Homework 1 Solutions

Homework 1 Solutions
January 29, 2007
1 Magnetic Field Inhomogeneity
A single proton is put into an NMR-like machine. The device can measure 〈σ z 〉,
〈σ x 〉, and 〈σ y 〉
1. At time t=0, the device measures 〈σ z 〉 = 〈σ y 〉 = 0 and 〈σ x 〉 = 1. What is ρ(0)
1.1 Solution
From class, we can just plug in the expected values to find the density matrix
ρ = 1 2 I + 1 2 (〈σ x〉σ x + 〈σ x 〉σ x + 〈σ x 〉σ x )
= 1 2 I + 1 2 σ x (1)
Alternate method: We know that if 〈σ x 〉 = 1 that σ x |ψ(0)〉 = |ψ(0)〉. Therefore,
|ψ(0)〉 = |+〉 = 1 √
2
(|↑〉 + |↓〉). (2)
Therefore
ρ = |+〉 〈+| (3)
= 1/2(|↑〉 〈↑| + |↑〉 〈↓| + |↓〉 〈↑| + |↓〉 〈↓|) (4)
2. In a uniform magnetic field, ⃗ B = B 0 ẑ, what is ρ(t) and 〈σ x (t)〉
1

1.2 Solution
First, we write down the Hamiltonian of a proton in a magnetic field
Ĥ = −⃗µ · ⃗B (5)
= − γ H
2 ⃗σ · ⃗B (6)
= − γ HB 0
σ z (7)
2
(8)
For covenience, we define ω = γ HB 0
¯h
We then write the unitary time evolution operator as either
or since σ z σ z = 1 we can write in terms of cos and sin
ˆ U(t) = exp(−iĤt/¯h)
= exp(i ωt
2 σ z) (9)
ˆ U(t) = exp(−iĤt/¯h)
First, we find ρ(t) using Eq. 1 and Eq. 10.
= cos( ωt )I + i sin(ωt
2 2 )σ z (10)
ρ(t) = U(t)ρ(0)U † (t)
= U(t)( 1 2 I + 1 2 σ x)U † (t)
= 1 2 I + 1 2 U(t)σ xU † (t) (11)
Now we need to find U(t)σ x U † . Recall that σ z σ x = −σ x σ z = iσ y
U(t)σ x U † = (cos( ωt
2 )I + i sin(γ HB 0 t
)σ z )σ x U † (t)
2
= cos( ωt
2 )σ x − i sin( γ HB 0 t
)σ x σ z U † (t)
2
= (cos 2 ( ωt
2 ) − sin2 ( ωt
2 ))σ x − (2 sin( ωt
2 ) cos(ωt 2 ))σ y
= cos(ωt)σ x − sin(ωt)σ y (12)
2

Substituting Eq. 12 into Eq. 11, we reach the answer
Alternate Method:
Using Eq. 9 and Eq. 2, we have
ρ(t) = 1 2 I + 1 2 (cos(ωt)σ x − sin(ωt)σ y ) (13)
|ψ(t)〉 =
=
Û(t) |ψ(0)〉
√
1
2 (eiωt/2 |↑〉 + e −iωt/2 |↓〉) (14)
We can then write
ρ(t) = |ψ(t)〉 〈ψ(t)| (15)
= 1 2 (|↑〉 〈↑| + eiωt |↑〉 〈↓| + e −iωt |↓〉 〈↑| + |↓〉 〈↓|) (16)
Finally we calculate 〈σ x (t)〉 = T r[σ x ρ(t)] = 〈ψ(t)| σ x |ψ(t)〉 = cos(ωt).
3. To measure 〈σ z 〉, the device performs a single measurement many times. Now
assume that every time the experiment is performed there is a slight deviation in
the magnetic field B ⃗ = (B 0 + B 1 )ˆx where B 1 = ξ and ξ is a random variable with a
probability distribution p(ξ) = √ 1
2πξ0
exp(− ξ2
).
2ξ0 2
Now what is ρ(t) What is 〈σ x (t)〉
These results are similar to the effect of an inhomogenous magnetic field in a real
NMR machnie.
1.3 Solution
As mentioned in class, if we are given a probability p i of having ω i the density matrix
is
ρ(t) = ∑ p i ρ(t, ω i )
i
(17)
In the case of a continuous probability distribution, we write
∫
ρ(t) = p(ξ)ρ(t, ω(ξ))dξ (18)
3

