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Modern Physics

One-Dimensional Linear Harmonic Oscillator in Quantum Mechanics

The one-dimensional linear harmonic oscillator describes a particle subject to a restoring force proportional to its displacement from an equilibrium position under quantum mechanics. Its potential energy is given by V(x) = &frac{1}{2}mω2x2, serving as a fundamental model for molecular vibrations and lattice dynamics.

UP TGT ScienceUP PGT PhysicsStudy Notes + Practice

Scope and prerequisites

TGT core

Understanding of energy eigenvalues, zero-point energy, basic wavefunction properties, and graphical representation of potentials for TGT level.

PGT extension

Rigorous solution of the time-independent Schrödinger equation using power series method (Hermite polynomials), creation and annihilation operators, and matrix mechanics for PGT level.

Prerequisites

Time-independent Schrödinger equation in one dimension
Operator algebra and commutation relations
Properties of differential equations and series solutions

Potential Energy Function and Hamiltonian

The classical potential for a linear harmonic oscillator is V(x) = &frac{1}{2}mω²x², where m is the mass and ω is the angular frequency.
The quantum mechanical Hamiltonian operator is $H = -\frac{hbar^{2}}{2m}\frac{d^{2}}{dx^{2}} + \frac{1}{2}m\omega^{2}x^{2}$.
This potential is symmetric about x = 0 and confines the particle to a finite region for any finite energy state, though quantum mechanically tunneling occurs.

Energy Eigenvalues and Zero-Point Energy

The allowed energy eigenvalues are quantized and given by E_n = (n + &frac{1}{2})ℏω, where n = 0, 1, 2, 3, ...
For the ground state (n = 0), the energy is E₀ = &frac{1}{2}ℏω, which is known as the zero-point energy.
The zero-point energy is a direct consequence of Heisenberg's uncertainty principle, preventing the particle from resting motionless at the potential minimum.

Wavefunctions and Hermite Polynomials

The spatial wavefunctions ψ_n(x) are expressed in terms of Hermite polynomials H_n(αx) multiplied by a Gaussian decay factor.
The wavefunctions alternate in parity: even n corresponds to even parity functions, and odd n corresponds to odd parity functions.
Adjacent wavefunctions satisfy the orthogonality condition over the domain from -∞ to +∞.

Quantum vs. Classical Turning Points

Classically, a particle cannot exist in regions where its total energy is less than its potential energy (E < V).
Quantum mechanically, the wavefunction penetrates into the classically forbidden region, leading to a non-zero probability density beyond the turning points.
In the classical limit (as n → ∞), the quantum probability distribution approaches the classical probability distribution.

Derivations

Derivation of Energy Eigenvalues using Series Solution

Conditions: Time-independent Schrödinger equation for V(x) = 0.5 m ω² x²

Write the time-independent Schrödinger equation: -&frac{hbar²}{2m}&frac{d²ψ}{dx²} + &frac{1}{2}mω²x²$\psi = E\psi$
Introduce dimensionless coordinates $\xi = \alphax$ where $\alpha = \sqrt{\frac{m\omega}{hbar}}$ and dimensionless energy $\lambda = \frac{2E}{hbar\omega}$.
Transform the differential equation to &frac{d²ψ}{dξ²} + (λ - ξ²)$\psi = 0$.
Assume a solution of the form ψ(ξ) = h(ξ)exp(-ξ²/2) and substitute into the differential equation to obtain Hermite's differential equation for h(ξ).
Apply the condition that the series must terminate to prevent the wavefunction from diverging at infinity, yielding $\lambda = 2n + 1$.

Result: $E_n = (n + 1/2)\hbar\omega$

Derivation of Ground State Wavefunction ψ0(x)

Conditions: For n = 0 using the differential equation and normalization condition

For n = 0, the Hermite polynomial is H_0(ξ) = 1.
The trial wavefunction for the ground state is ψ₀(ξ) = N_0 exp(-ξ²/2), where N_0 is the normalization constant.
Apply the normalization condition ∫_-∞^∞ |ψ₀(x)|² dx = 1.
Evaluate the Gaussian integral ∫_-∞^∞ exp(-α²x²) d$x = \sqrt{\pi}/\alpha$.
Solve for the normalization constant $N_0 = (\frac{\alpha}{\pi})^{1/4}$.

