**Quantum Mechanics and Momentum**

Part 5: Momentum Eigenstates

For any observable with operator , we recall that any state with a definite value for the observable has a wavefunction that is an eigenfunction of the operator, with the value of the observable given by the corresponding eigenvalue.

Now, we have that the momentum operator in one dimension is . What, then, are the states of specific momentum (in the position basis)? We set up the eigenvalue equation:

Now, using our operator, this becomes:

This is a simple differential equation (first-order separable), with solution:

where A is a constant (our constant of integration) and is the wavenumber corresponding to the wavefunction momentum *p*_{0} via our de Broglie relation .

Note that for all *x*, so fixed momentum requires equal probability for all positions (when Δ*p*→0, Δ*x*→∞, as required by the Heisenberg uncertanty principle). This, however, means that the wavefunction is not properly normalizable; pure momentum states represent an ultimately non-physical, idealized limit of real wavefunctions as the uncertanty in momentum shrinks.

Similar to the above arguements, we find in three dimensions that the wavefunction for a momentum is

where .

## Archive for November, 2008

### Physics Friday 48

November 28, 2008### Truly Epic

November 27, 2008This is simply unbelievable, and legendary:

### Happy Thanksgiving!

November 27, 2008### Monday Math 47

November 24, 2008I imagine that quite a few of you have previously encountered string art, where curves are approximated by a number of straight lines spaced along some pattern (see also here). Most string art is two dimensional, but I’ve also encountered some three dimensional cases. Here, we instead have the lines of string approximating a surface. Such a surface is a ruled surface; through each point on the surface there exists a straight line that lies on the surface. The set of such lines is called a ruling of the surface.

Let us consider a pair of parallel, congruent circles perpendicular to the axis connecting their centers. Let us choose our coordinates so that we have circles of radius R, parallel to the xy-plane with centers on the z axis at z=±h. If we connect corresponding points on the circles with straight lines, then we have a cylinder. Suppose, instead, that we then “twist” the circles, so that we thus have lines connecting points on the circles offset by some angle, which we will call θ. We noted that θ=0 gives a cylinder; one should also see that if θ=π, we have a double cone. So we consider 0<θ<π.

Picking one of the lines (one with a symmetrically placed position), we choose the

line through the points and on the circles. Describing this line parametrically, with parameter t such that we have p_{1} at t=-1 and p_{2} at t=1:

.

Now, via symmetry, we see the other lines consist of the image of this line under a rotation about the z axis. So, rotating this line by an angle φ about the z axis, we get

So, now, we can use this equation to define a parametric surface, with u (a renamed t) and φ as the parameters:

, with -∞<u<∞ and 0≤φ<2π.

We want to find a non-parametric equation for the surface; to do so, we seek to relate x, y, and z so as to eliminate u and φ. First, we use x(u,φ) and y(u,φ) and solve for cos(φ) and sin(φ):

If we square these and sum them, we obtain

Now, using z=hu, u=z/h, and we have:

This is a one-sheeted circular hyperboloid:

with and (as 0<θ<π, a,c>0)

Thus the circular hyperboloid of one sheet is a ruled surface. A suitable (non-uniform) scaling of our y coordinate will show that the general hyperboloid of one sheet x^2/a^2+y^2/b^2-z^2/c^2=1 (the elliptical hyperboloid when a≠b) also has the same property.

Now, note that using a ‘twist’ angle of -θ in our procedure generates the same hyperboloid as θ. Thus, the hyperboloid of one sheet is a doubly ruled surface (not just the circular, but the general hyperboloid of one sheet), a surface with two distinct rulings. It is one of only three such surfaces; does anyone care to guess the other two?

This property of hyperboloids of one sheet leads to them being used in structures, particularly towers, far more than one might realize. For example the classic shape for cooling towers of nuclear power plants is that of a segment of a hyperboloid. For other examples of hyperboloid structures, see here.

