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Modern Physics
Particle in a One-Dimensional Box (Infinite Potential Well)
A particle in a box is a fundamental quantum mechanical model describing a non-relativistic particle constrained to move in a one-dimensional region of constant potential bounded by infinite potential barriers. It demonstrates spatial quantization, energy quantization, and the absence of a zero-point energy state equal to zero.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Energy eigenvalues, wavefunctions, normalization, and fundamental probability interpretation for a 1D infinite potential well.
PGT extension
Expectation values, parity of wavefunctions, orthogonal properties, 3D generalization, and degeneracy of energy states.
Prerequisites
Time-independent Schrödinger wave equation
Boundary conditions for acceptable wavefunctions
De Broglie wavelength relation and momentum operator
Potential Energy Function and Boundaries
The potential energy V(x) is defined as zero for 0 < x < L and infinity for x <= 0 and x >= L.
The infinite potential barriers imply that the probability of finding the particle outside the box is identically zero.
Consequently, the wavefunction must vanish at the boundaries x = 0 and x = L due to continuity.
Quantization of Energy and Wavefunction
Solving the Schrödinger equation yields discrete energy levels proportional to n², where n is the principal quantum number (n = 1, 2, 3...).
The lowest possible energy state (n = 1) is called the zero-point energy or ground state energy and is strictly non-zero.
The wavefunctions alternate between symmetric and antisymmetric properties about the center of the box ($x = L/2)$ for even and odd quantum numbers.
Probability Density and Nodes
The probability density of finding the particle at position x is given by the square of the absolute value of the normalized wavefunction, |ψn(x)|².
The number of internal nodes (points where the wavefunction and probability density are zero) is given by n - 1.
For large values of n, the quantum probability distribution approaches the classical uniform probability distribution in accordance with the correspondence principle.
Orthogonality and Completeness
Wavefunctions corresponding to different energy states (m ≠ n) are mutually orthogonal over the interval from 0 to L.
The set of all particle-in-a-box wavefunctions forms a complete set, allowing any physically acceptable arbitrary wavefunction within the box to be expanded as a linear combination of these eigenstates.
Expectation values of observables are evaluated using standard operator integration over the spatial domain of the box.
Derivations
Derivation of Energy Eigenvalues and Wavefunctions
Conditions: Particle of mass m in 1D region 0 < x < L with V = 0 inside and $V = \infty outside$.
Write the time-independent Schrödinger equation inside the box: -(h-bar² / 2m) (d²ψ / dx²) = Eψ
Let k$^{2} = 2mE / h-bar^{2}$, transforming the differential equation into d²ψ / dx² + k²$\psi = 0$
Apply general solution ψ(x) = A sin(kx) + B cos(kx)
Apply boundary condition ψ(0) = 0, which yields B = 0, leaving ψ(x) = A sin(kx)
Apply boundary condition ψ(L) = 0, which yields A sin(kL) = 0. Since A ≠ 0, k$L = n\pi$ where n = 1, 2, 3...
Express wave vector as $k = n\pi / L$ and substitute to obtain energy En = (n² h²) / (8mL²)
Normalize the wavefunction by setting the integral of |ψ(x)|² dx from 0 to L equal to 1, yielding $A = \sqrt(2/L)$
\langle p^2 \rangle = mean square momentum kg^2 m^2 s^{-2}
Stationary state n
Frequency of Photon Emitted in Transition
$$\nu = \frac{E_{n_2} - E_{n_1}}{h}$$
\nu = frequency, E = energy levels Hertz (Hz)
Transition between states n_2 and n_1
3D Degenerate Energy Expression (Cube)
$$E = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$$
$L_x = L_y = L_z = L$ Joules (J)
Cubical box of side L
Graphs and experiments
Wavefunction Plot ψ<sub>n</sub>(x)
Draw: Sinusoidal curves for n = 1, 2, 3 showing amplitudes within 0 and L, with zero crossings at integer multiples of L/n.