Result: ψ_0(x) = (&frac{mω}{pihbar})^1/4 exp(-&frac{mωx²}{2hbar})

Formula sheet

Formula / law	Equation	Variables and units	Conditions
Potential Energy	$$V(x) = \frac{1}{2}momega^2x^2$$	m = mass, omega = angular frequency, x = displacement Joules (J)	Classical and quantum harmonic approximation
Hamiltonian Operator	$$H = -\frac{hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}momega^2x^2$$	hbar = reduced Planck's constant, m = mass, x = position coordinate Joules (J)	One-dimensional conservative system
Energy Eigenvalues	$$E_n = \left(n + \frac{1}{2}\right)hbaromega$$	n = quantum number (0, 1, 2, ...), omega = angular frequency Joules (J)	Bound states of harmonic oscillator
Zero-Point Energy	$$E_0 = \frac{1}{2}hbaromega$$	hbar = reduced Planck's constant, omega = angular frequency Joules (J)	Ground state (n = 0)
Dimensionless Scale Parameter	$$\alpha = \sqrt{\frac{momega}{hbar}}$$	m = mass, omega = angular frequency, hbar = reduced Planck's constant m^-1	Spatial coordinate scaling
Ground State Wavefunction	$$\psi_0(x) = \left(\frac{momega}{pihbar}\right)^{1/4} exp\left(-\frac{momega x^2}{2hbar}\right)$$	x = position, m = mass, omega = angular frequency m^-1/2	n = 0 state
Creation Operator	$$a^dagger = \sqrt{\frac{momega}{2hbar}} \left( x - \frac{i}{momega} p \right)$$	x = position operator, p = momentum operator Dimensionless operator	Raising operator for energy states
Annihilation Operator	$$a = \sqrt{\frac{momega}{2hbar}} \left( x + \frac{i}{momega} p \right)$$	x = position operator, p = momentum operator Dimensionless operator	Lowering operator for energy states

Graphs and experiments

Parabolic Potential Well and Energy Levels

Draw: A U-shaped parabola representing V(x) = &frac{1}{2}mω²x² with horizontal equidistant lines inside representing E₀, E₁, E₂, etc.

Interpret: Shows that energy levels are equally spaced by Δ$E = \hbar\omega$ and the ground state sits above the bottom of the well by &frac{1}{2}ℏω.

Exam use: Frequently asked in TGT/PGT to identify level spacing and zero-point energy concept.

Wavefunction and Probability Density for n = 0, 1, 2

Draw: Curves showing Hermite polynomial oscillations multiplied by decaying exponentials, alongside |ψ_n(x)|² curves.

Interpret: Illustrates nodes in wavefunctions (n nodes for n-th state) and penetration into classically forbidden regions.

Exam use: Tested via conceptual questions regarding number of nodes and probability density peaks.

Comparison of Classical vs. Quantum Probability Distribution

Draw: A bathtub-shaped classical probability curve plotted against the oscillating quantum probability curve for high n.

Interpret: Demonstrates the correspondence principle: as quantum number n increases, the quantum probability density mirrors the classical average distribution.

Exam use: Tested in PGT sections on correspondence principle and classical limits.

Solved numericals

TGT core

Calculate the zero-point energy of a linear harmonic oscillator whose frequency is 1.0 × 10¹³ Hz. Given $h = 6.63 \times 10^{-34} J s$.

Identify the formula for zero-point energy: $E_0 = 0.5 \hbar \omega = 0.5 h f$.
Substitute the given values: $h = 6.63 \times 10^{-34} J s$ and $f = 1.0 \times 10^{13} Hz$.
$E_0 = 0.5 \times (6.63 \times 10^{-34}) \times (1.0 \times 10^{13})$.
$E_0 = 3.315 \times 10^{-21} J$.

Answer: 3.315 \times 10^{-21} J

PGT extension

For a one-dimensional harmonic oscillator of mass 9.1 × 10^-31 kg and angular frequency 2.0 × 10¹⁴ rad/s, find the root-mean-square displacement √⟨x²⟩ for the ground state.

Recognize that for the ground state, ⟨x$⟩ = 0$, so √⟨x²$⟩ = \sqrt{⟨x^{2}⟩}$.
Use the expectation value relation for position in the ground state: ⟨x²$⟩ = \frac{hbar}{2momega}$.
Substitute the values: $\hbar = 1.054 \times 10^{-34} J s$, $m = 9.1 \times 10^{-31} kg$, $\omega = 2.0 \times 10^{14} rad/s$.
⟨x²$⟩ = \frac{1.054 \times 10^{-34}}{2 \times (9.1 \times 10^{-31}) \times (2.0 \times 10^{14})} = 2.895 \times 10^{-20} m^{2}$.
Take the square root: √⟨x²$⟩ = \sqrt{2.895 \times 10^{-20}} = 1.7 \times 10^{-10} m$.

