Linear Harmonic Oscillator

The linear harmonic oscillator is a fundamental model in physics that describes systems where a particle is displaced from an equilibrium position and experiences a restoring force proportional to the displacement. This force leads to oscillatory motion. A classic example is a mass attached to a spring obeying Hooke's Law.

Classical Harmonic Oscillator

For a classical one-dimensional harmonic oscillator, the potential energy is given by:

$V(x) = \frac{1}{2}kx^2$

where $k$ is the spring constant and $x$ is the displacement from equilibrium. The restoring force is $F = -kx$. The equation of motion is given by Newton's second law:

$m\frac{d^2x}{dt^2} = -kx$

This is a second-order linear differential equation with the solution:

$x(t) = A \cos(\omega_0 t + \phi)$

where $A$ is the amplitude, $\phi$ is the phase constant, and $\omega_0 = \sqrt{\frac{k}{m}}$ is the angular frequency of oscillation. The energy of the classical oscillator can range from zero to infinity.

Quantum Harmonic Oscillator

In quantum mechanics, the harmonic oscillator is described by the Hamiltonian operator, which includes the kinetic and potential energy operators:

$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega_0^2\hat{x}^2$

where $\hat{p}$ is the momentum operator ($p = -i\hbar\frac{d}{dx}$) and $\hat{x}$ is the position operator. The time-independent Schrödinger equation for the quantum harmonic oscillator is:

$\hat{H}\psi(x) = E\psi(x)$

Solving this equation yields quantized energy levels. Unlike the classical oscillator, the quantum harmonic oscillator can only have discrete energy values.

Energy Eigenvalues

The energy eigenvalues for the quantum harmonic oscillator are given by:

$E_n = \left(n + \frac{1}{2}\right)\hbar\omega_0$, where $n = 0, 1, 2, 3, \dots$

Here, $n$ is the quantum number. The lowest possible energy, for $n=0$, is $E_0 = \frac{1}{2}\hbar\omega_0$. This is known as the zero-point energy. It implies that even in its ground state, the oscillator has some residual energy and is never completely at rest. This is a direct consequence of the Heisenberg uncertainty principle.

Ladder Operators (Creation and Annihilation Operators)

A more elegant way to solve the quantum harmonic oscillator and understand its spectrum is by using ladder operators, also known as creation and annihilation operators. These operators allow us to move between energy eigenstates without explicitly solving the differential equation.

The annihilation operator, $\hat{a}$, is defined as:

$\hat{a} = \sqrt{\frac{m\omega_0}{2\hbar}}\hat{x} + \frac{i}{\sqrt{2m\hbar\omega_0}}\hat{p}$

The creation operator, $\hat{a}^\dagger$, is the Hermitian conjugate of $\hat{a}$:

$\hat{a}^\dagger = \sqrt{\frac{m\omega_0}{2\hbar}}\hat{x} - \frac{i}{\sqrt{2m\hbar\omega_0}}\hat{p}$

These operators satisfy the commutation relation:

$[\hat{a}, \hat{a}^\dagger] = \hat{a}\hat{a}^\dagger - \hat{a}^\dagger\hat{a} = 1$

The Hamiltonian can be expressed in terms of these operators:

$\hat{H} = \hbar\omega_0 \left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right)$

The operator $\hat{N} = \hat{a}^\dagger\hat{a}$ is called the number operator. Its eigenvalues are the non-negative integers $n$. The energy eigenvalues are then $E_n = \hbar\omega_0(n + \frac{1}{2})$.

The action of the ladder operators on the energy eigenstates $|\psi_n\rangle$ is:

$\hat{a}|\psi_n\rangle = \sqrt{n}|\psi_{n-1}\rangle$ (Annihilation operator lowers the energy by one quantum)

$\hat{a}^\dagger|\psi_n\rangle = \sqrt{n+1}|\psi_{n+1}\rangle$ (Creation operator raises the energy by one quantum)

Applying $\hat{a}$ to the ground state ($n=0$) results in zero: $\hat{a}|\psi_0\rangle = 0$. This shows that $|\psi_0\rangle$ is the lowest energy state.

