Quantum Mechanics II

1. Brackets are just matrices

Learning goal: Turn Dirac notation into matrix elements and expectation values.

In a two-level system, kets are column vectors and operators are matrices. The expectation value of an operator A in a normalized state is : multiply the conjugate-transpose row by the matrix by the column. For spin-1/2 in the Sz basis, Sx = (hbar/2) times the matrix with 0 on the diagonal and 1 off the diagonal.

For psi = (|up> + |down>)/sqrt 2 the vector is (1, 1)/sqrt 2. Then = (hbar/2) x (1/2)(1 x 1 + 1 x 1) = hbar/2. For psi = (|up> + i|down>)/sqrt 2, the conjugate row is (1, -i)/sqrt 2 and = 0. Phases between components matter.

The three spin-1/2 components follow the same pattern. The matrix for Sy is (hbar/2) times the matrix with 0 on the diagonal, -i in the upper right and i in the lower left. The state (|up> + i|down>)/sqrt 2 is the eigenstate of Sy with eigenvalue +hbar/2, so a measurement of Sy gives +hbar/2 with certainty, while the expectation of Sx in that state is 0. A useful test of any matrix you write down is that it is Hermitian, which means equal to its conjugate transpose, so its eigenvalues are real.

Worked example

Find <Sx> for psi = (|up> + |down>)/sqrt 2 using Sx = (hbar/2)[[0,1],[1,0]].

1. Write the vector (1, 1)/sqrt 2.
2. Apply the matrix: [[0,1],[1,0]](1,1) = (1,1), so Sx psi = (hbar/2)(1,1)/sqrt 2.
3. Multiply by the bra (1,1)/sqrt 2: (hbar/2)(1/2)(1 + 1).
4. <Sx> = hbar/2.

Mental model: Expectation = sum of eigenvalue times probability.

Common trap: Forgetting to conjugate the bra.

2. A small nudge to a two-level system

Learning goal: Apply second-order nondegenerate perturbation theory and compare with the exact answer.

For H = H0 + V with nondegenerate levels, the first-order energy shift is the diagonal element . The second-order shift is the sum over other states m of ||^2 / (E_n - E_m). Notice the sign: the ground state is pushed down by higher states.

Take H0 = diag(0, 2) and V with off-diagonal g = 0.1 (all energies in one unit). The first-order shift of the ground state is V00 = 0. The second-order shift is 0.01 / (0 - 2) = -0.005. The exact ground eigenvalue is 1 - sqrt(1 + g^2) = -0.0049876, so the approximation is close.

The state also changes. The first-order correction mixes in the excited state with coefficient V10 / (E0 - E1). With g = 0.1 and a gap of 2 this is 0.1 / (0 - 2) = -0.05, so the ground state becomes about |0> - 0.05|1>. The norm changes only at second order, by about 0.05 squared, or 0.25 percent. A check: the energy correction and the state correction use the same denominators, so a mistake in one usually shows in the other.

Worked example

H0 = diag(0, 2), off-diagonal g = 0.1. Find the ground-state energy to second order and compare with the exact value.

1. First order: V00 = 0.
2. Second order: g^2 / (E0 - E1) = 0.01 / (0 - 2) = -0.005.
3. Exact: eigenvalues of [[0, 0.1],[0.1, 2]] are 1 +/- sqrt(1 + 0.01); ground = 1 - 1.004988 = -0.004988.
4. The two agree to about 0.00001.

Mental model: Second order = sum of |coupling|^2 over (E_n - E_m).

Common trap: Using E_m - E_n.

3. How small is small

Learning goal: Judge the validity of perturbation theory by comparing coupling to the gap.

Perturbation theory is a series in the ratio of the coupling to the energy gap. If g is much smaller than the gap, low-order terms are accurate. If g is comparable to the gap, the series may converge slowly or fail. Check the ratio before trusting an answer.

With gap 1 (levels 0 and 1) and coupling g = 0.5, the second-order shift is -0.25 but the exact ground energy is 0.5 - sqrt(0.5) = -0.2071. The error is 0.043, about 21 percent of the exact shift. With gap 2 and g = 0.1 the error was below 0.3 percent. A small ratio is a requirement, not a nicety.

