polar coordinates and Prüfer's transformation

One of the most insightful visual techniques is to introduce polar coordinates into the phase plane; it is called Prüfer's transformation

$$\begin{array}{cccc} \begin{array}{ccc} \rho & = & \surd(x... ...\theta \\ y & = & \rho\sin\theta. \end{array} \end{array}$$

with variables which satisfy the differential equations,

$$\frac{d\rho}{ds} \frac{1}{2}\frac{1}{\surd(x^2 + y^2)}(2x\frac{dx}{ds} + 2y\frac{dy}{ds})$$ = (285)

$$\frac{x(cy+dx) + y(ay+bx)}{\surd(x^2+y^2)}$$ = (286)

$$\rho(a \sin^2\theta + (b+c) \sin\theta\cos\theta + d \cos^2\theta)$$ = (287)

for $\rho$, and

$$\frac{d\theta}{ds} \left( \frac{1}{1+\frac{y^2}{x^2}} \right) (y\frac{dx^{-1}}{ds}+\frac{dy}{ds}x^{-1}),$$ = (288)

$$\left( \frac{1} {1+\frac{y^2}{x^2}}\right) \frac{(x\frac{dy}{ds}-y\frac{dx}{ds})}{x^2}$$ = (289)

= x (ay + bx) - y (cy + dx) (290)

$$- c \sin^2\theta + (a - d) \sin\theta\cos\theta + b \cos^2\theta$$ = (291)

for $\theta$. In summary,

$$\frac{1}{\rho} \frac{d\rho}{ds} d \cos^2\theta + (b+c) \sin\theta\cos\theta + a \sin^2\theta$$ = (292)

$$\frac{d\theta}{ds} b \cos^2\theta + (a - d) \sin\theta\cos\theta - c \sin^2\theta,$$ = (293)

the form of which suggests using the double angle formulas:

$$\frac{1}{\rho} \frac{d\rho}{ds} \frac{a+d}{2} +\frac{a-d}{2}\cos2\theta +\frac{b+c}{2}\sin2\theta$$ = (294)

$$\frac{d\theta}{ds} \frac{b-c}{2} + \frac{a-d}{2}\sin2\theta + \frac{b+c}{2}\cos2\theta.$$ = (295)

An interesting comparison and check results from setting V = 0 in the matrix of the Schrödinger equation, Eq. (283). Then a = d = 0, b=-E and c=1. The Prüfer equations are

$$\frac{1}{\rho} \frac{d\rho}{ds} (1 - E) \sin\theta\cos\theta$$ = (296)

$$\frac{d\theta}{ds} - E \cos^2\theta - \sin^2\theta,$$ = (297)

whose solution for $\theta$ can be put in one or the other ot two forms:

$$\theta(s) - \theta(0) (1 - E)\ {\rm arctan}\ s - s$$ = (298)

$$(1 - E)\ {\rm arccot}\ s - E s,$$ = (299)

leaving the corresponding quadrature for $\rho$ still to be performed.

Often the Prüfer equations are not pursued beyond the $\theta$-equation because the sign and magnitude of the derivative establish the oscillatory nature of the solutions. If the derivative is of constant sign and bounded away from zero, the solutions will necessarily oscillate indefinitely with a period implied by the bound. Moreover, given two equations -- say the same equation with different values of the eigenvalue parameter E -- if the derivative of one of them is always greater than that of the other, conclusions about the interlacing of nodes can be drawn.

Even in the simple example shown, there is an unexpected jitter in the angle $\theta$, when the result is compared to the more familiar solution consisting of sines and cosines derived from uniform motion around the circumference of a circle, at an angular velocity which is the square root of E, not E itself. However, the result shown is not incorrect, and can be used to illustrate the advantages of preparing a system of differential equations for the optimal interpretation of results.

In this case, the coefficient of the differential equation is not a multiple of the quaternion ${\bf i}$ whose exponential produces a simple rotation, but contains and admixture of the quaternion ${\bf j}$ which would generate hyperbolic motion. The combination leads to an ellipse in the phase plane, rather than a circle. Furthermore, the expected $\surd E$ is the quaternion norm of the coefficient, whilst terms such as $(E\pm1)/2$ are the coefficients of the individual unit quaternions.

An appropriate preparation for the Prüfer transformation, actually one close to the self-adjoint form which is the version originally proposed by Prüfer, is to revise the matrix equation by writing

$$\left[ \begin{array}{cc} 1/\surd E & 0 \\ 0 & 1 \end{array} \... ...ht] \left[ \begin{array}{c} y(s) \\ x(s) \end{array} \right].$$

which, due to the constancy of E, takes the form

$$\frac{d}{ds} \left[ \begin{array}{c} y(s)/\surd E \\ x(s) \en... ...ft[ \begin{array}{c} y(s)/\surd E \\ x(s) \end{array} \right].$$

for which the appropriate Prüfer transformation would be

$$\begin{array}{cccc} \begin{array}{ccc} \rho & = & \surd\o... ... y & = & - \surd E\ \rho\sin\theta. \end{array} \end{array}$$

It then follows that

$$\frac{d\rho}{ds}$$ = 0 (300)

$$\frac{d\theta}{ds}$$ = -1 (301)

which is the expected result of high symmetry and great simplicity.