Quantum Computing with dartboards

Ishaan Ganti
Mission San Jose Highschool, Fremont, CA
Srinivasan S. Iyengar
Email<EMAIL_ADDRESS>Department of Chemistry, and the Indiana University Quantum Science and Engineering Center (IU-QSEC),
Indiana University, Bloomington, IN-47405

We present a physically appealing and elegant picture for quantum computing using rules constructed for a game of darts. A dartboard is used to represent the state space in quantum mechanics and the act of throwing the dart is shown to have close similarities to the concept of measurement, or collapse of the wavefunction in quantum mechanics. The analogy is constructed in arbitrary dimensional spaces, that is using arbitrary dimensional dartboards, and for for such arbitrary spaces this also provides us a “visual” description of uncertainty. Finally, connections of qubits and quantum computing algorithms is also made opening the possibility to construct analogies between quantum algorithms and coupled dart-throw competitions.

§ INTRODUCTIONS

The promise of solving exponentially complex problems efficiently using quantum technologies[Preskill, 2018, Nielsen and Chuang, 2000, Chia et al., 2020] and the development of the associated software is a rapidly evolving research frontier. While we are in the early stages of this emerging quantum revolution, there is already a diverse set of problems that can benefit from such developments. However, true progress can only be achieved by rigorous study facilitated by the development of a competitive quantum workforce. As a result, research and workforce development in the strongly inter-disciplinary area of quantum information sciences has been noted by the US National Science Foundation (NSF) as one of the “Big Ideas” and recognized through the introduction of the National Quantum Initiative (NQI) from the White House.
Furthermore, the 2019 National Academies Report entitled Quantum Computing: Progress and Prospects [National Academies of Sciences, Engineering, and Medicine, 2019] observes that “[a]dvances in QC theory and devices will require contributions from many fields beyond physics, including mathematics, computer science, materials science, chemistry, and multiple areas of engineering.”

By contrast, regarding the present state of the Quantum Information Science (QIS) workforce, the September 2018 National Science and Technology Council report [National Science and Technology Council, 2018] notes that “America’s current educational system typically focuses on discrete disciplinary tracks, rarely emphasizing cross-disciplinary study that equips graduates for complex modern questions and challenges, prominently including QIS.” Emphasizing this point, Jeremy Hilton,
Quantum Engineering Lead at Google AI,

wrote in Forbes [Hilton, 2019] that “one issue has everyone united: There’s a shortage of quantum computing talent. This shortage has a significant impact on the future of the industry. A trained, well-rounded quantum workforce is key to realizing the full practical value of quantum computing. And yet, pundits describe time and again the difficulty in recruiting talent. There isn’t a direct pipeline from universities, and there’s fierce competition for the limited workforce that is available…To continue expanding the quantum ecosystem, we need to grow the number of quantum-literate experts — now…”

Given these critical challenges it is vital that we develop new paradigms for quantum education that is accessible at multiple levels of pedagogy. Strongly influenced by other similar initiatives[Rudolph, 2017, Economou et al., 2020], here we present an approach where we utilize a game of darts to discover the complexities of quantum mechanics and eventually quantum computing. The first author in this article is a high-school senior who has deeply benefited from this interdisciplinary initiative. In fact, the goal of this effort can be labeled as “Quantum computing without physics." We make the case that it may be possible to introduce quantum theory without much background in physics at all.

This paper is organized as follows. We will begin by discussing the Schrodinger's cat problem along with a two-dimensional game of darts with rules constructed to reproduce the cat-state problem. Then, we will generalize this game of darts to multiple dimensions and eventually to a Fourier space, which yields one visualization of uncertainty. Finally, we  connect the game of darts to measurements performed on a set of qubits states, and thus quantum computing, and analyze a simple quantum circuit by constructing an analogy to a game of darts.

§ FROM CATS TO DARTBOARDS: SETTING THE STAGE

Schrödinger’s cat is a central thought experiment detailing the following scenario: a cat is trapped in a box with a lethal device for a certain duration of time. During this time, the device has some probability of activation. Then, by the end of a fixed duration of time, the box is
opened and the state of the cat (alive or dead) is observed.

Schrödinger’s cat is as famous as it is because according to the principles of quantum mechanics, at the moment just
before the box is opened, the cat is in  a “super-position” state, that is, both dead and alive. And somehow, when the box is opened, the cat “collapses” to one of the two, dead or alive states. Of
course, this sounds extremely odd. Even Schrödinger thought so. Yet, if we consider this scenario as an analogy to the behavior of
subatomic particles, this thought experiment accurately describes and presents the basis for much of contemporary non-relativistic physics, and also provides us with an algorithm to computationally inspect almost all phenomena in chemistry, biochemistry and material science.

To rationalize this idea and to set the stage for more general prescriptions, we may begin with a two-dimensional space where one axis is labeled as “alive”, and the other is labeled as “dead”. Any point in
this two-dimensional space represents the unobserved state of the cat prior to the box being opened. Thus we may say,

\begin{align}
cat = \alpha \cdot (alive) + \beta \cdot (dead)
\end{align}

or  formally,

\begin{align}
\ket{cat} = \alpha \ket{alive} + \beta \ket{dead}
\label{catstate}
\end{align}

This can be equivalently represented in a two-dimensional vector space as

\begin{align}
\begin{pmatrix}
\alpha \\
\beta
\end{pmatrix}
\label{cat-vector}
\end{align}

assuming the first axis is the “alive” axis and the second is the “dead” axis.
The only additional condition on choosing the specific point on the two-dimensional space is that the sum of the squares of the “alive” and
“dead” components, that is, the length of the vector in Eq. (<ref>)
must equal 1. In the Coppenhagen interpretation of quantum mechanics, this implies the probability that the cat is alive or dead is equal to 1.

In other words, the probabilities of the cat being found “alive” or “dead”

when the box is opened
correspond to $\alpha^2$ and $\beta^2$, respectively, and we have

\begin{align}
\alpha^2+\beta^2=1.
\label{norm}
\end{align}

§.§ A Dartboard for Our Cat

Imagine a game of darts for the above scenario. We will define the rules of the game as follows:

Rule 1 The action of throwing the dart is designed to correspond to the action of opening the box in Schrödinger’s scenario.
We thus require that the dart can only land directly on one of the axes that defines our space. While this requirement appears odd at this stage, we will see later, that this corresponds to a normal game of darts when the number of dimensions grow.

Rule 2 When the dart lands on a certain axis, the distance from the origin is noted. This distance, given Eq. (<ref>), represents one measurement of the quantity  $\alpha$ or $\beta$ depending on the axis the dart lands on. Indeed, as per the discussion that follows Eq. (<ref>), the probability that the dart lands on any given axis is proportional to the length $\alpha$ (or $\beta$).

Rule 3 Given Eq. (<ref>), the net probability of the dart landing on either axis is 1.

Rule 4 Whether the dart lands on the horizontal or vertical axes is determined by the magnitude of $\alpha$ and $\beta$. The greater a particular value is, more likely that the dart lands on that specific axis. More specifically, these likelihoods are $\alpha^2$ and $\beta^2$ as noted above.

Without loss of generality, our axes were chosen as the horizontal and vertical axis, and we labeled these as “dead” and “alive”. This is consistent
with the original thought experiment where, when we observe the cat, it must be either dead or alive.
Thus, our dart throw must also provide us with one of the two results.

the result of a single throw of dart is completely random and can result in a dead state or an alive state, with probabilities $\alpha^2$ and $\beta^2$. Thus, with a single dart throw, we don’t know much about the state of the cat prior to the throw.
However, with multiple such dart throws,

the ratio of the number of darts landing
on the “alive” axis to the number of darts landing on the “dead” axis would approach the ratio of the probabilities,
namely, $\alpha^2/\beta^2$.

§.§ Dartboards in Higher Dimensions

Now, what if we wanted to generalize this dartboard game to a situation involving more than two dimensions? First, we'd need more than 2 axes–more than two outcomes that can be thought of as “orthogonal." Thus, like our original “dead” and “alive” outcomes, these new possible set of outcomes cannot overlap with each other. One example of such a scenario could be a game of dice. We know that six outcomes are  likely from a single throw. However, while the die remains “unmeasured," it is some probabilistic jumble of these 6 values. This probabilistic jumble, readers may notice, then becomes a point in a six-dimensional space. Additionally, if we were to create a state equation for this scenario like in Eq. (<ref>), the square of the coefficients associated with each outcome would again equal 1 by rules (3) and (4).

Yet, in most real-world scenarios, we find that we need far more than two or even six dimensions to adequately represent a given situation. This calls for higher dimensional dartboards! We will proceed by understanding how these larger dartboards can be constructed mathematically, and then by demonstrating their significance. We will show that functions are a natural vehicle we may use to achieve this goal.

§.§ Game of darts in $N$-dimensions may be interpreted as a Monte Carlo problem

We may
express a function by enumerating its values at every single point on the number-line as

\begin{align}
\begin{pmatrix}
y_1 \\
y_2 \\
\vdots \\
\end{pmatrix}
\label{n-vector}
\end{align}

where the value, $y_i$, represents the function’s value at $x_i$. This infinitely long vector is not dissimilar to that in Eq. (<ref>).

An alternative imagery of the dartboard problem for Eq. (<ref>). The probability of a dart landing at $x_i$ is proportional to $y_i$.

The rules for the dartboard game in Section <ref> when applied to this vector space, leads to an appealing interpretation.
In our game of darts discussed above, each value along the vector in Eq. (<ref>) corresponds to one specific component that dictates the probability that the dart lands on a given axis. The same set of rules when applied to Eq. (<ref>) yields a game of darts with $n$ orthogonal dimensions where the probability of the dart landing on an axis labeled as $x_i$, is give by $y_i$.
However, since Eq. (<ref>) also represents some function,

we can re-express this multidimensional game of darts as being played on a two-dimensional coordinate system (see Figure <ref>), where

every throw of darts can land anywhere along the horizontal axis, with the probability of a given throw landing on a specific point along the horizontal axis being given by the square of the vertical axis value at that point as shown in Figure <ref>! This leads us to a natural interpretation that appears similar to a Monte Carlo problem.