A rule of thumb for a two-level system with gap D and coupling g: second-order shift is -g^2 / D, and the exact shift is (D/2)(1 - sqrt(1 + 4 g^2 / D^2)). Expand the square root for small g / D and the first term is -g^2 / D, which confirms the formula. The next term is +g^4 / D^3, so the relative error of the second-order estimate is about g^2 / D^2. For g / D = 0.5 that is about 25 percent, close to the 21 percent seen above. For g / D = 0.05 it is 0.25 percent.

Worked example

For levels 0 and 1 (gap 1) with off-diagonal coupling 0.5, compare the second-order ground shift to the exact one.

1. Second order: g^2/(0 - 1) = 0.25/(-1) = -0.25.
2. Exact: eigenvalues of [[0, 0.5],[0.5, 1]] are 0.5 +/- sqrt(0.25 + 0.25); ground = 0.5 - 0.7071 = -0.2071.
3. Error: 0.25 - 0.2071 = 0.043.
4. The ratio g/gap = 0.5 is too large for reliable second-order accuracy.

Mental model: Compare the coupling with the gap before trusting a series.

Common trap: Assuming perturbation theory is always accurate.

4. When the denominator is zero

Learning goal: Handle degenerate levels by diagonalizing the perturbation within the degenerate subspace.

If two unperturbed states share an energy, the second-order formula divides by zero, so the nondegenerate method fails. The fix is to write the perturbation as a matrix within the degenerate subspace and diagonalize it. Its eigenvalues are the first-order shifts, and its eigenvectors are the correct starting states.

For two degenerate levels at energy 1 with V = [[0, g],[g, 0]], the eigenvalues are +g and -g, so the levels split to 1 + g and 1 - g with eigenvectors (1, 1)/sqrt 2 and (1, -1)/sqrt 2. With g = 0.2 the splitting is 0.4. A perturbation that couples the levels removes the degeneracy at first order.

A physical example is the hydrogen atom in a weak electric field for principal quantum number n = 2. The 2s and 2p states with m = 0 have the same energy, so the field's perturbation must be diagonalized in that subspace. The result is a linear shift of plus or minus 3 e a0 E, where a0 is the Bohr radius, and two states remain unshifted. The 2s and 2p with m = 0 mix equally, so those states have permanent dipole moments, unlike the ground state.

Worked example

Two degenerate levels at energy 1 are coupled by V = [[0, 0.2],[0.2, 0]]. Find the first-order energies and eigenvectors.

1. The subspace matrix is V itself.
2. Eigenvalues solve lambda^2 - 0.04 = 0, so lambda = +0.2 and -0.2.
3. Energies: 1.2 and 0.8.
4. Eigenvectors: (1, 1)/sqrt 2 for +0.2 and (1, -1)/sqrt 2 for -0.2.

Mental model: Diagonalize the perturbation inside a degenerate subspace.

Common trap: Applying the nondegenerate formula.

5. Adding two spins

Learning goal: Count states and allowed total angular momentum values when adding two angular momenta.

When two angular momenta j1 and j2 are added, the total j takes every value from |j1 - j2| to j1 + j2 in steps of 1. The total number of states must be conserved: (2 j1 + 1)(2 j2 + 1) = sum of (2 j + 1) over allowed j.

For j1 = 1 and j2 = 1/2: j = 3/2 or 1/2, with 4 + 2 = 6 states, matching 3 x 2. For j1 = 2 and j2 = 1: j = 3, 2, 1 with 7 + 5 + 3 = 15 states, matching 5 x 3. Always run this dimension check, since it catches a missing or extra j value.

The singlet-triplet case is the most common. Two spin-1/2 particles give total spin 1, which has three states, and total spin 0, which has one state, for 4 in all and matching 2 x 2. The triplet is symmetric under exchange of the two spins, and the singlet is antisymmetric. Because the total wave function must be antisymmetric for two electrons, a symmetric spatial part goes with the singlet and an antisymmetric spatial part with the triplet.

Worked example

Add j1 = 2 and j2 = 1. List the allowed j and check the state count.

1. Smallest j is |2 - 1| = 1, largest is 2 + 1 = 3.
2. Allowed j: 1, 2, 3.
3. States: (2 x 1 + 1) + (2 x 2 + 1) + (2 x 3 + 1) = 3 + 5 + 7 = 15.
4. Product space: 5 x 3 = 15, so the list is complete.

Mental model: Dimension of the product space equals the sum over j.

Common trap: Missing the smallest j.

6. Splitting by total angular momentum

Learning goal: Compute energy levels from a spin-orbit-type coupling in a toy model.

