Magnetic Field Induced by Orbital Angular Momentum of a Carrier Trapped in a Quantum Dot

Problem Statement

Consider a semiconductor quantum dot formed by a spherically symmetric isotropic harmonic potential $V(r) = \frac{1}{2}m^*\omega^2 r^2$V(r)=21​m∗ω2r2 (where $r = \sqrt{x^2+y^2+z^2}$r=x2+y2+z2​) centered at the origin in a three-dimensional Cartesian coordinate system $(x, y, z)$(x,y,z).

A single carrier (electron) with effective mass $m^*$m∗ and charge $-q$−q ($q > 0$q>0) is confined in this quantum dot.

Among the stationary states of the one-particle Schrödinger equation, consider the state $|\psi\rangle$∣ψ⟩ with principal quantum number $N = 1$N=1 (one energy level above the ground state $N = 0$N=0) and with $z$z-component of orbital angular momentum $\hat{L}_z$L^z​ equal to $+\hbar$+ℏ. The wave function $\psi(\boldsymbol{r})$ψ(r) of this state is a normalized complex linear combination of the three first-excited Cartesian states (each with one quantum of excitation along $x$x, $y$y, or $z$z).

The carrier creates a probability current density $\boldsymbol{j}(\boldsymbol{r})$j(r) in space. In a stationary state, this gives rise to a steady current density $\boldsymbol{i}(\boldsymbol{r}) = -q\boldsymbol{j}(\boldsymbol{r})$i(r)=−qj(r).

Using the Biot–Savart law, this distributed steady current induces a magnetic flux density $B$B at the origin. The contribution from the carrier's spin magnetic moment is neglected.

Find $B$B and give the answer in the specified format.

Constraints

• Effective mass of carrier: $m^* = 2.0 \times 10^{-31}\,\text{kg}$m∗=2.0×10−31kg
• Carrier charge: $q = 3.0 \times 10^{-19}\,\text{C}$q=3.0×10−19C
• Angular frequency of the harmonic potential: $\omega = 1.0 \times 10^{15}\,\text{rad/s}$ω=1.0×1015rad/s
• Reduced Planck constant: $\hbar = 1.0 \times 10^{-34}\,\text{J}\cdot\text{s}$ℏ=1.0×10−34J⋅s
• Vacuum permeability: $\mu_0 = 4\pi \times 10^{-7}\,\text{T}\cdot\text{m/A}$μ0​=4π×10−7T⋅m/A

Input Format

Using the magnetic flux density $B\,[\text{T}]$B[T] at the origin, compute:

$$\pi B^2 \times 10^4$$πB2×104

and give the answer as a positive integer.