Can we draw parallels between function operations and vector operations? Some operations, such as the four
basic arithmetic operations seem to be straightforward. For example, if we know

\begin{align}
f(x) =
\begin{pmatrix}
q_1 \\
q_2 \\
\cdots \\
\end{pmatrix}
\end{align}

\begin{align}
g(x) =
\begin{pmatrix}
k_1 \\
k_2 \\
\cdots \\
\end{pmatrix}
\end{align}

\begin{align}
f(x) + g(x) =
\begin{pmatrix}
q_1+k_1 \\
q_2+k_2 \\
\cdots \\
\end{pmatrix}
\end{align}

But what about for more general operations, such as the dot product?

§.§ The Function Dot Product

Algebraically, the dot product of two vectors is the cumulative sum of the products of corresponding components within those two vectors:

\begin{align}
\vec{a}\cdot\vec{b} = \sum_i a_i b_i
\label{adotb}
\end{align}

But what would this look like with functions? Given the earlier vector representations,

the dot product, or inner product, of two functions over some
interval is the sum of all the products of the corresponding function values in that interval. Thus,

\begin{align}
\bra{f}\ket{g} = \int_a^b dx f(x) g(x)
\label{adotb-braket}
\end{align}

where the left side is simply a generalized notation for a dot product.

An extremely important insight is that when the dot product of two functions (or vectors, as they are interchangeable in our context) is 0,
the functions are orthogonal. However it is a subtle matter that vectors that represent the functions in Eq. (<ref>) in fact reside in infinite dimensional spaces. This Eq. (<ref>) is an infinite dimensional “dot” product.

This is useful for defining n-dimensional spaces. For example, given the orthogonal vectors
0 \\ 1
\end{pmatrix}$
1 \\ 0
\end{pmatrix}$, we can define a
two-dimensional space.

§ FUNCTIONS IN FOURIER SPACE LEAD TO COMPLEMENTARY FOURIER DARTBOARDS

Now, consider two real functions: $\cos(kx)$ and $\sin(mx)$, such that $k$ and  $m$ are integers. Since we’ve established above that we can treat these functions as vectors, as in Eq. (<ref>), that is,

\begin{align}
\cos(kx) \equiv
\begin{pmatrix}
\cos(kx_1) \\
\cos(kx_2) \\
\vdots \\
\cos(kx_n)
\end{pmatrix}
\label{cos-vector}
\end{align}

and similarly

\begin{align}
\sin(mx) \equiv
\begin{pmatrix}
\sin(mx_1) \\
\sin(mx_2) \\
\vdots \\
\sin(mx_n)
\end{pmatrix}
\label{sin-vector}
\end{align}

we can the express the dot product of these vectors as a sum of products, that is,

\begin{align}
\int_{-\pi}^{\pi} dx \cos(kx) \sin(mx)
\label{cosdotsin}
\end{align}

which is identically zero. Thus, we may say that the sine and cosine functions are essentially vectors that are orthogonal to each other. A similar statement can be made for any two Cosine or any two Sine functions as well, that is

\begin{align}
\int_{-\pi}^{\pi} dx \cos(kx) \cos(mx) = \delta_{k,m}
\label{cosdotsin}
\end{align}

\begin{align}
\int_{-\pi}^{\pi} dx \sin(kx) \sin(mx) = \delta_{k,m}
\label{cosdotsin}
\end{align}

where $\delta_{k,m}$ is the Kroneckar delta function.

As a result, we may use a family of Sine and Cosine functions, with different values for $k$ and $m$ in Eqs. (<ref>) and (<ref>)
to create a $n$-dimensional vector space where each axis is labelled using Eq. (<ref>) or (<ref>).

For example, $\cos(x)$ and $\cos(2x)$

define a two-dimensional vector space.

If we add a third function, such as $\cos(3x)$, we have a three-dimensional
space since $\cos(3x)$ is orthogonal to both $\cos(x)$ and $\cos(2x)$ and so on.

With $n$ such functions, we have now created an $n$-dimensional space and as we let $n\rightarrow\infty$, we

have an infinite-dimensional vector space.

Thus, in a manner that is complementary to Eq. (<ref>) we may express any function as a linear combination of the vectors that make up our new infinite dimensional basis, which are the
sinusoidal functions in the form $\cos(kx)$ and $\sin(kx)$, where $k \in \mathbb{Z}$:

\begin{align}
f(x) = \sum_{i=0}^\infty \left[ A_i \cos(k_i x) + B_i \sin(k_i x) \right]
\label{n-wave-function}
\end{align}

or to present this in a fashion complementary to Eq. (<ref>):

\begin{align}
\begin{pmatrix}
A_0 \\  A_1 \\
\vdots \\
A_n \\  B_0 \\ B_1 \\
\vdots \\
\end{pmatrix}
\label{n-wave-vector}
\end{align}

which exemplifies the many findings we just made. The coefficients $\left\{ A_n; B_n \right\}$ are evaluated by using the “dot” product:

\begin{align}
A_m = \int_{-\pi}^{\pi} dx f(x) \cos(mx)
\label{cosdotf}
\end{align}

As $f(x)$ is made up of a
certain amount of each wave, and we have shown that these waves are independent of one another (their dot product is 0), taking the dot product of $f(x)$
and one wave component, $\cos(nx)$ or $\sin(nx)$, will yield a numerical value corresponding to the amount of that wave in $f(x)$.

The dartboard problem for Eq. (<ref>). The probability of a dart landing at $k_i$ is proportional to $A_i$. A similar figure may be constructed for the Sine functions.

§.§ A complementary $N$-Dimensional Dartboard in Fourier space

The rules for our game of dart discussed in Section <ref>
when applied to Eq. (<ref>), led to a Monte Carlo interpretation in Section <ref>.
But we may apply the same rules to Eq. (<ref>) as represented by the Figure <ref>.
Based on “Rule 4” above, this leads to the interpretation that
the probability of the dart landing on any given axis of our new basis space (or any given point in the horizontal axis of Figure <ref>) is dictated by the extent to which a specific frequency captured by the sine or cosine wave is present in the function.

Thus, rather than having just two axes (the “dead” and “alive”
axes from the first example), in Section <ref> and here, we now have an infinite number of orthogonal axes. Each axis is labeled here by a wave
—specifically, its frequency. A point in this space, like in the first example, is simply a linear combination of these axes. From the discussion
of Fourier components, we know that every function can be uniquely represented as a point in this space.

For example, given the function

\begin{align}
f(x) = a \cos(x) + b \cos(2x) + c \sin(5x)
\end{align}

the probability of our dart throw landing along the $\cos(x)$ axis, or at a point with $k_i=1$ in Figure <ref>, would be $\frac{a^2}{a^2+b^2+c^2}$. Similarly, the probability of our dart throw landing along the $\cos(x)$ axis, or  at a point with $k_i=2$ in Figure <ref>, would be $\frac{b^2}{a^2+b^2+c^2}$, and so on. Notice how in the cat scenario, the calculation of the probabilities for landing on a specific axis is the same as the calculation above.
The “alive” probability would be $\frac{\alpha^2}{\alpha^2+\beta^2}$, and the “dead” probability would be $\frac{\beta^2}{\alpha^2+\beta^2}$. However, due to the normalization condition, where $\alpha^2+\beta^2=1$, this simplifies to $\alpha^2$ and $\beta^2$ respectively.

§ THE UNCERTAINTY DILEMMA BETWEEN THESE COMPLEMENTARY DARTBOARDS

Let us go back to Figure <ref> and imagine that our function $f(x)$, represented by Eq. (<ref>), is an exact Cosine wave, say $\cos(x)$. In such a situation, Eq. (<ref>) would have exactly one value, $A_1=1$, and all other values in Eq. (<ref>) would be identically zero. Consequently, while Figure <ref> would display a wave with a single frequency, where the dart may land anywhere along the horizontal axis, the complementary form of Figure <ref> would yield a single peak at $k_1=1$, zero everywhere else, thus yielding a probability of "1" that the dart lands at one precise point. Thus, while the outcome of every throw of dart is precisely dictated, with certainty in Figure <ref>, the corresponding outcome of a single throw of darts may land anywhere along the horizontal axis in Figure <ref>, and hence is uncertain!

Of course the description above is general, and does not really only apply to quantum mechanics. But if we were equipped with De Broglie's wave-particle duality, where the momentum of a particle is dictated by the frequency associated to its wave-nature, it is clear from the above discussion that the function, $\cos(x)$ has a precisely defined momentum. However, given that the game of darts in Figure <ref> may land anywhere along the horizontal axis, the corresponding position is infinitely uncertain!
Thus one may say, that using the game of darts, we have picturized in Figure <ref>, a momentum space, and in Figure <ref>, a complementary position space. But our development above did not need the principles of quantum mechanics and are hence expected to be more general. Indeed, it is well-known that a similar time-frequency uncertainty exists in signal processing[Press et al., 1992].

learn a fact about particles.
From the particle-wave duality, we know that a quantum object (which a particle is) possesses properties of both a wave and a particle, including
frequency and momentum. The precise relationship between a particle’s frequency and momentum is

\begin{align}
p = \frac{h}{\lambda}
\end{align}

where $p$ is momentum, is $h$ Planck’s constant, and $\lambda$ is the wavelength (note that wavelength is inversely proportional to frequency). The significance of this equation
stems from the following: a particle’s momentum and frequency are directly proportional.