A coupling of the form (A/2)[j(j+1) - l(l+1) - s(s+1)] depends on the total j. This is a toy form that shows the structure of fine-structure-type splitting: states with the same l and s but different j have different energies. The constant A sets the scale and is not derived here.

For l = 1 and s = 1/2: j = 3/2 gives (A/2)(3.75 - 2 - 0.75) = A/2, and j = 1/2 gives (A/2)(0.75 - 2 - 0.75) = -A. The splitting between the two levels is 1.5 A. Notice that the number-weighted average, (4 x A/2 + 2 x (-A))/6 = 0, so the coupling shifts levels without changing the total energy sum.

Try a different case, l = 2 and s = 1/2. Then j = 5/2 gives (A/2)(8.75 - 6 - 0.75) = A and j = 3/2 gives (A/2)(3.75 - 6 - 0.75) = -1.5 A. The splitting is 2.5 A. The weighted average with weights 6 and 4 is (6 x A + 4 x (-1.5 A)) / 10 = 0, as before. This confirms the pattern: the splitting is A x (l + 1/2), so it grows with l, and the weighted-average rule is a quick check that you have not mixed up the j values.

Worked example

In the toy coupling, find the energies and splitting for l = 1 and s = 1/2.

1. j = 3/2: j(j+1) = 3.75, l(l+1) = 2, s(s+1) = 0.75; (A/2)(3.75 - 2 - 0.75) = A/2.
2. j = 1/2: j(j+1) = 0.75; (A/2)(0.75 - 2 - 0.75) = -A.
3. Splitting = A/2 - (-A) = 1.5 A.
4. The weighted average with weights 4 and 2 is zero.

Mental model: Energy depends on j through j(j+1).

Common trap: Treating all j as degenerate.

7. Two fermions in a box

Learning goal: Build the ground state of identical fermions and count the energy.

For noninteracting particles in a one-dimensional infinite well, the single-particle energies are n^2 E1 for n = 1, 2, 3, .... Identical spin-polarized fermions cannot share a state, so they fill levels one each. Two such fermions have ground energy (1 + 4) E1 = 5 E1.

Their spatial wavefunction is antisymmetric: [phi1(x1) phi2(x2) - phi2(x1) phi1(x2)]/sqrt 2, and it vanishes if both particles are at the same place with the same spin. Two spinless bosons could both occupy n = 1 for 2 E1. Opposite-spin fermions are a different case, because the spin part can be antisymmetric and the spatial part symmetric.

Add more fermions. For three spin-polarized fermions the levels n = 1, 2, 3 are filled and the ground energy is (1 + 4 + 9) E1 = 14 E1. In general N fermions give E1 times N(N + 1)(2N + 1) / 6, which is the sum of the first N squares. Compare spin-polarized bosons, which would all sit in n = 1 for N E1. The difference between 14 E1 and 3 E1 is the exclusion pressure in a very simple model.

Worked example

Find the ground energy of two noninteracting spin-polarized fermions in a 1D infinite well (units of E1) and the wavefunction.

1. Single-particle energies are n^2 E1; the two lowest distinct states are n = 1 and n = 2.
2. Energy = (1 + 4) E1 = 5 E1.
3. Wavefunction: [phi1(x1) phi2(x2) - phi2(x1) phi1(x2)] / sqrt 2.
4. Norm: (1 + 1 - 0 - 0)/2 = 1 by orthonormality.

Mental model: Statistics decides which levels are filled.

Common trap: Putting all particles in the lowest level.

8. Counting allowed states

Learning goal: Count symmetric and antisymmetric two-particle states from single-particle levels.

With N single-particle levels, two distinguishable particles have N^2 joint states. Two identical bosons use symmetric states: N(N + 1)/2. Two identical spin-polarized fermions use antisymmetric states: N(N - 1)/2. The totals satisfy N^2 = N(N+1)/2 + N(N-1)/2.

For N = 3 these are 9, 6 and 3. For N = 4 they are 16, 10 and 6. The counting shows the exclusion principle at work: fermions lose the N doubly occupied states. These are counts of spatial occupation patterns for spin-polarized particles in a simple model.

Add spin. For two electrons with N spatial levels and spin 1/2, the allowed states are the spin singlet with a symmetric spatial part, giving N(N + 1)/2 states, plus the spin triplet with an antisymmetric spatial part, giving 3 x N(N - 1)/2 states. For N = 2 that is 3 + 3 = 6, which equals the number of ways to choose 2 of 4 spin-orbitals. This is the same total as the antisymmetric-state count for four spin-orbitals, which is a useful check.

