## Archive for January, 2009

### Physics Friday 57

January 30, 2009

Consider a single particle wavefunction (in the position basis) . Then for a volume V, the probability P that the particle will be measured to be in V is
.
The time derivative of this is:
.

Now, we recall that , where * indicates the complex conjugate. Thus, via the product rule,

So


Now, consider the time dependent Schrödinger equation:

Solving for the time derivative, we get

And taking the complex conjugate of that:
.
(Note that the potential  is real).
Thus
,
and


(the potential energy terms cancel).
So
.

Here, we need to use some vector calculus; namely, the product rule for the divergence operator : for scalar valued function φ and vector field , the divergence of their product is given by

Now, if our vector field is itself the gradient of a scalar function ψ, then

Swapping φ and ψ,

And taking the difference, we find
.
Putting in our wavefunction and it’s conjugate for φ and ψ,

So
.
Let us call the vector-valued function that is the argument of the divergence in the integrand :

Then

Now, recalling that the probability P is the integral over V of the probability density 
So in terms of the probability density 

and as this holds for all V, the integrand must vanish:


Now, this should look familiar to some of you. For any conserved quantity ρ with a flux given by the function , and no sources or sinks, the quantity and flux obey the continuity equation
.
[For example, if ρ is electric charge density, then the conservation of electric charge gives
,
where  is the electric current density.]
Now, as the probability density for the particle must always integrate over all space to unity, we similarly expect it to be a conserved quantity. Thus the above result is our continuity equation for quantum probability, and so  as defined above is our probability current (or probability flux). It has units of probability/(area × time), or probability density times velocity.

Note that the continuity equation tells us that for a stationary state, the divergence of the probability current must be zero. However, this does not mean that the current itself must be zero. Consider the three-dimensional plane wave .
Then


So

So
.
Note that as  is the particle’s velocity, the probability current of the plane wave, a stationary state, is the amplitude squared times the particle velocity.

Lastly, consider a wavefunction which has the same complex phase for all locations at any given time; that is , where  is a real-valued function. Then , and , and so we see the probability current is zero for such wavefunctions (one example of which are solutions to the particle in a one-dimensional box).

### Monday Math 56

January 26, 2009

Consider a sequence of polynomials Pn(x), where Pn is an nth degree polynomial. Let us also have the relationship that . Such a sequence is known as an Appell sequence, and includes a number of notable polynomial sequences (including the Hermite polynomials mentioned in a recent Physics Friday post). Note that the sequence Pn(x)=xn is a trivial example.

Now, we can obtain from  the relation
, or taking the definite integral of both sides, .
Letting our integral limits become x and x+1, we have:
. Now, note that Pn+1(x+1) is an n+1 degree polynomial with the same coefficient on the xn+1 term as Pn(x), and thus  must be a polynomial of degree no higher than n; in fact, one can check and find it must be a degree n polynomial.

So let us then choose the Appell sequence defined by

We can compute the first few terms immediately:
 gives us
B0(x)=1
Similarly, we can find




and so on.
Note that by putting x=0 in our defining equation, we have
.
This sequence is known as the Bernoulli polynomials. Amongst its properties is the symmetry relation
,
which tells us that
.
By differentiating the defining equation with respect to x, we get:


Now, consider the integral , for m a positive integer. Breaking up the region of integration into unit intervals, we find

and using our defining relation , we find that

And thus we see:
.
So

This is Faulhaber’s formula for the sum of the nth power of the first m integers in terms of the Bernoulli polynomials (see here).

The values of the Bernoulli polynomials at x=0 (the constant term of the polynomial) are known as the Bernoulli numbers:
.
The first few values are











For odd n>1, Bn=0;  is the only non-zero Bernoulli number for odd n. The even Bernoulli numbers, as one may observe from the above examples, alternate in sign.

Now, we recall that

Thus the coefficient of the xk term in Bn(x) is  times the coefficient of the xk-1 term of Bn-1(x); thus the coefficient of x in Bn(x) is equal to . Similarly, the coefficient of x2 in Bn(x) is equal to ; the x3 coefficient is , and so on.
Thus we see:
.
(proof by induction can verify this formula).
Therefore:

From this, and our symmetry, we find a recursive relation for the Bernoulli numbers:
,
which can be solved for Bn in terms of B0, B1, B2, …, Bn-1:
,
which with B0=1, can also be used to define the Bernoulli sequence.

### Happy 25th Birthday…

January 24, 2009

…to the Apple Macintosh.

### Physics Friday 56

January 23, 2009

Consider a rigid spherical body of density ρ. Now, imagine a test mass m located inside this body at a distance r from the center of the sphere. It will experience a gravitational force , with the negative sign indicating a force opposite in direction to the displacement r from the sphere center (an inward force). We see the force varies linearly with r. Thus, the gradient of the gravity is
, a constant. Here, the negative sign indicates that the gradient of the sphere’s gravitational self-attraction compresses the object.

