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Particle in a Box

Classical vs Quantum particle in a box

PIB-wiki

Figure 1:Classical (A) versus quantum (B-F) behavior of a particle in a box. The horizontal axis is position; the vertical axis shows the real (blue) and imaginary (red) parts of the wavefunction ψn(x)\psi_n(x). States B, C, D are the n=1,2,3n=1,2,3 eigenfunctions of the Hamiltonian, while E and F are not.

Solving the Schrödinger Equation for the Particle in a Box (PIB)

Particle in a box

Figure 2:Particle in a box subject to infinitely high potential walls.

The Schrödinger equation for a particle in a box (PIB) is defined by a Hamiltonian operator that incorporates a potential energy which is infinitely large at the boundaries of the box and zero inside. This potential confines the particle within the box, where it can only possess kinetic energy.

V(x)={∞x=0 or x=L00<x<LV(x) = \begin{cases} \infty & x = 0 \text{ or } x = L \\ 0 & 0 < x < L \end{cases}
(1)
ψ(0)=ψ(L)=0\psi(0) = \psi(L) = 0
(2)
H^=K^=−ℏ22md2dx2\hat{H} = \hat{K} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}
(3)
H^ψ(x)=Eψ(x)\hat{H} \psi(x) = E \psi(x)
(4)
−ℏ22md2dx2ψ(x)=Eψ(x)-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) = E \psi(x)
(5)
ψ′′(x)=−k2ψ(x)\psi''(x) = -k^2 \psi(x)
(6)
k2=2mEℏ2k^2 = \frac{2mE}{\hbar^2}
(7)

Solution and Boundary Conditions

ψ′′(x)=−k2ψ(x)\psi''(x) = -k^2 \psi(x)
(8)
ψ(x)=c1eikx+c2e−ikx=Acos⁡(kx)+Bsin⁡(kx)\psi(x) = c_1 e^{ikx} + c_2 e^{-ikx} = A \cos(kx) + B \sin(kx)
(9)
ψ(x)=Bsin⁡(kx)\psi(x) = B \sin(kx)
(10)
Bsin⁡(kL)=0B \sin(kL) = 0
(11)
kL=nπork=nπLkL = n\pi \quad \text{or} \quad k = \frac{n\pi}{L}
(12)
ψ(x)=Bsin⁡(nπLx)\psi(x) = B \sin\left(\frac{n\pi}{L}x\right)
(13)
En=n2h28mL2E_n = \frac{n^2 h^2}{8mL^2}
(14)

Wavefunctions Must Be Normalized

$$ \int_0^L \psi_n(x)^2 , dx = 1

$$

Bn2∫0Lsin⁡2(nπxL)dx=Bn22∫0L[1−cos⁡(2nπxL)]dx=1B_n^2 \int_0^L \sin^2\left(\frac{n\pi x}{L}\right) dx = \frac{B_n^2}{2} \int_0^L \left[ 1 - \cos\left(\frac{2n\pi x}{L}\right) \right] dx =1
(15)
Bn22⋅L=1\frac{B_n^2}{2} \cdot L = 1
(16)
Bn=2LB_n = \sqrt{\frac{2}{L}}
(17)

PIB Eigenfunctions and Eigenvalues

Source
import numpy as np
import matplotlib.pyplot as plt

# Parameters
L = 1.0  # Length of the box
x = np.linspace(0, L, 1000)  # Position array
n_max = 5  # Maximum quantum number to display

# Constants
hbar = 1.0545718e-34  # Reduced Planck's constant (J·s)
m = 9.10938356e-31  # Mass of an electron (kg)

# Functions
def psi_n(n, x, L):
    """Wavefunction for a particle in a 1D box."""
    return np.sqrt(2 / L) * np.sin(n * np.pi * x / L)

def psi_n_squared(n, x, L):
    """Probability density (wavefunction squared)."""
    return psi_n(n, x, L) ** 2

def energy_levels(n, L):
    """Energy level for the particle in the box."""
    return (n**2 * np.pi**2 * hbar**2) / (2 * m * L**2)

# Energy formula string
energy_formula = r"$E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2}$"

# Plotting
fig, axs = plt.subplots(1, 2, figsize=(8, 6), sharey=True)

# Plot wavefunctions in the first subplot
for n in range(1, n_max + 1):
    wavefunction = psi_n(n, x, L)
    axs[0].plot(x, wavefunction + n**2, label=f'n={n}')
    axs[0].hlines(n**2, 0, L, colors='gray', linestyles='dashed')  # Energy level lines

# Wavefunction plot labels and title
axs[0].set_xlabel('Position (x)', fontsize=12)
axs[0].set_ylabel(r'$\psi_n(x)$', fontsize=12)
axs[0].legend(loc='upper right', bbox_to_anchor=(1.15, 1))

# Plot wavefunction squared (probability densities) in the second subplot
for n in range(1, n_max + 1):
    wavefunction_squared = psi_n_squared(n, x, L)
    axs[1].plot(x, wavefunction_squared + n**2, label=f'n={n}')
    axs[1].hlines(n**2, 0, L, colors='gray', linestyles='dashed')  # Energy level lines

# Probability density plot labels
axs[1].set_xlabel('Position (x)', fontsize=12)
axs[1].set_ylabel(r'$|\psi_n(x)|^2$', fontsize=12)

# Display the energy formula and levels in the second subplot
for n in range(1, n_max + 1):
    energy_value = energy_levels(n, L)
    axs[1].text(L, n**2, f"$E_{n}$ = {energy_value:.2e} J", verticalalignment='bottom', fontsize=12)

# Overall title for the figure
fig.suptitle(f'Wavefunctions and Energies for a 1D Particle in a Box (n=1 to n={n_max})')

# Make room for labels outside plot and improve layout
plt.tight_layout(rect=[0, 0, 1, 0.95])
plt.show()
<Figure size 800x600 with 2 Axes>

Discrete energy levels and zero point energy

Quantum particles are never at rest

E1=h2/8mL2E_1 = h^2/8mL^2
(22)
En+1−En=(2n+1)h28mL2E_{n+1} - E_n = (2n+1)\frac{h^2}{8mL^2}
(23)

Increasing Box size leads to more classical behavior

Source
import numpy as np
import matplotlib.pyplot as plt

# Constants (set h^2/8m = 1 for simplicity)
L1 = 1.0
L2 = 2.0
n_max = 5

def energy_levels(L, n_max):
    n = np.arange(1, n_max+1)
    E = (n**2) / (L**2)  # scaled energy
    return n, E

# Compute for two box sizes
n, E1 = energy_levels(L1, n_max)
_, E2 = energy_levels(L2, n_max)

fig, axs = plt.subplots(1, 2, figsize=(12, 5), sharey=True)

# Left: small box (L=1)
for i, E in enumerate(E1, start=1):
    axs[0].hlines(E, 0.9, 1.1, colors="tab:blue", linewidth=3)
    axs[0].text(1.15, E, f"n={i}", va="center")
axs[0].set_title("Small box (L=1)\nLarger spacing")
axs[0].set_xlim(0.8, 1.5)
axs[0].set_ylabel("Energy (scaled units)")