Answer: 1.7 \times 10^{-10} m

PGT extension

Find the energy difference between the fourth excited state and the ground state for a harmonic oscillator with vibrational frequency $f = 3.0 \times 10^{12} Hz$.

Identify the quantum numbers: ground state n = 0, fourth excited state n = 4.
Energy formula: $E_n = (n + 0.5) h f$.
Calculate $E_0 = (0 + 0.5) h f = 0.5 h f$.
Calculate $E_4 = (4 + 0.5) h f = 4.5 h f$.
Find the difference Δ$E = E_4 - E_0 = 4.0 h f = 4 \times (6.63 \times 10^{-34}) \times (3.0 \times 10^{12})$.
Δ$E = 7.956 \times 10^{-21} J$.

Answer: 7.956 \times 10^{-21} J

Assumptions and limitations

Assumption: The restoring force strictly obeys Hooke's law ($F = -kx) indefinitely$ for all displacements. Limitation: Real molecular potentials (e.g., Morse potential) deviate from harmonic behavior at large amplitudes due to anharmonicity and dissociation.

Common misconceptions

Common error: The minimum energy of a quantum harmonic oscillator is zero.

Correct: The minimum energy is $E_0 = 0.5 \hbar \omega (zero-point energy)$.

A zero energy state would violate the Heisenberg uncertainty principle by allowing simultaneous exact localization and zero momentum.

Common error: A quantum harmonic oscillator particle cannot cross the classical turning points.

Correct: The particle has a non-zero probability of being found in the classically forbidden region.

Quantum wavefunctions decay exponentially rather than dropping abruptly to zero at the turning points.

Quick revision

Energy levels are equally spaced by Δ$E = \hbar\omega with ground state energy \frac{1}{2}\hbar\omega$.
Potential Energy: V(x) = \frac{1}{2}momega^2x^2
Hamiltonian Operator: $H = -\frac{hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}momega^2x^2$

Must remember

The potential energy for a one-directional linear harmonic oscillator is given by $V(x) = \frac{1}{2}m\omega^2x^2$.
The allowed energy eigenvalues are quantized as $E_n = \left(n + \frac{1}{2}\right)\hbar\omega$ for $n = 0, 1, 2, \dots$.
The zero-point energy of the oscillator is $E_0 = \frac{1}{2}\hbar\omega$, which is a direct consequence of Heisenberg's uncertainty principle.
The spatial wavefunctions $\psi_n(x)$ are expressed using Hermite polynomials multiplied by a Gaussian decay factor.
Quantum mechanically, a particle can penetrate into the classically forbidden region where $E < V$, unlike in classical mechanics.

Glossary

Hamiltonian OperatorTotal energy operator for the quantum system given by $H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2}m\omega^2x^2$.

Zero-Point EnergyThe lowest possible energy $E_0 = \frac{1}{2}\hbar\omega$ that a quantum mechanical harmonic oscillator can possess.

Hermite PolynomialsOrthogonal polynomials $H_n(\alpha x)$ that form part of the spatial wavefunctions of the harmonic oscillator.

Creation OperatorAn operator $a^dagger$ that raises the energy state of the harmonic oscillator by one quantum.

Annihilation OperatorAn operator $a$ that lowers the energy state of the harmonic oscillator by one quantum.

Classical vs Quantum Harmonic Oscillator

Feature	Classical Harmonic Oscillator	Quantum Harmonic Oscillator
Energy Values	Continuous energy values from zero upwards	Discretized energy values $E_n = \left(n + \frac{1}{2}\right)\hbar\omega$
Ground State Energy	Zero ($E = 0$ is allowed)	Non-zero zero-point energy ($E_0 = \frac{1}{2}\hbar\omega$)
Turning Points	Strict boundaries where $E = V$; motion strictly prohibited for $E < V$	Wavefunction penetrates classically forbidden region ($E < V$) with non-zero probability

Practice Quiz

Choose an option to reveal the explanation.

1. What is the classical potential energy function for a one-dimensional linear harmonic oscillator of mass $m$ and angular frequency $\omega$?

Explanation: The classical potential energy for a linear harmonic oscillator is defined as $V(x) = \frac{1}{2}m\omega^2x^2$.