Worked example

Count the two-particle states for N = 3 levels: distinguishable, bosons, fermions, and check the sum.

1. Distinguishable: 3 x 3 = 9.
2. Bosons: 3 x 4/2 = 6 (three doubly occupied plus three pairs).
3. Fermions: 3 x 2/2 = 3 (the three pairs).
4. Check: 6 + 3 = 9.

Mental model: Counting states depends on exchange symmetry.

Common trap: Using N^2 for identical particles.

9. Guess, then improve

Learning goal: Use the variational principle to bound the ground energy from above.

The variational principle says that for any normalized trial state, the expectation value of H is at least the ground-state energy. So a trial gives an upper bound, and lowering the trial energy gets closer from above. The method works best when the trial state shares the symmetry of the true ground state.

For H = [[0, 0.3],[0.3, 2]], the trial (1, 0) gives E = 0, an upper bound. The trial (1, -0.1)/sqrt(1.01) gives E = (2 x 0.01 + 2 x (1)(-0.1)(0.3))/1.01 = -0.0396. The exact ground energy is 1 - sqrt(1.09) = -0.0440. Both trials stay above it, as the principle requires.

The variational method improves by adding parameters. A trial with one adjustable parameter can only lower the best energy when a second parameter is added, since the old family is a special case. The linear-variation method, which uses a trial that is a combination of fixed basis states, reduces to finding the lowest eigenvalue of H in that basis. Adding more basis states can only push the lowest eigenvalue down toward the exact one, never up.

Worked example

For H = [[0, 0.3],[0.3, 2]], evaluate the energy of the trial state (1, -0.1) and compare with the exact ground energy.

1. Unnormalized norm: 1 + 0.01 = 1.01.
2. <H> numerator = 0 x 1 + 2 x 0.01 + 2 x (1)(-0.1)(0.3) = 0.02 - 0.06 = -0.04.
3. E = -0.04 / 1.01 = -0.0396.
4. Exact: 1 - sqrt(1 + 0.09) = -0.0440. Our bound -0.0396 lies above it.

Mental model: Always evaluate the Rayleigh quotient with a normalized state.

Common trap: Forgetting to divide by the norm.

10. Probabilities from amplitudes

Learning goal: Compute measurement probabilities and uncertainty for spin-1/2 states.

A state a|up> + b|down> has measurement probabilities |a|^2 and |b|^2 for Sz = +hbar/2 and -hbar/2. The expectation is the probability-weighted sum, and the uncertainty is sqrt( - ^2). Because Sz^2 = hbar^2/4 for spin-1/2, = hbar^2/4 in any state.

For theta = pi/6 with state cos theta |up> + sin theta |down>, the probabilities are 0.75 and 0.25. = (0.75 - 0.25) hbar/2 = 0.25 hbar. Then the uncertainty is sqrt(0.25 - 0.0625) hbar = 0.433 hbar. A definite Sz state has uncertainty zero; an equal superposition has the maximum, hbar/2.

A cross-check for a normalized spin-1/2 state is that the squares of the three expectation values add to (hbar/2)^2. For cos theta |up> + sin theta |down> with theta = pi/6, the expectation of Sz is 0.25 hbar and the expectation of Sx is hbar cos theta sin theta = 0.433 hbar. Then 0.25^2 + 0.433^2 = 0.0625 + 0.1875 = 0.25, which equals (1/2)^2 in units of hbar squared. That confirms the state points in a definite direction.

Worked example

For cos(theta)|up> + sin(theta)|down> with theta = pi/6, find <Sz> and the uncertainty in units of hbar.

1. Probabilities: cos^2(pi/6) = 0.75, sin^2(pi/6) = 0.25.
2. <Sz> = (0.75 - 0.25)/2 = 0.25.
3. <Sz^2> = 0.25.
4. Uncertainty = sqrt(0.25 - 0.0625) = 0.433.

Mental model: Expectation is the probability-weighted sum of eigenvalues.

Common trap: Forgetting the factor of 1/2.

11. A weak magnetic field and the Lande g-factor

Learning goal: Compute the splitting of a level with total angular momentum j in a weak magnetic field.

In a weak magnetic field, the fine-structure levels split further because states of different m_j have different energies. To first order the shift is g_J times mu_B times B times m_j, where mu_B is the Bohr magneton, about 5.788 x 10^-5 eV per tesla.