# Right: larger box (L=2)
for i, E in enumerate(E2, start=1):
    axs[1].hlines(E, 0.9, 1.1, colors="tab:green", linewidth=3)
    axs[1].text(1.15, E, f"n={i}", va="center")
axs[1].set_title("Larger box (L=2)\nSmaller spacing")
axs[1].set_xlim(0.8, 1.5)

# Add zero-point energy annotation
axs[0].annotate("Zero-point energy\n(n=1)", xy=(1.0, E1[0]), xytext=(1.3, E1[0]+2),
                arrowprops=dict(arrowstyle="->"))

for ax in axs:
    ax.set_xticks([])
    ax.set_xlabel("Particle in a box")
    ax.grid(True, axis="y", alpha=0.3)

plt.suptitle("Discrete Energy Levels and Zero-Point Energy", fontsize=14)
plt.tight_layout()
plt.show()
<Figure size 1200x500 with 2 Axes>

Non-uniform probabilities and nodes

  1. Nodes imply zero probability to find an electron in the box.

    • ∣ψn∣2=0|\psi_n|^2=0 at nodal points, which means zero probability.

    • This sharply contradicts classical mechanics, which bases its prediction on purely particle-like motion of the electron.

    • The existence of nodes implies a wave-like character of the electron.

  2. There are n−1n-1 nodes for quantum state nn

Source
import numpy as np
import matplotlib.pyplot as plt

# Parameters
n = 4
L = 1.0
Nsamples = 6000
bins = 60

# Grid and eigenstate
x = np.linspace(0, L, 1000)
psi = np.sqrt(2/L) * np.sin(n*np.pi*x/L)   # normalized
prob = psi**2

# Node locations (exclude 0 and L)
nodes = np.arange(1, n) * L / n

# Monte Carlo sampling via rejection
rng = np.random.default_rng(0)
samples = []
while len(samples) < Nsamples:
    u = rng.random(20000)
    x_prop = L * rng.random(20000)
    accept = u < (np.sin(n*np.pi*x_prop/L)**2)
    samples.extend(x_prop[accept].tolist())
samples = np.array(samples[:Nsamples])

# Plot
fig, axs = plt.subplots(1, 3, figsize=(14, 4))

# 1) psi_n(x)
axs[0].plot(x, psi, linewidth=2)
for xn in nodes:
    axs[0].axvline(xn, linestyle="--", linewidth=1)
axs[0].set_title(rf"$\psi_{n}(x)$ with {n-1} nodes")
axs[0].set_xlabel("x")
axs[0].set_ylabel(r"$\psi_n(x)$")
axs[0].set_xlim(0, L)
axs[0].grid(True, alpha=0.3)

# 2) |psi|^2
axs[1].plot(x, prob, linewidth=2)
for xn in nodes:
    axs[1].axvline(xn, linestyle="--", linewidth=1)
axs[1].set_title(rf"$|\psi_{n}(x)|^2$ (zero at nodes)")
axs[1].set_xlabel("x")
axs[1].set_ylabel(r"$|\psi_n(x)|^2$")
axs[1].set_xlim(0, L)
axs[1].grid(True, alpha=0.3)

# 3) Histogram vs analytic |psi|^2
axs[2].hist(samples, bins=bins, range=(0, L), density=True, alpha=0.35, label="Samples from $|\psi|^2$")
axs[2].plot(x, prob, linewidth=2, label="Analytic $|\psi|^2$")
for xn in nodes:
    axs[2].axvline(xn, linestyle="--", linewidth=1)
axs[2].set_title("Non-uniform probability & nodes")
axs[2].set_xlabel("x")
axs[2].set_ylabel("Probability density")
axs[2].set_xlim(0, L)
axs[2].legend()
axs[2].grid(True, alpha=0.3)

plt.suptitle("Non-uniform probabilities and nodes in a 1D box (n=4)", fontsize=14)
plt.tight_layout()
plt.show()
<>:52: SyntaxWarning: invalid escape sequence '\p'
<>:53: SyntaxWarning: invalid escape sequence '\p'
<>:52: SyntaxWarning: invalid escape sequence '\p'
<>:53: SyntaxWarning: invalid escape sequence '\p'
/tmp/ipykernel_3150/3569254680.py:52: SyntaxWarning: invalid escape sequence '\p'
  axs[2].hist(samples, bins=bins, range=(0, L), density=True, alpha=0.35, label="Samples from $|\psi|^2$")
/tmp/ipykernel_3150/3569254680.py:53: SyntaxWarning: invalid escape sequence '\p'
  axs[2].plot(x, prob, linewidth=2, label="Analytic $|\psi|^2$")
<Figure size 1400x400 with 3 Axes>

On time-dependence

Pure states

∣ψn(x,t)∣2=ψn(x,t)∗⋅ψn(x,t)=ψ(x)e−iEnt/ℏ⋅ψ(x)e+Ent/ℏ=∣ψ(x)∣2|\psi_n(x,t)|^2 = \psi_n(x,t)^{*}\cdot\psi_n(x,t) = \psi(x)e^{-iE_nt/\hbar}\cdot \psi(x)e^{+E_nt/\hbar} = |\psi(x)|^2
(24)
Source
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

# Parameters
L = 1.0  # Length of the box
hbar = 1.0  # Set hbar = 1 for simplicity
m = 1.0  # Set mass = 1 for simplicity

# Define spatial and time arrays
x = np.linspace(0, L,  200)  
t = np.linspace(0, 10, 200)  

n = 2

# Define energy for nth state
def energy_n(n, L):
    return (n**2 * np.pi**2 * hbar**2) / (2 * m * L**2)

# Define wavefunction for nth state
def psi_total(n, x, L, t):
    E_n = energy_n(n, L)
    return np.sqrt(2 / L) * np.sin(n * np.pi * x / L) * np.exp(-1j * E_n * t / hbar)


# Set up the figure and axis
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(11, 8))

# Set limits for the wavefunction and wavefunction squared
ax1.set_xlim(0, L)
ax1.set_ylim(-2, 2)
ax2.set_xlim(0, L)
ax2.set_ylim(0, 4)

# Set labels for the wavefunction plot
ax1.set_xlabel('Position (x)', fontsize=14)
ax1.set_ylabel(r'Re[$\psi(x,t)$]', fontsize=14)
ax1.set_title('Time Evolution of Wavefunction', fontsize=16)

# Set labels for the squared wavefunction plot
ax2.set_xlabel('Position (x)', fontsize=14)
ax2.set_ylabel(r'$|\psi(x,t)|^2$', fontsize=14)
ax2.set_title('Probability Density', fontsize=16)

### Animation settings 

# Initialize the wavefunction and its square plots
line_real, = ax1.plot([], [], lw=2)
line_prob, = ax2.plot([], [], lw=2)

# Initialization function for the plot
def init():
    line_real.set_data([], [])
    line_prob.set_data([], [])
    return line_real, line_prob

# Animation function which updates the plot each frame
def animate(i):
    t_val = t[i]
    psi = psi_total(n, x, L, t_val)
    psi_real = np.real(psi)
    psi_squared = np.abs(psi)**2
    line_real.set_data(x, psi_real)
    line_prob.set_data(x, psi_squared)
    return line_real, line_prob

# Create the animation
ani = FuncAnimation(fig, animate, init_func=init, frames=len(t), interval=150, blit=True)

plt.close(fig)  # Prevents static display of the last frame
HTML(ani.to_jshtml())
Loading...