Angular Momentum

Angular momentum is the rotational analogue of linear momentum. It is a vector quantity that describes the amount of rotation an object has. In classical mechanics, for a point particle, angular momentum $\vec{L}$ is defined as the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p}$:

$\vec{L} = \vec{r} \times \vec{p}$

In quantum mechanics, angular momentum is also a vector operator, and its components do not commute with each other. This non-commutativity has profound implications for the quantization of angular momentum.

Angular Momentum Operators in Quantum Mechanics

For a particle at position $\vec{r} = (x, y, z)$ with momentum $\vec{p} = (p_x, p_y, p_z)$, the angular momentum operators are:

$\hat{L}_x = \hat{y}\hat{p}_z - \hat{z}\hat{p}_y$

$\hat{L}_y = \hat{z}\hat{p}_x - \hat{x}\hat{p}_z$

$\hat{L}_z = \hat{x}\hat{p}_y - \hat{y}\hat{p}_x$

In spherical coordinates, these operators become more complex but are often more useful for problems with spherical symmetry.

Commutation Relations

The components of the angular momentum operator do not commute. The fundamental commutation relations are:

$[\hat{L}_x, \hat{L}_y] = i\hbar\hat{L}_z$

$[\hat{L}_y, \hat{L}_z] = i\hbar\hat{L}_x$

$[\hat{L}_z, \hat{L}_x] = i\hbar\hat{L}_y$

These relations imply that we cannot simultaneously know the precise values of all three components of angular momentum. It is conventional to choose $\hat{L}_z$ as the component that can be simultaneously diagonalized with the total angular momentum squared operator.

Total Angular Momentum Squared Operator

The operator for the square of the total angular momentum is:

$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$

The operator $\hat{L}^2$ commutes with each component of $\hat{L}$:

$[\hat{L}^2, \hat{L}_x] = [\hat{L}^2, \hat{L}_y] = [\hat{L}^2, \hat{L}_z] = 0$

This means that we can find a set of simultaneous eigenstates for $\hat{L}^2$ and any one component, say $\hat{L}_z$.

Eigenvalues of Angular Momentum

The eigenvalues of $\hat{L}^2$ and $\hat{L}_z$ are quantized. For a given angular momentum quantum number $l$, the eigenvalues of $\hat{L}^2$ are:

$\hat{L}^2 |\psi_{l,m}\rangle = l(l+1)\hbar^2 |\psi_{l,m}\rangle$

where $l$ is a non-negative integer ($l = 0, 1, 2, \dots$).

For a given $l$, the eigenvalues of $\hat{L}_z$ are:

$\hat{L}_z |\psi_{l,m}\rangle = m\hbar |\psi_{l,m}\rangle$

where $m$ can take integer values from $-l$ to $+l$: $m = -l, -l+1, \dots, l-1, l$.

The quantum number $l$ determines the magnitude of the angular momentum, while $m$ determines the projection of the angular momentum onto the z-axis. The total number of possible $m$ values for a given $l$ is $2l+1$.

Spin Angular Momentum

In addition to orbital angular momentum (due to motion), particles like electrons possess intrinsic angular momentum called spin. Spin is a purely quantum mechanical phenomenon and does not have a classical analogue.

Spin angular momentum operators ($\hat{S}_x, \hat{S}_y, \hat{S}_z, \hat{S}^2$) obey the same commutation relations as orbital angular momentum. However, the quantum numbers for spin are different. For an electron, the spin quantum number is $s = 1/2$.

The eigenvalues of $\hat{S}^2$ are $s(s+1)\hbar^2$. For $s=1/2$, this is $\frac{1}{2}(\frac{1}{2}+1)\hbar^2 = \frac{3}{4}\hbar^2$.

The eigenvalues of $\hat{S}_z$ are $m_s\hbar$, where $m_s$ can take values from $-s$ to $+s$. For $s=1/2$, $m_s$ can be $+1/2$ (spin up) or $-1/2$ (spin down).

Addition of Angular Momenta

In many physical systems, particles interact, and their individual angular momenta combine to form a total angular momentum for the system. For example, an electron has orbital and spin angular momentum. Two particles might interact, and their individual angular momenta add up. The process of combining angular momenta requires a specific mathematical formalism.

The Problem

Suppose we have two systems, System 1 with angular momentum $\vec{J}_1$ and System 2 with angular momentum $\vec{J}_2$. The total angular momentum of the combined system is $\vec{J} = \vec{J}_1 + \vec{J}_2$. We want to find the possible values of the total angular momentum quantum numbers.

