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We now simply integrate (using Mathematica or a book of integrals or knowldege<br />

about the Fourier transform of a Gaussian distribution)<br />

∫<br />

ρ(t) = p(ξ)( 1 2 I + 1 2 (cos ω(1 + ξ/B 0)tσ x − sin ω(1 + ξ/B 0 )tσ y ))dξ (19)<br />

= 1 2 I + 1 ∫<br />

p(ξ)(cos ω(1 + ξ/B 0 )tσ x − sin ω(1 + ξ/B 0 )tσ y )dξ (20)<br />

2<br />

= 1 2 I + 1 2 exp(−(ωξ 0t) 2<br />

)(cos ωtσ x − sin ωtσ y ) (21)<br />

2B 2 0<br />

The 〈σ x (t)〉 = exp(− (ωξ 0t) 2<br />

) cos ωt.<br />

2B0<br />

2<br />

Note that ρ(t) rapidly approaches the totally mixed state, ρ(∞) = 1 I, with the<br />

2<br />

rate determined by the ratio of the width of the inhomogeneity ξ 0 and the applied<br />

field B 0 . We will see later that NMR is possible because B 0 is much larger than ξ 0 .<br />

2 Density Matrices and Harmonic Oscillators<br />

Alice and Bob are trying to build a device that generates I 2 in the ground electronic<br />

state, a translational velocity of 300 m/s, and a vibrational state of |ψ〉 = 1/2 |0〉 +<br />

√<br />

1/2 |1〉 + 1/2 |3〉. (For this problem, assume the molecule has a classical velocity<br />

along ẑ, and the molecular axis is fixed to point along ˆx. We can then treat I 2 as<br />

a 1d harmonic oscillator). Alice builds three boxes and Bob sets up a detector z=1<br />

meter away. Bob’s detector can measure the velocity of the molecule, the number<br />

state, and 〈X〉 and 〈P x 〉. (Hint: Write ˆX and ˆP x in terms of a † and a).<br />

Alice makes three boxes that satisfy 〈N〉=1.25 (where ˆN = a † a) and the translational<br />

velocity of I 2 is 300 m/s.<br />

1. a. For |ψ〉 what is 〈X〉<br />

b. If one of Alice’s boxes has produced |ψ〉 at the box, how does 〈X〉 change with<br />

the distance to Bob’s detector How does 〈P x 〉 change<br />

2.1 Solution<br />

Remember that ˆX =<br />

√ ¯h<br />

2MΩ (a† + a) = x 0 (a † + a) Define |ψ ′ 〉 = a |ψ〉<br />

First, we calculate 〈X〉<br />

〈X〉 = 〈ψ| ˆX |ψ〉<br />

= x 0 〈ψ| (a † + a) |ψ〉<br />

4

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