Interpret: Shows spatial distribution of phase and amplitude; demonstrates alternating symmetry (even/odd parity about center).
Exam use: Direct identification of nodes and nodal positions for given quantum numbers.
Probability Density Plot |ψ<sub>n</sub>(x)|²
Draw: Curves showing positive lobes only, with n peaks for the n-th state and n-1 points touching zero.
Interpret: Represents spatial probability distribution; points of zero probability represent regions where the particle is never found.
Exam use: Explaining classical vs quantum correspondence at high quantum numbers.
Energy Level Diagram
Draw: Horizontal energy rungs spaced unevenly, with spacing increasing as n increases (proportional to n²).
Interpret: Highlights quantization of energy and non-zero ground state energy E1.
Exam use: Calculating absorption/emission spectral frequencies during quantum transitions.
Solved numericals
TGT core
Calculate the lowest three energy levels for an electron confined in a one-dimensional box of length 0.2 nm. (Given $h = 6.63 x 10^{-34} J s$, $m = 9.1 x 10^{-31} kg)$
Identify formula: $E_n = \frac{n^2 h^2}{8mL^2}$
Convert units: $L = 0.2 nm = 0.2 x 10^{-9} m$, $m = 9.1 x 10^{-31} kg$
Calculate ground state energy ($n=1): E_1 = \frac{1^2 \times (6.63 \times 10^{-34})^2}{8 \times (9.1 \times 10^{-31}) \times (0.2 \times 10^{-9})^2}$
An electron is in the ground state of an infinite potential box of length L. What is the probability of finding the electron in the region from x = 0 to $x = L/4?$
Identify ground state wavefunction: \ps$i_1(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)$
Set up probability integral: $P = \int_{0}^{L/4} |\psi_1(x)|^2 dx = \frac{2}{L} \int_{0}^{L/4} \sin^2\left(\frac{\pi x}{L}\right) dx$
Use trigonometric identity \si$n^2(\theta) = \frac{1 - \cos(2\theta)}{2}: P = \frac{2}{L} \int_{0}^{L/4} \frac{1 - \cos(2\pi x / L)}{2} dx$
Test integer combinations of (n_x, n_y, n_z) where n >= 1
Check 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14$
List all distinct permutations of (1, 2, 3): (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)
Count total number of valid distinct states, which is 6
Answer: Degeneracy is 6
Assumptions and limitations
Assumption: Potential energy inside the box is strictly zero and outside is strictly infinite, forming perfectly rigid walls. Limitation: Real physical potential wells always have finite barrier heights, allowing quantum mechanical tunneling into and beyond the walls.
Common misconceptions
Common error: The lowest energy state of a particle in a box can be zero.
Correct: The lowest energy state (ground state) is always strictly greater than zero.
A zero energy state would violate Heisenberg's Uncertainty Principle, as uncertainty in position is bounded by box length L, requiring a non-zero minimum momentum.
Common error: Nodes in a 1D box wavefunction include the boundary points x = 0 and x = L.
Correct: Nodes are defined strictly as interior points where the wavefunction vanishes (0 < x < L).
Boundary zeros are boundary conditions, not internal nodes; the n-th state has n-1 internal nodes.
Quick revision
Ground state energy is $E_1 = h^2 / (8mL^2)$ and is never zero.
The energy eigenvalues for a particle in a 1D box are given by $E_n = \frac{n^2 h^2}{8mL^2}$, where $n = 1, 2, 3, \dots$
The ground state energy ($n=1$) is never zero; it represents the zero-point energy required by the Heisenberg Uncertainty Principle.
The normalized wavefunction for a 1D infinite well is $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$ for $0 \le x \le L$.
The number of internal nodes (where the wavefunction vanishes strictly inside the box) for the $n$-th state is $n - 1$.
The expectation value of the position $\langle x \rangle$ for any stationary state in a symmetric 1D box is always $\frac{L}{2}$.
Wavefunctions corresponding to different quantum numbers ($m \neq n$) are mutually orthogonal over the domain $[0, L]$.