2. The allowed energy eigenvalues for a quantum harmonic oscillator are given by:

Explanation: The energy eigenvalues are quantized as $E_n = \left(n + \frac{1}{2}\right)\hbar\omega$ where $n = 0, 1, 2, \dots$.

3. What is the zero-point energy of a one-dimensional quantum harmonic oscillator?

Explanation: For the ground state ($n = 0$), the energy is $E_0 = \frac{1}{2}\hbar\omega$, which is known as the zero-point energy.

4. The zero-point energy in a quantum harmonic oscillator is a direct consequence of:

Explanation: The zero-point energy prevents the particle from resting motionless at the potential minimum, directly arising from Heisenberg's uncertainty principle.

5. What is the correct expression for the Hamiltonian operator $H$ of a 1D linear harmonic oscillator?

Explanation: The Hamiltonian includes the kinetic energy operator $-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ and the potential energy term $\frac{1}{2}m\omega^2x^2$.

6. The spatial wavefunctions $\psi_n(x)$ of a harmonic oscillator are expressed using Hermite polynomials multiplied by which decay factor?

Explanation: Wavefunctions are formed by Hermite polynomials $H_n(\alpha x)$ multiplied by a Gaussian decay factor.

7. What is the parity of the ground state wavefunction $\psi_0(x)$ of a harmonic oscillator?

Explanation: Even quantum numbers $n$ correspond to even parity functions. For the ground state $n = 0$, it is even ($n=0$ is even).

8. In the context of harmonic oscillator scaling, what is the dimension of the dimensionless scale parameter $\alpha = \sqrt{\frac{m\omega}{\hbar}}$?

Explanation: The parameter $\alpha$ has units of inverse meters ($\mathrm{m}^{-1}$) to scale the spatial coordinate $x$.

9. Classically, a particle cannot exist in regions where:

Explanation: Classically, kinetic energy cannot be negative, meaning the particle cannot enter regions where total energy is less than potential energy ($E < V$).

10. Quantum mechanically, what happens to the particle's probability density beyond the classical turning points?

Explanation: The wavefunction penetrates into the classically forbidden region ($E < V$), leading to a non-zero probability density beyond the turning points.

11. As the quantum number $n \to \infty$, what does the quantum probability distribution of a harmonic oscillator approach?

Explanation: According to the correspondence principle, in the classical limit ($n \to \infty$), the quantum probability distribution approaches the classical probability distribution.

12. Which operator acts as the raising operator (creation operator) for the energy states of a harmonic oscillator?

Explanation: The creation operator $a^\dagger$ raises the energy state by one quantum.

13. What is the function of the lowering operator (annihilation operator) $a$?

Explanation: The annihilation operator $a$ lowers the energy state, reducing the energy eigenvalue by $\hbar\omega$.

14. Adjacent wavefunctions of the harmonic oscillator over the domain from $-\infty$ to $+\infty$ satisfy which condition?

Explanation: Adjacent wavefunctions satisfy the orthogonality condition over the entire domain from $-\infty$ to $+\infty$.

15. What is the spatial parity of the first excited state ($n = 1$) wavefunction of a linear harmonic oscillator?

Explanation: Wavefunctions alternate in parity: even $n$ corresponds to even parity, and odd $n$ corresponds to odd parity. Thus, $n=1$ is odd.

PYQ Patterns

These are pattern-based practice questions. Authentic year labels are only used when verified from a source.

1. The energy of the ground state of a one-dimensional simple harmonic oscillator is $\frac{1}{2}\hbar\omega$. This implies that:

Explanation: The minimum energy is non-zero (zero-point energy), which prevents simultaneous exact localization and zero momentum, complying with Heisenberg's uncertainty principle.

2. What is the energy difference $\Delta E$ between two adjacent energy levels $E_n$ and $E_{n+1}$ of a quantum harmonic oscillator?

Explanation: Since $E_n = \left(n + \frac{1}{2}\right)\hbar\omega$, the difference is $E_{n+1} - E_n = \left(n + 1 + \frac{1}{2}\right)\hbar\omega - \left(n + \frac{1}{2}\right)\hbar\omega = \hbar\omega$.

3. Which of the following represents the correct normalized ground state wavefunction $\psi_0(x)$ for a 1D harmonic oscillator with parameter $\alpha = \sqrt{\frac{m\omega}{\hbar}}$?

Explanation: Substituting $\alpha = \sqrt{\frac{m\omega}{\hbar}}$, the ground state wavefunction expression transforms to $\left(\frac{\alpha^2}{\pi}\right)^{1/4} e^{-\frac{\alpha^2 x^2}{2}}$.

