Representer theorem

The solution to the Schrödinger equation, the wavefunction, describes the quantum mechanical properties of a particle on microscopic scales. Measurable quantities such as position, momentum and energy are all derived from the wavefunction.

$\mathrm{i}\hslash \frac{\partial }{\partial t}\psi (x,t)=-\frac{{\hslash }^2}{2m}\frac{{\partial }^2}{\partial x^2}\psi (x,t)+V(x)\psi (x,t),$ where $\hslash$ is the reduced Planck constant, $m$ is the mass of the particle, $\mathrm{i}$ is the imaginary unit and $t$ is time.

One peculiar potential that can be solved exactly is when the electric quadrupole moment is the dominant term of an infinitely long cylinder of charge.

It can be shown that the Schrödinger equation is solvable for a cylindrically symmetric electric quadrupole, thus indicating that the quadrupole term of an infinitely long cylinder can be quantized. In the physics of classical electrodynamics, it can be easily shown that the scalar potential and associated mechanical potential energy of a cylindrically symmetric quadrupole is as follows:

${\mathbf{V}}_{\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{d}}=\frac{\lambda d^{2}Cos[2\varphi ]}{4\pi {ϵ}_0{s}^2}$ (SI units)

${\mathbf{V}}_{\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{d}}=\frac{Q\lambda d^{2}Cos[2\varphi ]}{4\pi {ϵ}_0{s}^2}$ (SI units)

Cylindrical symmetry should be used when solving the equation. The time independent Schrödinger equation becomes the following in cylindrical symmetry.

$E\psi (x)=-\frac{{\hslash }^2}{2ms}\frac{\partial }{\partial s}(s\frac{\partial }{\partial s})\psi (s,\varphi )-\frac{{\hslash }^2}{2m{s}^2}\frac{{\partial }^2}{\partial {\varphi }^2}\psi (s,\varphi )+\frac{Q\lambda d^{2}Cos[2\varphi ]}{4\pi {ϵ}_0{s}^2}\psi (s,\varphi ),$

Using the technique Separation of Variables, the above equation can be written as two ordinary differential equations in both the radial and azimuthal directions. The radial equation is Bessel's equation as can be seen below. If one changes variables to $x=ks$, Bessel's equation is exactly obtained.

$\frac{1}{x}\frac{\partial }{\partial x}(x\frac{\partial }{\partial x})S(x)+(1-\frac{{\nu }^2}{x^2})S(x)=0$

Azimuthal equation

The azimuthal equation is Mathieu equation, is as follows:

$\frac{{\partial }^2}{\partial {\varphi }^2}\mathrm{\Phi }(\varphi )+({\nu }^2-\frac{\lambda qm{d}^2}{2\pi {ϵ}_0\hslash }Cos[2\varphi ])\mathrm{\Phi }[\varphi ]=0$

Since the canonical form of Mathieu's equation can be written as follows, it can be shown that ${\nu }^2$ corresponds to a and $\frac{\lambda qm{d}^2}{2\pi {ϵ}_0\hslash }$ corresponds to q.

$\frac{d^{2}y}{d{x}^2}+[a-2q\cos (2x)]y=0.$

The solution of the Mathieu equation is expressed in terms of Mathieu cosine $C(a,q,x)$ Mathieu sine $S(a,q,x)$ for a unique a and q. This indicates that the quadrupole moment can be quantized in order of the Mathieu characteristic values $a_n$ and $b_n$.

In general, Mathieu functions are not periodic. The term q must be that of a characteristic value in order for Mathieu functions to be periodic. Immediately, it can be shown that the solution of the radial equation highly depends on what characteristic values are seen in this case.

References

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External links

• - Cylindrical Multipole Moments
• MULTIPOLE EXPANSION
• The nonvanishing coefficients of the dipole moment expansion in axially symmetric molecules