GATE Physics Quantum Mechanics Syllabus

Give the approximate transmission coefficient for tunneling through a wide/high rectangular barrier of width $L$ and height $V_{0}$ (with $E < V_{0}$).

$$T \approx e^{-2\kappa L}, \qquad \kappa = \frac{\sqrt{2m(V_{0}-E)}}{\hbar}.$$ Transmission falls off exponentially with barrier width and with the square root of $(V_{0}-E)$.

State the WKB (Gamow) tunneling formula for transmission through an arbitrary smooth barrier.

$$T \approx \exp\!\left( -\frac{2}{\hbar}\int_{x_{1}}^{x_{2}} \sqrt{2m\,[V(x)-E]}\;dx \right),$$ where $x_{1}$ and $x_{2}$ are the classical turning points where $V(x)=E$.

Write the energy eigenvalues for a particle in a one-dimensional infinite square well of width $L$.

$$E_{n} = \frac{n^{2}\pi^{2}\hbar^{2}}{2mL^{2}}, \qquad n = 1,2,3,\dots$$ Energy is quantized and there is no $n=0$ state (zero state would violate the uncertainty principle).

Write the normalized eigenfunctions for a 1D infinite square well of width $L$ (from $0$ to $L$).

$$\psi_{n}(x) = \sqrt{\frac{2}{L}}\,\sin\!\left(\frac{n\pi x}{L}\right), \qquad n = 1,2,3,\dots$$ with $\psi_{n}(0)=\psi_{n}(L)=0$.

For a 1D infinite square well, what is the ground-state (zero-point) energy and why can't it be zero?

Ground state ($n=1$): $$E_{1} = \frac{\pi^{2}\hbar^{2}}{2mL^{2}}.$$ It cannot be zero because $\psi \equiv 0$ is not normalizable, and a zero-energy localized state would violate the uncertainty principle $\Delta x\,\Delta p \geq \hbar/2$.

Write the energy levels of a 3D cubic infinite well of side $L$ and give the degeneracy reasoning.

$$E_{n_x n_y n_z} = \frac{\pi^{2}\hbar^{2}}{2mL^{2}}\left(n_x^{2}+n_y^{2}+n_z^{2}\right), \quad n_i = 1,2,\dots$$ Different sets $(n_x,n_y,n_z)$ giving the same $n_x^2+n_y^2+n_z^2$ have the same energy, producing degeneracy.

Write the energy eigenvalues of the 1D quantum harmonic oscillator.

$$E_{n} = \left(n + \tfrac{1}{2}\right)\hbar\omega, \qquad n = 0,1,2,\dots$$ The levels are equally spaced by $\hbar\omega$, and the zero-point energy is $E_{0} = \tfrac{1}{2}\hbar\omega$.

Define the ladder (raising and lowering) operators for the harmonic oscillator and give their commutator.

$$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right), \quad \hat{a}^{\dagger} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right),$$ with $[\hat{a},\hat{a}^{\dagger}] = 1$. The Hamiltonian is $\hat{H} = \hbar\omega(\hat{a}^{\dagger}\hat{a} + \tfrac{1}{2})$.

State the action of the raising and lowering operators on the number state $|n\rangle$.

$$\hat{a}|n\rangle = \sqrt{n}\,|n-1\rangle, \qquad \hat{a}^{\dagger}|n\rangle = \sqrt{n+1}\,|n+1\rangle.$$ Also $\hat{a}^{\dagger}\hat{a}|n\rangle = n|n\rangle$ (number operator) and $\hat{a}|0\rangle = 0$.

Write the ground-state wavefunction of the 1D harmonic oscillator.

$$\psi_{0}(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \exp\!\left(-\frac{m\omega x^{2}}{2\hbar}\right).$$ It is a Gaussian with no nodes.

Define degeneracy in quantum mechanics and give its usual physical origin.

Degeneracy occurs when two or more linearly independent eigenstates share the same energy eigenvalue. The number of such states is the degree of degeneracy. It usually arises from a symmetry of the Hamiltonian (e.g. rotational symmetry giving $(2l+1)$-fold degeneracy in $m_l$).