4. In quantum mechanics, the turning points for a harmonic oscillator of energy $E_n$ are given by the points where:

Explanation: Turning points occur where the total energy equals the potential energy, i.e., $E_n = V(x) = \frac{1}{2}m\omega^2x^2$.

5. The parity of the wave function $\psi_n(x)$ of a one-dimensional harmonic oscillator is:

Explanation: The wavefunctions of a harmonic oscillator alternate in parity, being even when $n$ is even and odd when $n$ is odd.

6. What is the commutator $[x, p]$ between the position operator $x$ and the momentum operator $p$?

Explanation: The canonical commutation relation in quantum mechanics is $[x, p] = i\hbar$.

Timed Topic Test

Attempt these questions under a ten-minute limit using a separate timer.

1. For a harmonic oscillator, if the angular frequency $\omega$ is doubled while keeping mass constant, the zero-point energy becomes:

Explanation: Zero-point energy is $E_0 = \frac{1}{2}\hbar\omega$. Since $E_0 \propto \omega$, doubling $\omega$ doubles the zero-point energy.

2. The Hermite polynomial $H_0(\alpha x)$ for the ground state of a harmonic oscillator is equal to:

Explanation: The zeroth Hermite polynomial is identically equal to 1 ($H_0(y) = 1$).

3. What is the Hermite polynomial $H_1(\alpha x)$ for the first excited state?

Explanation: The first-order Hermite polynomial is $H_1(y) = 2y$, so $H_1(\alpha x) = 2\alpha x$.

4. Which of the following operators can be expressed in terms of creation and annihilation operators to write the position operator $x$?

Explanation: The position operator in terms of ladder operators is $x = \sqrt{\frac{\hbar}{2m\omega}}(a + a^\dagger)$.

5. The number operator $N$ is defined as $N = a^\dagger a$. What is the eigenvalue of $N$ for the state $\psi_n$?

Explanation: The number operator $N = a^\dagger a$ counts the quanta of excitation, yielding $N\psi_n = n\psi_n$.

6. What is the expectation value of the potential energy $\langle V \rangle$ for the ground state of a harmonic oscillator?

Explanation: By the virial theorem for harmonic potentials, the average kinetic energy equals the average potential energy, each being half of the total ground state energy $\frac{1}{2}E_0 = \frac{1}{4}\hbar\omega$.

7. The classical frequency of a harmonic oscillator is $\nu = \frac{\omega}{2\pi}$. What is the energy spacing between adjacent quantum states in terms of $\nu$?

Explanation: Since $\hbar = \frac{h}{2pi}$ and $\omega = 2\pi\nu$, the energy spacing is $\hbar\omega = \left(\frac{h}{2\pi}\right)(2\pi\nu) = h\nu$.

8. What is the effect of applying the creation operator $a^\dagger$ on the energy eigenstate $|n\rangle$?

Explanation: The creation operator action is given by $a^\dagger |n\rangle = \sqrt{n+1} |n+1\rangle$.

9. Which of the following quantum states of a harmonic oscillator represents the ground state wavefunction profile?

Explanation: The ground state wavefunction $\psi_0(x)$ is proportional to $e^{-\frac{m\omega x^2}{2\hbar}}$, which is a bell-shaped Gaussian curve.

10. Why is quantum mechanical tunneling possible in a harmonic oscillator potential?

Explanation: Tunneling occurs because the quantum wavefunction penetrates into the classically forbidden region ($E < V$) with a non-zero amplitude before decaying.

One-Dimensional Linear Harmonic Oscillator in Quantum Mechanics

Scope and prerequisites

TGT core

PGT extension

Prerequisites

Potential Energy Function and Hamiltonian

Energy Eigenvalues and Zero-Point Energy

Wavefunctions and Hermite Polynomials

Quantum vs. Classical Turning Points

Derivations

Derivation of Energy Eigenvalues using Series Solution

Derivation of Ground State Wavefunction &psi;<sub>0</sub>(x)

Formula sheet

Graphs and experiments

Parabolic Potential Well and Energy Levels

Wavefunction and Probability Density for n = 0, 1, 2

Comparison of Classical vs. Quantum Probability Distribution

Solved numericals

Calculate the zero-point energy of a linear harmonic oscillator whose frequency is 1.0 × 1013 Hz. Given $h = 6.63 \times 10^{-34} J s$.