What if we then Fourier transformed this function? We would find that our graph in the frequency space would have equal sized spikes at
every single frequency, since the Fourier transform of a delta fucntion is simple a contant ($1/2\pi$).

This implies that a corresponding game of darts in the frequency domain would yield every single frequency with equal probability!

Since frequency and momentum are directly proportional, the same goes for
momentum. No matter how many darts we threw at this Fourier space dartboard, we will most certainly get a different result for the momentum of the particle every for every through, yet the position is known with infinite certainty.

In short, certainty of the particle’s position leads to infinite uncertainty of the particle’s momentum and this can be seen from our little game of darts.

What if the opposite occurs? Say we know the exact momentum of the particle. In the frequency/momentum domain, this would be
represented as a delta function at the known momentum. In the frequency/momentum dartboard space, the point representing this situation would
lie on an axis. In turn, every throw of our dart will land on the aforementioned axis, yielding the same momentum no matter what. But what about
the particle’s position? It would just be an endless sine wave with frequency $\frac{pc}{h}$, where is the momentum of the particle. In other words, we will
be infinitely uncertain as to where the particle resides.

§ SCHRODINGER’S CAT AS A QUBIT, DARTS AS MEASUREMENTS

Now, let us move on to the connection to quantum bits. Quantum bits, also known as qubits, are similar to the Schrödinger’s cat dartboard
example we began with earlier. This time, instead of our two states being “dead” and “alive”, we label the states as $\ket{0}$ and $\ket{1}$, which are again defined to be orthogonal to each other as before. If we define  $\ket{0}$ along the horizontal
axis and $\ket{1}$ along the vertical axis, then, our qubit is similar to our cat-state from Section <ref>.
We can
express it in an analogous fashion as in Eq. (<ref>), and

\begin{align}
\ket{\psi} = \alpha \ket{0} + \beta \ket{1}
\label{qubitstate}
\end{align}

where $\ket{\psi}$ represents a qubit state with components $\alpha$ and $\beta$ which dictate
the probabilities of “collapsing” the state to the $\ket{0}$ and $\ket{1}$ directions, respectively, exactly as in the game of darts.

Note how qubits differ from standard bits. Standard bits can only be a 0 or a 1. Qubits, on the other hand, can exist in a superposition as in Eq. (<ref>). But, when measured, equivalent to our earlier dart throws, these qubits collapse to a $\ket{0}$ or $\ket{1}$ state.

Now consider two qubits. Since these can both independently reside on their respective $\ket{0}$ and $\ket{1}$ states, the two-qubit system is a four-dimensional space with axes labeled as,

$\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$. How would we
be able to express this pair of qubits? The key is in recognizing that these four states are, similar to (a) the $\ket{0}$ and $\ket{1}$ states for a single qubit, (b) the basis states for the function in Eq. (<ref>), and (c) the states in Eqs. (<ref>) and (<ref>), that is, the states $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$ are
orthogonal. Four orthogonal states implies a four-dimensional space, and a state,

\begin{align}
\ket{q} = q_{00} \ket{00} + q_{01} \ket{01} + q_{10} \ket{10} + q_{11} \ket{11}
\label{2qubitstate}
\end{align}

is a point in this four dimensional space. Again all our discussions on the dartboards apply to this state.
The square of the coefficients, as usual, corresponds to the probability of the pair of qubits collapsing to one specific dimension in this four-dimensional space.
But, upon adding an extra qubit, we have an expanded game of darts. If we were
to add a third then we have eight different possible basis states and in general we have $2^N$ orthogonal dimensions for $N$-qubits. In this fashion the complexity and the ability to store (and process) information grows exponentially from a family of qubits.

We may also construct an analogy between Eq. (<ref>) and the state corresponding to two cats, that may be independently “dead” or “alive”. Thus,

\begin{align}
\ket{2cats} =& q_{alive_1 alive_2} \ket{alive_1 alive_2} + \nonumber \\ & q_{alive_1 dead_2} \ket{alive_1 dead_2} + \nonumber \\ & q_{dead_1 alive_2} \ket{dead_1 alive_2} + \nonumber \\ & q_{dead_1 dead_2} \ket{dead_1 dead_2}
\label{2qubit-catstate}
\end{align}

where for example the probability of recovering both cats as living is proportional to $q_{alive_1 alive_2}$.

$\ket{0}$     H      1

$\ket{0}$          X
A quantum circuit to create the Bell state in Eq. (<ref>). An initial state, $\ket{0}\otimes\ket{0}$ is propagated through a Hadamard transform of qubit one to yield, $\left\{ \frac{1}{\sqrt{2}}\left[\ket{0}+\ket{1}\right]\otimes\ket{0}\right\}$. Following this, application of a CNOT gate yields the state in Eq. (<ref>).

§ BELL STATES USING “ENTANGLED” DARTS

We now introduce a sophisticated concept in quantum information known as entanglement, which is used in the context of quantum teleportation. We develop a dartboard analogue to this concept. For this we first introduce a Bell state[Nielsen and Chuang, 2000] which is created from a subset of the four vectors in Eq. (<ref>), namely

\begin{align}
\ket{q_B} = \frac{1}{\sqrt{2}} \left[ \ket{00} + \ket{11} \right]
\label{2qubitBellstate}
\end{align}

Thus by comparison with Eq. (<ref>), when cat-1 is recovered to be alive, this is also the case for cat-2 and vice versa. Thus, in some sense, the outcomes of the throw of darts corresponding to the two cats, has not become “entangled”. That is the outcome from the first dart throw, also dictates and influences the second dart.
A quantum circuit that creates such a state is shown in Figure <ref>. How does one create a picture of this  using our dartboard analogy?

Let us imagine two darts, that as we will discover, behave in a very peculiar fashion. The two wires in Figure <ref> represent the two different dartboards where these darts are supposed to land. We will also assumes, as done in Eq. (<ref>), that there exists two cats, whose lives are influenced by the outcome from these two darts and their respective boards.
Dartboard-1, the top wire in Figure <ref>, is pre-processed through a transformation given by the Hadamard transform $H$ in Figure <ref>. This Hadamard transform simply rotates the state that dictates the dart landing probabilities for Dartboard-1 by 45 degrees. So our Dartboard-1 will now present dart-1 with equal probabilities of landing along the horizontal and vertical axis. Thus, in a sense, going back to our cat states in Eq. (<ref>), the Hadamard transform has created a state that looks like, $\frac{1}{\sqrt{2}} \left[\ket{alive} \pm \ket{dead} \right]$ and notice then by comparison with Eq. (<ref>), that both $\alpha^2$ and $\beta^2$ are equal. Thus after the Hadamard transform, Dart-1 is free to land on either axis, with no bias.

Let us now presume that Dart-2 begins, with a dartboard where the dart throws are dictated by the probability, using the cat-state analogy, $\alpha=0$ and $\beta=1$. Thus we know with certainty that the state that dictates Dart-2 is, for example, the $\ket{0}$ state, or the cat represented here is, for example $\ket{alive}$ with probability 1. Thus without influence from Dart-1, Dart-2 will always provide an outcome where the cat is “alive”.

Now something very strange happens as a result of the operation

in Figure <ref>. The two darts are not independent anymore. When Dart-1 lands on the state $\ket{0}$, that is finds that the corresponding cat is “alive”, Dart-2, dictated by the probabilities that control Dartboard-2 discussed in the previous paragraph, recovers an an $\ket{alive}$ or $\ket{0}$ too. But, when Dart-1 lands on the state $\ket{1}$ that is finds that the corresponding cat is “dead”, Dart-2, promptly flips the probabilities in the state that dictates Dartboard-2 and also recovers a $\ket{dead}$ or $\ket{1}$ state. Thus the behavior or Dart-2 is not independent, from that of Dart-1, and this property is called entanglement.

This bizarre result is one of the major hallmarks of quantum mechanics and is thought to be central to a presumed advantage that a computer created from the principles of quantum mechanics may have over those that we currently use.

§ CONCLUSION

We develop an approach to discuss the fundamental underpinnings of quantum mechanics and quantum computing using the concept of dart throws onto the two axis of a two-dimensional space. The concept of qubits and measurement is related to these dart throws and the process, when generalized, yields visual descriptions of uncertainty and Fourier transforms. We follow the prescriptions of this game and develop a way to relate these concepts to fundamental notions in quantum information storage and processing, including analogues to quantum circuits and Bell states.

This relationship of uncertainty between a particle’s momentum and its position is known as the Heisenberg Uncertainty Principle. The
formal definition of the principle is as follows: $\Delta x \Delta p \geq \frac{h}{4 \pi}$. Hopefully, readers now generally understand why this inverse relationship exists
between the uncertainty of position and momentum (or, more generally, frequency). However, the Heisenberg Uncertainty Principle is just in the
case of subatomic particles. Dennis Gabor made the connection that the uncertainty principle existing in quantum systems also
existed in signal processing. This is due to the previously mentioned wave particle duality, which essentially makes signals and subatomic
particles analogous.
The purpose of this paper was to demonstrate how the underlying reason for this uncertainty is really due to the general mathematical
uncertainty principle. As we’ve seen, it shows up many fields in different forms, and it is often at the heart of the field which it belongs to. This is
the case in quantum mechanics. Hopefully, this paper has left you all with a glimpse of the workings of uncertainty, and maybe has you left you
pondering as to how it can be treated as a tool rather than as a limitation.

§ ACKNOWLEDGMENTS
This research was supported by the National Science
Foundation grants OMA-1936353 and CHE-2102610 to SSI.