Let the quantum numbers for System 1 be $(j_1, m_1)$ and for System 2 be $(j_2, m_2)$. These describe the eigenvalues of $\vec{J}_1^2$, $\vec{J}_{1z}$, $\vec{J}_2^2$, and $\vec{J}_{2z}$. The basis states for the combined system in this uncoupled representation are $|j_1, m_1; j_2, m_2\rangle$.

We are interested in the eigenvalues of the total angular momentum squared operator $\vec{J}^2 = (\vec{J}_1 + \vec{J}_2)^2$ and its z-component $\vec{J}_z = \vec{J}_{1z} + \vec{J}_{2z}$. The operator $\vec{J}_z$ commutes with $\vec{J}_{1z}$ and $\vec{J}_{2z}$, so its eigenvalues are simply the sum of the individual eigenvalues:

$\hat{J}_z |j_1, m_1; j_2, m_2\rangle = (m_1 + m_2)\hbar |j_1, m_1; j_2, m_2\rangle$

The operator $\vec{J}^2$ involves cross terms like $\vec{J}_1 \cdot \vec{J}_2$, which do not commute with $\vec{J}_{1z}$ or $\vec{J}_{2z}$. Therefore, the states $|j_1, m_1; j_2, m_2\rangle$ are generally not eigenstates of $\vec{J}^2$. We need to find a new set of basis states, the coupled representation, $|j, m; j_1, j_2\rangle$, which are simultaneous eigenstates of $\vec{J}^2$ and $\vec{J}_z$.

Clebsch-Gordan Coefficients

The transformation from the uncoupled basis to the coupled basis is given by:

$|j, m; j_1, j_2\rangle = \sum_{m_1, m_2} \langle j_1, m_1; j_2, m_2 | j, m; j_1, j_2 \rangle |j_1, m_1; j_2, m_2\rangle$

The coefficients $\langle j_1, m_1; j_2, m_2 | j, m; j_1, j_2 \rangle$ are called Clebsch-Gordan coefficients, denoted by $C^{j,m}_{j_1,m_1; j_2,m_2}$.

Rules for Combining Angular Momenta

When adding two angular momenta $j_1$ and $j_2$, the resulting total angular momentum quantum number $j$ can take values ranging from $|j_1 - j_2|$ to $j_1 + j_2$ in integer steps.

$j = |j_1 - j_2|, |j_1 - j_2| + 1, \dots, j_1 + j_2 - 1, j_1 + j_2$

For each possible value of $j$, the possible values of $m$ range from $-j$ to $j$ in integer steps:

$m = -j, -j+1, \dots, j-1, j$

The total number of states in the uncoupled basis is $(2j_1+1)(2j_2+1)$. The total number of states in the coupled basis is the sum of $(2j+1)$ for all possible $j$ values. These two numbers must be equal.

Addition of Orbital and Spin Angular Momentum

For a single electron, its orbital angular momentum $\vec{L}$ and spin angular momentum $\vec{S}$ combine to form the total angular momentum $\vec{J} = \vec{L} + \vec{S}$.

If an electron has orbital quantum numbers $(l, m_l)$ and spin quantum numbers $(s, m_s)$, the total angular momentum quantum number $j$ can take values:

$j = |l - s|, \dots, l + s$. For an electron, $s=1/2$.

So, $j = l \pm 1/2$ (if $l > 0$). If $l=0$, then $j=1/2$.

The quantum number $m_j$ ranges from $-j$ to $+j$.

Perturbation Theory

Perturbation theory is a powerful mathematical method used in quantum mechanics to find approximate solutions to the Schrödinger equation for systems that are "close" to exactly solvable systems. It's used when the Hamiltonian of the system can be split into a solvable part and a small, "perturbing" part.

