Problem 1:
The figure below shows one of the possible energy eigenfunctions ψ(x) for a particle bouncing freely back and forth between impenetrable walls located at x = -a and x = +a. The potential energy equals zero for |x| < a. If the energy of the particle is 2 eV when it is in the quantum state associated with this eigenfunction, find the energy when it is in quantum state of lowest possible energy.
Solution:
 | Concepts, principles, relations that apply to the
problem: The infinite square well |
| Why do they apply? We are asked to identify which eigenfunction of the infinite square well is shown in the figure. |
| How do they apply? The eigenfunctions of H are yn(x) = (1/a)1/2sin(np(x+a)/(2a)) with eigenvalues En = n2p2h2/(8ma2). y2(x) is shown in the figure. E2 = 2 eV, E1 = ½ eV is the lowest possible energy eigenvalue. |
| Details of the calculation: None |
Problem 2:
A particle is in the state |y> that has angular momentum j and angular momentum projection on the z-axis m such that
J2|y> = h2(j(j+1)|y>, Jz|y> = mh|y>.
Find expectation values of the angular momentum components Jx and Jy in this state.
Solution:
| Concepts, principles, relations that apply to the
problem: The operators J± |
| Why do they apply? We define the operators J+ = Jx + iJy and J- = Jx - iJy. We then have Jx = (1/2)(J+ + J-) and Jy = (-i/2)(J+ - J-) . The operators J± operating on the basis states {|k,j,m>} yield J±|k,j,m> = [j(j+1)-m(m±1)]1/2h|k,j,m±1>. |
| How do they apply? |y> = |k,j,m>. <Jx> = ½<k,j,m|(J+ + J-)|k,j,m> = 0. <Jy> = (-i/2)<k,j,m|(J+ - J-)|k,j,m> = 0. |
| Details of the calculation: None |
Problem 3:
The Heisenberg Hamiltonian representing the “exchange
interaction” between two spins (S1 and S2) is given by H =
-2f(R)S1×S2,
where f(R) is the so-called exchange coupling constant and R is the spatial
separation between the two spins. Find the eigenstates and eigenvalues of the
Heisenberg Hamiltonian describing the exchange interaction between two
electrons.
HINT: The total spin operator is S =
S1
+ S2.
Solution:
| Concepts, principles, relations that apply to the
problem: The eigenstates and eigenvalues of S1×S2 |
| Why do they apply? H is proportional to S1×S2 . |
| How do they apply? S1×S2 = ½(S2 – S12 – S22). The eigenstates of S1×S2 are the singlet state and the triplet states, {|S, Ms>}, S = 0, 1. For the singlet state we have S1×S2 |0, 0> = -(3/4)h2. For the triplet states we have S1×S2 |1, Ms> = (1/4)h2. The eigenstates of H are |0, 0> with eigenvalue (3/2)f(R)h2 and |1, 1>, |1, 0>, |1, -1> with eigenvalue -(1/2)f(R)h2 (3-fold degenerate). |
| Details of the calculation: None |
Problem 4:
A collection of non-interacting undistinguishable spin 1/2 particles is trapped in a 3-dimensional, isotropic harmonic-oscillator potential. How many particles can occupy the three lowest lying energy states?
Solution:
| Concepts, principles, relations
that apply to the problem: The energy levels of the 3D harmonic oscillator, the Pauli exclusion principle. |
| Why do they apply? The energy levels of the 3D harmonic oscillator are degenerate. |
| How do they apply? The energy levels of the isotropic, 3D harmonic oscillator are E = (nx + ny + nz + 3/2)hw, nx, ny, nz = 0, 1, 2, ..., (Cartesian coordinate) or E = (2n + l + 3/2)hw, n, l - 0, 1, 2, ..., (spherical coordinates). E0 = (3/2)hw, two undistinguishable spin 1/2 particles can occupy this state (nx, ny, nz = 0). E1 = (1 + 3/2)hw, 6 undistinguishable spin 1/2 particles can occupy this state (nx or ny or nz = 1 with the other ni = 0). E2 = (2 + 3/2)hw, 12 undistinguishable spin 1/2 particles can occupy this state (nx or ny or nz = 2 with the other ni = 0, or two of the ni = 1 and one of the ni = 0). |
| Details of the calculation: None |
Problem 5:
The La line of the characteristic X-ray spectra of heavy atoms consists of several components of different frequencies corresponding to the various allowed transitions from levels with n = 3 to levels with n = 2. Predict the number of different frequencies to be observed, on the basis of the selection rules Dl = ±1, Dj = 0, ±1.
