**Problem 1:**

Consider a hydrogenic atom with nuclear charge Zq_{e} in a **strong** magnetic field **B** = B**k**. Assume
that the Zeeman slitting is much larger than the spin-orbit splitting of the
energy levels, so that to first order the spin-orbit interaction can be ignored.

(a) Find the energies of the 2s and 2p energy levels in the strong magnetic
field. Is the degeneracy completely removed by the Zeeman interaction?

(b) Estimate the magnitude of the magnetic field, B, required to give a Zeeman
splitting in the hydrogenic atom comparable to the binding energy of the ground
state of the hydrogen atom. Can such a magnetic field be created in the
laboratory?

Solution:

Concepts: A magnetic moment in a magnetic field | |

Reasoning: We can use first order stationary perturbation theory to find the energy levels of the perturbed hydrogenic atom. | |

Details of the calculation: (a) H = H _{0} + H_{z}, H_{0} = p^{2}/(2μ) – Ze^{2}/r,
e^{2} = (4πε_{0}), μ = m_{e} = reduced mass of the electron.The eigenstates of H _{0} are {|n, l, m, m_{s}>}, characterized
by the 4 quantum numbers n, l, m, m_{s}, with eigenvalues are E_{n}
= -Z^{2}*13.6 eV/n^{2}. All n = 2 levels have energy E_{2}
= -Z^{2}*3.4 eV, the level is 8-fold degenerate. The 2s level is
two-fold degenerate and the 2p level is 6-fold degenerate.H _{Z} = -μ∙B = -μ_{z}B.
μ_{z} = (-q_{e}/(2m_{e}))(L_{z} + 2S_{z})
= -μ_{B}(L_{z} + 2S_{z})/ħ.H _{0} commutes with L^{2} an L_{z}, so H_{0}
commutes wit H_{Z}. The eigenstates of H are {|n, l, m, m_{s}>}
and the matrix of H_{z} in that basis is diagonal.The first order corrections to the energy eigenvalues therefore is ΔE _{n,l,m,ms }= (q_{e}ħB/(2m_{e}))(m
+ 2m_{s}) = μ_{B}B(m + 2m_{s}), where μ_{B} is
the Bohr magneton.The eigenvalues are E _{n} = -Z^{2}*13,6 eV/n^{2} +
μ_{B}B(m + 2m_{s}).The 2s level splits in into two and the 2p levels into five sublevels. The energy difference between the sublevels is ΔE = μ _{B}B.For the 2s level we have E _{2} = -Z^{2}*3.4
eV ± μ_{B}B, and for the 2p level we have E _{2} = -Z^{2}*3.4 eV ± m_{t}μ_{B}B,
where m_{t} = 0, ±1, ±2The degeneracy of the 2s level is completely removed, the degeneracy of the 2p level is only partially lifted. The m = -1, m _{s} = ½ and the m = -1, m_{s} = -½ are
not shifted and still degenerate.If we neglect all other interactions, then the n = 2 level splits into 5 sublevels, of which 3 are still two-fold degenerate. (b) For μ _{B}B =
13.6 eV, we need B = 13.6 eV/ 5.79*10^{-5} eV/T ~ 2*10^{5} T.The largest magnetic field created in the laboratory is less than 100 T. |

**
Problem 2:**

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

∮p dq = (n - ¼)h.

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: The WKB approximation | |

Reasoning: We are instructed to use the WKB approximation. For this problem the WKB approximation requires that ∫ _{q1}^{q2}
pdq = (n - ¼)(h/2) or ∮pdq = ∮ħkdq = (n - ¼)h. Here ∮denote an integral over one complete cycle of the classical motion and k ^{2}
= (2m/ħ^{2})(E - V(q)). | |

Details of the calculation: For stationary bound states we want ∫ _{qmin}^{qmax}
ħkdq = (n - ¼)(πħ), n = 1, 2, ... .In our problem k ^{2} = (2m/ħ^{2})(E - mgq). (Here q is the vertical coordinate. The q-axis points upward and the origin lies at the surface.) We need (2m) ^{½}∫_{0}^{qmax}
(E - mgq)^{½}dq = (n - ¼)(πħ).Here mgq _{max} = E, q_{max} = E/(mg).Therefore (2m) ^{½}∫_{0}^{E/(mg)} (E - mgq)^{½}dq =
(n - ¼)(πħ). |

**Problem 3:**

In one dimension, the potential energy of an electron as a function of x is
given by

U(x) = -30 eV exp(-x^{2}/(4 Å^{2})).

Use the variational method to find the energy of the ground state in units of
eV.