The Lande factor depends on the quantum numbers: g_J = 1 + [j(j + 1) + s(s + 1) - l(l + 1)] / [2 j(j + 1)]. For l = 1, s = 1/2 and j = 3/2 this gives 1 + (3.75 + 0.75 - 2) / 7.5 = 4/3. For j = 1/2 it gives 1 + (0.75 + 0.75 - 2) / 1.5 = 2/3. For l = 0 the factor is 2.

A level with total angular momentum j splits into 2j + 1 equally spaced sublevels, and the spacing is g_J mu_B B. For j = 3/2 in a 1 tesla field the spacing is (4/3) x 57.9 micro-eV, about 77 micro-eV. This is small compared with the fine-structure gap, which is why the weak-field assumption holds.

The approximation fails when the field is strong enough that the magnetic energy is comparable to the fine-structure splitting. Then l and s decouple and the better quantum numbers are m_l and m_s. Always compare the two energy scales before you pick the method.

Worked example

For l = 1, s = 1/2, j = 3/2, what is g_J? Use j(j+1) = 3.75.

1. Numerator 3.75 + 0.75 - 2 = 2.5.
2. Denominator 2 x 3.75 = 7.5.
3. 2.5 / 7.5 = 1/3.
4. g = 1 + 1/3 = 4/3.

Mental model: g_J = 1 + [j(j+1) + s(s+1) - l(l+1)] / [2 j(j+1)]. Spacing = g_J mu_B B.

Common trap: Using the weak-field formula in a strong field.

12. Hyperfine structure and the 21 cm line

Learning goal: Add the electron and nuclear spins and connect the splitting to a measured line.

The nucleus has its own spin I and the electron has s = 1/2. They couple to give total F running from |I - s| to I + s. For hydrogen I = 1/2, so F = 1 or F = 0, with 3 + 1 = 4 states matching 2 x 2.

The magnetic interaction between the two spins splits the ground-state level into F = 1 and F = 0. Transitions between them give a spectral line of frequency about 1420.4 MHz, which is a wavelength near 21.1 cm. Radio astronomers use this line to map neutral hydrogen.

The wavelength follows from lambda = c / f. With c = 299,792,458 m/s and f = 1,420,405,752 Hz, lambda is 0.2111 m. The photon energy is h f, about 5.87 x 10^-6 eV, which is tiny compared with an optical photon of about 2 eV.

The counting rule stays the same for other nuclei: for deuterium, I = 1 gives F = 3/2 and 1/2, with 4 + 2 = 6 states matching 3 x 2. Always run the dimension check.

Worked example

For deuterium I = 1 and s = 1/2. How many states in F = 3/2?

1. 2F + 1.
2. F = 3/2.
3. 2 x 3/2 + 1.
4. 4 states.

Mental model: F runs from |I - s| to I + s. Dimension check: (2I + 1)(2s + 1) = sum of (2F + 1).

Common trap: Forgetting the nuclear spin.

13. A trial with a kink: the exponential versus the Gaussian

Learning goal: Compare two variational trials for the harmonic oscillator and see both stay above the exact energy.

Use units with hbar = m = omega = 1. The exact ground energy of the harmonic oscillator is 1/2. A variational trial gives an upper bound, so any trial energy must be at least 1/2.

For the Gaussian trial exp(-a x^2) the energy is E(a) = a/2 + 1/(8a). Setting the derivative to zero gives a = 1/2 and E = 1/2, exact, because the true ground state is a Gaussian of that width.

For the exponential trial exp(-b |x|) the kinetic energy is b^2/2 and the potential energy is 1/(4 b^2), so E(b) = b^2/2 + 1/(4 b^2). The minimum is at b^4 = 1/2, so b = 0.841 and E = 0.7071. This is above 1/2 by about 41 percent, because the trial has a kink at the origin and the wrong tail.

The lesson is that the energy error is a measure of how wrong the trial shape is. A one-parameter family cannot fix a wrong shape, but it tells you the best it can do, and it never goes below the truth.

Worked example

Gaussian trial with a = 0.5. E(a) = a/2 + 1/(8a). Compute.

1. a/2 = 0.25.
2. 1/(8a) = 1/4 = 0.25.
3. Sum.
4. 0.5.

Mental model: Trial energy is an upper bound. The error reveals how wrong the shape is.

Common trap: Treating a lower trial energy as proof the trial is exact.