Solution:
| Concepts, principles,
relations that apply to the problem: Addition of angular momentum, selection rules |
| Why do they apply? For a single hole in a shell the possible quantum numbers l, s, and j are the same as those for a single electron in the shell. |
| How do they apply? If we have a hole in the n = 2 shell, the shell angular momentum quantum numbers are l = 0 or l = 1 and j = l ± 1/2. We can have the following energy levels: 2s1/2, 2p1/2, 2p3/2. If an electron from the n = 3 shell fills that hole we are left with a hole in the n = 3 shell. We then can have the following energy levels: 3s1/2, 3p1/2, 3p3/2, 3d3/2, 3d5/2. The allowed transitions are indicated by a + while the forbidden transitions are indicated by a ´.  There are seven allowed transitions. |
| Details of the
calculation: None |
Problem 6:
The weak interactions (for example, beta decay) are mediated by massive particles called intermediate vector bosons, which are observed in accelerator experiments to have masses in the range mc2 ~ 80-90 x 109 electron-volts. Assuming the weak interactions to occur because of the quantum-mechanical exchange of a virtual intermediate vector boson between two particles, estimate the maximum range of the weak force.
Solution:
| Concepts, principles, relations that apply to the
problem: The uncertainty principle: Dt ~ h/DE |
| Why do they apply? Dt ~ h/DE ~h/mc2. DE ~ mc2, since the particle either exists or does not exist. The virtual particle can propagate a distance no larger than R = cDt in a time interval Dt. If we insert Dt from above, we have R ~ h/mc. |
| How do they apply? R ~ hc/(mc2) ~ 10-18 m. |
| Details of the calculation: None |
Problem 7:
Let us consider a carbon atom whose electrons are in the following configuration (1s)2 (2s)2 2p 3p. List all the expected terms 2S+1LJ on the basis of the L-S (Russell-Sanders) coupling scheme.
Solution:
| Concepts, principles,
relations that apply to the problem: Atomic spectra, addition of angular momentum |
| Why do they apply? The Pauli exclusion principle states that no two identical fermions can have exactly the same set of quantum numbers. We use it and the rules for adding angular momentum to find the allowed terms 2S+1LJ |
| How do they apply? (1s)2 couples to L = 0, S = 0. (Pauli exclusion principle) (2s)2 couples to L = 0, S = 0. (Pauli exclusion principle) The 2p and 3p electrons have different principal quantum numbers n. We do not have to worry about the exclusion principle when adding their angular momenta. Both electrons have l = 1, s = 1/2. This can lead to L = 2, 1, 0, S = 0, 1. We therefore can form the following terms: 1D2, 3D123, 1P1, 3P012, 1S0, 3S1. |
| Details of the
calculation: None |
Problem 8:
Consider a two-dimensional infinite potential square well
of width L,
(V = 0 for 0 < x,y < L, V = infinite everywhere else) with an added perturbation
(a) Calculate the first order perturbation to the ground
state energy eigenvalue.
(b) Calculate the first order perturbation to the first
excited state energy eigenvalue
Solution:
| Concepts, principles, relations that apply to the
problem: First order perturbation theory for non-degenerate states, first order perturbation theory for degenerate states. |
| Why do they apply? The ground state of the 2D infinite square well is non-degenerate, the first excited state is degenerate. |
| How do they apply? H = H0 + H’. The eigenfunctions of H0 are fnl(x,y) = (2/L)sin(npx/L)sin(lpy/L), with eigenvalues E = (n2 + l2)p2h2/(2mL2). (a) For the ground state n = l = 1. The ground state is not degenerate. The unperturbed ground state energy is E00 = h2/(mL2). The first order perturbation correction is E10 = <f11|H’|f11> = (g4/L2)ò0Lsin2(px/L) sin2(py/ L)sin(2px/L) sin(2py/L)dxdy. = (4g/L2)ò0Lsin2(px/L) sin(2px/L)dx)ò0Lsin2(py/L) sin(2py/L)dy = 0. (b) The first excited state is two-fold degenerate. We can have n = 2, l = 1, or n = 1, l = 2. We have to diagonalize the matrix of H’ in the subspace spanned by these two degenerate states.
<f12|H’|f12>
= (4g/L2)ò0Lsin2(px/L)sin(2px/L)dx)ò0Lsin2(2py/L)
sin(2py/L)dy = 0. |
| Details of the calculation: None |
Problem 9:
A spin 1/2 particle is in an external magnetic field B
= B0(êx
+ êy).