Mixed states

ψ(x,t)=c1ψ1(x)e−iE1t/ℏ+c2ψ2e−iE2t/ℏ\psi(x,t) = c_1 \psi_1(x) e^{-iE_1t/\hbar} +c_2 \psi_2 e^{-iE_2t/\hbar}
(25)
Step-by-Step illustration of time-dependence

  1. The complex conjugate of ψ(x,t)\psi(x,t) is:

ψ∗(x,t)=c1∗ψ1(x)eiE1t/ℏ+c2∗ψ2(x)eiE2t/ℏ\psi^*(x,t) = c_1^* \psi_1(x) e^{iE_1t/\hbar} + c_2^* \psi_2(x) e^{iE_2t/\hbar}
(26)

  1. The squared wavefunction ∣ψ(x,t)∣2|\psi(x,t)|^2 is:

∣ψ(x,t)∣2=(c1ψ1(x)e−iE1t/ℏ+c2ψ2(x)e−iE2t/ℏ)(c1∗ψ1(x)eiE1t/ℏ+c2∗ψ2(x)eiE2t/ℏ)|\psi(x,t)|^2 = \left( c_1 \psi_1(x) e^{-iE_1t/\hbar} + c_2 \psi_2(x) e^{-iE_2t/\hbar} \right) \left( c_1^* \psi_1(x) e^{iE_1t/\hbar} + c_2^* \psi_2(x) e^{iE_2t/\hbar} \right)
(27)

  1. Expanding the product:

∣ψ(x,t)∣2=∣c1∣2∣ψ1(x)∣2+∣c2∣2∣ψ2(x)∣2+c1c2∗ψ1(x)ψ2∗(x)e−i(E1−E2)t/ℏ+c1∗c2ψ1∗(x)ψ2(x)ei(E1−E2)t/ℏ|\psi(x,t)|^2 = |c_1|^2 |\psi_1(x)|^2 + |c_2|^2 |\psi_2(x)|^2 + c_1 c_2^* \psi_1(x) \psi_2^*(x) e^{-i(E_1 - E_2)t/\hbar} + c_1^* c_2 \psi_1^*(x) \psi_2(x) e^{i(E_1 - E_2)t/\hbar}
(28)

Final Expression

∣ψ(x,t)∣2=∣c1∣2∣ψ1(x)∣2+∣c2∣2∣ψ2(x)∣2+2 Re[c1c2∗ψ1(x)ψ2∗(x)e−i(E1−E2)t/ℏ]|\psi(x,t)|^2 = |c_1|^2 |\psi_1(x)|^2 + |c_2|^2 |\psi_2(x)|^2 + 2 \, \text{Re} \left[ c_1 c_2^* \psi_1(x) \psi_2^*(x) e^{-i(E_1 - E_2)t/\hbar} \right]
(29)

This consists of:

  • The probability densities of ψ1(x)\psi_1(x) and ψ2(x)\psi_2(x) with coefficients ∣c1∣2|c_1|^2 and ∣c2∣2|c_2|^2.

  • A time-dependent interference term that oscillates with the frequency (E1−E2)/ℏ(E_1 - E_2)/\hbar.

Source
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation
from IPython.display import HTML

# Parameters
L = 1.0  # Length of the box
hbar = 1.0  # Set hbar = 1 for simplicity
m = 1.0  # Set mass = 1 for simplicity

n_states = [1, 2]  # Quantum numbers for the superposition
coeffs = [1, 1]  # Coefficients for the superposition
coeffs = np.array(coeffs) / np.sqrt(np.sum(np.array(coeffs) ** 2))  # Normalize coefficients

# Define spatial and time arrays
x = np.linspace(0, L,  200)  
t = np.linspace(0, 10, 200)  

# Define wavefunction for nth state
def psi_n(n, x, L):
    return np.sqrt(2 / L) * np.sin(n * np.pi * x / L)

# Define energy for nth state
def energy_n(n, L):
    return (n**2 * np.pi**2 * hbar**2) / (2 * m * L**2)

# Define time-dependent wavefunction
def psi_total(x, t, L, coeffs, n_states):
    psi_t = np.zeros_like(x, dtype=complex)
    for idx, n in enumerate(n_states):
        psi_x = psi_n(n, x, L)
        E_n = energy_n(n, L)
        time_factor = np.exp(-1j * E_n * t / hbar)
        psi_t += coeffs[idx] * psi_x * time_factor
    return psi_t

# Set up the figure and axis
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(11, 8))

# Set limits for the wavefunction and wavefunction squared
ax1.set_xlim(0, L)
ax1.set_ylim(-2, 2)
ax2.set_xlim(0, L)
ax2.set_ylim(0, 4)

# Set labels for the wavefunction plot
ax1.set_xlabel('Position (x)', fontsize=14)
ax1.set_ylabel(r'Re[$\psi(x,t)$]', fontsize=14)
ax1.set_title('Time Evolution of Wavefunction', fontsize=16)

# Set labels for the squared wavefunction plot
ax2.set_xlabel('Position (x)', fontsize=14)
ax2.set_ylabel(r'$|\psi(x,t)|^2$', fontsize=14)
ax2.set_title('Probability Density', fontsize=16)

### Animation settings 

# Initialize the wavefunction and its square plots
line_real, = ax1.plot([], [], lw=2)
line_prob, = ax2.plot([], [], lw=2)

# Initialization function for the plot
def init():
    line_real.set_data([], [])
    line_prob.set_data([], [])
    return line_real, line_prob

# Animation function which updates the plot each frame
def animate(i):
    t_val = t[i]
    psi = psi_total(x, t_val, L, coeffs, n_states)
    psi_real = np.real(psi)
    psi_squared = np.abs(psi)**2
    line_real.set_data(x, psi_real)
    line_prob.set_data(x, psi_squared)
    return line_real, line_prob

# Create the animation
ani = FuncAnimation(fig, animate, init_func=init, frames=len(t), interval=150, blit=True)

plt.close(fig)  # Prevents static display of the last frame
HTML(ani.to_jshtml())
Loading...

Quantum PIB in 3D

pib1

Figure 3:Particle in a 3D box subject to infinitely high potential walls.