Solution:

Concepts: The variational method | |

Reasoning: The variational method often yields a very good estimate for the ground state energy of a system. | |

Details of the calculation: The Hamiltonian is H = (-ħ ^{2}/2μ)(∂^{2}/∂x^{2}) -
V_{0}exp(-x^{2}/a^{2})).Here V _{0} = 30 eV and a = 2 Å and μ = m_{e}.Choose ψ(x) = Aexp(-x ^{2}/b^{2}). Here b is
an adjustable parameter.ψ(x) = 0 as x --> ±∞, and dψ/dx must be continuous at all x. A ^{2}∫_{-∞}^{∞}exp(-2x^{2}/b^{2})dx = 1, A^{2} = (2/πb^{2})^{½}.<H> = A ^{2}∫_{-∞}^{∞} exp(-x^{2}/b^{2})((-ħ^{2}/2μ)(∂^{2}/∂x^{2}) -
V_{0}exp(-x^{2}/a^{2}))exp(-x^{2}/b^{2})dx= 2A ^{2}(-ħ^{2}/2μ)∫_{0}^{∞}
[-2/b^{2} + 4x^{2}/b^{4}]exp(-2x^{2}/b^{2})dz
- 2A^{2}V_{0}∫_{0}^{∞} exp(-x^{2}(2/b^{2} - 1/a^{2}))dx= 2A ^{2}[(-ħ^{2}/2μ)(2/b^{2})^{½}
- V_{0}(a^{2}b^{2}/(2a^{2 }+ b^{2})^{½}]∫_{0}^{∞} exp(-x^{2})dx- 2A ^{2}[(2ħ^{2}/μb^{4})(b^{2}/2)^{3/2}∫_{0}^{∞}
x^{2}exp(-x^{2})dx.∫ _{0}^{∞} exp(-r^{2})dr = π^{½}/2.
∫_{0}^{∞}
r^{2}exp(-r^{2})dr = π^{½}/4.<H> = ħ ^{2}/(2μb^{2})
- 2^{½}aV_{0}/(2a^{2 }+ b^{2})^{½}.d<H>/db = 0 --> -ħ ^{2}/(μb^{3})
+ 2^{½}abV_{0}/(2a^{2}+b^{2})^{3/2}
= 0.Let V _{0}' = V_{0}μa^{2}/ħ^{2}.Then a ^{2}/b^{2} - 2^{½}ab^{2}V_{0}'/(2a^{2 }+ b^{2})^{3/2}
= 0, a^{4}/b^{4} = 2(a^{2}b^{4})V_{0}'^{2}/(2a^{2 }+
b^{2})^{3},a ^{4}/b^{4} = 2(a^{2}/b^{2})V_{0}'^{2}/(2a^{2}/b^{2
}+ 1)^{3}.Let y = a ^{2}/b^{2}.Then y ^{2} = 2yV_{0}'^{2}/(2y^{ }+ 1)^{3}.
y(2y + 1)^{3} = 2V_{0}'^{2}.Here V _{0}'^{ }= 30 eV
m_{e} a^{2}/ħ^{2}
= 17.5.y = 2.6, b ^{2} = a^{2}/2.6.E = 2.6ħ ^{2}/(2μa^{2})
- V_{0}(0.916) = -25.2 eV. |

**Problem 4:**

A proton and a neutron are confined by a three-dimensional
potential. For this problem assume that the proton and neutron do not interact
with each other, and neglect spin-orbit interactions. Both particles have spin
½. Including the spins, the ground state is four-fold degenerate. To this
system we now add the interaction between the magnetic dipole moments of the
particles described by the interaction Hamiltonian

H' =
k** S**_{p}**∙S**_{n},

where **S**_{p}, **S**_{n} are the spin operators of the proton and neutron,
respectively, and k is a positive constant.

(a) Consider the following operators:

S_{p}^{2}, S_{n}^{2}, S_{pz}, S_{nz},
S^{2}, and S_{z},

where **S** = **S**_{p}
+ **S**_{n}. State which of these operators commute with H'.

(b) Into how many distinct energy levels does the original ground state split
in the presence of H'? Calculate the corresponding energies and state their
degeneracy.

(c) We now place the
system into a uniform external magnetic field, which points in the positive
z-direction, **B** = B **k**. The spin-spin interaction described by H'
continues to be present and the additional interaction Hamiltonian is

H'_{B} = b(S_{pz}
+ S_{nz})B_{z},_{
}where b is a positive constant.

Calculate
the corrections to the energies of the states identified in part (b) due to the
presence of the magnetic field.

(d) Sketch a graph of the energy levels as a function of the external magnetic
field strength, B, including the effects of both H' and H'_{B}.
Identify the curves with the corresponding states identified in part (b).