Glossary
Zero-point EnergyThe minimum possible energy ($E_1$) of a quantum mechanical system, which is strictly non-zero.
Internal NodeA point strictly between $x = 0$ and $x = L$ where both the wavefunction and probability density vanish.
Expectation ValueThe probabilistic average outcome of a large number of measurements of an observable on identical systems.
OrthogonalityA mathematical property where the integral of the product of two distinct eigenstates over the domain is zero.
Quantum vs Classical Particle in a Box
Feature
Quantum Mechanical Model
Classical Mechanical Model
Energy States
Quantized ($E_n \propto n^2$)
Continuous and can be zero
Position Probability
Determined by $|\psi_n(x)|^2$, varies across box
Uniform probability across the entire box length
Ground State Velocity
Non-zero due to zero-point energy
Can be exactly zero
Practice Quiz
Choose an option to reveal the explanation.
1. What is the expression for the ground state energy of a particle of mass $m$ in a 1D box of length $L$?
Explanation: Substituting $n=1$ into the energy eigenvalue formula $E_n = \frac{n^2 h^2}{8mL^2}$ yields $E_1 = \frac{h^2}{8mL^2}$.
2. How many internal nodes does the wavefunction $\psi_n(x)$ have for a particle in a 1D box in the state $n = 4$?
Explanation: The number of internal nodes (excluding boundaries) for the $n$-th state is given by $n - 1$. For $n = 4$, the number of internal nodes is $4 - 1 = 3$.
3. What is the expectation value of position $\langle x \rangle$ for a particle in a 1D box of length $L$ in any stationary state $n$?
Explanation: By symmetry, the probability density $|\psi_n(x)|^2$ is symmetric about the midpoint $x = L/2$, leading to an average position of $\langle x \rangle = \frac{L}{2}$.
4. Why is the lowest possible energy state ($n=1$) of a particle in a box non-zero?
Explanation: If energy were zero, momentum would be zero with certainty, which violates the Heisenberg Uncertainty Principle given that position is confined within box length $L$.
5. What is the probability density $P(x)$ of finding the particle at position $x$ for the state $\psi_n(x)$?
Explanation: The probability density is the square of the absolute value of the normalized wavefunction: $P(x) = |\psi_n(x)|^2 = \left|\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right|^2 = \frac{2}{L} \sin^2\left(\frac{n\pi x}{L}\right)$.
6. Which of the following statements is true regarding the wavefunctions $\psi_m(x)$ and $\psi_n(x)$ for $m \neq n$ in a 1D box?
Explanation: Eigenfunctions of a Hermitian operator (such as the Hamiltonian) corresponding to distinct eigenvalues are mutually orthogonal, meaning $\int_{0}^{L} \psi_m^*(x) \psi_n(x) dx = 0$ for $m \neq n$.
7. What is the expected value of momentum squared $\langle p^2 \rangle$ for a particle in a 1D box in state $n$?
Explanation: Applying the momentum squared operator $-\hbar^2 \frac{d^2}{dx^2}$ to the wavefunction yields the expectation value $\langle p^2 \rangle = \frac{n^2 \pi^2 \hbar^2}{L^2}$.
8. For a particle in a 3D cubical box of side $L$, what is the energy eigenvalue expression?
Explanation: Separation of variables in a 3D cubical box leads to energy eigenvalues that are the sum of independent 1D energies: $E = \frac{h^2}{8mL^2}(n_x^2 + n_y^2 + n_z^2)$.
9. How do the wavefunctions alternate in terms of symmetry about the center of the box ($x = L/2$)?
Explanation: When the coordinate origin is chosen at the center of the box, odd quantum numbers yield symmetric wavefunctions (even parity), while even quantum numbers yield antisymmetric wavefunctions (odd parity).
10. What happens to the quantum probability distribution of a particle in a box as the principal quantum number $n$ becomes very large?