Now, let us consider a mass M with its center of gravity a distance d from the center of our sphere. Let us consider a displacement r from the center of the sphere on the line between the centers of the sphere and M, with positive r being toward the mass M. Then the gravitational force on our test mass due to our external mass is
,
where the positive sign indicates the force is in the direction of positive r.
This, in turn, has a gradient


Note here that the tidal stretching increases as one approaches the mass M, while the self-attraction compression is constant. In fact, the sum of the two gradients becomes positive, and the tidal force dominates, when

Where  is the radius of a sphere of density ρ with a mass 2M.

Thus, note that if d<deff, we see that the tidal forces overcome the self-attraction down to the center of the sphere, and thus a stable rigid body held together by gravitational self-attraction alone cannot exist. If our body of mass M is a sphere of radius R, it has density . Thus, writing deff in terms of ρM and R,
.

A real body would also have a material tensile strength to help hold it together, and also would deform from spherical into tidal bulges under the tidal force; however, there still remains a similar limit, inside which a self-gravitating celestial body (our sphere) orbitting a larger body (the mass M) will disintegrate under the tidal forces. This is called the Roche Limit.

### Monday Math 55

January 19, 2009

How might one perform the unit-square integral
?
First, let us swap the order of integration:
.
Next, we approach the inner integral by performing the u-substitution
 → 
The limits become y=0 → u=∞, y=1 → u=-ln(x).

So, with this substitution, we find , and


for 0<x<1, ln(x)<0, so -ln(x)>0; -ln(x)→∞ as x→0, and -ln(1)=0, so the integration region can be seen as between  and x=1, u>0. So, reversing the order of the double integral,


We found previously that , and the gamma function is defined as ; this means


This is called Hadjicostas’s formula, and holds for s any complex number with .
Note that as the pole of ζ(s) at s=1 is of order one: the Laurent series is of the form ,
where the  are called the Stieltjes constants, and , the Euler-Mascheroni constant; the singularity in the term  is thus removable, with 

Thus we have
.

Similarly, with the related integral
,
we can use the same subtitution and change of order:
.

What if we try instead try the integral
?
Performing the same u-substitution:
.

Reversing the order of the double integration, and performing the inner integral,


We found previously that , so then

(for )
and, in analogy to I(s), we can also find that.


### Physics Friday 55

January 16, 2009

Classically, a one-dimensional harmonic oscillator is a system with a mass under a restoring force proportional to displacement from the equilibrium position: F=-kx. The energy is , and the equation of motion  has solution , where.

Analogously, a one-dimensional quantum harmonic oscillator is a particle with Hamiltonian , where here p is the quantum momentum operator, and x the position operator. In the position basis, this is then
.
A cursory examination of the expectation for energy

shows that we can expect our energies to be non-negative.

Most quantum mechanics books and courses I have encountered address the energy eigenstates by means of the raising and lowering operators. Instead, here we will use an analytical method of finding the eigenvalues to the time-independent Schrödinger equation
.
Let us replace x and E with corresponding dimensionless variables: we define the dimensionless variables , . Then , 
the Schrödinger equation becomes
.

To find a solution to this differential equation, we consider asymptotic behavior. Namely, when the energy is small, the ξ2 dominates over ε, and our equation approaches , which hints at a solution that behaves like a Gaussian. Adopting, then, a test form
, and plugging this into our differential equation , we get
,
or, eliminating the common Gaussian factor,

Applying the power series method, we plug in  to find

Which gives recursion relation .

Now, we return to the physics of the problem, namely that the wavefunctions have to be normalized. This means that ψ→0 as x→±∞. Now, considering that
, and comparing to the series expansion of the gaussian, we see our series diverges too fast for ψ to be normalized, unless the series terminates. This requires  for some n, which gives , and as E≥0, we see that the ground state has energy ; this is known as the ground state energy, or zero-point energy.

Solving for the wavefunctions themselves requires more mathematics, and gives solutions involving the Hermite polynomials.

### Monday Math 54

January 12, 2009

Let us consider an integral of the form , with parameter λ>0, and where f(x) has a single local minimum in the interval (a,b). The method of steepest decent, also known as saddle-point integration, is a useful method for approximating such an integral in the small λ limit. Namely, if the local minimum occurs at x=x0, then we can expand f(x) in the Taylor series about this point; as it is a local minimum, f‘(x0)=0 and f”(x0)≥0. Here, we assume that this second derivative is nonzero. Thus, our Taylor series is
.
Now, we note that as λ→0+, the term in the exponent becomes ever more negative, and the integrand approaches zero; the local minimum at x0 is the “slowest” to approach, and thus the function near that point comes to dominate the rest of the integral; and so

and

Now, we see that the last integral is part of a gaussian integral; the peak of our gaussian is in the region of integration, and as λ→0+, the gaussian’s width goes to zero as well, so that the tails become negligible in the limit, and
.
Now, recall that the integral of the gaussian over all real numbers is given by
.
Thus ,
and so
.

