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L.J. Sham December 11, 2007
Physics 130B Final Examination
PRINT your name and attach only the front page to the front of your bluebook.
Name:
Do problem 1 and choose ONLY THREE PROBLEMS from problem 2 to problem 5. (Cross out the problem you do NOT wish to be graded.)
Problem 1 2 3 4 5 Total
Grade
Out of 40 20 20 20 20 100
Useful information:
Pauli spin matrices
I =
[
]
, σx =
[
]
σy =
[
0 −i i 0
]
σz =
[
]
Simple Harmonic Oscillator
H^ ˆ = pˆ
2 2 m
mω^2 xˆ^2 = ℏω
ˆc†cˆ +
, Hˆ|n〉 = ℏω(n +
)|n〉, n = 0, 1 , 2 ,... (2)
x ˆ = a(ˆc + ˆc†), pˆ =
i 2 a
(ˆc − cˆ†), where a =
2 mω
ˆc|n〉 =
n|n − 1 〉, ˆc†|n〉 =
n + 1|n + 1〉, [ˆc, ˆc†] = 1. (4)
Angular momentum states ℓ = 1 : In polar coordinates (θ, φ),
Y 1 , 1 = −
√^1
sin θeiφ; Y 1 , 0 = cos θ; Y 1 ,− 1 =
√^1
sin θe−iφ. (5)
- Quickies. Each answer must be supported by brief reasoning.
(a) An electron in the hydrogen atom has at a certain instant the same form of wave func- tion as the ground state except the Bohr radius is replaced by a length which is larger by 1%. The mean energy of the electron at that instant is different from -1 Ry by (I) 1%; (II) 10%; (III) 0.01%. Choose the correct one and reason. (b) Let Tˆ be the operator representing a rotation of the ammonia molecule, NH 3 , through 120 ◦^ about the nitrogen axis normal to the plane of the hydrogen atoms. Find the eigenvalues of the operator Tˆ.
Figure 1: The top view from the nitrogen atom (circle) above the normal axis to the centroid of the equilateral triangle of hydrogen atoms (black dots). (c) A hydrogen atom is excited by white light with the polarization vector (i.e., the electric field) in the x direction. Consider the possible transitions induced between energy eigenstates of the electron motion relative to the proton, denoted by the usual notation |n, ℓ, m, ms〉 for the principal n, orbital angular momentum (ℓ, m) and spin ms quantum numbers. Out of the eight states: | 2 , 0 , 0 , ms〉, | 2 , 1 , 1 , ms〉, | 2 , 1 , 0 , ms〉, | 2 , 1 , − 1 , ms〉, with ms = ±^12 , select the ones to which the transition from the ground state | 1 , 0 , 0 , +^12 〉 is possible. (It might or might not be helpful to look at Problem 3 first but p. 1 is.) (d) A beam of light with a spectral distribution centered at the resonance frequency of two atomic levels is shone on the atom. Explain succinctly (i) the reason that the Fermi Golden Rule gives the same probability per unit time for the excitation from the lower to high energy level and for the reverse process from the higher to the lower; (ii) the reason that we are not happy with this result. (e) A beam of neutrons is sent through a Stern-Gerlach apparatus. The detector screen shows two lines of equal intensity. The starting condition is known to be one of the only two mutually exclusive possibilities: i. half of the neutrons in the beam are in the spin-up state |+〉 along the direction of the vertical magnetic field (the z axis) and half in the spin-down state |−〉; ii. every neutron in the beam is in the spin state along the positive x axis which is normal to the plane of motion of the neutrons (y − z plane), i.e.,
| + x〉 =
Design an experiment to distinguish between only these two possibilities.
(b) The Hamiltonian of the particle in a uniform static magnetic field B is given by
H = μB B · Lˆ, (12)
where μB is the Bohr magneton. If the magnetic field is fixed along the y direction, find the expectation value of Lˆy at time t with the particle initial state given by Eq. (11). (c) If the magnetic field is fixed along the z direction since t = 0, find the expectation value of Lˆy at time t with the particle initial state given by Eq. (11).
- An immobilized proton was prepared with its spin pointing vertically up, thus in state | + z〉, in zero magnetic field. We wanted to flip the spin to point vertically down, i.e., to flip to the state | − z〉. An electromagnet was installed to provide a horizontal magnetic field along the x axis with time dependence B(t) which may be controlled at will within a range of ±b.
(a) Find the Hamiltonian for the proton spin with the installed magnet. (b) Design the time dependence of B(t) which will bring the proton spin to the down state and to stop there. (c) Unfortunately, the professor bumped into the magnet and knocked the field axis to an angle φ from the x axis in the x − y plane. Find the Hamiltonian for the proton spin with the displaced magnet. (d) Moreover, exactly half way through the application of the B(t) field using your design, there was a blackout, stopping the experiment. What was the final state of the spin, in the basis of the two states | ± z〉?