§ A BRIEF ASIDE ON FOURIER TRANSFORMS,  IMAGES AND JPG FILES AS CONNECTED TO THE DARTBOARDS IN SECTION <REF>

The Fourier transform takes in a function, $g(x)$, as input and produces another function, $\hat{g}(\omega)$ that
represents $g(x)$ in the frequency domain, represented by $\omega$. What does this mean? Well, when we take the dot product of our function, $g(x)$ with sine and cosine
waves of differing frequencies to find out the coefficients of our Fourier series, we are essentially figuring out how significant each frequency is
in constructing $g(x)$. If the dot product of $g(x)$ and $\cos(6\pi x)$ is greater the dot product of $g(x)$ and $\cos(8\pi x)$, then 3 Hz is a more important
frequency in $g(x)$ than 4 Hz. The Fourier transform does this for every frequency present in the function over a specified interval, revealing the
most and least important frequencies. The Fourier transform of a function is

\begin{align}
\hat{g}(\omega) = \int dx g(x) \exp{\imath \omega x}
\end{align}

where generally $\omega=2\pi f$.
One extremely important usage of the Fourier transform is in image processing. An image really is just a rectangular grid, where each square
in the grid is a pixel, and where each pixel has a color. So, we’ll treat an image as a sort of coordinate plane. Then, we can split the image up into
columns, such that each column is one pixel wide. By doing this, we essentially split the larger image into narrow sub-images that only have one
dimension. This means that we can express them as functions of one variable. The input for each sub-image function will be the position on the
sub-image, and the output will be the color of the pixel at that position. Now if we convert these sub-image functions to the frequency space, we’ll be able to see the important frequencies in the image. Because
visible light is an electromagnetic wave, and differing frequencies within the range of visible light correspond to different colors, we will then be
able see what the important colors in the image are. This connection is crucial.

§ THE GEOMETRIC DOT PRODUCT

§ EVALUATING THE COSINE DOT PRODUCT

§ FINDING THE FOURIER SERIES OF A KNOWN FUNCTION

§ FOURIER TRANSFORM ON A CLASSICAL WAVE (SOUNDWAVES)

[Preskill, 2018]
Preskill, J. Quantum Computing in the NISQ era and beyond. Quantum
2018, 2, 79
[Nielsen and Chuang, 2000]
Nielsen, M. A.; Chuang, I. L. Quantum computation and quantum information;
Cambridge University Press: New York, NY, USA, 2000
[Chia et al., 2020]
Chia, N.-H.; Chung, K.-M.; Lai, C.-Y. On the Need for Large Quantum Depth.
Proceedings of the 52nd Annual ACM SIGACT Symposium on Theory of Computing.
STOC 2020; New York, NY, USA, Association for Computing Machinery: New York,
NY, USA, 2020; p 902–915
[National Academies of Sciences, Engineering, and Medicine, 2019]
National Academies of Sciences, Engineering, and Medicine, Quantum Computing:
Progress and Prospects. The National Academies Press 2019,
[National Science and Technology Council, 2018]
National Science and Technology Council, National Strategic Overview For
Quantum Information Science. subcommittee on quantum information science
[Hilton, 2019]
Hilton, J. Building the Quantum Workforce of the Future. Forbes 2019,

[Rudolph, 2017]
Rudolph, T. Q is for Quantum; Terrence Rudolph, 2017
[Economou et al., 2020]
Economou, S. E.; Rudolph, T.; Barnes, E. Teaching quantum information science
to high-school and early undergraduate students. 2020;
[Press et al., 1992]
Press, W. H.; Teukolsky, S. A.; Vetterling, W. T.; Flannery, B. P. Numerical
Recipes in C; Cambridge University Press, New York, 1992


# Phenomenological models of baryon

Do Quoc Tuan
Dept. of Theoretical Physics, College of Science, VNU, Hanoi, Vietnam

Abstract

In this paper, I present almost my works performed during my time at VNU. I
was interested in the composite Skyrme model proposed by H. Y. Cheung and F.
Gursey. The expanding for this model based on results obtained from the
original Skyrme model given by T. H. R. Skyrme in 1961 was general and
interesting.

Acknowledgments

I want to show deep gratitude to my supervisor Dr. Pham Thuc Tuyen (Department
of Theoretical Physics, College of Science (another name is Hanoi University
of Science), Vietnam National University) because of everything that he has
been teaching and supporting me during a student’s time! I also thank
Department of Theoretical Physics and Faculty of Physics for giving me the
best conditions to completing the thesis! Finally, and most of all, I want to
give my thesis to my parent, my beloved as a special gift for the supports
over a long period!

Ha noi, 6/2007

Do Quoc Tuan111email<EMAIL_ADDRESS>
###### Contents

  1. 1 Why is the Skyrme model
    1. 1.1 The weak point of QCD
    2. 1.2 The idea of Skyrme
  2. 2 The massless composite Skyrme model
    1. 2.1 The formalism
    2. 2.2 Calculations for static quantities of baryon
  3. 3 The composite Skyrme model with pion’s mass
    1. 3.1 The term of pion’mass
    2. 3.2 The field equation and numerical results
  4. 4 The supersymmetric composite Skyrme model
  5. 5 Appendix
    1. 5.1 Some notations for Chapter 2 and 3
    2. 5.2 Some notations for Chapter 4

##  Chapter 1 Why is the Skyrme model

###  1.1 The weak point of QCD

In physically, we can divide the nature in two types of particles: matter
particles and field particles. Matter particles interact together by fields
concluding transfer interaction particles. There are four fields corresponding
four interactions (four forces) and their particles. The graviton is a
transfer interaction particle of gravitational force, the photon is a transfer
interaction particle of electromagnetic force, $W^{\pm}$ and $Z^{o}$ are
transfer interaction particle of weak force, the gluon is transfer interaction
particle of strong force. It is known well, quarks are interact together by
the strong force. In other word, the interaction of quarks is the exchange of
gluons. And the theory that describes quarks and their interaction is $QCD$
(Quantum Chromo Dynamics). In mathematically, Gell-Mann and Neuman proposed
that quarks make up a multiple which represents the gauge $SU(3)$ being the
group of $3\times 3$ unitary matrices with determinant 1. Gell-Mann firstly
gives generators of $SU(3)$, therefore, they are called Gell-Mann matrices
$\lambda$ . They satisfy the Lie algebra

$\left[{\lambda_{i},\lambda_{j}}\right]=2if_{ijk}\lambda_{k},$ (1.1)

where $f_{ijk}$ is the structure constant. It is totally antisymmetric

$f_{ijk}=-f_{jik}.$ (1.2)

To describe the interaction between quarks, we need using the gluon field
$A_{\mu}^{\alpha}$ . And now, the Lagrangian density is the Yang-Mills
Lagrangian density

${\cal
L}=-\frac{1}{4}F_{\mu\nu}^{\alpha}F^{\mu\nu\alpha}+i\bar{\psi}\gamma^{\mu}(\partial_{\mu}+igA_{\mu})\psi+M\bar{\psi}\psi,$
(1.3) $A_{\mu}=A_{\mu}^{\alpha}\lambda^{\alpha},$ (1.4)
$F_{\mu\nu}^{\alpha}=\partial_{\mu}A_{\nu}^{\alpha}-\partial_{\nu}A_{\mu}^{\alpha}-gf^{\alpha\beta\gamma}A_{\mu}^{\beta}A_{\nu}^{\gamma},$
(1.5)

where g is a constant of interaction. At the high energy scale ($>TeV$), $QCD$
is very good to use. But, when the energy lower (in $GeV$), it meets with some
problems in which describe quantities of quarks and gluons. The main cause is
the high value of g, it obstructs an expansion of interaction terms of hadron.
To repair this problem, t’Hooft proposed that $1/N_{c}$ plays as g($N_{c}$ is
color number in a gauge theory). And, he gave that in large-$N_{c}$ limit,
$QCD$ becomes equivalent to an effective field theory of meson.

###  1.2 The idea of Skyrme

In 1961, Skyrme gave the new model in which baryons may be obtained from the
theory of meson [1]. He began with the Lagrangian density of pion

${\cal
L}=\frac{1}{2}\partial_{\mu}\vec{\pi}\partial^{\mu}\vec{\pi}-\frac{{m^{2}}}{2}\vec{\pi}\vec{\pi},$
(1.6)

in which $\pi$’s make up a three dimensional isotopic space. After, he
expanded the isotopic $SU\left(2\right)$ becomes a new group with the algebra

$\left[{V_{i},V_{j}}\right]=2i\varepsilon_{ijk}V_{k},$ (1.7)
$\left[{A_{i},A_{j}}\right]=2i\varepsilon_{ijk}V_{k},$ (1.8)
$\left[{V_{i},A_{j}}\right]=2i\varepsilon_{ijk}A_{k},$ (1.9)

where generators $V_{i}$ are vector charges, generators $A_{i}$ are axial
charges. And now, an isotopic space have four components
$\left\\{{\vec{\pi}}\right\\}\to\left\\{{\vec{\varphi},\varphi_{4}}\right\\}$
transforming in a rule

$\left[{V_{i},\varphi_{j}}\right]=2i\varepsilon_{ijk}\varphi_{k},$ (1.10)
$\left[{V_{i},\varphi_{4}}\right]=0,$ (1.11)
$\left[{A_{i},\varphi_{j}}\right]=-2i\varepsilon_{ij}\varphi_{4},$ (1.12)
$\left[{A_{i},\varphi_{4}}\right]=2i\varphi_{i}.$ (1.13)

Particles described by $\varphi_{i}\left({i=1,2,3}\right)$ are pseudo- scalars
and making isotopic vectors. A particle described by $\varphi_{4}$ has a
positive parity, a spin and an isotopic spin equals zero, and it is a vacuum.
If replace $V_{i}$ and $A_{i}$ by $L_{i}$ and $R_{i}$ defined