Let the magnetic moment associated with the spin be
m = -gS, where
g is a constant and is known as the
gyromagnetic ratio. Use the eigenstates of Sz, |+> and |–>,
as basis kets.
(a) What is the interaction Hamiltonian of the particle
with the magnetic field? Express in matrix form.
(b) The most general spin state of the particle is |c(t)>
= a(t) |+> +
b(t) |–>.
Write down the coupled time dependent Schroedinger
equations for a(t) and
b(t).
(c) Suppose at t = 0 we have |c(0)>
= | + >. Find |c(t)>.
(d) Evaluate <Sz(t)> for the
c(t) obtained in (c).
Solution:
| Concepts, principles, relations that apply to the
problem: The two dimensional state space of a spin 1/2 particle, the evolution of the state vector |
| Why do they apply? We are given detailrd instructions on how to find the state vector at time t given the state vector at t = 0. |
| How do they apply? (a) H = -m×B = gB0(Sx + Sy). In matrix form
(b)
(c) Assume
a(t) = Acos(wt), b(t)
= Bsin(wt). (d) <Sz(t)> =
|
| Details of the calculation: None |
Problem 10:
A spinless particle of mass M is scattered by a central potential of finite range such that the wave function at low energies and for large r is well described by
,
where d is a phase
shift.
(a) Find the total low-energy cross section in terms of
the quantities defined.
(b) Find the phase shift
d if the slow particle scatters off a central potential of the form
V(r) = V0, for r < a, V(r) = 0, for r > a.
Solution:
| Concepts, principles, relations that apply to the
problem: Elastic scattering, the method of partial waves, s-wave scattering. |
| Why do they apply? For very slow particles or very short-range potentials the method of partial waves is the preferred method of calculating the scattering cross section, because only s-waves need to be considered. |
| How do they apply? (a) For a central potential of finite range, i.e. V(r) --> 0 as r --> ¥ faster than 1/r, stationary state solutions of the eigenvalue equation Hfk(r) = Ekfk(r) have the asymptotic form fk(r) = exp(ikz) + fk(q)exp(ikr)/r. The scattering amplitude is given by fk(q) = (1/k)Sl=0¥(2l+1)exp(idl)sindlPl(cosq). and the differential scattering cross section is dsk/dW = |fk(q)|2 = (1/k2)|Sl=0¥(2l+1)exp(idl)sindlPl(cosq)|2, Here fk(q) = (1/k)sind, dsk/dW = (1/k2)sin2d, and the total scattering cross section is s = (4p/k2)sin2d. |
| Details of the calculation: (b) (¶2/¶r2 + k2 - U(r) - l(l+1)/r2)ukl(r) = 0 is the radial equation for stationary states. Here E = h2k2/(2m), V(r) = h2U(r)/(2m). We need ukl(0) = 0, ukl(r) and ¶ukl(r)/¶r to be continuous at r = a, and ukl(r) to stay finite at infinity. To calculate d = d0 we must solve the radial equation for l = 0. (¶2/¶r2 + k2 + U0)uk0(r) = 0, uk0(0) = 0, for r < a. Therefore uk0(r) = C1sin((k2 + U0)1/2r). (¶2/¶r2 + k2)uk0(r) = 0, for r > a. uk0(r) = C2sin(kr + d). We need uk0(r) and ¶uk0(r)/¶r to be continuous at r = a. This yields C1sin((k2 + U0)1/2a) = C2sin(ka + d) and (k2 + U0)1/2C1cos((k2 + U0)1/2a) = kC2cos(ka + d) or tan(ka + d) = (k/(k2 + U0)1/2)tan((k2 + U0)1/2a) = C. We have d = np + tan-1(C) - ka. |
Problem 11:
In the WKB approximation, find the allowed energies that a ball of mass m bouncing due to gravity on a perfectly reflecting surface, can have. You can use the fact that for this problem the WKB approximation gives
where p(q) is the momentum of the ball at the height q and the integral is over a full periodic path. You may leave your answer as an integral equation which could be solved to yield the energy levels.
Solution:
| Concepts, principles, relations that
apply to the problem: The WKB approximation |
| Why do they apply? We are instructed to use the WKB approximation. For this problem the WKB approximation requires that òq1q2 pdq = (n + 1/4)(h/2) or  pdq
=
hkdq
= (n + 1/4)h. Here denote
an integral over one complete cycle of the classical motion and k2
= (2m/h2)(E - V(q)). |
|
How do they apply? |
|
Details of the calculation: |
.
.