Explanation: According to the Bohr correspondence principle, for very large $n$, the quantum mechanical probability distribution matches the classical expectation of a uniformly distributed particle.
11. What are the boundary conditions for the wavefunction $\psi(x)$ of a particle in an infinite potential well of length $L$ ($0 \le x \le L$)?
Explanation: Due to infinite potential barriers at $x = 0$ and $x = L$, the probability of finding the particle outside is zero, requiring the continuous wavefunction to vanish at the boundaries: $\psi(0) = 0$ and $\psi(L) = 0$.
12. What is the unit of the normalized wavefunction $\psi_n(x)$ in a 1D box?
Explanation: Since the probability density $|\psi_n(x)|^2$ integrated over length ($dx$) must equal 1, the integral $\int |\psi_n(x)|^2 dx = 1$ implies that $|\psi_n(x)|^2$ has units of $\text{m}^{-1}$, making $\psi_n(x)$ have units of $\text{m}^{-1/2}$.
13. What is the frequency $\nu$ of a photon emitted when a particle transitions from energy level $n_2$ to $n_1$?
Explanation: By conservation of energy, the energy difference between the two states is released as a photon of energy $h\nu = E_{n_2} - E_{n_1}$, giving $\nu = \frac{E_{n_2} - E_{n_1}}{h}$.
14. Which of the following represents a degenerate state in a 3D cubical box of side $L$?
Explanation: Different sets of quantum numbers that yield the exact same sum of squares ($1^2 + 2^2 + 3^2 = 3^2 + 2^2 + 1^2 = 14$) correspond to the same energy eigenvalue, representing degeneracy.
15. What is the value of the normalization constant for the wavefunction $\psi_n(x) = A \sin\left(\frac{n\pi x}{L}\right)$?
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If the length of a 1D box containing a particle is halved, how does the ground state energy of the particle change?
Explanation: Energy is inversely proportional to the square of the box length ($E_1 \propto \frac{1}{L^2}$). If $L$ becomes $L/2$, then $E_1$ becomes proportional to $\frac{1}{(L/2)^2} = \frac{4}{L^2}$, increasing by a factor of 4.
2. What is the probability of finding a particle in the middle third of a 1D box ($L/3 \le x \le 2L/3$) when it is in the ground state ($n = 1$)?
Explanation: Integrating the ground state probability density from $L/3$ to $2L/3$: $\int_{L/3}^{2L/3} \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) dx = \left[ \frac{x}{L} - \frac{\sin(2\pi x / L)}{2\pi} \right]_{L/3}^{2L/3} = \frac{1}{3} + \frac{\sqrt{3}}{2\pi}$
3. For a particle in a 1D infinite potential well, the ratio of the energy of the third excited state to that of the ground state is:
Explanation: The ground state corresponds to $n = 1$. The first excited state is $n=2$, the second is $n=3$, and the third excited state is $n = 4$. The energy ratio is $\frac{E_4}{E_1} = \frac{4^2}{1^2} = 16$.
4. Which operator expectation value for a stationary state $n$ of a particle in a 1D box equals zero?
Explanation: Since the particle has equal probability of moving left or right with equal momenta magnitudes, the expectation value of linear momentum $\langle p \rangle$ is identically zero.
5. What is the degeneracy of the energy level corresponding to $E = \frac{50 h^2}{8mL^2}$ for a particle in a 3D cubical box?
Explanation: We need combinations of $(n_x^2 + n_y^2 + n_z^2) = 50$. The possible sets of positive integers are $(5, 5, 0)$ not allowed since $n \ge 1$, so integers such as $(5^2 + 5^2 + 0^2)$ fail. Checking valid permutations: $50 = 6^2 + 3^2 + 3^2$ ($36+9+9=54$ - no); $50 = 5^2 + 5^2 + 0^2$ (invalid); let's check $5^2 + 4^2 + 3^2 = 25 + 16 + 9 = 50$. The distinct permutations of $(5, 4, 3)$ are $(5,4,3), (5,3,4), (4,5,3), (4,3,5), (3,5,4), (3,4,5)$, giving a degeneracy of 6? Wait, let's re-verify: $5^2+4^2+3^2 = 25+16+9 = 50$. All three integers are distinct ($5, 4, 3$), so there are $3! = 6$ permutations. Let's check options: 6 is an option.
