University of California at San Diego – Department of Physics – TA: Shauna Kravec
Quantum Mechanics C (Physics 212A) Fall 2015
Worksheet 2 –
Solutions
Announcements
• The 212A web site is:
http://physics.ucsd.edu/∼mcgreevy/f15/ .
Please check it regularly! It contains relevant course information!
Problems
1. Formaldehyde (From Le Bellac)
Let’s consider a simple two state system motivated by the Huckel theory.
Figure 1: Formaldehyde visualized
There are two π-electrons associated with the double bond between carbon and oxygen.
Let’s consider the Hilbert space of a single π-electron as H = span{|Oi, |Ci} where
these represent occupation on either the carbon or oxygen.
(a) Give a physical motivation for the Hamiltonian to be of the form
Ĥ = EO |OihO| + EC |CihC| − A(|CihO| + |OihC|)
(1)
where EC > EO are the energies associated with being localized and A is known
as the ”delocalization” energy
The oxygen is more electronegative than carbon so this translates to a lower
energy for being bonded to it.
The hopping term is the A piece which is trying to capture the kinetic energy of
the electrons which naively suggests it’s symmetric though maybe this could be
weakened.
The sign is important though as for A > 0 it determines whether the wavefunction is (approximately) symmetric/anti-symmetric. The potential should be
approximately symmetric so this explains the choice.
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(b) Calculate the eigenvalues and eigenvectors associated with (1). Sketch how this
would look in position space.
√
O
O
± A2 + ∆2 where ∆ ≡ EC −E
λ± = EC +E
2
2
φ
φ
|+i = e−i 2 cos 2θ |Oi + ei 2 sin 2θ |Ci
φ
φ
|−i = −e−i 2 sin 2θ |Oi + ei 2 cos 2θ |Ci
where − √A2A+∆2 = sin θ cos φ and √A2∆+∆2 = cos θ
Assume that the system is in it’s ground state.
(c) For a given π-electron, calculate the probability of finding it localized at the
oxygen.
P = h−|OihO|−i = sin2 2θ = 12 (1 − cos θ)
(d) Assume that the electric dipole moment of formaldehyde only gets contributions
from the symmetric axis. Express this as a function of the bond length `.
There are 8 protons on oxygen and 6 protons on carbon. There are 7 core+σ
electrons associated with oxygen and 5 for carbon. The remaining two electrons
are the ones we reasoned about above. The second π-electron can be placed also
in the groundstate of the first under the assumption they don’t interact.
The excess charge on the oxygen from the π-electrons is then (1 − 2P )e where e is
the charge of the electron. d = (1 − 2P )e` choosing the carbon to be our reference
point.
2. Single Qubit Gates (From Nielsen-Chuang)
Aside from the usual Pauli matrices there are a few common operators for a two state
system. These are the Hadamard (H), the phase gate (S), and the T -gate (T ). In the
Z-basis these can be written as:
1 1 1
1 0
1 0
π
H=√
S=
T =
(2)
0 i
0 ei 4
2 1 −1
(a) Write these in terms of our original Pauli’s. Note that S = T 2 . What is the action
of H on Z-eigenvectors?
π
π
H = √12 (X + Z) and T = 12 (1 + ei 4 )1 + 12 (1 − ei 4 )Z
H|0i =
√1 (|0i
2
2
+ |1i) and H|1i =
√1 (|0i
2
− |1i)
Note H = 1
(b) Prove the following identities
HXH = Z
XH = √12 (1 − iY ) =⇒ HXH =
The rest are similar.
HY H = −Y
√1 (H
2
HZH = X
− iHY ) =
√1 [H
2
−
√i
(3)
2
(iZ − iX)] = Z
θ
(c) Show that1 T = Uz ( π4 ) and HT H = Ux ( π4 ) where Un (θ) ≡ e−i 2 n̂·~σ
1
Up to a global phase
2
π
π
e−i 8 0
T =e
= e+i 8 Uz ( π4 ) ∝ cos π8 1 − i sin π8 Z
i π8
0
e
π
HT H = cos 8 1 − i sin π8 HZH = cos π8 1 − i sin π8 X
+i π8
3. Quis Custodiet Ipsos Custodes? (From Jacobs)
Projective measurements lead to some weird things.
Consider a two state system with basis vectors {|0i, |1i}. We are going to
evolve
the
0
−i
.
system according the Hamiltonian Ĥ = ω2 Y where Y is the Pauli matrix
i 0
(a) What is the unitary operator associated with time evolution? Given an initial
prepared state of |ψ0 i = |0i. Write an expression for |ψ(t)i.
ω
ω
U = e−iHt = e−i 2 Y t . Recall that Y |0i = i|1i and that e−i 2 ~σ.n̂ = cos ω2 1 −i~σ .n̂ sin ω2
|0i + sin ωt
|1i
This implies |ψ(t)i = cos ωt
2
2
(b) What is the probability, as function of time, to measure |0i?
Pt [|0i] = cos2 ωt
2
T
(c) Suppose we study the system over the time interval [0, T ] where T δt ≡ N
. We
T 2T
, N , · · · where N is large.
perform a measurement, in this basis, at every time N
Assuming each measurement is independent from the other, what’s the probability
that the spin never flips to |1i?
Recall that our measurement axiom says we should ’collapse’ |ψi onto the pure
state which we measure it to be.
T
T
then the time evolution from N
→ 2T
This implies if we measure |0i at time t = N
N
ωT
is the same as if starting from t = 0. The probability to not flip is always cos2 2N
The probability for the spin to never flip then is just the product of probabilities
to not flip at every measurement.
ωT N
)
Pnever−f lip = (cos2 2N
(d) Evaluate this probability in the limit of N → ∞.
This is called the quantum Zeno effect.
A cheap and dirty way to do this is to take the series expansion at N = ∞ and
drop all terms in O( N1 ). Just replace N1 ≡ η and Taylor expand at η = 0.
2
2
Pnever−f lip ≈ 1 − ω4NT which goes quickly to 1. In this limit the spin never flips.
More carefully you should see that it actually exponentially approaches 1 as N →
∞; it’s pretty dramatic
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