$L_{i}=\frac{1}{2}\left({V_{i}-A_{i}}\right),$ (1.14)
$R_{i}=\frac{1}{2}\left({V_{i}+A_{i}}\right),$ (1.15)

then obtaining a new algebra

$\left[{L_{i},L_{j}}\right]=\left[{\frac{1}{2}\left({V_{i}-A_{i}}\right),\frac{1}{2}\left({V_{j}-A_{j}}\right)}\right]$
,

$=\frac{1}{4}\left[{V_{i},V_{j}}\right]-\frac{1}{4}\left[{V_{i},A_{j}}\right]-\frac{1}{4}\left[{A_{i},V_{j}}\right]+\frac{1}{4}\left[{A_{i},A_{j}}\right]$
,

$=\frac{1}{2}i\varepsilon_{ijk}V_{k}-\frac{1}{2}i\varepsilon_{ijk}A_{k}+\frac{1}{2}i\varepsilon_{jik}A_{k}+\frac{1}{2}i\varepsilon_{ijk}V_{k}$
,

$=i\varepsilon_{ijk}V_{k}-i\varepsilon_{jik}A_{k}$ ,

$\bf{\Rightarrow\left[{L_{i},L_{j}}\right]=2i\varepsilon_{ijk}L_{k}}$ (1.16)

and

$\left[{R_{i},R_{j}}\right]=\left[{\frac{1}{2}\left({V_{i}+A_{i}}\right),\frac{1}{2}\left({V_{j}+A_{j}}\right)}\right]$
,

$=\frac{1}{4}\left[{V_{i},V_{j}}\right]+\frac{1}{4}\left[{V_{i},A_{j}}\right]+\frac{1}{4}\left[{A_{i},V_{j}}\right]+\frac{1}{4}\left[{A_{i},A_{j}}\right]$
,

$=\frac{1}{2}i\varepsilon_{ijk}V_{k}+\frac{1}{2}i\varepsilon_{ijk}A_{k}-\frac{1}{2}i\varepsilon_{jik}A_{k}+\frac{1}{2}i\varepsilon_{ijk}V_{k}$
,

$=i\varepsilon_{ijk}V_{k}+i\varepsilon_{jik}A_{k}$ ,

$\bf{\Rightarrow\left[{R_{i},R_{j}}\right]=2i\varepsilon_{ijk}R_{k}}$ (1.17)

and

$\left[{L_{i},R_{j}}\right]=\left[{\frac{1}{2}\left({V_{i}-A_{i}}\right),\frac{1}{2}\left({V_{j}+A_{j}}\right)}\right]$
,

$=\frac{1}{4}\left[{V_{i},V_{j}}\right]+\frac{1}{4}\left[{V_{i},A_{j}}\right]-\frac{1}{4}\left[{A_{i},V_{j}}\right]-\frac{1}{4}\left[{A_{i},A_{j}}\right]$
,

$=\frac{1}{2}i\varepsilon_{ijk}V_{k}+\frac{1}{2}i\varepsilon_{ijk}A_{k}+\frac{1}{2}i\varepsilon_{jik}A_{k}-\frac{1}{2}i\varepsilon_{ijk}V_{k}$
,

$\bf{\Rightarrow\left[{L_{i},R_{j}}\right]=0}.$ (1.18)

This is an algebra of the chiral group $SU\left(2\right)\times
SU\left(2\right)$. Thus, the expanding group is the chiral group
$SU\left(2\right)\times SU\left(2\right)$. Continuously, he proposed that
$\left\\{{\vec{\varphi},\varphi_{4}}\right\\}$ form a three dimensional sphere

$\vec{\varphi}^{2}+\varphi_{4}^{2}=1.$ (1.19)

Recently, $\left\\{{\vec{\varphi},\varphi_{4}}\right\\}$ are associated with
$2\times 2$ unitary $U\left({x,t}\right)$ matrix

$U\left({x,t}\right)=\exp i\vec{\tau}\vec{\pi}\left({x,t}\right)$ ,

$=\exp i\vec{\tau}\frac{{\vec{\pi}}}{\pi}\pi$ ,

$=\cos\pi+i\vec{\tau}\frac{{\vec{\pi}}}{\pi}\sin\pi,$ (1.20)

where $\pi=\sqrt{\vec{\pi}^{2}}$, $\tau$’s are Pauli matrices. Pion fields now
are associated with $\vec{\pi}$ fields

$\varphi_{4}\Leftrightarrow\cos\pi,$ (1.21)
$\vec{\varphi}\Leftrightarrow\frac{{\vec{\pi}}}{\pi}\sin\pi.$ (1.22)

Really

$\vec{\varphi}^{2}+\varphi_{4}^{2}=\frac{{\vec{\pi}^{2}}}{{\pi^{2}}}\sin^{2}\pi+\cos^{2}\pi=1.$
(1.23)

The Lagrangian density proposed by Skyrme [1] is

${\cal
L}=\frac{{F_{\pi}^{2}}}{{16}}Tr\left({\partial_{\mu}U\partial_{\mu}U^{\dagger}}\right)+\frac{1}{{32e^{2}}}Tr\left[{\left({\partial_{\mu}U}\right)U^{\dagger},\left({\partial_{\nu}U}\right)U^{\dagger}}\right]^{2},$
(1.24)

here $U$ is an $SU\left(2\right)$ matrix transforming as $U\to AUB^{-1}$ under
an chiral $SU\left(2\right)\times SU\left(2\right)$. He proved that the model
has a soliton solution (called hedgehog111firstly called by Polyakov, see at
L. Ryder, Quantum Field Theory (second edition), Cambridge University Press
solution)

$U_{o}=\exp\left({i\vec{\tau}\frac{{\vec{r}}}{r}F\left(r\right)}\right),$
(1.25)

where $F\left(r\right)$ is the chiral angle, baryons are obtained from the
quantization of soliton solution(see chapter 2).

It is well known that in the large-$N$ limit, $QCD$ may be regarded as the
theory of effective meson fields. E. Witten argued that baryons may be
regarded as solitons of this effective meson theory without an further
reference to their quark content [2]. We know that solitons have not a spin.
But, by quantizing around solitons, we obtain the special topological
construct of these solitons which makes a spin. On the other hand, soliton
solutions have a limited spatial distribution, therefore, they are correlative
to particles that have a limited spatial dimension [2,15].

Figure 1.1: The configuration of hedgehog, arrows on it are isotopic vectors
of $\vec{c}$

##  Chapter 2 The massless composite Skyrme model

###  2.1 The formalism

In 1983, G. S. Adkins, C. R. Nappi and E. Witten used the Skyrme’s idea to
computed the static properties of baryons on the chiral group
$SU\left(2\right)\times SU\left(2\right)$ and found that the model gives
predictions that agree well with experiments [2]. However, this model has a
problem in which the situation of particles at low energy are not interested
but it regards that particles in free states, and this point can reduce the
exactitude of theoretical datum. In 1989, this problem is treated by P. T.
Tuyen and N. A. Viet [3]. They considered that at low energy particles are not
free but they would shrink together into the cluster. Therefore, the group
that describes them is $SU\left(2\right)\times
SU\left(2\right)/SU\left(2\right)_{diag}$. This model gives few datum better
than the sigma model. In 1990, H. Y. Cheung, F. Gursey [4] gave the general
model in which the sigma model and the V-T model are two special cases
corresponding with $n=1$ and $n=2$. They introduced a composite _SU(2)_
configuration $V_{n}=UPUP...PU$, where _U(x,t)_ is the _SU(2)_ matrix and
$P=p_{o}+i\vec{\tau}.\vec{p}$ is the constant _SU(2)_ matrix, and they
proposed the composite Skyrme lagrangian density

${\cal
L}_{n}=-\frac{{F_{\pi}^{2}}}{{16n^{2}}}Tr\left({\partial_{\mu}V^{-n}\partial^{\mu}V^{n}}\right)+\frac{1}{{32e^{2}n^{4}}}Tr\left({\left[{V^{-n}\partial_{\mu}V^{n},V^{-n}\partial_{\nu}V^{n}}\right]^{2}}\right),$
(2.1)

with $F_{\pi}$ is the pion decay constant, _e_ is a dimensionless parameter.

Under the chiral group $SU\left(2\right)\times SU\left(2\right)$, _U_
transforms as $U\to AUB^{\dagger}$, _P_ transforms as $P\to BPA^{\dagger}$,
and so $V_{n}$ transforms as $V_{n}\to AV_{n}B^{\dagger}$. In the limit $P\to
I$, $V_{n}\to U_{n}$, the lagrangian (2.1) becomes approximated by the
lagrangian density

${\cal
L}_{n}=-\frac{{F_{\pi}^{2}}}{{16n^{2}}}Tr\left({\partial_{\mu}U^{-n}\partial^{\mu}U^{n}}\right)+\frac{1}{{32e^{2}n^{4}}}Tr\left({\left[{U^{-n}\partial_{\mu}U^{n},U^{-n}\partial_{\nu}U^{n}}\right]^{2}}\right).$
(2.2)

Now, using the hedgehog soliton solution gave by Skyrme
$U_{0}\left(r\right)=\exp\left[{i\tau.\hat{r}F\left(r\right)}\right]$, where
$\tau$’s are Pauli’s matrices and $\hat{r}=\vec{r}/r$, the lagrangian density
(2.2) is

${\cal
L}_{n}=\frac{{e^{2}F_{\pi}^{4}}}{8}\left[{\left({\frac{{dF}}{{d\tilde{r}}}}\right)^{2}+2\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right]+\frac{{e^{2}F_{\pi}^{4}\sin^{2}nF}}{{2n^{2}\tilde{r}^{2}}}\left[{2\left({\frac{{dF}}{{d\tilde{r}}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right],$
(2.3)

where $\tilde{r}=eF_{\pi}r$ is the dimensionless variable. And the static
energy of hedgehog is