H^ψ(x,y,z)=Eψ(x,y,z)\hat{H}\psi(x,y,z) = E\psi(x,y,z)
(30)
−ℏ22m(∂2ψ∂x2+∂2ψ∂y2+∂2ψ∂z2)=Eψ{-\frac{\hbar^2}{2m}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) = E\psi}
(31)
∫−∞∞∫−∞∞∫−∞∞∣ψ(x,y,z)∣2dxdydz=1{\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}\left|\psi(x,y,z)\right|^2dxdydz = 1}
(32)
−ℏ22mΔψ=Eψwith ψ(a,y,z)=ψ(x,b,z)=ψ(x,y,c)=0and ψ(0,y,z)=ψ(x,0,z)=ψ(x,y,0)=0{-\frac{\hbar^2}{2m}\Delta\psi = E\psi} \\ {\textnormal{with }\psi(a,y,z) = \psi(x,b,z) = \psi(x,y,c) = 0} \\ {\textnormal{and }\psi(0,y,z) = \psi(x,0,z) = \psi(x,y,0) = 0}
(33)
ψ(x,y,z)=X(x)Y(y)Z(z){\psi(x,y,z) = X(x)Y(y)Z(z)}
(34)
−ℏ22m[1X(x)d2X(x)dx2+1Y(y)d2Y(y)dy2+1Z(z)d2Z(z)dz2]=E{-\frac{\hbar^2}{2m}\left[\frac{1}{X(x)}\frac{d^2X(x)}{dx^2} + \frac{1}{Y(y)}\frac{d^2Y(y)}{dy^2} + \frac{1}{Z(z)}\frac{d^2Z(z)}{dz^2}\right] = E}
(35)
−ℏ22m[1X(x)d2X(x)dx2]=Ex with X(0)=X(a)=0−ℏ22m[1Y(y)d2Y(y)dy2]=Ey with Y(0)=Y(b)=0−ℏ22m[1Z(z)d2Z(z)dz2]=Ez with Z(0)=Z(c)=0{-\frac{\hbar^2}{2m}\left[\frac{1}{X(x)}\frac{d^2X(x)}{dx^2}\right] = E_x\textnormal{ with }X(0) = X(a) = 0}\\ {-\frac{\hbar^2}{2m}\left[\frac{1}{Y(y)}\frac{d^2Y(y)}{dy^2}\right] = E_y\textnormal{ with }Y(0) = Y(b) = 0}\\ {-\frac{\hbar^2}{2m}\left[\frac{1}{Z(z)}\frac{d^2Z(z)}{dz^2}\right] = E_z\textnormal{ with }Z(0) = Z(c) = 0}
(36)
X(x)=2asin⁡(nxπxa)Y(y)=2bsin⁡(nyπyb)Z(z)=2csin⁡(nzπzc){X(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n_x\pi x}{a}\right)}\\ {Y(y) = \sqrt{\frac{2}{b}}\sin\left(\frac{n_y\pi y}{b}\right)}\\ {Z(z) = \sqrt{\frac{2}{c}}\sin\left(\frac{n_z\pi z}{c}\right)}
(37)
Source
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D  # noqa: F401

# -------- Parameters --------
L = 1.0
nx, ny, nz = 1, 2, 3        # quantum numbers
N = 60                      # grid points per axis

# -------- Wavefunction on a 3D grid --------
x = np.linspace(0, L, N)
y = np.linspace(0, L, N)
z = np.linspace(0, L, N)
X, Y, Z = np.meshgrid(x, y, z, indexing='ij')

norm = (2.0/np.sqrt(L**3))
psi = norm * np.sin(nx*np.pi*X/L) * np.sin(ny*np.pi*Y/L) * np.sin(nz*np.pi*Z/L)
rho = psi**2  # probability density

# -------- Create illustrative views --------
# 1) Three orthogonal central slices: xy|_{z=L/2}, xz|_{y=L/2}, yz|_{x=L/2}
z_mid = np.argmin(np.abs(z - L/2))
y_mid = np.argmin(np.abs(y - L/2))
x_mid = np.argmin(np.abs(x - L/2))

slice_xy = rho[:, :, z_mid]
slice_xz = rho[:, y_mid, :]
slice_yz = rho[x_mid, :, :]

# 2) "Isosurface-like" point cloud: plot points above a probability threshold
# Choose threshold so that top ~2% of voxels are shown
flat = rho.ravel()
thresh = np.quantile(flat, 0.98)
mask = rho >= thresh
pts = np.column_stack((X[mask], Y[mask], Z[mask]))

# Downsample if there are too many points
max_pts = 8000
if pts.shape[0] > max_pts:
    idx = np.random.choice(pts.shape[0], size=max_pts, replace=False)
    pts = pts[idx]

# -------- Plot --------
fig = plt.figure(figsize=(12, 9))

# XY slice
ax1 = fig.add_subplot(2, 2, 1)
im1 = ax1.imshow(slice_xy.T, origin='lower', extent=[0, L, 0, L], aspect='equal')
ax1.set_title(rf"$|\psi(x,y,z=L/2)|^2$  (n=({nx},{ny},{nz}))")
ax1.set_xlabel("x")
ax1.set_ylabel("y")
fig.colorbar(im1, ax=ax1, shrink=0.8)

# XZ slice
ax2 = fig.add_subplot(2, 2, 2)
im2 = ax2.imshow(slice_xz.T, origin='lower', extent=[0, L, 0, L], aspect='equal')
ax2.set_title(rf"$|\psi(x,y=L/2,z)|^2$")
ax2.set_xlabel("x")
ax2.set_ylabel("z")
fig.colorbar(im2, ax=ax2, shrink=0.8)

# YZ slice
ax3 = fig.add_subplot(2, 2, 3)
im3 = ax3.imshow(slice_yz.T, origin='lower', extent=[0, L, 0, L], aspect='equal')
ax3.set_title(rf"$|\psi(x=L/2,y,z)|^2$")
ax3.set_xlabel("y")
ax3.set_ylabel("z")
fig.colorbar(im3, ax=ax3, shrink=0.8)

# 3D point cloud (isosurface-like)
ax4 = fig.add_subplot(2, 2, 4, projection='3d')
ax4.scatter(pts[:, 0], pts[:, 1], pts[:, 2], s=2, alpha=0.4)
ax4.set_title("High-probability region (point cloud)")
ax4.set_xlabel("x")
ax4.set_ylabel("y")
ax4.set_zlabel("z")
ax4.set_xlim(0, L); ax4.set_ylim(0, L); ax4.set_zlim(0, L)