Let us perform an example: . This does not fit the form as it is, but can be transformed to do so. First, let us make the substitution . Then we have:
. Second, we note that for positive u, .
Thus, we have
,
and the integral above fits our form with  and ; thus our approximation will be for n→∞. Now, , which is zero for u=1. , so , and . Thus, our first saddle-point approximation says

and so

which you may recognize as Stirling’s approximation (see here).

### Physics Friday 54

January 9, 2009

Quantum Mechanics and Momentum
Part 11: Angular Momentum Eigenfunctions

In our previous part, we introduced the ladder operators, and showed that they generate a “ladder” of simulataneous eigenstates to L2 and Lz. Now, we examine the angular momentum operators in spherical coordinates (r,θ,φ).
Using our conversions and chain rule, we find the linear momentum operator components to be:



From these, we find
,
,
.
Note that all of these are fully independent of the radial coordinate.
Summing the squares of these, we can find that
.

This then gives us our eigenvalue relation  as being
.
Compare this to the angular portion of Laplace’s equation in spherical coordinates. We see then that the eigenfunctions of L2 are (any radial function times) spherical harmonics:
,
and to make our wavefunctions 2π periodic in φ, we require that j be an integer. This means that the eigenvalues of Lz should be integer multiples of ℏ, ranging from –jℏ to jℏ.

Applying  to , we find
,
and so
.
So the spherical harmonic  is the simultaneous eigenfunction for L2 and Lz, with eigenvalues  and , where j and m are integers with j≥0, –jmj.

Lastly, we note that it can be shown, via more that one method, that if the potential depends purely on the radial coordinate, V(r), so that the problem is spherically symmetric, then the quantum hamiltonian  commutes with the angular momentum operators L2 and Lz, so that one may simultaneously specify the energy, the square magnitude of the angular momentum, and the projection onto one axis of the angular momentum. This is very useful in determining the electron wavefunctions for the hydrogen atom.

### Monday Math 53

January 5, 2009

Previously, we introduced the spherical harmonics. We note that as written, the spherical harmonic  is real for m=0, but complex for m≠0. However, one can find (normalized) linear combinations of  and  for m≠0 that are real, namely
 for m>0 and  for m<0. With , we have that the first few real spherical harmonics  are:










Let us examine where these are zero. First,  is a non-zero constant, and so is zero nowhere. For l=1, we have
, which is zero when , which is the plane ; , which is zero when , which is the plane ; and , which is zero when , which is the plane . Thus, when l=1, we will have a single surface (a plane) on which the harmonic is zero; such a surface is called a node.
We see that for , we have , which gives the cones  and  ().
 when , or on the planes x=0 and z=0.
 when , or on the planes y=0 and z=0.
 becomes , which becomes the planes x=y and x=-y.
Lastly,  becomes , which becomes the planes x=0 and y=0.
In fact, the real harmonics  always have l nodes, all of which are vertical planes (constant φ), the plane z=0 (θ=π/2), or cones with the z-axis as their axis (constant θ≠π/2).

### Physics Friday 53

January 2, 2009

Quantum Mechanics and Momentum

Last week, we introduced the quantum angular momentum operator L, and found some properties of its magnitude and components. Most notably, that only one component may be fixed at a time, and that component, usually the z component, and the magnitude squared may be fixed simultaneously. So now, let us look at the common eigenfunctions of L2 and Lz. We want eigenfunction Y so that  and .

Now, let us define two “ladder operators” L+ and L by

Let us examine the operation of L+ on our Lz eigenvalue equation:

Now, from the definition of the commutator,
,
and from our definition of L+,
.
So , and
,
which tells us that  is also an eigenstate of Lz, with “raised” eigenvalue . Thus, we call L+ the “raising operator.” Repeated application of this shows that for n applications of the raising operator,

Similarly, we can show that  gives
,
and so it is called the “lowering operator.”
Now, let us examine how L2 behaves on these raised an lowered eigenstates. The commutators of L2 with our raising and lowering operators are
,
and so
,
and the ladder of eigenvalues …b-2ℏ, b-ℏ, b, b+ℏ b+2ℏ,… generated by the raising and lowering operators are all eigenstates of L2 with eigenvalue a.
We have eigenfunctions , and , with eigenvalues .
Thus, .
Subtracting this from , and recalling that , we find:
.
Note that the middle term above corresponds to a non-negative physical quantity; thus we expect that  cannot be negative: . This, in turn, means that our ladder must be bounded both above and below; there exists  and  such that  and .
So then we have . But
.
Therefore:

Analogously, using , and , one finds
.
Taking the difference of these two equations, and thus cancelling a, one finds:
,
as .
Combine this with the knowledge that all eigenvalues on the ladder are separated by units of ℏ, we see , for some integer n, and so
, ,
giving us our eigenvalue ladder .
And as , we find
.