${\cal E}_{n}=\frac{{F_{\pi}}}{e}A_{n}=M_{n},$ (2.4)
$A_{n}=\int\limits_{0}^{\infty}{4\pi\tilde{r}^{2}\left\\{{\frac{1}{8}\left[{\left({\frac{{dF}}{{d\tilde{r}}}}\right)^{2}+2\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right]+\frac{{e^{2}F_{\pi}^{4}\sin^{2}nF}}{{2n^{2}\tilde{r}^{2}}}\left[{2\left({\frac{{dF}}{{d\tilde{r}}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right]}\right\\}}d\tilde{r}.$
(2.5)

We obtain the non-linear differential equation of _F(r)_ from minimum
condition of the hedgehog’s energy

$\delta_{F}{\cal E}_{n}=0,$ (2.6)

or

$\frac{d}{{dr}}\frac{{\delta{\cal E}_{n}}}{{\delta
F^{\prime}}}-\frac{{\delta{\cal E}_{n}}}{{\delta F}}=0,(Euler-Lagrange-
equation)$ (2.7)
$\left({\frac{{\tilde{r}^{2}}}{4}+\frac{2}{{n^{2}}}\sin^{2}nF}\right)F^{\prime\prime}+\frac{{\tilde{r}}}{2}F^{\prime}+\frac{1}{n}\sin
2nF\left({F^{\prime
2}-\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}-\frac{1}{4}}\right)=0.$ (2.8)

To solve this equation we need boundary conditions for the _F(r)_. The first
condition of _F(r)_ that relates to the limit of hedgehog’s energy when
$r\to\infty$ is $F\left(\infty\right)=0$. The second condition relating to the
exist of baryonic number B is $F\left(0\right)=\pi$. This equation can not
give us the analytic solution. So we will solve it by the numerical method.

Figure 2.1: The numerical solution of (2.8) for cases of n=2 and n=4 [22, 25].

Now we will quantize the soliton by collective coordinates
$A\left(t\right)=a_{o}\left(t\right)+i\vec{a}\left(t\right)\vec{\tau}$ with
$a_{\mu}\left(t\right)$ $\left({\mu=0,1,2,3}\right)$ are collective
coordinates, $A\left(t\right)$ is $SU(2)$ matrix. From properties of $SU(2)$
matrix, we have

$A\left(t\right)A^{\dagger}\left(t\right)=1$ ,

$\left\\{{a_{o}\left(t\right)+i\vec{a}\left(t\right)\vec{\tau}}\right\\}\left\\{{a_{o}\left(t\right)-i\vec{a}\left(t\right)\vec{\tau}}\right\\}=1$
,

$\bf{a_{\mu}a_{\mu}=1}.$ (2.9)

By time derivative, resulting

$\bf{a_{\mu}\dot{a}_{\mu}=0}$ (2.10)

By this way, solitons to be particles. Different quantization take different
particles. Thus, it takes us to the new idea about the origin of particles in
nature in which they are made by a basic particle (called soliton). With the
_A(t)_ matrix, the $U^{n}\left({x,t}\right)$ will transform as

$U^{n}\left({x,t}\right)=A\left(t\right)U_{0}^{n}\left(x\right)A^{-1}\left(t\right)$

, and we put it into (2.2), the lagrangian now is

$L_{n}=-{\cal
E}_{n}+\Gamma_{n}Tr\left[{\partial_{0}A\left(t\right)\partial_{0}A^{-1}\left(t\right)}\right]$
,

$=-{\cal E}_{n}+2\Gamma_{n}\sum\limits_{\mu=0}^{3}{\dot{a}_{\mu}^{2}},$ (2.11)

with

$\Gamma_{n}=\frac{1}{{e^{3}F_{\pi}}}\Phi_{n},$ (2.12)
$\Phi_{n}=\frac{{2\pi}}{{3n^{2}}}\int\limits_{0}^{\infty}{\tilde{r}^{2}\sin^{2}nF\left\\{{1+4\left[{\left({\frac{{dF}}{{d\tilde{r}}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right]}\right\\}}d\tilde{r}.$
(2.13)

The conjugate momenta are obtained from the derivative of lagrangian following
$\dot{a}_{\mu}(t)$

$\pi_{\mu}=\frac{{\partial L_{n}}}{{\partial\dot{a}_{\mu}}}$ ,

$=\frac{{\partial\left[{-M_{n}+2\Gamma_{n}\sum\limits_{\mu=0}^{3}{\dot{a}_{\mu}^{2}}}\right]}}{{\partial\dot{a}_{\mu}}}$
,

$=4\Gamma_{n}\dot{a}_{\mu}.$ (2.14)

We go to the hamiltonian

$H_{n}=\pi_{\mu}\dot{a}_{\mu}-L_{n}$ ,

$=4\Gamma_{n}\dot{a}_{\mu}\dot{a}_{\mu}-L_{n}$ ,

$=M_{n}+2\Gamma_{n}\dot{a}_{\mu}\dot{a}_{\mu}$ ,

$=M_{n}+\frac{{\left({\sum\limits_{\mu=0}^{3}{\pi_{\mu}^{2}}}\right)}}{{8\Gamma_{n}}}.$
(2.15)

Canonical quantizing momenta $\pi_{\mu}=-i\frac{\partial}{{\partial a_{\mu}}}$
let the hamiltonian as

$H_{n}=M_{n}+\frac{1}{{8\Gamma_{n}}}\sum\limits_{\mu=0}^{3}{\left({\frac{{-\partial^{2}}}{{\partial
a_{\mu}^{2}}}}\right)}.$ (2.16)

Now, we will define the isotopic vector $\vec{c}$ following the way

$\dot{A}^{\dagger}A=\left({\dot{a}_{0}-i\dot{a}_{i}\tau_{i}}\right)\left({a_{0}+ia_{i}\tau_{i}}\right)$
,

$=\dot{a}_{0}a_{0}+\dot{\vec{a}}\vec{a}-i\left({a_{0}\dot{\vec{a}}-\dot{a}_{0}\vec{a}+\vec{a}\times\dot{\vec{a}}}\right)\vec{\tau}$
,

note (2.10), we have

$\dot{A}^{\dagger}A=-i\vec{c}\vec{\tau},$ (2.17)

with

$\vec{c}=\left({a_{0}\dot{\vec{a}}-\dot{a}_{0}\vec{a}+\vec{a}\times\dot{\vec{a}}}\right).$
(2.18)

The square isotopic vector $\vec{c}$ is

$\vec{c}^{2}=\dot{a}_{0}^{2}\vec{a}^{2}+\dot{\vec{a}}^{2}a_{0}^{2}+\vec{a}^{2}\dot{\vec{a}}^{2}-(\vec{a}\dot{\vec{a}})^{2}-2\dot{a}_{0}a_{0}\dot{\vec{a}}\vec{a}$
,

$=\dot{a}_{0}^{2}\vec{a}^{2}+\dot{\vec{a}}^{2}a_{0}^{2}+\vec{a}^{2}\dot{\vec{a}}^{2}-(\vec{a}\dot{\vec{a}})^{2}+2(\vec{a}\dot{\vec{a}})^{2}$
,

$=\dot{a}_{\mu}\dot{a}_{\mu}.$ (2.19)

In the configuration of hedgehog, since the isotopic vector and the radius
vector have the same direction, rotation in the co-ordinate space can make an
isospin of particle, and we assume that a spin equals to an isospin. The
canonical momenta that correspond with collective coordinates are isospins. We
can homogenize $\vec{c}^{2}$ with the square of a spin and an isospin

$\vec{c}^{2}=\frac{1}{{16\Gamma_{n}^{2}}}\vec{J}^{2}=\frac{1}{{16\Gamma_{n}^{2}}}\vec{T}^{2}.$
(2.20)

Thus, we have

$H_{n}=M_{n}+\frac{1}{{8\Gamma_{n}}}\vec{J}^{2}=M_{n}+\frac{1}{{8\Gamma_{n}}}\vec{T}^{2},$
(2.21)

with $\vec{J}$ is a spin and $\vec{T}$ is an isospin. Eigenvalues of
hamiltonian (2.21) are

$E_{n}=M_{n}+\frac{{l\left({l+2}\right)}}{{8\Gamma_{n}}},$ (2.22)

with _l=2J_ and _J_ is the quantum number of spin. So, the nucleon
$\left({J=1/2}\right)$ and the delta $\left({J=3/2}\right)$ masses are given
by

$M_{N}=M_{n}+\frac{3}{{8\Gamma_{n}}},$ (2.23)
$M_{\Delta}=M_{n}+\frac{{15}}{{8\Gamma_{n}}}.$ (2.24)

Note (2.12), (2.13), (2.23), (2.24) we have

$F_{\pi}=\sqrt[4]{{\frac{{\left({5M_{N}-M_{\Delta}}\right)^{3}\left({M_{\Delta}-M_{N}}\right)\Phi_{n}}}{{384A_{n}^{3}}}}},$
(2.25) $e=\frac{{4A_{n}F_{\pi}}}{{\left({5M_{N}-M_{\Delta}}\right)}}.$ (2.26)

###  2.2 Calculations for static quantities of baryon

After constructing the formalism of massless composite model, we apply to the
baryon (nucleon). Since this model is constructed depend on some ideas of
theory of topology then we can start at the topological current to
calculations for static quantities of baryon [2, 3, 4, 15]. The topological
current (a baryonic current) is defined

$B^{\mu}=\frac{{\varepsilon^{\mu\nu\alpha\beta}}}{{24\pi^{2}n}}Tr\left[{\left({U^{{\dagger}n}\partial_{\nu}U^{n}}\right)\left({U^{{\dagger}n}\partial_{\alpha}U^{n}}\right)\left({U^{{\dagger}n}\partial_{\beta}U^{n}}\right)}\right],$
(2.27)

with $\varepsilon_{0123}=-\varepsilon^{0123}=1$ (totally antisymmetry). If we
put
$U^{n}\left({x,t}\right)=A\left(t\right)U_{0}^{n}\left(x\right)A^{-1}\left(t\right)$
into (2.27), we will have