L.J. Sham December 11, 2007
Physics 130B Final Examination Solutions
1 Correction of Frequently Made Errors
- (a) Use the variational principle. (b) (c) Find the selection rules from the matrix elements of x between two states. The mag- netic field term is weak by comparison with the electric field term but its consideration was given credit. (d) (e) It is encouraging that a surprising number of students figured out the right answer.
- The state was given as a tensor product of three vector spaces x, y, z:
|nx, ny , nz〉 ≡ |nx〉|ny〉|nz〉.
Then, the operator in the x space acts only on the vector in x space, e.g.,
p ˆx|nx, ny, nz 〉 = [ˆpx|nx〉]|ny〉|nz〉,
so that, 〈mx, my , mz |pˆx|nx, ny, nz 〉 = 〈mx|pˆx|nx〉〈my|ny 〉〈mz|nz 〉.
- Be careful to pick the variational wave function which will contain information on the added term in the Hamiltonian. In this case, the SHO Hamiltonian in Eq. (2) is symmetry under mirror operation x → −x and so is the ground state wave function but the additional term, V^ ˆ , is odd under the mirror operation. So an even variational wave function would give zero, which means not that the effect is zero but that your variational wave function contains no possibility to include the effect of the perturbation. Remember that for 〈Ψ| given by Eq. (11), take the complex conjugate of the elements.
- A little nitpicking: the magnetic moment factor for proton is μp.
Useful information:
Pauli spin matrices
I =
[
]
, σx =
[
]
σy =
[
0 −i i 0
]
σz =
[
]
(c) A hydrogen atom is excited by white light with the polarization vector (i.e., the electric field) in the x direction. Consider the possible transitions induced between energy eigenstates of the electron motion relative to the proton, denoted by the usual notation |n, ℓ, m, ms〉 for the principal n, orbital angular momentum (ℓ, m) and spin ms quantum numbers. Out of the eight states: | 2 , 0 , 0 , ms〉, | 2 , 1 , 1 , ms〉, | 2 , 1 , 0 , ms〉, | 2 , 1 , − 1 , ms〉, with ms = ±^12 , select the ones to which the transition from the ground state | 1 , 0 , 0 , +^12 〉 is possible. (It might or might not be helpful to look at Problem 3 first but p. 1 is.) Solution – Consider the electric perturbation only, He = eE ˆx. From Eq.(5), we could make,
|Y 1 ,± 1 〉 =
√^1
(|x〉 ± i|y〉, and |Y 1 , 0 〉 = |z〉. (8)
The the selection rules from the consideration of the matrix elements between | 1 , 0 , 0 , +^12 〉 and the excited states are the change ∆ms = 0 (no spin-dependent perturbation is in- volved), ∆ℓ = 0, and ∆m = ± 1. Hence, the only possible final states are | 2 , 1 , ± 1 , 12 〉.
(d) A beam of light with a spectral distribution centered at the resonance frequency of two atomic levels is shone on the atom. Explain succinctly (i) the reason that the Fermi Golden Rule gives the same probability per unit time for the excitation from the lower to high energy level and for the reverse process from the higher to the lower; (ii) the reason that we are not happy with this result. Solution – i. Consider the three terms in the Golden rule, they are all the same for the absorption and the emission process. ii. The higher energy state can spontaneously decay to the lower energy state. This is not taken into account by the Fermi Golden rule. (e) A beam of neutrons is sent through a Stern-Gerlach apparatus. The detector screen shows two lines of equal intensity. The starting condition is known to be one of the only two mutually exclusive possibilities: i. half of the neutrons in the beam are in the spin-up state |+〉 along the direction of the vertical magnetic field (the z axis) and half in the spin-down state |−〉; ii. every neutron in the beam is in the spin state along the positive x axis which is normal to the plane of motion of the neutrons (y − z plane), i.e.,
| + x〉 =
Design an experiment to distinguish between only these two possibilities. Solution – Repeat the experiment with the magnetic axis rotated about the path through π/ 4 radians. Scenario (i) still yields two lines on the detector screen but Scenario (ii) now yields only one line.
- A harmonic oscillator of mass m and frequency ω in three dimensions has the Hamiltonian,
H 0 =
2 m
(p^2 x + p^2 y + p^2 z ) +
mω^2 (x^2 + y^2 + z^2 ). (10)
Its energy eigenstates are given by the number states |nx, ny, nz 〉 ≡ |nx〉|ny〉|nz〉, where nx, ny, nz = 0, 1 , 2 ,.. .. The states | 1 , 0 , 0 〉, | 0 , 1 , 0 〉, | 0 , 0 , 1 〉, are degenerate.