We now simply integrate (using Mathematica or a book of integrals or knowldege
about the Fourier transform of a Gaussian distribution)
∫
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)
= 1 2 I + 1 ∫
p(ξ)(cos ω(1 + ξ/B 0 )tσ x − sin ω(1 + ξ/B 0 )tσ y )dξ (20)
2
= 1 2 I + 1 2 exp(−(ωξ 0t) 2
)(cos ωtσ x − sin ωtσ y ) (21)
2B 2 0
The 〈σ x (t)〉 = exp(− (ωξ 0t) 2
) cos ωt.
2B0
2
Note that ρ(t) rapidly approaches the totally mixed state, ρ(∞) = 1 I, with the
2
rate determined by the ratio of the width of the inhomogeneity ξ 0 and the applied
field B 0 . We will see later that NMR is possible because B 0 is much larger than ξ 0 .
2 Density Matrices and Harmonic Oscillators
Alice and Bob are trying to build a device that generates I 2 in the ground electronic
state, a translational velocity of 300 m/s, and a vibrational state of |ψ〉 = 1/2 |0〉 +
√
1/2 |1〉 + 1/2 |3〉. (For this problem, assume the molecule has a classical velocity
along ẑ, and the molecular axis is fixed to point along ˆx. We can then treat I 2 as
a 1d harmonic oscillator). Alice builds three boxes and Bob sets up a detector z=1
meter away. Bob’s detector can measure the velocity of the molecule, the number
state, and 〈X〉 and 〈P x 〉. (Hint: Write ˆX and ˆP x in terms of a † and a).
Alice makes three boxes that satisfy 〈N〉=1.25 (where ˆN = a † a) and the translational
velocity of I 2 is 300 m/s.
1. a. For |ψ〉 what is 〈X〉
b. If one of Alice’s boxes has produced |ψ〉 at the box, how does 〈X〉 change with
the distance to Bob’s detector How does 〈P x 〉 change
2.1 Solution
Remember that ˆX =
√ ¯h
2MΩ (a† + a) = x 0 (a † + a) Define |ψ ′ 〉 = a |ψ〉
First, we calculate 〈X〉
〈X〉 = 〈ψ| ˆX |ψ〉
= x 0 〈ψ| (a † + a) |ψ〉
4

= x 0 (〈ψ ′ |ψ〉 + 〈ψ |ψ ′ 〉)
= x 0 (2Re(〈ψ |ψ ′ 〉)) (22)
where Re(g) is the real part of g. If g = a + ib, Re(g)=a, Im(g)=b
Now we calculate 〈ψ |ψ ′ 〉 recall a |n〉 = √ n |n − 1〉.
〈ψ |ψ ′ 〉 = (1/2 〈0| + 1/ √ 2 〈1| + 1/2 〈3|)(1/ √ 2 |0〉 + 3/2 |2〉) (23)
= (1/2) 3/2 (24)
Plugging into Eq. 22, we get 〈X〉 = x 0 / √ 2
b. As stated in the problem, the propagation of the distance between Alice and
Bob, z, is classical. Therefore, we only need to worry about the time, t = z/v.
We can then define Û(t) = exp(−iω(a† a + 1/2)t). Note Û(t) |n〉 = exp(−iω(n +
1/2)t) |n〉.
Therefore
|ψ(t)〉 =
as we showed on the first day of class.
Furthermore, we can define
Û(t) |ψ〉 (25)
= e −iωt/2 /2 |0〉 + e −i3ωt/2 / √ 2 |1〉 + e −i7ωt/2 /2 |3〉 (26)
|ψ ′ (t)〉 = |ψ(t)〉 (27)
= e −i3ωt/2 / √ 2 |0〉 + 3e −i7ωt/2 /2 |2〉 (28)
Making the same calculation as Eq. 22 and substituting t = z/v, One can show
that
A similar calculation shows that
where P x0 =
〈X(z)〉 = 2x 0 Re(〈ψ(z/v) |ψ ′ (z/v)〉) (29)
= (x 0 / √ 2) cos( ωz
v ) (30)
〈P x (z)〉 = 2P x0 Im(〈ψ(z/v) |ψ ′ (z/v)〉) (31)
= −(P x0 / √ 2) sin( ωz
v ) (32)
√
Mω¯h/2.
5

