The quantum cylindrical quadrupole is a solution to the Schrödinger equation,
i ℏ ∂ ∂ t ψ ( x , t ) = − ℏ 2 2 m ∂ 2 ∂ x 2 ψ ( x , t ) + V ( x ) ψ ( x , t ) , {\displaystyle \mathrm {i} \hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),}
where ℏ {\displaystyle \hbar } is the reduced Planck constant, m {\displaystyle m} is the mass of the particle, i {\displaystyle \mathrm {i} } is the imaginary unit and t {\displaystyle 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 shown that the scalar potential and associated mechanical potential energy of a cylindrically symmetric quadrupole is as follows:
V q u a d = λ d 2 C o s [ 2 ϕ ] 4 π ϵ 0 s 2 {\displaystyle \mathbf {V} _{\mathrm {quad} }={\frac {\lambda d^{2}Cos[2\phi ]}{4\pi \epsilon _{0}s^{2}}}} (SI units)
V q u a d = Q λ d 2 C o s [ 2 ϕ ] 4 π ϵ 0 s 2 {\displaystyle \mathbf {V} _{\mathrm {quad} }={\frac {Q\lambda d^{2}Cos[2\phi ]}{4\pi \epsilon _{0}s^{2}}}} (SI units) Using cylindrical symmetry, the time independent Schrödinger equation becomes the following:
E ψ ( x ) = − ℏ 2 2 m s ∂ ∂ s ( s ∂ ∂ s ) ψ ( s , ϕ ) − ℏ 2 2 m s 2 ∂ 2 ∂ ϕ 2 ψ ( s , ϕ ) + Q λ d 2 C o s [ 2 ϕ ] 4 π ϵ 0 s 2 ψ ( s , ϕ ) . {\displaystyle E\psi (x)=-{\frac {\hbar ^{2}}{2ms}}{\frac {\partial }{\partial s}}(s{\frac {\partial }{\partial s}})\psi (s,\phi )-{\frac {\hbar ^{2}}{2ms^{2}}}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\psi (s,\phi )+{\frac {Q\lambda d^{2}Cos[2\phi ]}{4\pi \epsilon _{0}s^{2}}}\psi (s,\phi ).}
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