$B^{0}=-\frac{1}{{2\pi^{2}}}\frac{{\sin^{2}nF}}{{r^{2}}}F^{\prime},$ (2.28)
$B^{i}=\frac{{i\varepsilon^{{\rm{ijk}}}}}{{2\pi^{2}}}\frac{{\sin^{2}nF}}{{r^{2}}}F^{\prime}x_{k}Tr\left[{\dot{A}^{\dagger}A\tau_{j}}\right].$
(2.29)

Note (2.17) and (2.20), then (2.29) is now

$B^{i}=\frac{{\varepsilon^{{\rm{ijk}}}}}{{2\pi^{2}}}\frac{{\sin^{2}nF}}{{r^{2}}}F^{\prime}x_{k}\frac{{J_{j}}}{{2\Gamma_{n}}}.$
(2.30)

The baryonic number is

$B=\int\limits_{0}^{\infty}{4\pi r^{2}}B^{0}\left(r\right)dr=1.$ (2.31)

The iso-scalar electric square radius is

$\left\langle{r^{2}}\right\rangle_{E,I=0}=\int\limits_{0}^{\infty}{4\pi
r^{2}}B^{0}\left(r\right)r^{2}dr.$ (2.32)

The iso-scalar magnetic square radius is

$\left\langle{r^{2}}\right\rangle_{M,I=0}=\frac{3}{5}\frac{{\int\limits_{0}^{\infty}{4\pi
r^{2}B^{0}\left(r\right)r^{4}}}}{{\left\langle{r^{2}}\right\rangle_{E,I=0}}}.$
(2.33)

We have the time’s component of vector current $\vec{V}$

$V_{i}^{0}=\frac{i}{3}\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}\left({F_{\pi}^{2}+\frac{4}{{e^{2}}}\left[{\left({\frac{{dF}}{{dr}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}}\right]}\right)Tr\left({\dot{A}A^{\dagger}\tau_{i}}\right),$
(2.34)

note (2.17)

$V_{i}^{0}=\frac{{\sin^{2}nF}}{{3n^{2}r^{2}}}\left({F_{\pi}^{2}+\frac{4}{{e^{2}}}\left[{\left({\frac{{dF}}{{dr}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}}\right]}\right)\frac{{T_{i}}}{{2\Gamma_{n}}}.$
(2.35)

The density of charge of nucleon is defined by Gell-Mann - Nishijima

$Q=\int\limits_{0}^{\infty}{\rho_{nucleon}}\left(r\right)dr=I_{3}+\frac{1}{2}B,$
(2.36)

and

$I_{3}=\int\limits_{0}^{\infty}{r^{2}}V_{3}^{0}d\Omega
dr=\int\limits_{0}^{\infty}{\frac{{4\pi\sin^{2}nF}}{{3n^{2}}}\left({F_{\pi}^{2}+\frac{4}{{e^{2}}}\left[{\left({\frac{{dF}}{{dr}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}}\right]}\right)\frac{{T_{3}}}{{2\Gamma_{n}}}}dr.$
(2.37)

So, we have

$\rho_{nu}\left(r\right)=\frac{{4\pi\sin^{2}nF}}{{3n^{2}}}\left({F_{\pi}^{2}+\frac{4}{{e^{2}}}\left[{\left({\frac{{dF}}{{dr}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}}\right]}\right)\frac{{T_{3}}}{{2\Gamma_{n}}}-\frac{{\sin^{2}nF}}{\pi}F^{\prime}.$
(2.38)

With $\bf{T_{3}=\frac{1}{2}}$ and $\bf{T_{3}=-\frac{1}{2}}$, we have the
charge’s densities of proton and neutron.

After integrating, we obtain charges of a proton and a neutron

$Q_{pro}=\int\limits_{0}^{\infty}{\rho_{pro}}\left(r\right)dr=1,$ (2.39)
$Q_{neu}=\int\limits_{0}^{\infty}{\rho_{neu}}\left(r\right)dr=0.$ (2.40)

The iso-vector electric square radius and the iso-vector magnetic square
radius are

$\left\langle{r^{2}}\right\rangle_{M,I=1}=\left\langle{r^{2}}\right\rangle_{E,I=1}=\int\limits_{0}^{\infty}{r^{2}V_{3}^{0}}dr=\infty.$
(2.41)

However, in the experiment the iso-vector electric radius is 0.88, the iso-
vector magnetic radius is 0.8. This is the weak point of the massless
composite Skyrme model! The iso-scalar magnetic moment is defined

$\vec{\mu}_{I=0}=\frac{1}{2}\int{\left({\vec{r}\times\vec{B}}\right)d^{3}x},$
(2.42)

and the result is

$\vec{\mu}_{I=0}=\frac{1}{{3\Gamma_{n}}}\left\langle{r^{2}}\right\rangle_{I=0}\vec{J}.$
(2.43)

The connection between the spin and the magnetic moment is

$\vec{\mu}=\frac{g}{{2M_{N}}}\vec{J}.$ (2.44)

Note (2.23, 2.24), we have

$g_{I=0}=g_{proton}+g_{neutron}=\frac{4}{9}M_{N}\left({M_{\Delta}-M_{N}}\right)\left\langle{r^{2}}\right\rangle_{I=0}.$
(2.45)

The iso-vector magnetic moment is defined

$\vec{\mu}_{I=1}=\frac{1}{2}\int{\left({\vec{r}\times\vec{V}_{3}}\right)d^{3}}x,$
(2.46)

where $\vec{V}_{3}$ is the vector current. After few calculations, the result
is

$\vec{\mu}_{I=1}=\frac{1}{3}\Gamma_{n}\vec{J}.$ (2.47)

And we also have

$g_{I=1}=g_{proton}-g_{neutron}=\frac{{2M_{N}}}{{M_{\Delta}-M_{N}}}.$ (2.48)

Using (2.44) and (2.47), we can find

$\mu_{pro}=\frac{{g_{pro}}}{2}=\frac{1}{4}\left[{\frac{4}{9}M_{N}\left({M_{\Delta}-M_{N}}\right)\left\langle{r^{2}}\right\rangle_{I=0}+\frac{{2M_{N}}}{{M_{\Delta}-M_{N}}}}\right],$
(2.49)
$\mu_{neu}=\frac{{g_{neu}}}{2}=\frac{1}{4}\left[{\frac{4}{9}M_{N}\left({M_{\Delta}-M_{N}}\right)\left\langle{r^{2}}\right\rangle_{I=0}-\frac{{2M_{N}}}{{M_{\Delta}-M_{N}}}}\right].$
(2.50)

The axial coupling is defined

$g_{A}=-\frac{\pi}{{3e^{2}}}D_{n},$ (2.51)

with

$D_{n}=\int\limits_{0}^{\infty}{\tilde{r}^{2}}\left\\{{\frac{{\sin
2nF}}{{n\tilde{r}}}\left[{1+4\left({F^{\prime
2}+\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right)}\right]+F^{\prime}\left({1+\frac{{8\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}}\right)}\right\\}d\tilde{r}.$
(2.52)

Figure 2.2: Plots of density of neutron charge and proton charge for a case of
$n=2$ [25]. Figure 2.3: Plots of density of neutron charge and proton charge
for a case of $n=4$ [25].

NUMERICAL RESULTS TABLE FOR THE CASE OF N=4 [25]

Quantity Prediction Experiment $e$ 1.10 - $F_{\pi}$(MeV) 168 186
$\left\langle{r^{2}}\right\rangle_{E,I=0}^{1/2}$(fm) 1.10 0.72
$\left\langle{r^{2}}\right\rangle_{M,I=0}^{1/2}$(fm) 1.93 0.81
$\left\langle{r^{2}}\right\rangle_{E,I=1}^{1/2}$(fm) $\infty$ 0.88
$\left\langle{r^{2}}\right\rangle_{M,I=1}^{1/2}$(fm) $\infty$ 0.8
$\mu_{pro}$(mag) 2.55 2.79 $\mu_{neu}$(mag) -0.66 -1.91
$\left|{\frac{{\mu_{pro}}}{{\mu_{neu}}}}\right|$ 3.86 1.46 $g_{A}$ 3.52 1.23

##  Chapter 3 The composite Skyrme model with pion’s mass

###  3.1 The term of pion’mass

When Skyrme constructed a model from the theory of pi-meson, he did not let
the term of mass of pion because of its spontaneous symmetry breaking.
However, with $n=1,2,3,4$, datum of prediction are not really exact. One of
causes that we do not let the term of pion’s mass into the lagrangian [2, 3].
Therefore, we will propose the composite lagrangian density with the pion’s
mass [22, 24]. The lagrangian density now is

${\cal L}_{n}^{pion}={\cal
L}_{n}+\frac{1}{{8n}}m_{\pi}^{2}F_{\pi}^{2}\left[{Tr\left({U^{n}}\right)-2}\right].$
(3.1)

Now using the hedgehog form gave by Skyrme
$U_{o}\left(r\right)=\exp\left[{i\tau.\hat{r}F\left(r\right)}\right]$, where
$\tau$’s are Pauli’s matrices and
$\hat{r}=\frac{{\vec{r}}}{{\left|{\vec{r}}\right|}}$, the lagrangian density
is

${\cal L}_{n}^{pion}={\cal
L}_{n}+\frac{1}{{4n}}m_{\pi}^{2}F_{\pi}^{2}\left({\cos nF-1}\right),$ (3.2)
${\cal
L}_{n}=\frac{{F_{\pi}^{2}}}{8}\left[{\left({\frac{{dF}}{{dr}}}\right)^{2}+2\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}}\right]+\frac{{\sin^{2}nF}}{{2e^{2}n^{2}r^{2}}}\left[{2\left({\frac{{dF}}{{dr}}}\right)^{2}+\frac{{\sin^{2}nF}}{{n^{2}r^{2}}}}\right].$
(3.3)

The energy of hedgehog is

${\cal
E}_{n}^{pion}=\frac{{F_{\pi}}}{e}A_{n}+\frac{1}{{e^{3}F_{\pi}}}\Theta_{n},$
(3.4)

with

$A_{n}=\int\limits_{0}^{\infty}{4\pi\tilde{r}^{2}\left\\{{\left[{\frac{{F^{\prime
2}}}{8}+\frac{{\sin^{2}nF}}{{4n^{2}\tilde{r}^{2}}}}\right]+\frac{{e^{2}F_{\pi}^{4}\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}\left[{F^{\prime
2}+\frac{{\sin^{2}nF}}{{2n^{2}\tilde{r}^{2}}}}\right]}\right\\}}d\tilde{r},$
(3.5)

and

$\Theta_{n}=\int\limits_{0}^{\infty}{\frac{{\pi\tilde{r}^{2}m_{\pi}^{2}}}{n}}\left({\cos
nF-1}\right)d\tilde{r},$ (3.6)

$\tilde{r}=eF_{\pi}r$ is a dimensionless variable.