The other thing to do is realize that 〈E〉 is invariant with time so mω 2 〈X(z)〉 2 /2+
〈P (z)〉 2 /(2m) is also invariant with time.
For I 2 , ω/(2π) is 30 GHz. For a velocity of 300 m/s, this means that moving
the detector 10 nm along Z that 〈X〉 will oscillate in value between (x 0 / √ 2) and
−(x 0 / √ 2)
2. Bob measures that for B1 and B3 〈X〉 = answer of 1a and that 〈P x 〉=0
For B2, the expected value for both quantitites is 0.
Bob then looks at the measured values of N. For B1 and B2, the values are 0,
1, and 3. For B3, the values are 0, 1, and 2. Based on the data, construct possible
density matrices (limit yourself to 3X3 density matrices using only the eigenvalues
measured for each case).
Caclulate the minimum and maximum of T r[|ψ〉 〈ψ| ρ]. (This is one way to measure
how close two density martices are.)
2.2 Solution
A density matrix contains all the information we know about a quantum system.
Therefore, it is equivalent to knowing the values of a set of measurements.
For example, with a spin 1/2 system all density matrices can be written as
ρ = 1/2I + 1/2(〈σ x 〉σ x + 〈σ y 〉σ y + 〈σ z 〉σ z ) (33)
There are constraints on the density matrix (see Shankar pg. 133 and 144). For
spin 1/2, these constraints mean 〈⃗σ · ⃗σ〉 ≤ 1
We will use the following constraints
ρ = ρ † (34)
T r[ρ] = 1 (35)
1 ≥ ρ ii ≥ 0 (36)
|ρ ij | ≤ | √ ρ ii ρ jj | (37)
Eq. 36 means that the probability of measuring the state i is non-negative. This
implies Eq. 37.
(For example, say ρ = 1/2I +1/2(1+2ɛ)σ x . I can of course redfine my coordinates
so that I now label x as z and z as -x. This yields ρ = 1/2I + 1/2(1 + 2ɛ)σ z =
1 + ɛ |↑〉 〈↑| − ɛ |↓〉 〈↓|. Please convince yourself this is true for any size matrix.)
Throught this problem ρ nm = 〈n| ρ |m〉 and ρ = ∑ m,n ρ nm |n〉 〈m|
6
(38)

step 1: use N measurement and T r[ρ] = 1
For B1 and B2, we know that 〈N〉=1.25 and that the possible values of N are 0,
1, and 3. That gives us two equations
ρ 00 + ρ 11 + ρ 33 = 1 (39)
ρ 11 + 3ρ 33 = 1.25 (40)
Notice that this measurement and constraint does not distinguish between the
two boxes. Three unknowns and two equations means that there is a one parameter
family of solutions. We denote the density matrix associated with B1 as ρ 1 and the
density matrix associated with B2 as ρ 2 . Since at this stage we cannot distinguish
between B1 and B2, I use the shorthand ρ 1,2 For ρ 1,2 we have
where 7/24 ≥ ξ ≥ 0
For B3 we measure values of N 0,1, and 2
This also yields two solutions For ρ 3 we have
where 3/8 ≥ ξ ≥ 0
step 2: use 〈X〉 and 〈P x0 〉 to refine our assesment.
We will use 〈X〉/x 0 and 〈P x 〉/P 0
First for B1 and B3, we can write
Applying the trace we find
ρ 1,2
00 = 7/12 − 2ξ (41)
ρ 1,2
11 = 3ξ (42)
ρ 1,2
33 = 5/12 − ξ (43)
ρ 3 00 + ρ 11 + ρ 22 = 1 (44)
ρ 3 11 + 2ρ 22 = 1.25 (45)
ρ 3 00 = 3/8 − ξ (46)
ρ 3 11 = 2ξ (47)
ρ 3 22 = 5/8 − ξ (48)
ˆX/x 0 = a † + a = 〈1| |0〉 + 〈0| |1〉 (49)
ˆP x /P x0 = i(a † − a) = i 〈1| |0〉 − i 〈0| |1〉 (50)
〈X〉/x 0 = ρ 1,3
01 + ρ 1,3
10 = 2Re(ρ 1,3
10 ) (51)
〈P x 〉/P x0 = iρ 1,3
01 − iρ 1,3
10 = 2Im(ρ 1,3
10 ) (52)
7