What is the total degeneracy of the $n$th energy level of the hydrogen atom (ignoring spin, then including spin)?

Ignoring spin: $\sum_{l=0}^{n-1}(2l+1) = n^{2}$. Including electron spin (factor 2): $2n^{2}$. This degeneracy in $l$ is a special ('accidental') symmetry of the pure Coulomb potential.

Write the energy eigenvalues of the hydrogen atom and the value for the ground state.

$$E_{n} = -\frac{13.6\ \text{eV}}{n^{2}} = -\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}}, \qquad n = 1,2,3,\dots$$ Ground state: $E_{1} = -13.6\ \text{eV}$.

List the four quantum numbers of the hydrogen atom and their allowed ranges.

Principal $n = 1,2,3,\dots$; orbital (azimuthal) $l = 0,1,\dots,n-1$; magnetic $m_{l} = -l,\dots,0,\dots,+l$; spin $m_{s} = \pm\tfrac{1}{2}$. Energy depends only on $n$ (Coulomb degeneracy).

Write the value of the Bohr radius $a_{0}$ and the hydrogen ground-state radial wavefunction.

Bohr radius: $a_{0} = \frac{4\pi\varepsilon_{0}\hbar^{2}}{m e^{2}} \approx 0.529\ \text{\AA}$. Ground-state wavefunction: $$\psi_{100}(r) = \frac{1}{\sqrt{\pi}\,a_{0}^{3/2}}\,e^{-r/a_{0}}.$$

State the eigenvalues of $\hat{L}^{2}$ and $\hat{L}_{z}$ for orbital angular momentum.

$$\hat{L}^{2}|l,m\rangle = l(l+1)\hbar^{2}|l,m\rangle, \qquad \hat{L}_{z}|l,m\rangle = m\hbar\,|l,m\rangle,$$ with $l = 0,1,2,\dots$ and $m = -l,\dots,+l$.

Write the commutation relations among the angular momentum components.

$$[\hat{L}_{x},\hat{L}_{y}] = i\hbar \hat{L}_{z}, \quad [\hat{L}_{y},\hat{L}_{z}] = i\hbar \hat{L}_{x}, \quad [\hat{L}_{z},\hat{L}_{x}] = i\hbar \hat{L}_{y},$$ compactly $[\hat{L}_{i},\hat{L}_{j}] = i\hbar\,\epsilon_{ijk}\hat{L}_{k}$. Also $[\hat{L}^{2},\hat{L}_{i}] = 0$.

Define the angular momentum raising and lowering operators and their action on $|l,m\rangle$.

$\hat{L}_{\pm} = \hat{L}_{x} \pm i\hat{L}_{y}$, with $$\hat{L}_{\pm}|l,m\rangle = \hbar\sqrt{l(l+1)-m(m\pm 1)}\;|l,m\pm 1\rangle.$$ They raise/lower $m$ by one unit while leaving $l$ fixed.

Give the spin operators for a spin-1/2 particle in terms of the Pauli matrices, and the eigenvalues of $\hat{S}^{2}$ and $\hat{S}_{z}$.

$\hat{\vec{S}} = \frac{\hbar}{2}\vec{\sigma}$ with $$\sigma_{x}=\begin{pmatrix}0&1\\1&0\end{pmatrix},\ \sigma_{y}=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\ \sigma_{z}=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$ $\hat{S}^{2}$ eigenvalue $= \frac{3}{4}\hbar^{2}$; $\hat{S}_{z}$ eigenvalues $= \pm\frac{\hbar}{2}$.

State the Pauli matrices' key algebraic properties.

Each squares to the identity: $\sigma_{i}^{2} = I$. They anticommute: $\{\sigma_{i},\sigma_{j}\} = 2\delta_{ij}I$. Commutators: $[\sigma_{i},\sigma_{j}] = 2i\,\epsilon_{ijk}\sigma_{k}$. They are Hermitian, traceless, and unitary with $\det\sigma_i = -1$.

