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Blog - Visualization of diamond defect field response (Rev #6)

This is a blog article in progress, written by Jacob Biamonte. See also the quantum network theory series. To see discussions of the article as it was being written, visit the Azimuth Forum. For the final polished version, go to the Azimuth Blog.

NOTE: Here’s Part I

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Visualization of the quantum energies of a defect in diamond

Last time we talked about how defects in diamonds exhibit controllable quantum degrees of freedom. We showed that the energy levels of an elementary model of such a system resulted in a cubic equation. Today we are going to use a few oldtime tricks to visualize these energy levels.

Cubic equations embrace a long and twisted tale stretching back to the early days of mathematics. They lead to the eventual discovery of complex numbers, and even group theory.

Sometimes on Babylonian cuneiform tablets from the 20th to 16th centuries BC, people have noticed tables for calculating cubes and cube roots. The Babylonians could have used the tables to solve cubic equations, no one can provide evidence that they did.

The tale started to get more lively in the 16th century when del Ferro found a method for solving a class of cubic equations, namely those of the form x 3+mx=nx^3 + mx = n.

As we now know, all cubic equations can be reduced to this form if we allow m and n to be negative, but negative numbers were not known to del Ferro at that time!

Del Ferro kept his result to himself, until just before his death. He showed it to his student Fiore.

In 1530, Tartaglia entered the game. In fact, he was bold enough to let everyone know that he was able to solve two problems in cubic equations.


At that time, it seems the thing to do was to offer a challenge, and this is exactly what Fiore did. Each contestant put up money and proposed a number of problems for his rival. The winner would solve the most problems within 30 days. Sounds pretty reasonable.

Tartaglia received questions in the form x 3+mx=nx^3 + mx = n, for which he had discovered a general method. Fiore questions came as x 3+mx 2=nx^3 + mx^2 = n,

Later, Tartaglia was persuaded by Cardano to reveal his secret for solving cubic equations and did so under some strict conditions of secrecy, which had a few loops holes in them which Cardano exploited to publish and credit Tartaglia for his results some six years after Tartaglia told him.

This led to a challenge to Cardano by Tartaglia, which Cardano didn’t have the courage to accept. The challenge was eventually accepted by Cardano’s student Ferrari. Ferrari did better than Tartaglia in the competition, and Tartaglia is known to have lost both his prestige and income.

So I guess the moral of the story is not to gamble your life away on mathematics. This tale is not over: it has an even more complex ending.

Cardano noticed that Tartaglia’s method sometimes required him to extract the square root of a negative number.

He even included a calculation with these complex numbers in his famous Ars Magna, but he did not really understand it. Bombelli studied it in detail and is hence, often considered to have discovered complex numbers.

Sensing the magnetic field

Initially the system has a degenerate energy space configuration. This is called the zero-field splitting. In this case, the Hamiltonian is simply

(1)H 0=ΔS z 2 H_0 = \Delta S_z^2

With a static external magnetic field applied at angle θ \theta with respect to the NV-symmetry axis and with strength β \beta, one then arrives at the Hamiltonian together with a so called Zeeman splitting.

(2)H=H 0+β(cosθS x+sinθS z) H = H_0 + \beta (\cos \theta S_x + \sin \theta S_z)

When probing the optical properties of the NV-center by driving the system with microwave pulses and detecting optical emissions, one can determine two resonant peaks.

We can consider some relationships of the eigenvalues of the Hamiltonian. The first is that

(3)Tr(H)=2Δ=λ 0+λ 1+λ 2 \text{Tr}(H) = 2 \Delta = \lambda_0 + \lambda_1 + \lambda_2

We know that the first transition occurs at the energy gap between the first excited energy state λ 1 \lambda_1, and the ground state λ 0 \lambda_0. In other words we know that the magic frequency is

(4)f 1=λ 1λ 0 f_1 = \lambda_1 - \lambda_0

The system is initially in the low-energy state corresponding to λ 0 \lambda_0. So the transitions occur with respect to this state. The second transition is then

(5)f 2=λ 2λ 0 f_2 = \lambda_2 - \lambda_0

Combining this information into one equation gives us a value for λ 0 \lambda_0 provided we know the zero-field splitting Δ \Delta and the transition frequencies from the ground state into the first and also second excited states.

