Homework 1 Solutions

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B3 purity 0.05 0.7 0.1 0.65 ξ 0.15 0.2 0.6 0.25 0.55 0.3 0.5 0.35 0.05 0.1 0.15 0.2 0.25 0.3 0.35 η 0.45 3 Single Molecule NMR In Problem 1, we assumed a single proton spin NMR Device. Here we look at whether or not that is possible. We assume in this problem that the H-NMR resonance is at 500 MHz. 1. A magnetic dipole produces a magnetic field B ⃗ = µ 0 ⃗µ . In QM, we replace 4π r 3 ⃗µ with 〈ˆ⃗µ〉. In NMR, 〈ˆµ(t)〉 oscillates in time creating a sinusoidal signal. As a toy model, assume that the pickup coil transforms the magnetic field into a current by the following equation I(t) = 2π(1mm) µ 0 B(t). This current passes through a 1 Ohm resistor and the voltage drop over the resistor is measured. We will say that NMR is possible in this toy model when the voltage drop is larger than the thermal voltage due to Johnson noise. a. Assume that the spin is at T=0 K. (The resistor is still at 300 K). What distance r does the pickup coil need to be for NMR to be possible b. Assume that the spin is at T=300 K. How close does the pickup coil need to be now 16
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 17
Page 1 and 2: Homework 1 Solutions January 29, 20
Page 3 and 4: Substituting Eq. 12 into Eq. 11, we
Page 5 and 6: = x 0 (〈ψ ′ |ψ〉 + 〈ψ |ψ
Page 7 and 8: step 1: use N measurement and T r[
Page 9 and 10: For B2 we have ρ 2 00 = 7/12 − 2
Page 11 and 12: 0.25 B1 0.245 0.24 0.235 min Tr[|ψ
Page 13 and 14: maxminT r[|ψ〉〈ψ| ρ] = 1 2 (
Page 15: We now know the offdiagonal element

Homework 1 Solutions

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