The Setup

Consider a system with Hamiltonian $\hat{H}_0$ that is exactly solvable. Let its eigenvalues and eigenstates be $E_n^{(0)}$ and $|\psi_n^{(0)}\rangle$, respectively, satisfying:

$\hat{H}_0 |\psi_n^{(0)}\rangle = E_n^{(0)} |\psi_n^{(0)}\rangle$

Now, we introduce a small perturbation, represented by the Hamiltonian $\hat{H}'$. The total Hamiltonian of the perturbed system is:

$\hat{H} = \hat{H}_0 + \lambda\hat{H}'$

where $\lambda$ is a dimensionless parameter, $0 \le \lambda \le 1$, which indicates the strength of the perturbation. We assume that the exact solutions $\hat{H}$ can be expressed as power series in $\lambda$:

$E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \dots$

$|\psi_n\rangle = |\psi_n^{(0)}\rangle + \lambda |\psi_n^{(1)}\rangle + \lambda^2 |\psi_n^{(2)}\rangle + \dots$

Here, $E_n^{(k)}$ and $|\psi_n^{(k)}\rangle$ are the $k$-th order corrections to the energy and wavefunction, respectively.

Non-Degenerate Perturbation Theory

This applies when the energy levels of the unperturbed system are non-degenerate (i.e., each energy eigenvalue corresponds to only one unique state).

First-Order Energy Correction ($E_n^{(1)}$)

The first-order correction to the energy of the $n$-th state is simply the expectation value of the perturbation Hamiltonian in the unperturbed state:

$E_n^{(1)} = \langle\psi_n^{(0)}|\hat{H}'|\psi_n^{(0)}\rangle$

This means the first-order correction is the average value of the perturbation experienced by the unperturbed state.

First-Order Wavefunction Correction ($|\psi_n^{(1)}\rangle$)

The first-order correction to the wavefunction involves a sum over all other unperturbed states $|\psi_m^{(0)}\rangle$ (where $m \neq n$):

$|\psi_n^{(1)}\rangle = \sum_{m \neq n} \frac{\langle\psi_m^{(0)}|\hat{H}'|\psi_n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}} |\psi_m^{(0)}\rangle$

The term $\langle\psi_m^{(0)}|\hat{H}'|\psi_n^{(0)}\rangle$ is the matrix element of the perturbation between states $|\psi_n^{(0)}\rangle$ and $|\psi_m^{(0)}\rangle$. The denominator $E_n^{(0)} - E_m^{(0)}$ is the energy difference between the unperturbed states.

Second-Order Energy Correction ($E_n^{(2)}$)

The second-order correction to the energy is given by:

$E_n^{(2)} = \sum_{m \neq n} \frac{|\langle\psi_m^{(0)}|\hat{H}'|\psi_n^{(0)}\rangle|^2}{E_n^{(0)} - E_m^{(0)}}$

Notice that $E_n^{(2)}$ is always negative (since $E_n^{(0)} - E_m^{(0)}$ has the opposite sign to the numerator's square if $E_n^{(0)} > E_m^{(0)}$, and the numerator is squared, making it positive. However, the usual convention is that $E_n^{(0)} - E_m^{(0)}$ is in the denominator, and if $E_n^{(0)}$ is the higher energy, the denominator is positive. If $E_n^{(0)}$ is the lower energy, the denominator is negative, and the correction is negative). The second-order correction tends to lower the energy.

Degenerate Perturbation Theory

When the unperturbed energy level $E_n^{(0)}$ corresponds to more than one linearly independent state (i.e., it is degenerate), the non-degenerate formulas are insufficient. We must consider the subspace of degenerate states.

Let $E^{(0)}$ be a degenerate energy level with degeneracy $d$. Let the $d$ linearly independent unperturbed states corresponding to this energy be $|\psi_1^{(0)}\rangle, |\psi_2^{(0)}\rangle, \dots, |\psi_d^{(0)}\rangle$.

The perturbation $\hat{H}'$ will generally lift this degeneracy. We need to find the eigenvalues of the $d \times d$ matrix formed by the perturbation Hamiltonian within this subspace. Let this matrix be $H'_{ij} = \langle\psi_i^{(0)}|\hat{H}'|\psi_j^{(0)}\rangle$, where $i, j = 1, \dots, d$.

The corrected energies $E_n$ are the eigenvalues of this matrix $H'$. The corresponding corrected wavefunctions $|\psi_n\rangle$ are the eigenvectors of this matrix.

The first-order energy correction for a degenerate state is then one of the eigenvalues of the matrix $H'_{ij}$. Higher-order corrections can be calculated if needed, but they are more complex.

Transition Probabilities

Transition probabilities describe the likelihood that a quantum system will move from one state to another, typically due to the influence of an external time-dependent perturbation, such as electromagnetic radiation.