(6)λ 0=2Δ(f 1+f 2)3 \lambda_0 = \frac{2 \Delta - (f_1+f_2)}{3}

We then consider the characteristic equation of the Hamiltonian, written in variable λ \lambda

(7)x 32Δx 2+(Δ 2β 2)x+Δβsin 2(θ)=0 x^3 - 2 \Delta x^2 + (\Delta^2 - \beta^2)x + \Delta \beta \sin^2(\theta) = 0

There exists eigenvalues (roots) λ 0\lambda_0, λ 1\lambda_1, λ 2\lambda_2 such that the above polynomial can be expressed as

(xλ 0)(xλ 1)(xλ 2)=x 3x 2Tr(H)+x2(Tr(H) 2Tr(H 2))det(H) (x-\lambda_0)(x-\lambda_1)(x-\lambda_2) = x^3 - x^2 \text{Tr}(H) + \frac{x}{2}(\text{Tr}(H)^2 - \text{Tr}(H^2)) - \text{det}(H)

the coefficients of a monic polynomial are (up to sign) the elementary symmetric polynomials in the roots

The coefficients of indeterminate xx are of interest to us (as opposed to e.g. x 2x^2 or x 0x^0). By the way, the expression of these terms into traces as above, are a special case of the (Newton–Girard identies)[], that’s how old this sort of trick is. If you really wanted to have fun, you might notice that they look just list a dispersion and write them like this …

The point is that by equating coefficients from the equation

(8)12(Tr(H) 2Tr(H 2))=(Δ 2β 2) \frac{1}{2}(\text{Tr}(H)^2 - \text{Tr}(H^2)) = (\Delta^2 - \beta^2)

we arrive at an expression for the magnitude of the magnetic field β\beta in terms of the measured transition frequencies f 1f_1 and f 2f_2 as

(9)β 2=13(f 1 2+f 2 2f 1f 2Δ 2) \beta^2 = \frac{1}{3} (f_1^2 + f_2^2 - f_1f_2 - \Delta^2)

In other words, we can can use this NV to probe the external magnetic field! So it’s really a sensor, as we stated from the get go.


We then shift to the so called, center of mass of the polynomial. This can be thought of as adding a term proportional to the identity to make the Hamiltonian traceless. This amounts to the assignment

(10)κ i:=λ i2Δ3 \kappa_i := \lambda_i - \frac{2 \Delta}{3}

for i=0,1,2 i = 0, 1, 2 and from this we arrive at the depressed cubic

(11)κ 3+aκ+b=0 \kappa^3 + a \kappa + b = 0


(12)a:=(Δ 23+β 2) a := - \left( \frac{\Delta^2}{3} + \beta^2 \right)
(13)b:=Δ(227Δ 2+β 2(sin 2(θ)23)) b := \Delta\left( \frac{2}{27} \Delta^2 + \beta^2 \left( \sin^2(\theta) - \frac{2}{3}\right)\right)

We’ve already solved for λ 0 \lambda_0 in terms of the resonant frequencies f 1 f_1 and f 2 f_2. Changing this into an equation for κ 0 \kappa_0 we arrive at

(14)κ 0=(f 1+f 2) \kappa_0 = -(f_1 + f_2)

which we know is a root of κ 3+aκ+b=0 \kappa^3 + a \kappa + b = 0. In other words, there exists q,r,p q, r, p such that

(15)(κκ 0)(qκ 2+pκ+r)=κ 3+aκ+b=0 (\kappa - \kappa_0)(q \kappa^2 + p \kappa + r) = \kappa^3 + a \kappa + b = 0

The cubic has a neat feature, if the roots are real, they are known to be related to the triangle as follows

Let’s first understand that, and then equate the quantities we just determine with the length of the sides, and the orientation (that is the rotation) of the triangle.

  • The applied static magnetic field is related to the length of the sides
  • The angle of the triangle depends on both the applied field and also the angle of the NV symmetry axis


So there you have it. NV-centers in a nut shell, or at least in a rather long blog article. I certainly find this stuff interesting. As I mentioned at the start, the content here was inspired mainly from

There are a few more related references, including

For those that want to read another quick introduction, with more of an experimental focus, maybe check out

As I said, we’ve been working with Florian Dolde and others. So we have two of our own papers on NV-centers:

The sort of things I hear from experts in NV, about what will happen next, are mainly about scaling up the numbers of interacting NV-centers. Right now, the record is an impressive, but small handful of interacting particles (like 4 or 5), but with new the new methods being developed to create samples, people tell me this will likely double in the next two year. This would certainly open the door up to a lot more interesting applications. For example, say they are able to get 5 NV-centers to interact, this means we would then need to consider 59,049 energy levels. Modelling such a system using existing computing technologies will be challenging. Turning this on its head: such a system offers a computational resource, to at least, model itself.

Scrap yard

Among other things, quantum physics explains how structures, such as crystals, form. In principle, using quantum physics, we could understand many more things than we actually do about the nature of mater.

  • To understand the world around us, all we have to do is solve the quantum equations describing what we see in nature.

The problem is that these equations seem impossible to solve using any computer not governed by quantum physics! The sad fact is that quantum computers which exist today as far to small to do anything practical.

One of the fun aspects of this is that we’re actually going to give a geometric interpretation to some things my friends Florian and Ya are actually measuring in their Stuttgart lab.

category: experiments