Homework 1 Solutions

3.1 Solution
First calculate the Johsnon Noise, V therm = 4Rk b T ∆f. For our case ∆f = 500MHz,
yielding V therm = 2.9µV.
Now we calculate the V = IR.
V = IR (125)
= 1Ohm 〈µ〉(1mm)
2r 3 (126)
= 1Ohm γ H(1mm)
〈σ
4r 3 Z 〉 (127)
= 1Ohm γ H(1mm)
T r[σ
4r 3 z ρ(T )] (128)
where ρ(T ) is the thermal density matrix at temperature T and γ H = 5.6 Nuclear
Bohr Magnetons = 5.6 × 5.5 × 10 −26 J/T . We need to find r such that V = V therm
We now examine ρ(T ) at two temperatures. At 0 K, the thermal density matrix
is always the ground state. (Proton spin points along the magnetic field).
Consequently,T r[σ z ρ(T )] = 1. We can then calculate the distance to be r = 29
nm. This is already difficult.
For part b we look at the spin at temperatute T =300K. As we discussed in class
this results in a very small magnetic moment. T r[σ z ρ(T )] = 4x10 −5 . This leads to a
distance of r = 1.1nm.
Single spin measurements of electrons involving nm leads and devices at 4K has
been shown. A single spin measurement of a nuclei has been performed in atoms
using the hyperfine interaction to couple the nuclear magnetic dipole to the electron
magnetic dipole and then using the fine structure to couple the electron magnetic
dipole to the electric dipole of the atom.
2. Single molecule fluorescence experiments at room temperature are possible.
Why is this the case (qualitative and/or quantitative argument). Hint: These
experiments involve visible light and an electron dipole.
3.2 Solution
Single molecule NMR is hard at room temperature because the magnetic dipole of
proton is small and the residual or effective magnetic dipole at room temperature is
tiny.
Electric dipole of a molecule is much larger than the magnetic dipole of a nucleus
and, therefore, generates a larger electro-magnetic field when oscillating. See the
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