plt.suptitle("3D Particle-in-a-Box Orbital: n = (1, 2, 3)", y=0.98)
plt.tight_layout()
plt.show()
<Figure size 1200x900 with 7 Axes>
Table of energy levels for 3D box
Quantum Numbers (nx,ny,nz( n_x, n_y, n_z)Energy (Enx,ny,nz)( E_{n_x, n_y, n_z} )Degeneracy
(1,1,1)(3ℏ22mL2)( \frac{3\hbar^2}{2mL^2} ) 1
(1,1,2), (1,2,1), (2,1,1)(6ℏ22mL2)( \frac{6\hbar^2}{2mL^2} ) 3
(1,1,3), (1,3,1), (3,1,1)(11ℏ22mL2)( \frac{11\hbar^2}{2mL^2} ) 3
(1,2,2), (2,1,2), (2,2,1)(9ℏ22mL2)( \frac{9\hbar^2}{2mL^2} ) 3
(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)(14ℏ22mL2)( \frac{14\hbar^2}{2mL^2} )6
(1,3,3), (3,1,3), (3,3,1)(19ℏ22mL2)( \frac{19\hbar^2}{2mL^2} )3
(2,2,2)(12ℏ22mL2)( \frac{12\hbar^2}{2mL^2} )1
(2,2,3), (2,3,2), (3,2,2)(17ℏ22mL2)( \frac{17\hbar^2}{2mL^2} )3
(2,3,3), (3,2,3), (3,3,2)(22ℏ22mL2)( \frac{22\hbar^2}{2mL^2} )3
(3,3,3)(27ℏ22mL2)( \frac{27\hbar^2}{2mL^2} )1
(1,1,4), (1,4,1), (4,1,1)(18ℏ22mL2)( \frac{18\hbar^2}{2mL^2} )3
(1,2,4), (1,4,2), (2,1,4), (2,4,1), (4,1,2), (4,2,1)(21ℏ22mL2)( \frac{21\hbar^2}{2mL^2} )6
(1,3,4), (1,4,3), (3,1,4), (3,4,1), (4,1,3), (4,3,1)(26ℏ22mL2)( \frac{26\hbar^2}{2mL^2} )6
(1,4,4), (4,1,4), (4,4,1)(33ℏ22mL2)( \frac{33\hbar^2}{2mL^2} )3
(2,2,4), (2,4,2), (4,2,2)(24ℏ22mL2)( \frac{24\hbar^2}{2mL^2} )3
(2,3,4), (2,4,3), (3,2,4), (3,4,2), (4,2,3), (4,3,2)(29ℏ22mL2)( \frac{29\hbar^2}{2mL^2} )6
(2,4,4), (4,2,4), (4,4,2)(36ℏ22mL2)( \frac{36\hbar^2}{2mL^2} )3
(3,3,4), (3,4,3), (4,3,3)(34ℏ22mL2)( \frac{34\hbar^2}{2mL^2} )3
(3,4,4), (4,3,4), (4,4,3)(39ℏ22mL2)( \frac{39\hbar^2}{2mL^2} )3
(4,4,4)(48ℏ22mL2)( \frac{48\hbar^2}{2mL^2} )1

Note on Computing Average Properties from a Wave Function

Because of the probabilistic interpretation of the wave function, average properties can be computed from the wave function. The general formula is

⟨A⟩=∫ψ∗(x)A^ψ(x)dx\langle A \rangle = \int \psi^*(x)\hat{A}\psi(x)dx
(42)

where A^\hat{A} is any operator. This could be momentum, kinetic energy, and so on. Below are a few problems illustrating how to do such calculations.

To calculate the average position (or expectation value of position) for a particle in a 1D box, follow the steps shown in the example below.

Example: Calculate average (expectation) of position, ⟨x⟩\langle x\rangle

1. Wavefunction of the Particle in a 1D Box

The wavefunction for a particle in a 1D box of length LL with infinite potential walls at x=0x = 0 and x=Lx = L is given by:

ψn(x)=2Lsin⁡(nπxL)\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)
(43)

where:

  • nn is the quantum number (1, 2, 3, ...),

  • LL is the length of the box,

  • ψn(x)\psi_n(x) is the wavefunction.

2. Expectation Value of Position ⟨x⟩\langle x \rangle

The expectation value of the position xx for a particle is given by:

⟨x⟩=∫0Lx∣ψn(x)∣2 dx\langle x \rangle = \int_0^L x |\psi_n(x)|^2 \, dx
(44)

This integral gives the average position of the particle based on the probability density ∣ψn(x)∣2|\psi_n(x)|^2. For the wavefunction ψn(x)\psi_n(x), the probability density is:

∣ψn(x)∣2=(2Lsin⁡(nπxL))2=2Lsin⁡2(nπxL)|\psi_n(x)|^2 = \left( \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right) \right)^2 = \frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right)
(45)

3. Set Up the Integral

Substitute ∣ψn(x)∣2|\psi_n(x)|^2 into the expression for ⟨x⟩\langle x \rangle:

⟨x⟩=∫0Lx2Lsin⁡2(nπxL) dx\langle x \rangle = \int_0^L x \frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right) \, dx
(46)

This is the integral you need to solve to find the average position.

4. Solve the Integral

The integral can be simplified using known trigonometric identities. First, use the identity:

sin⁡2θ=12(1−cos⁡(2θ))\sin^2 \theta = \frac{1}{2} \left(1 - \cos(2\theta)\right)
(47)

So, the integral becomes:

⟨x⟩=2L∫0Lx(12(1−cos⁡(2nπxL)))dx\langle x \rangle = \frac{2}{L} \int_0^L x \left( \frac{1}{2} \left( 1 - \cos\left( \frac{2n\pi x}{L} \right) \right) \right) dx
(48)

Simplifying:

⟨x⟩=1L∫0Lx(1−cos⁡(2nπxL))dx\langle x \rangle = \frac{1}{L} \int_0^L x \left( 1 - \cos\left( \frac{2n\pi x}{L} \right) \right) dx
(49)

Now split this into two integrals:

⟨x⟩=1L(∫0Lx dx−∫0Lxcos⁡(2nπxL)dx)\langle x \rangle = \frac{1}{L} \left( \int_0^L x \, dx - \int_0^L x \cos\left( \frac{2n\pi x}{L} \right) dx \right)
(50)

5. Evaluate the Integrals

  • The first integral is straightforward:

∫0Lx dx=L22\int_0^L x \, dx = \frac{L^2}{2}
(51)

  • The second integral can be solved using integration by parts or by referring to standard integral tables. It turns out that this integral evaluates to 0 for any integer nn. Thus:

∫0Lxcos⁡(2nπxL)dx=0\int_0^L x \cos\left( \frac{2n\pi x}{L} \right) dx = 0
(52)

6. Final Result

Thus, the expectation value of position simplifies to:

⟨x⟩=1L×L22=L2\langle x \rangle = \frac{1}{L} \times \frac{L^2}{2} = \frac{L}{2}
(53)

7. Interpretation

For any quantum state nn, the average position ⟨x⟩\langle x \rangle of a particle in a 1D box is always:

⟨x⟩=L2\langle x \rangle = \frac{L}{2}
(54)

This result makes sense intuitively because, due to the symmetry of the problem, the particle is equally likely to be found on either side of the box, so its average position is right in the middle of the box at x=L2x = \frac{L}{2}.