6. The wavefunctions for a particle in a 1D box form a complete set. What does completeness allow us to do?
Explanation: Completeness of the orthonormal set of eigenfunctions $\{\psi_n\}$ guarantees that any arbitrary physical state wavefunction $\Psi(x)$ within the box can be expressed as $\Psi(x) = \sum c_n \psi_n(x)$بيه.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. What is the zero-point energy of an electron confined in a 1D box of length $1\text{ nm}$?
Explanation: Using $E_1 = \frac{h^2}{8mL^2}$, substituting $h = 6.626 \times 10^{-34}\text{ J s}$, $m = 9.11 \times 10^{-31}\text{ kg}$, and $L = 10^{-9}\text{ m}$, the energy in Joules converts to approximately $0.376\text{ eV}$.
2. For a particle in a 1D box, at which positions inside the box is the probability density maximum for the state $n = 3$?
Explanation: For $n=3$, the probability density $\sin^2(3\pi x / L)$ reaches its maxima where $\sin(3\pi x / L) = \pm 1$, which occurs at $3\pi x / L = \pi/2, 3pi/2, 5pi/2$, leading to $x = L/6, L/2, 5L/6$ (i.e., $L/6, 3L/6, 5L/6$).
3. What is the expectation value of position squared $\langle x^2 \rangle$ for a particle in the ground state ($n=1$) of a 1D box of length $L$?
4. In a 1D infinite potential well, if the potential energy inside ($0 < x < L$) is raised from $0$ to a constant positive value $V_0$, how do the energy eigenvalues change?
Explanation: Adding a constant potential energy $V_0$ everywhere inside the box simply shifts the zero point of the potential energy scale, adding $V_0$ to every energy eigenvalue ($E_n' = E_n + V_0$).
5. What is the parity of the wavefunction $\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)$ when the coordinate system is centered at the middle of the box ($x = 0$ at center)?
Explanation: With the origin at the center, substituting $x$ with $-x$ in $\sin(n\pi(x+L/2)/L)$ shows that odd $n$ yields a symmetric (even) function, while even $n$ yields an antisymmetric (odd) function.
6. What is the root-mean-square momentum $p_{\text{rms}}$ of a particle in the $n$-th state of a 1D box of length $L$?
Explanation: Since $\langle p \rangle = 0$ and $\langle p^2 \rangle = \frac{n^2 \pi^2 \hbar^2}{L^2}$, the root-mean-square momentum is $p_{\text{rms}} = \sqrt{\langle p^2 \rangle} = \frac{n \pi \hbar}{L}$.
7. Which of the following conditions ensures that a wavefunction $\psi(x)$ is physically acceptable for a particle in a box?
Explanation: Acceptable quantum mechanical wavefunctions must be single-valued, continuous, differentiable (except at infinite potential steps), and square-integrable (normalizable).
8. If a particle is in the ground state of a 1D box of length $L$, what is the probability of finding it in the region $0 \le x \le L/2$?
Explanation: By symmetry, the probability density $|\psi_1(x)|^2$ is symmetric about $x = L/2$, meaning exactly half of the total probability lies in the left half of the box ($0.50$).
9. What is the physical significance of the internal nodes in a wavefunction?
Explanation: An internal node is a point where both the wavefunction and the probability density $|\psi(x)|^2$ vanish, meaning the probability of finding the particle at that exact point is zero.
10. How does the spacing between consecutive energy levels ($E_{n+1} - E_n$) behave as the quantum number $n$ becomes very large?
Explanation: Since $E_n \propto n^2$, the difference $E_{n+1} - E_n \propto (2n + 1)$, which increases linearly with $n$ for large quantum numbers.