For a one-dimensional harmonic oscillator of mass 9.1 × 10-31 kg and angular frequency 2.0 × 1014 rad/s, find the root-mean-square displacement √⟨x2⟩ for the ground state.

Find the energy difference between the fourth excited state and the ground state for a harmonic oscillator with vibrational frequency $f = 3.0 \times 10^{12} Hz$.

Assumptions and limitations

Common misconceptions

Quick revision

Must remember

Glossary

Classical vs Quantum Harmonic Oscillator

Practice Quiz

1. What is the classical potential energy function for a one-dimensional linear harmonic oscillator of mass $m$ and angular frequency $\omega$?

2. The allowed energy eigenvalues for a quantum harmonic oscillator are given by:

3. What is the zero-point energy of a one-dimensional quantum harmonic oscillator?

4. The zero-point energy in a quantum harmonic oscillator is a direct consequence of:

5. What is the correct expression for the Hamiltonian operator $H$ of a 1D linear harmonic oscillator?

6. The spatial wavefunctions $\psi_n(x)$ of a harmonic oscillator are expressed using Hermite polynomials multiplied by which decay factor?

7. What is the parity of the ground state wavefunction $\psi_0(x)$ of a harmonic oscillator?

8. In the context of harmonic oscillator scaling, what is the dimension of the dimensionless scale parameter $\alpha = \sqrt{\frac{m\omega}{\hbar}}$?

9. Classically, a particle cannot exist in regions where:

10. Quantum mechanically, what happens to the particle's probability density beyond the classical turning points?

11. As the quantum number $n \to \infty$, what does the quantum probability distribution of a harmonic oscillator approach?

12. Which operator acts as the raising operator (creation operator) for the energy states of a harmonic oscillator?

13. What is the function of the lowering operator (annihilation operator) $a$?

14. Adjacent wavefunctions of the harmonic oscillator over the domain from $-\infty$ to $+\infty$ satisfy which condition?

15. What is the spatial parity of the first excited state ($n = 1$) wavefunction of a linear harmonic oscillator?

PYQ Patterns

1. The energy of the ground state of a one-dimensional simple harmonic oscillator is $\frac{1}{2}\hbar\omega$. This implies that:

2. What is the energy difference $\Delta E$ between two adjacent energy levels $E_n$ and $E_{n+1}$ of a quantum harmonic oscillator?

3. Which of the following represents the correct normalized ground state wavefunction $\psi_0(x)$ for a 1D harmonic oscillator with parameter $\alpha = \sqrt{\frac{m\omega}{\hbar}}$?

4. In quantum mechanics, the turning points for a harmonic oscillator of energy $E_n$ are given by the points where:

5. The parity of the wave function $\psi_n(x)$ of a one-dimensional harmonic oscillator is:

6. What is the commutator $[x, p]$ between the position operator $x$ and the momentum operator $p$?

Timed Topic Test

1. For a harmonic oscillator, if the angular frequency $\omega$ is doubled while keeping mass constant, the zero-point energy becomes:

2. The Hermite polynomial $H_0(\alpha x)$ for the ground state of a harmonic oscillator is equal to:

3. What is the Hermite polynomial $H_1(\alpha x)$ for the first excited state?

4. Which of the following operators can be expressed in terms of creation and annihilation operators to write the position operator $x$?

5. The number operator $N$ is defined as $N = a^\dagger a$. What is the eigenvalue of $N$ for the state $\psi_n$?

6. What is the expectation value of the potential energy $\langle V \rangle$ for the ground state of a harmonic oscillator?

7. The classical frequency of a harmonic oscillator is $\nu = \frac{\omega}{2\pi}$. What is the energy spacing between adjacent quantum states in terms of $\nu$?

8. What is the effect of applying the creation operator $a^\dagger$ on the energy eigenstate $|n\rangle$?

9. Which of the following quantum states of a harmonic oscillator represents the ground state wavefunction profile?

10. Why is quantum mechanical tunneling possible in a harmonic oscillator potential?

Derivation of Ground State Wavefunction ψ<sub>0</sub>(x)

Calculate the zero-point energy of a linear harmonic oscillator whose frequency is 1.0 × 10¹³ Hz. Given $h = 6.63 \times 10^{-34} J s$.

For a one-dimensional harmonic oscillator of mass 9.1 × 10^-31 kg and angular frequency 2.0 × 10¹⁴ rad/s, find the root-mean-square displacement √⟨x²⟩ for the ground state.