Solution:

Concepts: Addition of angular momentum, common eigenbasis for J _{1}^{2},
J_{2}^{2}, J^{2}, J_{z}. | |

Reasoning: We find the eigenstate and eigenvalues of H' and then the corrections to these energy eigenvalues due to H' _{B}. | |

Details of the calculation: (a) The operators S _{p}^{2}, S_{n}^{2},
S^{2}, and S_{z} all commute, and a common eigenbasis {|s_{p},s_{n};s,m_{s}>}
exists. We have the singlet state |00> = ½^{½}(|+-> -
|-+>) and the triplet states |11> = |++>, |10> = ½^{½}(|+->
+ |-+>), |1-1> = |-->.S_{p}∙S_{n
}= ½(S^{2} – S_{p}^{2} – S_{n}^{2}),
S_{pz} + S_{nz} = S_{z}, the vectors {|s_{1},s_{2}; s,m_{s}>}
are also eigenvectors of S_{p}∙S_{n }and S_{pz} + S_{nz}.So S _{p}^{2}, S_{n}^{2}, S^{2}, and S_{z}
commute with H', S_{pz} and S_{nz} do not commute with H'.(b) <s,m_{s}|H'|s,m_{s}>
= (k/2)ħ^{2}[s(s+1) – s_{p}(s_{p}+1)
– s_{n}(s_{n}+1)].For the singlet state <0,0|H|0,0> =
-(3k/4)ħ^{2}.For the triplet states <1,m_{s}|H|1,m_{s}> = (k/4)ħ^{2}.The original ground state split in into two energy levels. The level (3k/4)ħ ^{2}
is non-degenerate and the level -(k/4)ħ^{2}
is three-fold degenerate.(c) S _{pz}
+ S_{nz} = S_{z}, <1,m_{s}|H'_{B}|1,m_{s}>
= bB_{z}m_{s}ħ, m_{s}
= -1, 0, 1.The energy levels now are -(3k/4)ħ ^{2},
(k/4)ħ^{ 2} - bBħ,
(k/4)ħ^{ 2}, and (k/4)ħ^{
2} + bBħ.
(d) |

**Problem 5:**

A harmonic oscillator in two dimensions has the unperturbed Hamiltonian

H = ½m(p_{x}^{2 }+ p_{y}^{2}) + ½mω^{2}(x^{2
}+ y^{2}).

It is subjected to the perturbation H_{1 }= Δxy, where Δ is the strength
of the perturbation.

(a) Write the eigenstates of the unperturbed oscillator in terms of the
eigenstates of the 1-dimensional harmonic oscillator. What are the eigenvalues
of H?

(b) Evaluate the first order corrections to the energies of the three states
lowest in energy
when Δ > 0.

Solution:

Concepts: First order perturbation theory for non-degenerate and degenerate states | |

Reasoning: Separation of variables is possible. The 2D harmonic oscillator eigenfunctions are products of the of the 1D harmonic oscillator eigenfunctions. The ground state is non-degenerate, and the first excited state is two-fold degenerate. For the ground state ΔE = <0,0|H _{1}|0,0> is the first order energy
correction, using non-degenerate perturbation theory. The first order
corrections to the degenerate first excited state are found by diagonalizing H_{1}
in the subspace of degenerate states. | |

Details of the calculation: (a) Let |n> and |m> denote the eigenstate of the 1D harmonic oscillator in x and y, respectively. The eigenstates of the 2D oscillator are {|n,m>}. H = H _{x} + H_{y}. The operators p_{x}^{
}, x
and p_{y}, y, act in different subspaces. The energy of the state |n,m> is E _{nm} = (½ + n)ħω + (½ + m)ħω = (n + m
+ 1)ħω.For the ground state E _{00} = ħω. For the first excited state E_{10}
= E_{01} = 2ħω. (b) For a 1D Hamiltonian H _{x} = p_{x}^{2}/(2m) + mω^{2}x^{2}/2
define a = αx + iβp _{x}, a^{†} = αx - iβp_{x},
with α =√(mω/(2ħ)), β =1/√(2mωħ).Then [a, a ^{†}] = 1 and H_{x} = ħω (a^{†}a + ½).We have x = (a + a ^{†})/(2α), p_{x} = -i(a - a^{†})/(2β).For eigenstates |n> of H _{x} we have a^{†}|n> = (n + 1)^{½}|n
+ 1>, a|n> = n^{½}|n - 1>.Similarly for the y-direction we have y = (b + b ^{†})/(2α), p_{y} = -i(b - b^{†})/(2β).For eigenstates |m> of H _{y} we have b^{†}|m> = (m + 1)^{½}|m
+ 1>, b|m> = m^{½}|m - 1>.For the ground state ΔE = Δ<0,0|xy|0,0> = Δ<0 _{x}|x|0_{x}><0_{y}|y|0_{y}>
= 0. The first order energy correction for the ground state is zero. In the subspace spanned by {|1,0>, |0,1>} the matrix elements of xy are <1,0|xy|1,0> = <0,1|xy|0,1> = 0, <1,0|xy|0,1> = <1 _{x}|x|0_{x}><0_{y}|y|1_{y}>
= (1/(2α)^{2}) = <0,1|xy|1,0>.det(H' - ΔE) = ΔE ^{2} -Δ^{2}/(2α)^{4} = 0.The eigenvalues are ΔE = ±Δ/(2α) ^{2}.The first order energy corrections for the first excited states are ΔE = ±Δ/(2α) ^{2},The degeneracy is removed. |