Now using our measurements we can differentiate between B1 and B2.
For B1 we have
This puts additional constraints on ξ by Eq. 37.
We look at the squared equation
ρ 1 00 = 7/12 − 2ξ (53)
ρ 1 11 = 3ξ (54)
ρ 1 33 = 5/12 − ξ (55)
√
ρ 1 01 = ρ 1 10 = 1/8 (56)
√ √ √
1/8 ≤ 3ξ 7/12 − 2ξ (57)
1/8 ≤ 7/4ξ − 6ξ 2 (58)
First, we find values of ξ where it is equal. We find ξ = 1/8 and ξ = 1/6.
This means that 1/6 ≥ ξ ≥ 1/8
We have no information about the other elements. Therfore at this point our
density matrix is
ρ 1 00 = 7/12 − 2ξ (59)
ρ 1 11 = 3ξ (60)
ρ 1 33 = 5/12 − ξ (61)
√
ρ 1 01 = ρ 1 10 = 1/8 (62)
ρ 1 03 = ρ 1∗
30 = |ρ 1 03|e −iφ 03
(63)
ρ 1 13 = ρ 1∗
31 = |ρ 1 13|e −iφ 13
(64)
(65)
where
|ρ 1 03| ≤
|ρ 1 13| ≤
√
7/12 − 2ξ
√
3ξ
√
5/12 − ξ (66)
√
5/12 − ξ (67)
1/6 ≥ ξ ≥ 1/8 (68)
8

For B2 we have
ρ 2 00 = 7/12 − 2ξ (69)
ρ 2 11 = 3ξ (70)
ρ 2 33 = 5/12 − ξ (71)
ρ 2 01 = ρ 2 10 = 0 (72)
ρ 2 03 = ρ 2∗
30 = |ρ 1 03|e −iφ 03
(73)
ρ 2 13 = ρ 2∗
31 = |ρ 1 13|e −iφ 13
(74)
(75)
where
|ρ 2 03| ≤
|ρ 2 13| ≤
√
7/12 − 2ξ
√
3ξ
√
5/12 − ξ (76)
√
5/12 − ξ (77)
7/24 ≥ ξ ≥ 0 (78)
Now we repeat for B3
First we can write
ˆX/x 0 = a † + a = 〈1| |0〉 + 〈0| |1〉 + √ 2(〈2| |1〉 + 〈1| |2〉) (79)
√
ˆP x /P x0 = i(a † − a) = i 〈1| |0〉 − i 〈0| |1〉 + i (2)(〈2| |1〉 − 〈1| |2〉) (80)
Applying the trace we find
We can then write that
where
〈X〉/x 0 = 2Re(ρ 2 10) + 2 √ 2Re(ρ 2 21) (81)
〈P x 〉/P x0 = 2Im(ρ 2 10) + 2 √ 2Re(ρ 2 21) (82)
ρ 3 00 = 3/8 − ξ ρ 3 11 = 2ξ (83)
ρ 3 22 = 5/8 − ξ (84)
ρ 3 01 = ρ 3 10 = η (85)
√
ρ 3 12ρ 3 20 = ( 1/8 − η)/ √ 2 (86)
ρ 3 20 = ρ 3∗
21 = |ρ 3 11|e −iφ 03
(87)
√
1/8 ≥ η ≥ 0. We can now use η to constrain ξ
9

