|San José State University|
& Tornado Alley
The deuteron, the hydrogen 2 nucleus, is an especially interesting structure. The two nucleons, the proton and neutron, are held together by the nuclear forces alone.
The equations describing the motion of a system of two particles can be separated into one equation determining the location of the center of mass of the particles and another equation describing their motion with respect to that center of mass. It is only the second equation that is of interest for the physics of the deuteron nucleus.
The time-independent Schroedinger equation for two equal-mass particles in a potential field of U(r), where r is the separation of the two particles, can be factored into a radially-dependent factor and angle-dependent factors. The radially-dependent factor R(r) satisfies the equation
where l is an integer representing angular momentum and E is a constant.
from the rotation of the two particles about their center of mass, the
The substitution of u(r)/r for R(r) simplifies the equation to:
This equation can be solved for a limited set of potential functions. In general the equation has solutions satisfying certain boundary conditions only for discrete values of E, called eigenvalues. The potential that gives rise to a force proportional to exp(-r/a)/r2 is not one of the potential functions for which an analytical solution of the equation is possible.
It is possible to obtain a solution for a square well potential; i.e.,
The equation within the well reduces to:
The solution within the well is of the form:
Since R(r)=u(r)/r must be bounded a s r→0, B=0.
Outside of the well the equation to be satisfied is the same as the
above but the potential is zero so the corresponding value of
k'2 = -Em/
h is negative and so the
solution is of the form
and since u must be bounded as r→∞, C=0.
At the boundary, r=b, the two solutions must fit together smoothly; i.e., both their values and their slopes must match. This means that
These equations imply that
When k and k' are replaced by their values in terms of V0 and E the above equation becomes
For only certain values of E will the above conditions have a solution.
The equation to be satisfied for this case can be put into the form
for Deuteron Based Upon
a Square Well Potential
|Inside the Well||Outside the Well|
|l=1||A√k[sin(kr)/kr) - cos(kr)]/kr||iC√K((1/Kr)+1)exp(-Kr)/Kr|
|l=2||A√k[(3/(kr)2 - 1) - 3cos(kr)/kr)]/kr||C√k[3/(Kr)2 + 3/Kr + 1]exp(-Kr)/Kr|
In the above table
The above material gives an idea of the cumbersomeness of the analytic solutions to the equation. What is needed is a more general approach that gives some information about the solution even if it cannot be the solutions themselves. Such an approach is available in the Fourier transform method.
The Fourier transform of a function f(r) is
where i is the imaginary unit, the square root of −1.
Note that even if the function f(r) is real valued the Fourier transform Ff(ω) will be a complex-valued function.
The power of the transform method is that it converts the problem of the solution of a differential equation into the solution of an algebraic equation. One of the properties of the Fourier transform is that the Fourier transform of the derivative of a function can be expressed in terms of the Fourier transform of the function.
Thus the Fourier transform of the second derivative of f(r) is given by
Another important property of the Fourier transform is that there is a special formula for the Fourer transform of the product of two functions; i.e.,
The integral on the right is known as the convolution of the functions Ff and Fg, the Fourier transforms of f and g. The roles of f and g may be interchanged in the convolution integral.
Consider the differential equation
This is a previous equation with the
U(r) term and the
term collapsed into one term denoted as V(r).
When the Fourier transform is applied to this equation the result is
The left-hand side of the above equation is equivalent to a linear operator in the space of continuous function defined on the interval [−∞, ∞]. Call this operator A, then the above equation is
This is an eigenvalue equation. It has a solution only if the determinant of (A−EI) is zero, where I is the identity operator. Thus the energy eigenvalues must be discrete.
With the insight provided by the Fourier transform method it is now easy to see that the discreteness of the energy eigen values follows directly from the original equation. The second derivative is a linear operator that maps functions into functions. It can then be represented as an operator D so that the equation becomes
The multiplication of u(r) by V(r) is what might be described as a diagonal operator. The entire LHS can be represented as a linear operator H acting upon u(r) so the equation is then
This is an eigenvalue equation, which has solutions corresponding to the determinant of (H−EI) being zero.
(To be continued.)
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