### Physics Friday 47

November 21, 2008**Quantum Mechanics and Momentum**

Part 4: The Momentum Operator

Frequently, the quantum mechanical operator for momentum is simply presented to students without justification (that was my experience, at least). However, use of our previous developments in Fourier transforms and observables allow us to see a (non-rigorous) justification of the momentum operator for wavefunctions represented in position space.

Let us consider a one-dimensional problem, with a particle described by the (position domain) wavefunction *ψ*(*x*). As previously, we can define a Fourier transform of this to the (angular) wavenumber domain, *φ*(*k*):

Now, as we noted previously, the relation shows that the wavenumber domain represents momentum space.

Now, recall that for an observable with (Hermitian) operator and (position space) wavefunction *ψ*(*x*), the expectation value of the observable is

For example, the expectation value for the position is

,

where ^{*} indicates the complex conjugate.

Expectation of a function of momentum *g*(*p*) can be found in the wavefunction domain as

and so

(Just as multiplication by *x* is the position operator is position space, multiplication by is the momentum operator in momentum space.)

If we wish to write this in terms of *ψ*(*x*), we start from our Fourier transform:

.

Taking the complex conjugate,

Now, we know that the Fourier transform of the derivative of a function obeys the rule:

and thus we see that as , then

Combining these, and remembering the delta “function” relation we used in proving Plancherel’s Theorem, we have:

.

Comparing this to our operator expectation value

which implies that the operator is the momentum operator for wavefunctions in the position domain. Similar arguments can show that the position operator in the frequency domain is .

For three dimensions, we have momentum vector , which gives us vector-valued momentum operator

### A Mathematical Addendum

November 20, 2008**Quantum Mechanics and Momentum**

Part 3′ (Mathematical Addendum): Fourier Transforms and Derivatives.

Again, we have our Fourier transform . Now, let us consider the transform of the derivative of a function, . We can use integration by parts on this: with , ; we have , and , and so

.

Now, if we require (as any meaningful signal, or quantum wavefunction, must be), the first term on the right side of the above vanishes, and thus:

.

This will prove important in later posts.

### Monday Math 46

November 17, 2008Find .

First, we note via symmetry that

.

Now, recall that it was shown here, that

.

Thus

.

And so we find that

.

Now, we can use our previous results for the partial derivatives of the beta function.

We will need

,

,

,

and .

We have

,

and so

.

### Physics Friday 46

November 14, 2008**Quantum Mechanics and Momentum**

Part 3: The Fourier Transform, Dirac Delta, and Plancherel’s Theorem

Consider the Fourier transform in one dimension. There are several common conventions for defining it; here, we will use the unitary angular form frequently used in physics: for function *f*(*t*) the Fourier transform is

,

and the inverse transform is

These convert between the time domain and the angular frequency domain. For transforms on a function of a position variable *f*(*x*), we have

,

and

which convert between the position domain and the (angular) wavenumber domain.

For an *n* dimensional position space with position vector we find the transform of a function to be in the *n* dimensional wavenumber space with wavenumber vector

via

and

Now, consider the Fourier transform of the Dirac delta “function”:

and by examination of this, and the inverse transform, we obtain

.

(Note that this is not a mathematically rigorous equality, but is valid under integration with a function: .)

This relation is useful for proving a number of relations, one of which follows.

For a valid wavefunction of one spatial dimension *ψ*(*x*), let us define its Fourier transform *φ*(*k*).

Then

The complex conjugate of this is

where ^{*} indicates the complex conjugate.

Thus, let us now find :

This is known as Plancherel’s Theorem (or in some physics and engineering sources as Parseval’s Theorem), and we see that if the (position space) wavefunction *ψ*(*x*) is normalized, , then the corresponding “wavenumber space” wavefunction *φ*(*k*) is also normalized.

The reason we want to use such a Fourier transformed function goes back to our de Broglie relations, particularly . Using this, we see that our “wavenumber domain,” via this equation, represents the momentum space basis for our quantum state.

### Another Sad Story of Superstition at Work

November 13, 2008### Patent on ‘Patent Trolling’

November 13, 2008Over at Overlawyered, Walter Olson reports on the attempt by Halliburton patent attorney Clive D. Menezes to patent a form of patent trolling.