(a) Evaluate with the help of Eq. (3) in the basis of the three given degenerate states, the matrix elements of the perturbation, V = λxy. (11)
Solution – 〈mx, my , mz|xy|nx, ny , nz 〉 = 〈mx|x|nx〉〈my |y|ny〉〈mz|nz 〉. (12)
Use the creation and annihilation operators in Eq. (3) to evaluate the matrix elements as,
〈mx = 1|x|nx = 0〉 = a; 〈my = 0|y|ny = 1〉 = a. (13) Thus, the matrix elements of V are,
〈 1 , 0 , 0 |xy| 0 , 1 , 0 〉 = a^2 = 〈 0 , 1 , 0 |xy| 1 , 0 , 0 〉, (14) and the rest are zero. The matrix elements of V are shown next. (b) Find the new eigenstates to which the three degenerate eigenstates of the unperturbed Hamiltonian H 0 have changed and the corresponding change in energy eigenvalues. Solution – In the basis of the three degenerate states, | 1 , 0 , 0 〉, | 0 , 1 , 0 〉, | 0 , 0 , 1 〉, the matrix of V is
V = λa^2
Diagonalizing the matrix leads to the three eigenvalues, ±λa^2 , 0 , with the correspond- ing eigenstate vectors,
1 √ 2
- An ion of charge e and mass m moves in a one-dimensional harmonic potential with angular frequency ω. Its Hamiltonian is given by Eq. (2). The ion is placed in an electric field E which adds a term to the Hamiltonian given by V^ ˆ = −eE x,ˆ where |eEa| ≪ ℏω, (17)
a being defined in Eq. (3).
The exact solution. Note that the addition of a potential linear in x yields a displaced harmonic oscillator since the total potential is,
V =
mω^2 x^2 − eEx =
mω^2 (x − x 0 )^2 −
mω^2 x^20 , (25)
where
x 0 =
eE mω^2
The displacement SHO gives the same ground state. Thus, the net change in energy is
∆E 0 = −
mω^2 x^20 = −
(eE)^2 2 mω^2
The perturbation method. The first order change is zero by symmetry,
〈 0 |V | 0 〉 = 0. (28)
The second order perturbation contribution is
∆E 0 =
∑^ ∞
n=
|〈n|V | 0 〉|^2 E 0 − En
|〈 1 |V | 0 〉|^2
E 0 − E 1
(eE)^2 2 mω^2
(c) Compare the accuracy of your two answers. Solution – All three methods give the same answer. The perturbation result is correct because the exact result is a power series which terminates at second order in E. The variational method is a bit of a fluke. Its result before expansion is not exactly correct but the choice of the wave function leads to the correct O(E)^2 term.
- A spin-less particle is confined to be in an ℓ = 1 orbital angular momentum state. The orthonormal basis set is given by the spherical harmonics |Y 1 ,m〉, m = 1, 0 , − 1. The matrix representations of the angular momentum operators ˆL are given by
Lz = 1
(^) ; Lx = √^1 2
(^) ; Ly = √^1 2
0 −i 0 i 0 −i 0 i 0
(a) At time t = 0, the particle has the state vector
Ψ(t = 0) =
− 2 i √^1 i^2 2
Find the expectation value and uncertainty of Lˆy at t = 0.
Solution – Verify that LyΨ(t = 0) = Ψ(t = 0), i.e., the initial state is an eigenstate of L^ ˆy. Thus,
〈Ψ(0)| Lˆy|Ψ(0)〉 = 〈Ψ(0)|Ψ(0)〉 = 1, (32)
and the variation is zero. (b) The Hamiltonian of the particle in a uniform static magnetic field B is given by
H = μB B · Lˆ, (33)
where μB is the Bohr magneton. If the magnetic field is fixed along the y direction, find the expectation value of Lˆy at time t with the particle initial state given by Eq. (31). Solution – Since
H = μB B Lˆy, (34)
, the initial state is also an eigenstate of the Hamiltonian and thus at time t, it becomes
|Ψ(t)〉 = |Ψ(t = 0)〉e−iωt/ℏ, where ω =
μB B ℏ
The expectation value remains constant. (c) If the magnetic field is fixed along the z direction since t = 0, find the expectation value of Lˆy at time t with the particle initial state given by Eq. (31). Solution – From
H = μB B
the state at time t is
Ψ(t) =
− 2 i e−iωt √^1 i^2 2 e
iωt
Then
〈Ψ(t)| Lˆy|Ψ(t)〉 = Ψ(t)†Ly Ψ(t) = cos(ωt). (38)
- An immobilized proton was prepared with its spin pointing vertically up, thus in state | + z〉, in zero magnetic field. We wanted to flip the spin to point vertically down, i.e., to flip to the state | − z〉. An electromagnet was installed to provide a horizontal magnetic field along the x axis with time dependence B(t) which may be controlled at will within a range of ±b.