Summary

To find the average position of a particle in a 1D box for a general wavefunction:

  • Use the wavefunction ψn(x)\psi_n(x),

  • Set up the expectation value integral ⟨x⟩=∫0L⋅∣ψn(x)∣2 dx\langle x \rangle = \int_0^L \cdot |\psi_n(x)|^2 \, dx,

  • Solve the integral, which results in ⟨x⟩=L2\langle x \rangle = \frac{L}{2} for all nn.

Thus, the particle’s average position is always at the midpoint of the box, independent of the quantum number nn.

Source
# Visual proof that the positive and negative areas cancel:
# ∫_0^L x cos(2 n π x / L) dx = 0

import numpy as np
import matplotlib.pyplot as plt

# Parameters (you can change n or L for other examples)
n = 1
L = 1.0

# Define function
x = np.linspace(0, L, 1500)
f = x * np.cos(2 * np.pi * n * x / L)

# Numerical integral for confirmation
num_int = np.trapezoid(f, x)

# Create plot
plt.figure(figsize=(8, 4.5))
plt.plot(x, f, label=r"$f(x) = x\cos\!\left(\frac{2n\pi x}{L}\right)$")
# Shade positive and negative regions (same default color; annotated for clarity)
pos = f >= 0
neg = f < 0
p = plt.fill_between(x, 0, f, where=pos, alpha=0.3, label="positive area")
q = plt.fill_between(x, 0, f, where=neg, alpha=0.3, label="negative area")

# Add zero line and formatting
plt.axhline(0, linewidth=1)
plt.xlim(0, L)
plt.xlabel("x")
plt.ylabel("f(x)")
plt.title(r"Positive and negative areas cancel:  $\int_0^L x\cos\!\left(\frac{2n\pi x}{L}\right)\,dx = 0$")

# Annotate a positive lobe and a negative lobe
# Find a representative positive and negative region to place annotations
if np.any(pos):
    idx_pos = np.argmax(f)  # near first positive peak
    plt.annotate("positive area", xy=(x[idx_pos], f[idx_pos]), xytext=(x[idx_pos]+0.1*L, 0.7*np.max(f)),
                 arrowprops=dict(arrowstyle="->"))
if np.any(neg):
    idx_neg = np.argmin(f)  # near first negative through
    plt.annotate("negative area", xy=(x[idx_neg], f[idx_neg]), xytext=(x[idx_neg]-0.2*L, 0.7*np.min(f)),
                 arrowprops=dict(arrowstyle="->"))

# Show numerical check in a small textbox
plt.text(0.02*L, 0.9*np.max(f), f"Numerical integral ≈ {num_int:.2e}")

plt.legend(loc="upper right")
plt.tight_layout()
plt.show()
'vis '
<Figure size 800x450 with 1 Axes>
'vis '

Problems

Problem 1: Compute probability of finding particle somewhere

Compute the probability of observing the particle in a box in the domain a3≤x≤2a3\frac{a}{3} \leq x \leq \frac{2a}{3}.

Solution

Since the square of the wave function is a probability density, we can determine the probability of observing the particle in a particular domain using the relationship

Prob(x1≤x≤x2)=∫x1x2P(x)dx=∫x1x2ψ∗(x)ψ(x)dx\text{Prob}(x_1 \leq x \leq x_2) = \int_{x_1}^{x_2}P(x)dx = \int_{x_1}^{x_2} \psi^*(x)\psi(x)dx
(55)

We simply use the above equation together with the normalized particle-in-a-box wave function:

Prob(a3≤x≤2a3)=2a∫a32a3sin⁡2nπxadx\begin{align} \text{Prob}(\frac{a}{3} \leq x \leq \frac{2a}{3}) = \frac{2}{a}\int_{\frac{a}{3}}^{\frac{2a}{3}} \sin^2\frac{n\pi x}{a}dx \end{align}
(56)

We will use the definite integral of sin⁡2ax\sin^2ax from a table:

∫sin⁡2axdx=x2−sin⁡2ax4a\int\sin^2axdx = \frac{x}{2} - \frac{\sin2ax}{4a}
(57)

Perform a uu-substitution on the integral above to put it into the table form:

Prob(a3≤x≤2a3)=2a[x2−sin⁡2nπxa4nπa]a32a3=2a[x2−asin⁡2nπxa4nπ]a32a3=2a[a3−asin⁡4nπ34nπ−a6+asin⁡2nπ34nπ]=2[16+sin⁡2nπ3−sin⁡4nπ34nπ]\begin{align} \text{Prob}(\frac{a}{3} \leq x \leq \frac{2a}{3}) &= \frac{2}{a}\left[ \frac{x}{2} - \frac{\sin\frac{2n\pi x}{a}}{\frac{4n\pi}{a}}\right]_{\frac{a}{3}}^{\frac{2a}{3}} \\ &= \frac{2}{a}\left[ \frac{x}{2} - \frac{a\sin\frac{2n\pi x}{a}}{4n\pi}\right]_{\frac{a}{3}}^{\frac{2a}{3}} \\ &= \frac{2}{a}\left[ \frac{a}{3} - \frac{a\sin\frac{4n\pi}{3}}{4n\pi} - \frac{a}{6} + \frac{a\sin\frac{2n\pi }{3}}{4n\pi}\right] \\ &= 2\left[ \frac{1}{6} + \frac{\sin\frac{2n\pi }{3} - \sin\frac{4n\pi}{3}}{4n\pi}\right] \end{align}
(58)

Problem 2: Compute an expectation of x2x^2

Compute the average of x2x^2 for a particle in a box.

Solution

To compute the average value of x2x^2, we start by writing the integral expression:

⟨x2⟩=∫ψ∗(x)x2ψ(x)dx\langle x^2 \rangle = \int \psi^*(x) x^2 \psi(x)dx
(59)

For the particle in a box, we can limit the domain, and thus the bounds of integration, to 0≤x≤a0\leq x \leq a. We can also set ψn(x)=2asin⁡nπxa\psi_n(x) = \sqrt{\frac{2}{a}}\sin\frac{n\pi x}{a}.

Thus, for a particle in a 1D box of size aa, we get

⟨x2⟩=∫0a2asin⁡(nπxa)x22asin⁡(nπxa)dx=2a∫0ax2sin⁡2nπxadx\begin{align} \langle x^2 \rangle &= \int_0^a \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right) x^2 \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)dx \\ &= \frac{2}{a} \int_0^a x^2 \sin^2\frac{n\pi x}{a}dx \end{align}
(60)

From an integral table we find that

∫x2sin⁡2αxdx=x36−(x24α−18α3)sin⁡2αx−xcos⁡2αx4α2+C\int x^2\sin^2\alpha xdx = \frac{x^3}{6} - \left(\frac{x^2}{4\alpha} - \frac{1}{8\alpha^3}\right)\sin2\alpha x - \frac{x\cos 2\alpha x}{4\alpha^2} + C
(61)

We use this equation with α=nπa\alpha = \frac{n\pi}{a} and get:

⟨x2⟩=∫0a2asin⁡(nπxa)x22asin⁡(nπxa)dx=2a∫0ax2sin⁡2nπxadx=2a[x36−(x24α−18α3)sin⁡2αx−xcos⁡2αx4α2]0a=2a[a36−(a24α−18α3)sin⁡2αa−acos⁡2αa4α2]=2a[a36−(a24nπa−18(nπa)3)sin⁡2nπaa−acos⁡2nπaa4(nπa)2]=2a[a36−a3(2nπ)2]=a23−a22(nπ)2\begin{align} \langle x^2 \rangle &= \int_0^a \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right) x^2 \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)dx \\ &= \frac{2}{a} \int_0^a x^2 \sin^2\frac{n\pi x}{a}dx \\ &= \frac{2}{a}\left[ \frac{x^3}{6} - \left(\frac{x^2}{4\alpha} - \frac{1}{8\alpha^3}\right)\sin2\alpha x - \frac{x\cos 2\alpha x}{4\alpha^2}\right]_0^a \\ &= \frac{2}{a}\left[ \frac{a^3}{6} - \left(\frac{a^2}{4\alpha} - \frac{1}{8\alpha^3}\right)\sin2\alpha a - \frac{a\cos 2\alpha a}{4\alpha^2} \right] \\ &= \frac{2}{a}\left[ \frac{a^3}{6} - \left(\frac{a^2}{4\frac{n\pi}{a}} - \frac{1}{8\left(\frac{n\pi}{a}\right)^3}\right)\sin2\frac{n\pi}{a} a - \frac{a\cos 2\frac{n\pi}{a} a}{4\left(\frac{n\pi}{a}\right)^2} \right] \\ &= \frac{2}{a}\left[ \frac{a^3}{6} - \frac{a^3}{\left(2n\pi\right)^2} \right] \\ &= \frac{a^2}{3} - \frac{a^2}{2\left(n\pi\right)^2} \end{align}
(62)

This result, combined with the result for ⟨x⟩\langle x \rangle, can be used to determine σx\sigma_x, the standard deviation of the particle’s position:

σx=⟨x2⟩−⟨x⟩2=a2πnπ2n23−2\sigma_x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} = \frac{a}{2\pi n}\sqrt{\frac{\pi^2n^2}{3} -2}
(63)

Problem 3: Compute expectation of energy

Compute the average energy of a particle in a box.

Solution

The average energy of the particle in a box is a special case of computing an average quantity. We start by writing out the standard definition of an average computed from a wavefunction:

⟨E⟩=∫0aψn∗(x)E^ψn(x)dx\langle E \rangle = \int_0^a \psi_n^*(x)\hat{E}\psi_n(x)dx
(64)

where E^\hat{E} is the total energy operator. We know the total energy operator by another symbol, namely E^=H^\hat{E} = \hat{H}. We plug this into the above equation to get:

⟨E⟩=∫0aψn∗(x)H^ψn(x)dx.\langle E \rangle = \int_0^a \psi_n^*(x)\hat{H}\psi_n(x)dx.
(65)

We now recognize that the particle-in-a-box wavefunctions we are discussing were derived from the Schrödinger equation:

H^ψn(x)=Enψn(x)\hat{H}\psi_n(x) = E_n\psi_n(x)
(66)

where EnE_n is a scalar. Thus, for the average energy we get:

⟨E⟩=∫0aψn∗(x)H^ψn(x)dx=∫0aψn∗(x)Enψn(x)dx=En∫0aψn∗(x)ψn(x)dx=En\begin{align} \langle E \rangle &= \int_0^a \psi_n^*(x)\hat{H}\psi_n(x)dx \\ &=\int_0^a \psi_n^*(x)E_n\psi_n(x)dx \\ &=E_n\int_0^a \psi_n^*(x)\psi_n(x)dx \\ &= E_n \end{align}
(67)

The last equality holds because the wave functions are normalized.

Problem 4: Compute expectation of momentum

Compute the average momentum for a particle in a box.

Solution

To compute the average momentum of a particle in a 1D box, we start in the usual way:

⟨p⟩=∫0aψn∗(x)p^ψn(x)dx\langle p \rangle = \int_0^a \psi_n^*(x)\hat{p}\psi_n(x)dx
(68)

Recall that the momentum operator in one dimension is given by

p^x=−iℏddx\hat{p}_x = -i\hbar\frac{d}{dx}
(69)

We now substitute this into the above equation and solve:

⟨p⟩=∫0aψn∗(x)(−iℏddx)ψn(x)dx=−2iℏa∫0asin⁡(nπxa)ddx(sin⁡(nπxa))dx=−2iℏa∫0asin⁡(nπxa)nπacos⁡(nπxa)dx=−2inπℏa2∫0asin⁡(nπxa)cos⁡(nπxa)dx=0\begin{align} \langle p \rangle &= \int_0^a \psi_n^*(x)\left(-i\hbar\frac{d}{dx}\right)\psi_n(x)dx \\ &= -\frac{2i\hbar}{a}\int_0^a \sin\left(\frac{n\pi x}{a}\right)\frac{d}{dx}\left(\sin\left(\frac{n\pi x}{a}\right)\right)dx \\ &= -\frac{2i\hbar}{a}\int_0^a \sin\left(\frac{n\pi x}{a}\right)\frac{n\pi}{a}\cos\left(\frac{n\pi x}{a}\right)dx \\ &= -\frac{2in\pi\hbar}{a^2}\int_0^a \sin\left(\frac{n\pi x}{a}\right)\cos\left(\frac{n\pi x}{a}\right)dx \\ &= 0 \end{align}
(70)

where the last equality can be found in an integral table.

So the average momentum of a particle in a box is zero. This is because it is equally probable for the particle to be moving forward and backward.

Problem 5

Consider an electron in superfluid helium (4^4He) where it forms a solvation cavity with a radius of 18A˚18 \text{Å}. Calculate the zero-point energy and the energy difference between the ground and first excited states by approximating the electron by a particle in a 3-dimensional box.

Solution

The zero-point energy can be obtained from the lowest-state energy (n=1n = 1) with a=b=c=36A˚a = b = c = 36 \text{Å}. The first excited state is triply degenerate (E112E_{112}, E121E_{121}, and E211E_{211}).