Fermi's Golden Rule

Fermi's Golden Rule is a fundamental result that gives the transition rate from an initial state $|i\rangle$ to a continuum of final states $|f\rangle$ under the influence of a constant perturbation. It applies when the perturbation is weak and causes transitions between states that are very close in energy.

Let the initial state be $|i\rangle$ with energy $E_i$. The perturbation is $\hat{H}'$. We are interested in transitions to a dense set of final states $|f\rangle$ with energies $E_f$ around $E_i$. The density of states is denoted by $\rho(E_f)$.

The transition rate $W_{i \to f}$ (probability per unit time) is given by:

$W_{i \to f} = \frac{2\pi}{\hbar} |\langle f|\hat{H}'|i\rangle|^2 \rho(E_f)$

where $\langle f|\hat{H}'|i\rangle$ is the matrix element of the perturbation between the initial and final states. The density of states $\rho(E_f)$ accounts for the number of available final states per unit energy interval.

Transitions due to Constant Perturbations

If a system is in an eigenstate $|i\rangle$ of $\hat{H}_0$ and a constant perturbation $\hat{H}'$ is suddenly applied, Fermi's Golden Rule can be used to calculate the rate of transitions to other states.

For example, consider a particle in a box. If a small region of the box suddenly changes its potential (a constant perturbation), the particle might transition from its initial energy level to higher energy levels. The rate of these transitions is governed by Fermi's Golden Rule, provided the final states are in a continuum or effectively form one.

Transitions due to Harmonic Perturbations (Time-Dependent Perturbation Theory)

A very common type of perturbation is one that varies sinusoidally with time, such as an oscillating electric field (e.g., light). This is described by a time-dependent Hamiltonian $\hat{H}(t) = \hat{H}_0 + \hat{H}'(t)$.

Let $\hat{H}'(t) = \vec{d} \cdot \vec{E}(t)$, where $\vec{d}$ is the dipole moment operator and $\vec{E}(t)$ is the electric field. If the electric field is monochromatic, $\vec{E}(t) = \vec{E}_0 \cos(\omega t)$.

The time-dependent Schrödinger equation is:

$i\hbar \frac{\partial}{\partial t}|\Psi(t)\rangle = (\hat{H}_0 + \hat{H}'(t))|\Psi(t)\rangle$

If the system starts in an eigenstate $|i\rangle$ of $\hat{H}_0$ at $t=0$, the perturbation $\hat{H}'(t)$ can induce transitions to other states $|f\rangle$.

Absorption and Emission

For a harmonic perturbation $\hat{H}'(t)$ with frequency $\omega$, transitions are most likely to occur between states $|i\rangle$ and $|f\rangle$ if the energy difference $E_f - E_i$ is close to $\hbar\omega$ (absorption) or $-\hbar\omega$ (stimulated emission).

The transition probability per unit time from state $|i\rangle$ to state $|f\rangle$ induced by a harmonic perturbation $\hat{H}'(t) = \hat{V} e^{-i\omega t} + \hat{V}^\dagger e^{i\omega t}$ (where $\hat{V}$ is the time-independent part) is approximately:

For absorption ($E_f > E_i$): $W_{i \to f} \approx \frac{2\pi}{\hbar} |\langle f|\hat{V}^\dagger|i\rangle|^2 \delta(E_f - E_i - \hbar\omega)$

For stimulated emission ($E_f < E_i$): $W_{i \to f} \approx \frac{2\pi}{\hbar} |\langle f|\hat{V}|i\rangle|^2 \delta(E_f - E_i + \hbar\omega)$

The delta functions ensure energy conservation. These formulas are essentially Fermi's Golden Rule applied to the specific form of the harmonic perturbation. The terms $\langle f|\hat{V}^\dagger|i\rangle$ and $\langle f|\hat{V}|i\rangle$ are the matrix elements responsible for absorption and emission, respectively.

Spontaneous Emission

In addition to stimulated absorption and emission, systems can also undergo spontaneous emission, where a system in an excited state decays to a lower energy state by emitting a photon, even in the absence of an external field. This is a purely quantum effect related to the interaction of the atom with the vacuum fluctuations of the electromagnetic field. The rate of spontaneous emission is also calculable using more advanced quantum electrodynamics, but it is related to the same matrix elements involved in stimulated emission.