We have two inequalities
√
( 1/8 − η)/ √ 2 ≤
√
η ≤ 3/8 − ξ
√
2ξ
√
2ξ (88)
√
5/8 − ξ (89)
We will comeback to these.
step 3:Finally we calculate the min and max of T r[|ψ〉 〈ψ| ρ]
First, we note that if the measurements are consistent we reach the max when
ρ = |ψ〉 〈ψ|
To calculate the min and max generally we do the following: First, we write out
|ψ〉 〈ψ|
|ψ〉 〈ψ| = 1 2 |1〉 〈1|+1 1
(|0〉 〈0|+|3〉 〈3|+|0〉 〈3|+|0〉 〈3|)+
4 2 √ (|0〉 〈1|+|0〉 〈1|+|1〉 〈3|+|1〉 〈3|)
2
(90)
We can then write for any ρ
T r[|ψ〉 〈ψ| ρ] = 1 2 ρ 11 + 1 4 (ρ 00 + ρ 33 + 2Re(ρ 30 )) + 1
2 √ 2 (2Re(ρ 10) + 2Re(ρ 13 )) (91)
B1. For B1 the measurements are consistent with ρ 1 = |ψ〉 〈ψ| Therefore
maxT r[|ψ〉 〈ψ| ρ 1 ] = 1 (92)
To calculate the min we examine Eq. 91. To minimize this function we set
Re(ρ 30 ) and Re(ρ 10 ) to negative values with the largest magnitude allowed by the
constraints.
minT r[|ψ〉 〈ψ| ρ] = 1 √
2 (3ξ)+1 4 (1−3ξ−2 (7/12 − 2ξ)(5/12 − ξ))+ 1
2 √ 2 ( √ 1 √
−2 (3ξ)(5/12 − ξ))
2
(93)
Now we need to pick an ξ that minimizes this. I usually plot it in matlab,
mathematica, maple, or scilab. We can see from the espression that the function is
basically linear in ξ so we suspect that the answer will be either 1/6 or 1/8. For the
ρ = |ψ〉 〈ψ|, η = 1/6, so I would guess the minimum would occur for η = 1/8. This
is true as shown below
The min T r[|ψ〉 〈ψ| ρ] = 0.204
B2.
10

0.25
B1
0.245
0.24
0.235
min Tr[|ψ>

B2 measurements are not consistent with ρ 2 = |ψ〉 〈ψ|.
We write that
maxminT r[|ψ〉 〈ψ| ρ] = 1 √
2 (3ξ)+1 4 (1−3ξ±2 (7/12 − 2ξ)(5/12 − ξ))+ 1 √
2 √ 2 (±2 (3ξ)(5/12 − ξ))
(94)
Notice that this expression is the same as for B1 without a term 1
2 √ √1
2 2
= 1 4
Therefore the max is B1max-1/4=3/4.
The minimum is negative (≤ −0.46). This is why this measurement of overlap is
not good. The minimum should be zero (for orthogonal states).
We again simply plot with ξ between the bounds of zero and 7/24.
The min trace value is -0.0888
0.25
B2
0.2
0.15
min Tr[|ψ>

maxminT r[|ψ〉 〈ψ| ρ] = 1 2 (2ξ) + 1 1
(3/8 − ξ) +
4 2 √ (2η) (95)
2
Notice how all the terms vanish since |ψ〉 〈ψ| has no overlap with |2〉. Recall that
0 ≤ η ≤ 1/( √ 8) and 3/8 ≥ ξ ≥ 0 We could simply bound the min and max with
these bounds but we have more information.
Using Eq. √88, we need the quadratic problem to have real solutions. After some
algebra this 9/16 − 8η must be real. This means η ≤ 9/(8 · 16) can solve for ξ and
√
find ξ pm = 3/4± 9/16−8η
4
Maximum We choose the maximum ξ for each η and plot the trace. We see that
η = 0 and ξ = 3/8 is the best leading to a trace of 3/8 = 0.375
Minimum We √ set a minimum η=0. Then we need to refer to the other inequality
Eq. 89, 1/4 ≤ 5/4ξ − 2ξ 2 . The smallest value of ξ allowed is ξ = 1/16(5 − √ 17).
We then calculate the minimum Tr is 0.138
3. Bob buys a new detector. It can measure 〈a † a † + aa〉. The values obtained are
B1= √ 3 (ACTUAL HOMEWORK 6/ √ 2 WAS WRITTEN in ERROR THIS VALUE
MEANS THE MEASUREMENT WAS BAD), B2=0, and B3=0. What does this
measure in terms of position and momentum
Update the density matrices.
Calculate the min and max of T r[ρρ] for each case. (One way to measure the
purity. A pure state yields 1 and the most mixed state, 1/3 I, yields 1/3).
Calculate the min and max of T r[|ψ〉 〈ψ| ρ B1 ].
2.3 Solution
Another measurement givesus more insight into the density matrix
We further note
a † a † + aa = ( ˆX 2 /x 0 − ˆP 2 x /P x0 )/2 (96)
and
a † a † + aa = √ 2(|2〉 〈0| + |0〉 〈2|) √ 6(|3〉 〈1| + |1〉 〈3|) (97)
T r[(a † a † + aa)ρ] = 2 √ 2Re(ρ 02 ) + 2 √ 6(|3〉 〈1| + |1〉 〈3|) (98)
13

