GSO003 Problem 10

1. Determining the Wave Function

The ground-state spatial dependence of the 3D isotropic harmonic oscillator is proportional to $e^{-\alpha^2 r^2/2}$ , where $\alpha = \sqrt{m^*\omega/\hbar}$ . The first-excited level ( $N = 1$ ) is three-fold degenerate, with Cartesian basis states:

\psi_x = N_1 x e^{-\alpha^2 r^2/2}, \quad \psi_y = N_1 y e^{-\alpha^2 r^2/2}, \quad \psi_z = N_1 z e^{-\alpha^2 r^2/2}

To form an eigenstate of $\hat{L}_z = \frac{\hbar}{i}\frac{\partial}{\partial\phi}$ with eigenvalue $+\hbar$ , take the combination $\psi_x + i\psi_y$ . Using $x + iy = r\sin\theta\, e^{i\phi}$ :

\psi = N\, r\sin\theta\, e^{i\phi}\, e^{-\alpha^2 r^2/2}

This satisfies $\hat{L}_z\psi = +\hbar\psi$ .

2. Normalization Constant

From $\int|\psi|^2\,dV = 1$ :

|N|^2 \left(\int_0^\infty r^4 e^{-\alpha^2 r^2}\,dr\right)\left(\int_0^\pi \sin^3\theta\,d\theta\right)(2\pi) = 1

Using the Gaussian integral formula twice (differentiating with respect to $a$ ):

\int_0^\infty r^4 e^{-\alpha^2 r^2}\,dr = \frac{3\sqrt{\pi}}{8\alpha^5}

\int_0^\pi \sin^3\theta\,d\theta = \frac{4}{3}

Substituting:

|N|^2 \cdot \frac{3\sqrt{\pi}}{8\alpha^5} \cdot \frac{4}{3} \cdot 2\pi = |N|^2 \frac{\pi^{3/2}}{\alpha^5} = 1 \implies |N|^2 = \frac{\alpha^5}{\pi^{3/2}}

3. Probability Current Density and Steady Current

The probability current density is:

\boldsymbol{j} = \frac{\hbar}{m^*}\,\text{Im}(\psi^*\nabla\psi)

Writing $\psi = |\psi|e^{i\phi}$ and using $\nabla\phi = \frac{1}{r\sin\theta}\hat{\phi}$ :

\boldsymbol{j} = \frac{\hbar}{m^*}|\psi|^2\nabla\phi = \frac{\hbar}{m^*}|N|^2 r\sin\theta\, e^{-\alpha^2 r^2}\,\hat{\phi}

Steady current density (charge $-q$ ):

\boldsymbol{i} = -q\boldsymbol{j} = -\frac{q\hbar}{m^*}|N|^2 r\sin\theta\, e^{-\alpha^2 r^2}\,\hat{\phi}

4. Magnetic Flux Density at the Origin via Biot–Savart

The Biot–Savart law gives the contribution from volume element $dV$ at position $\boldsymbol{r}$ to the field at the origin. Integrating the $\phi$ -component over $\phi$ from $0$ to $2\pi$ eliminates $\hat{x}$ and $\hat{y}$ parts of $\hat{\theta}$ , leaving only the $z$ -component:

B_z = \frac{\mu_0}{2}\int_0^\infty dr\int_0^\pi d\theta\; i_\phi \sin^2\theta

Substituting $i_\phi$ :

B = |B_z| = \frac{\mu_0 q\hbar |N|^2}{2m^*}\left(\int_0^\infty r\,e^{-\alpha^2 r^2}\,dr\right)\left(\int_0^\pi \sin^3\theta\,d\theta\right)

Radial integral: $\int_0^\infty r\,e^{-\alpha^2 r^2}\,dr = \frac{1}{2\alpha^2}$ . Therefore:

B = \frac{\mu_0 q\hbar |N|^2}{2m^*} \cdot \frac{1}{2\alpha^2} \cdot \frac{4}{3} = \frac{\mu_0 q\hbar}{3m^*\alpha^2}|N|^2

Substituting $|N|^2 = \frac{\alpha^5}{\pi^{3/2}}$ and $\alpha^2 = \frac{m^*\omega}{\hbar}$ :

B = \frac{\mu_0 q}{3\pi^{3/2}}\sqrt{\frac{m^*\omega^3}{\hbar}}

5. Numerical Evaluation

\frac{m^*\omega^3}{\hbar} = \frac{2.0 \times 10^{-31} \times (10^{15})^3}{10^{-34}} = 2.0 \times 10^{48}

\sqrt{\frac{m^*\omega^3}{\hbar}} = \sqrt{2} \times 10^{24}

B = \frac{4\pi \times 10^{-7} \times 3.0 \times 10^{-19}}{3\pi^{3/2}} \times \sqrt{2} \times 10^{24} = \frac{4\sqrt{2}}{\sqrt{\pi}} \times 10^{-2}\,\text{T}

B^2 = \frac{32}{\pi} \times 10^{-4}

\pi B^2 \times 10^4 = \pi \cdot \frac{32}{\pi} \times 10^{-4} \times 10^4 = 32

Answer: 32

GSO003 Problem 10

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

Problem Statement

Constraints

Input Format

Solution

1. Determining the Wave Function

2. Normalization Constant

3. Probability Current Density and Steady Current

4. Magnetic Flux Density at the Origin via Biot–Savart

5. Numerical Evaluation