In the addition of two angular momenta $j_{1}$ and $j_{2}$, what values can the total angular momentum $j$ take?

$$j = |j_{1}-j_{2}|,\ |j_{1}-j_{2}|+1,\ \dots,\ j_{1}+j_{2},$$ in integer steps. For each $j$, $m = -j,\dots,+j$, and the total number of states is $(2j_{1}+1)(2j_{2}+1)$.

When adding two spin-1/2 particles, what total-spin states result and how are they classified?

They combine into a triplet ($s=1$, symmetric, 3 states: $m_s = +1,0,-1$) and a singlet ($s=0$, antisymmetric, 1 state). Dimension check: $2\times 2 = 3 + 1 = 4$ states.

What are Clebsch-Gordan coefficients and what do they accomplish?

Clebsch-Gordan coefficients $\langle j_{1}m_{1};j_{2}m_{2}|j\,m\rangle$ are the expansion coefficients relating the coupled basis $|j,m\rangle$ to the uncoupled product basis $|j_{1}m_{1}\rangle|j_{2}m_{2}\rangle$. They are nonzero only when $m = m_{1}+m_{2}$ and $|j_1-j_2|\le j\le j_1+j_2$.

In time-independent (non-degenerate) perturbation theory with $\hat{H} = \hat{H}_{0} + \lambda \hat{H}'$, write the first-order energy correction.

$$E_{n}^{(1)} = \langle \psi_{n}^{(0)} | \hat{H}' | \psi_{n}^{(0)} \rangle,$$ the expectation value of the perturbation in the unperturbed state $|\psi_{n}^{(0)}\rangle$.

Write the second-order energy correction and the first-order wavefunction correction in non-degenerate perturbation theory.

$$E_{n}^{(2)} = \sum_{m\neq n} \frac{|\langle \psi_{m}^{(0)}|\hat{H}'|\psi_{n}^{(0)}\rangle|^{2}}{E_{n}^{(0)} - E_{m}^{(0)}},$$ $$|\psi_{n}^{(1)}\rangle = \sum_{m\neq n} \frac{\langle \psi_{m}^{(0)}|\hat{H}'|\psi_{n}^{(0)}\rangle}{E_{n}^{(0)} - E_{m}^{(0)}}\,|\psi_{m}^{(0)}\rangle.$$

Why does non-degenerate perturbation theory fail for degenerate levels, and how is it fixed?

The energy denominators $E_{n}^{(0)} - E_{m}^{(0)}$ vanish for degenerate states, giving divergences. One uses degenerate perturbation theory: diagonalize the perturbation $\hat{H}'$ within the degenerate subspace; the eigenvalues of that matrix are the first-order corrections, and the perturbation may lift the degeneracy.

What is the Born approximation used for, and what is its central physical assumption?

The Born approximation is a method in scattering theory to compute the scattering amplitude $f(\theta)$ for a potential $V(\vec{r})$. It assumes the scattering potential is weak, so the incident wave is only slightly perturbed and can be approximated by a plane wave inside the integral (first-order treatment).

Write the first Born approximation for the scattering amplitude $f(\theta)$.

$$f(\theta) = -\frac{m}{2\pi\hbar^{2}} \int e^{-i\vec{q}\cdot\vec{r}}\, V(\vec{r})\, d^{3}r,$$ where $\vec{q} = \vec{k}' - \vec{k}$ is the momentum transfer with $|\vec{q}| = 2k\sin(\theta/2)$. The amplitude is essentially the Fourier transform of the potential.

How is the differential scattering cross-section related to the scattering amplitude?

$$\frac{d\sigma}{d\Omega} = |f(\theta)|^{2}.$$ The total cross-section is $\sigma = \int |f(\theta)|^{2}\,d\Omega$.

For a spherically symmetric potential, simplify the Born approximation scattering amplitude.

$$f(\theta) = -\frac{2m}{\hbar^{2}q}\int_{0}^{\infty} r\,V(r)\,\sin(qr)\,dr,$$ with $q = 2k\sin(\theta/2)$. The angular integration reduces the 3D Fourier transform to this radial form.