B1. For B1 this measurement of √ 3 almost tells us the whole density matrix. It
sets ρ 1 13 and as a consequence there is only one value of ξ, ξ = 1/6, that satsifies the
Eq. 37
where
ρ 1 00 = 1/4 (99)
ρ 1 11 = 1/2 (100)
ρ 1 33 = 1/4
√
(101)
ρ 1 01 = ρ 1 10 = 1/8 (102)
ρ 1 03 = ρ 1∗
30 = |ρ 1 13|e −iφ 13
(103)
√
ρ 1 13 = ρ 1 31 = 1/8 (104)
(105)
|ρ 1 03| ≤ 1/4 (106)
ρ 1 could still be a pure state so the max T r[ρ 1 ρ 1 ]=1 and max T r[|ψ〉 〈ψ| |ψ〉 〈ψ|]=1.
For the minimum purity, we set ρ 1 13=0 and find T r[ρ 1 ρ 1 ] = 1−2(1/16) = 7/8. For the
minimum overlap, we set ρ 1 13=-1/4 and find T r[|psi〉 〈psi| ρ 1 ] = 1 − 4(1/16) = 0.75.
B2.
For B2 we have
ρ 2 00 = 7/12 − 2ξ (107)
ρ 2 11 = 3ξ (108)
ρ 2 33 = 5/12 − ξ (109)
ρ 2 01 = ρ 2 10 = 0 (110)
ρ 2 03 = ρ 2∗
30 = |ρ 1 03|e −iφ 03
(111)
ρ 2 13 = −ρ 2∗
31 = ±i|ρ 1 13| (112)
(113)
where
|ρ 2 03| ≤
|ρ 2 13| ≤
√
7/12 − 2ξ
√
3ξ
√
5/12 − ξ (114)
√
5/12 − ξ (115)
7/24 ≥ ξ ≥ 0 (116)
14

We now know the offdiagonal element ρ 2 13 is pure imaginary.
Minimum purity occurs when the offdiagonal elements have minimum value.
Then
T r[ρ 2 ρ 2 ] = (7/12 − 2ξ) 2 + (3ξ) 2 + (5/12 − ξ) 2 (117)
= 74/144 − 3/2ξ + 14ξ 2 (118)
First, we check where the minimum of the quadratic function is. We find that
ξ min = 3/56. This is within the allowed bounds.
minT r[ρ 2 ρ 2 ] = 0.4737 (119)
For the maximum, we use a pure state that agrees with ξ = 0, |Ψ〉 = √ 7
√ |0〉 + 12
5
|3〉. The T r[|Ψ〉 〈Ψ| |Ψ〉 〈Ψ|] = 1.
12
(We said that 1’s were measured out of B2 so this answer is technically wrong.
However, since we didn’t mention the frequency of 1 we can assume that it was
measured only once out of our millions of measurements. The correct answer is
ρ 2 = (1 − ɛ) |Ψ〉 〈Ψ| + ɛ |1〉 〈1|. The max Tr measureing purity is then 1 − ɛ + ɛ 2 ).
B3.
We can then write that
ρ 3 00 = 3/8 − ξ ρ 3 11 = 2ξ (120)
ρ 3 22 = 5/8 − ξ (121)
ρ 3 01 = ρ 3 10 = η (122)
√
ρ 3 12 = ρ 3 12 = ( 1/8 − η)/ √ 2 (123)
ρ 3 20 = −ρ 3 20 = ±i|ρ 3 20| (124)
√
where 1/8 ≥ η ≥ 0.
The maximum purity we find numerically to be 0.74. The minimal purity we find
numerically to be 0.45.
Here’s a plot of the max purity. The boundaries defined by the inequalities are
denoted by the sudden jump in color.
15

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
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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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notes about EM waves. Equally valid are arguments about photon detectors at
visible wavelength but not at 500 MHz.
Since the transition is visible the energy spacing between the levels is large (h
time a few 100 THz). At room temperature most of the population is still in the
electronic ground state. Unlike the case of a nuclear spin, the expectation value for
the dipole element is approximately the same as what it would be at 0 K.
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