E111=h28me(nx2a2+ny2b2+nz2c2)E_{111} = \frac{h^2}{8m_e}\left(\frac{n_x^2}{a^2} + \frac{n_y^2}{b^2} + \frac{n_z^2}{c^2}\right)
(71)
=(6.626076×10−34 Js)28(9.109390×10−31 kg)(1(36×10−10 m)2+1(36×10−10 m)2+1(36×10−10 m)2)= \frac{(6.626076\times 10^{-34}\textnormal{ Js})^2}{8(9.109390\times 10^{-31}\textnormal{ kg})}\left(\frac{1}{(36\times 10^{-10}\textnormal{ m})^2} + \frac{1}{(36\times 10^{-10}\textnormal{ m})^2} + \frac{1}{(36\times 10^{-10}\textnormal{ m})^2}\right)
(72)
=1.39×10−20 J=87.0 meV= 1.39\times10^{-20}\textnormal{ J} = 87.0\textnormal{ meV}
(73)
E211=E121=E112=(6.626076×10−34 Js)28(9.109390×10−31 kg)E_{211} = E_{121} = E_{112} = \frac{(6.626076\times 10^{-34}\textnormal{ Js})^2}{8(9.109390\times 10^{-31}\textnormal{ kg})}
(74)
×(22(36×10−10 m)2+12(36×10−10 m)2+12(36×10−10 m)2)\times\left(\frac{2^2}{(36\times 10^{-10}\textnormal{ m})^2} + \frac{1^2}{(36\times 10^{-10}\textnormal{ m})^2} + \frac{1^2}{(36\times 10^{-10}\textnormal{ m})^2}\right)
(75)
=2.79×10−20 J=174 meV⇒ΔE=87 meV= 2.79\times 10^{-20}\textnormal{ J} = 174\textnormal{ meV} \Rightarrow \Delta E = 87\textnormal{ meV}
(76)

(Experimental value: 105 meV; Phys. Rev. B 41, 6366 (1990))

Problem 6: Energy Levels in a 3D Box

A particle is confined in a 3D box with side lengths Lx=Ly=Lz=LL_x = L_y = L_z = L. The energy levels for a particle in this box are given by the formula:

Enx,ny,nz=ℏ2π22mL2(nx2+ny2+nz2)E_{n_x, n_y, n_z} = \frac{\hbar^2 \pi^2}{2mL^2} \left( n_x^2 + n_y^2 + n_z^2 \right)
(77)

where nxn_x, nyn_y, nzn_z are the quantum numbers associated with the particle’s motion in the xx-, yy-, and zz-directions.

Calculate the energy levels for the quantum states with nx=1n_x = 1, ny=1n_y = 1, nz=2n_z = 2 and nx=2n_x = 2, ny=2n_y = 2, nz=1n_z = 1. Are these energy levels degenerate?

Solution

The energy for the state (nx,ny,nz)=(1,1,2)(n_x, n_y, n_z) = (1, 1, 2) is:

E1,1,2=ℏ2π22mL2(12+12+22)=ℏ2π22mL2(1+1+4)=ℏ2π22mL2×6E_{1,1,2} = \frac{\hbar^2 \pi^2}{2mL^2} \left( 1^2 + 1^2 + 2^2 \right) = \frac{\hbar^2 \pi^2}{2mL^2} (1 + 1 + 4) = \frac{\hbar^2 \pi^2}{2mL^2} \times 6
(78)

The energy for the state (nx,ny,nz)=(2,2,1)(n_x, n_y, n_z) = (2, 2, 1) is:

E2,2,1=ℏ2π22mL2(22+22+12)=ℏ2π22mL2(4+4+1)=ℏ2π22mL2×9E_{2,2,1} = \frac{\hbar^2 \pi^2}{2mL^2} \left( 2^2 + 2^2 + 1^2 \right) = \frac{\hbar^2 \pi^2}{2mL^2} (4 + 4 + 1) = \frac{\hbar^2 \pi^2}{2mL^2} \times 9
(79)

Since E1,1,2≠E2,2,1E_{1,1,2} \neq E_{2,2,1}, these energy levels are not degenerate.

Problem 7: Degeneracy of Energy Levels

Consider a particle confined in a cubic box with side lengths Lx=Ly=Lz=LL_x = L_y = L_z = L. The energy levels are given by the same formula as in Problem 1.

Solution

Part 1: We want to find the quantum numbers that satisfy:

nx2+ny2+nz2=14n_x^2 + n_y^2 + n_z^2 = 14

We can check different combinations of nxn_x, nyn_y, and nzn_z:

  • For nx=3n_x = 3, ny=2n_y = 2, and nz=1n_z = 1:

32+22+12=9+4+1=143^2 + 2^2 + 1^2 = 9 + 4 + 1 = 14
(80)

  • Other permutations of these quantum numbers will give the same energy:

    • (3,2,1)(3, 2, 1)

    • (3,1,2)(3, 1, 2)

    • (2,3,1)(2, 3, 1)

    • (2,1,3)(2, 1, 3)

    • (1,3,2)(1, 3, 2)

    • (1,2,3)(1, 2, 3)

Part 2: The energy level corresponding to nx2+ny2+nz2=14n_x^2 + n_y^2 + n_z^2 = 14 has 6 degenerate states, since the quantum number triplets are (3,2,1)(3, 2, 1), (3,1,2)(3, 1, 2), (2,3,1)(2, 3, 1), (2,1,3)(2, 1, 3), (1,3,2)(1, 3, 2), and (1,2,3)(1, 2, 3).

Problem 8: Degeneracy of the Ground State

Solution

Part 1: The ground state corresponds to the quantum numbers nx=ny=nz=1n_x = n_y = n_z = 1. The energy for this state is:

E1,1,1=ℏ2π22mL2(12+12+12)=ℏ2π22mL2×3E_{1,1,1} = \frac{\hbar^2 \pi^2}{2mL^2} \left( 1^2 + 1^2 + 1^2 \right) = \frac{\hbar^2 \pi^2}{2mL^2} \times 3
(81)

There is only one combination of quantum numbers that gives this energy, so the degeneracy of the ground state is 1.

Part 2: The ground state is non-degenerate because there is only one way to assign the quantum numbers nx=ny=nz=1n_x = n_y = n_z = 1. Degeneracy arises when multiple different sets of quantum numbers give the same energy, which is not the case for the ground state.

Problem 9: Higher Energy Degeneracy

Consider a particle in a cubic box. The energy levels are quantized as:

Enx,ny,nz=ℏ2π22mL2(nx2+ny2+nz2)E_{n_x, n_y, n_z} = \frac{\hbar^2 \pi^2}{2mL^2} \left( n_x^2 + n_y^2 + n_z^2 \right)
(82)
Solution

Part 1: We want to solve:

nx2+ny2+nz2=9n_x^2 + n_y^2 + n_z^2 = 9

Possible combinations of nxn_x, nyn_y, and nzn_z:

  • nx=3n_x = 3, ny=0n_y = 0, nz=0n_z = 0 gives 32+02+02=93^2 + 0^2 + 0^2 = 9.

  • nx=2n_x = 2, ny=2n_y = 2, nz=1n_z = 1 gives 22+22+12=4+4+1=92^2 + 2^2 + 1^2 = 4 + 4 + 1 = 9.

  • Permutations of (2,2,1)(2, 2, 1) are:

    • (2,2,1)(2, 2, 1)

    • (2,1,2)(2, 1, 2)

    • (1,2,2)(1, 2, 2)

Part 2: The degeneracy for the energy level corresponding to nx2+ny2+nz2=9n_x^2 + n_y^2 + n_z^2 = 9 is 4 (the state (3,0,0)(3,0,0) and the 3 permutations of (2,2,1)(2,2,1)).

Chapter 3: The Schrödinger Equation
The Wavefunction
Chapter 3: The Schrödinger Equation
Applications of the Particle in a Box