that are smaller than a molecule. Although we can’t have water below one molecule we can still consider 1/2 a water molecule providing 1/2 the volume. But the nature of a molecule no longer really has a continuous nature to its content so this is really an approximate treatment that works. This provides an example of the relationship of discrete and continuous in terms of thinking about how one approximates fluids as continuous media.
2. The mathematics of continuity defines what we mean by an integral or a derivative. And considering a volume in the limit that we divide into infinitesimal elements leads to the definition of the integral {or the derivative}.

The problem is a bit more complicated for the plane wave states because we transition to the continuum by taking the limit .

When the sums in the above relationships go over to integrals the variable of interest is the momentum and not the index n. We need to replace the index with the momentum in keeping track of the sum.

The change is going to zero because

We keep the subscript on p to remind us that this is still a valid summation but we need the nth momentum and coefficient A for the nth momentum state. This is just the infinite sum of numbers times a differential element. Clearly the number which represents the contribution or amount of the state with momentum could be consider as the amount per unit step. Whether you treat this as a number or a density doesn’t effect the sum. The in the sum implies that we treat the as the amount per step in the sum. With the distinction is irrelevant but it clarifies that the number per step goes over to the number per momentum region .

All of the discussion leads to a factor of which now makes the limit

The initial expression is defined and evaluated based on a series of momenta and as a get larger and larger the amount of particle at each location gets smaller and smaller. The state is spread out until the amount at any location is zero. The second state is obtained by transitioning from . The amplitude at each location x is discrete however when you express this an amplitude per unit dp which itself is small the density function remains finite. The product of the differential and the amplitude approaches 0.

In the limit the above equality holds.

Care will always be required in dealing with the idealization of a state that has a singular location or momentum. Consider calculating

The first approach tells you that the amplitude is very small where as the second approach defines a factor that seems to imply a finite amount at each value of x. This is an incorrect interpretation.

To get real physical content one might need to build a wave packet that combines a range of momenta. This state can be built so that it possesses mathematical behavior.

For discrete states

This conditions will allow one to pick out the jth state via the inner product. Thus the state

has a sensible interpretation in terms of normalized basis vectors.

For continuous states

the normalization condition requires the states to have a value of when you choose a specific state. The usual normalization for states compared to continuous states are

standard states	continuous

The requirement for the delta function is that when you take the specific state and examine its normalization it must be .

So in order to include the orthonormality for basis states one implements a delta function that requires an infinite normalization. The infinity is required in order to pick out the state in a weighted sum.

This discussion is repetitive and circular. It describes how in the limit of large a for the box normalization of . One conclusion is that this normalization seems to go to zero goes to . This conclusion is reached when the approach to zero is absorbed in the differential dp. Thus mathematically one can see the step size is getting smaller as a gets larger but the density amplitude/length or probability/length remains finite and leads to an infinite value for the states normalization (required by the delta function definition). So there is not really a different value for the normalization when taking the limit of the periodic states (periodic boundary condition for plane wave states). The mathematical complexity is then better viewed in terms of the need to define a workable value for when momentum is continuous. Thus the definition is the central mathematical piece and the normalization from the periodic states just reaches this result. If this is the required condition then an integral over a state of definite momentum will yield a finite result for a finite ranged integral. Thus the normal probability density remains finite. The integral therefore over all x will be infinite and the amount of the state at any location in z appears to be non-zero. This contradicts our understanding that spreading of a state over all space will make it have a amplitude that approaches zero for any finite volume.

These contradictory results are only truly resolved when the actual states used are a convolution of the idealized plane wave states. Restricting your states to a finite region by weighting some regions of position or momentum space with zero amplitude eliminates the singularity. Thus normal traditional amplitudes and probability densities are present in the physical systems that do not extend to the ends of the universe. The wavelike relationship between position and momentum or frequency and time has introduced the basic conundrum that a tuning fork must play for an infinite amount of time in order to play a singular frequency.