After some steps of calculation as chapter 2, we go to the hamiltonian

$H_{n}={\cal E}_{n}^{pion}+\frac{1}{{8\Gamma_{n}}}\vec{J}^{2}={\cal
E}_{n}^{pion}+\frac{1}{{8\Gamma_{n}}}\vec{T}^{2},$ (3.7)

with $\vec{J}$ is a spin and $\vec{T}$ is an isospin. Eigenvalues of
hamiltonian (3.7) are

$M_{n}=E_{n}=\frac{{F_{\pi}}}{e}A_{n}+\frac{1}{{e^{3}F_{\pi}}}\Theta_{n}+\frac{{l\left({l+2}\right)}}{{8\Gamma_{n}}},$
(3.8)

$l=2J$ and $J$ is the quantum number of spin. So, the nucleon
$\left({J=1/2}\right)$ and delta $\left({J=3/2}\right)$ masses are given by

$M_{N}=\frac{{F_{\pi}}}{e}A_{n}+\frac{1}{{e^{3}F_{\pi}}}\Theta_{n}+\frac{{3e^{3}F_{\pi}}}{{8\Phi_{n}}},$
(3.9)
$M_{\Delta}=\frac{{F_{\pi}}}{e}A_{n}+\frac{1}{{e^{3}F_{\pi}}}\Theta_{n}+\frac{{15e^{3}F_{\pi}}}{{8\Phi_{n}}}.$
(3.10)

From (3.9) and (3.10), we obtain

$F_{\pi}=\frac{{2\left({M_{\Delta}-M_{N}}\right)\Phi_{n}}}{{3e^{3}}},$ (3.11)
$e=\sqrt[4]{{\frac{{4\left({M_{\Delta}-M_{N}}\right)^{3}\Phi_{n}^{2}A_{n}}}{{6M_{N}\left({M_{\Delta}-M_{N}}\right)^{2}\Phi_{n}-9\Theta_{n}-3\left({M_{\Delta}-M_{N}}\right)^{2}\Phi_{n}}}}}.$
(3.12)

###  3.2 The field equation and numerical results

We obtain the nonlinear differential equation of $F\left(r\right)$ from
minimum condition of the hedgehog’s energy

$\delta_{F}{\cal E}_{n}^{pion}=0,$ (3.13)

or

$\frac{d}{{dr}}\frac{{\delta{\cal E}_{n}^{pion}}}{{\delta
F^{\prime}}}-\frac{{\delta{\cal E}_{n}^{pion}}}{{\delta F}}=0(Euler-Lagrange-
equation),$ (3.14)
$\left({\frac{{\tilde{r}^{2}}}{4}+\frac{2}{{n^{2}}}\sin^{2}nF}\right)F^{\prime\prime}+\frac{{\tilde{r}}}{2}F^{\prime}+\frac{{\sin
2nF}}{n}\left({F^{\prime
2}-\frac{{\sin^{2}nF}}{{n^{2}\tilde{r}^{2}}}-\frac{1}{4}}\right)-\beta\tilde{r}^{2}\sin
nF=0,$ (3.15)

where

$\beta=\frac{{m_{\pi}^{2}}}{{4e^{2}F_{\pi}^{2}}}.$ (3.16)

We see that (3.16) contains the constant of $e$ and $F_{\pi}$ but they are not
known because they are must found after solving (3.15). Thus, we will solve
(3.15) by the way is taking the value of $\beta$ in spite of $e$ and $F_{\pi}$
and raising it until it equals approximately the $\beta$ that defined by
(3.11, 3.12, 3.16). In this case $n=4$, we need the value of $\beta$ is 0.0025
[24].

Figure 3.1: Plots of numerical solution for a case of $n=4$ [24]. Figure 3.2:
Plots of density of neutron charge and proton charge for a case of $n=4$ [24].

Following results that are obtained above by us (chapter 2), quantities of
baryon (proton and neutron) are described below [24]

NUMERICAL RESULTS TABLE FOR THE CASE OF N=4

Quantity Prediction Experiment $e$ 2.87 - $F_{\pi}$(MeV) 446.56 186
$\left\langle{r^{2}}\right\rangle_{E,I=0}^{1/2}$(fm) 0.21 0.71
$\left\langle{r^{2}}\right\rangle_{E,I=1}^{1/2}$(fm) 2.26 0.88
$\left\langle{r^{2}}\right\rangle_{M,I=0}^{1/2}$(fm) 1.0 0.81
$\left\langle{r^{2}}\right\rangle_{M,I=1}^{1/2}$(fm) 2.26 0.8 $\mu_{pro}$(mag)
1.64 2.79 $\mu_{neu}$(mag) -1.57 -1.91
$\left|{\frac{{\mu_{pro}}}{{\mu_{neu}}}}\right|$ 1.04 1.46 $g_{A}$ 1.22 1.23

##  Chapter 4 The supersymmetric composite Skyrme model

The fundamental theories of particles consider that elementary particles are
points. They all have the spin half-one and combining them together will make
particles which have the structural space. But, the model gave by Skyrme is
quite otherwise. He consider that there are solitons. In these models, one
consider that soliton solutions have the structural space and the spin zero.
By quantizing these solitons, we obtain real particles that have the spin non-
zero. Since working with solitons, these models are all nonlinear. There is an
important point that these models work at the ideal case (a pion’s mass is
zero). That makes datum of prediction are not really exact. Thus, we need to
let the term of pion’s mass in these models. But, with only pion, the models
would not be exact, because we also have other mesons. Must let all of mesons
into the model. Thus, we extend the model onto the _SU(3)_ group. But then
will be an another cause that makes the model is not really exact. In general
we must consider the contributions of the particle partners of the mesons. On
the other hand, each meson has a ”superpartner” (make up from quark
superpartners). When all of these contributions are counted in the model, it
may be exact [23].

In 1984, E. A. Bergshoeff, R. I. Nepomachie, H. J. Schnitzer proposed Skyrmion
of four-dimensional supersymmetric non-linear sigma model [5]. The composite
Skyrme model proposed by H. Y. Cheung and F. Gursey in 1990. It’s the general
model in which the sigma model and $V-T$ model are two special cases
corresponding $n=1,2$. Thus, we will calculate the composite model in the
supersymmetry by the way proposed by E. A. Bergshoeff et al [26].

We recall the Lagrangian density of composite Skyrme model

${\cal
L}_{n}=-\frac{{f_{\pi}^{2}}}{{16n^{2}}}Tr\left({\partial_{\mu}U^{-n}\partial^{\mu}U^{n}}\right)+\frac{1}{{32e^{2}n^{4}}}Tr\left({\left[{U^{-n}\partial_{\mu}U^{n},U^{-n}\partial_{\nu}U^{n}}\right]^{2}}\right),$
(4.1)

with $f_{\pi}$ is the pion decay constant, _e_ is a dimensionless parameter.
Replacing ordinary derivatives in the Lagrangian density into covariant
derivatives

$D_{\mu}U^{n}=\partial_{\mu}U^{n}-iV_{\mu}U^{n}\tau_{3},$ (4.2)

the Lagrangian density (4.1) is now

${\cal
L}_{n}=-\frac{{f_{\pi}^{2}}}{{16n^{2}}}Tr\left({D_{\mu}U^{-n}D^{\mu}U^{n}}\right)+\frac{1}{{32e^{2}n^{4}}}Tr\left({\left[{U^{-n}D_{\mu}U^{n},U^{-n}D_{\nu}U^{n}}\right]^{2}}\right).$
(4.3)

and is invariant under local $U\left(1\right)_{R}$ and global
$SU\left(2\right)_{L}$ transformations

$U^{n}\left(r\right)\to
AU^{n}\left(x\right)e^{i\lambda\left(r\right)\tau_{3}},A\in
SU\left(2\right)_{L},$ (4.4) $V_{\mu}\left(r\right)\to
V_{\mu}\left(r\right)+\partial_{\mu}\lambda\left(r\right).$ (4.5)

Really, we have

$\left({D_{\mu}U^{-n}}\right)^{\prime}=\left[{\partial_{\mu}+i\left({V_{\mu}+\partial_{\mu}\lambda}\right)\tau_{3}}\right]\left({U^{-n}e^{-i\lambda\left(x\right)\tau_{3}}}\right)A^{\dagger}$
,

$=\left({\partial_{\mu}U^{-n}}\right)e^{-i\lambda\tau_{3}}A^{\dagger}-i\left({\partial_{\mu}\lambda}\right)\tau_{3}U^{-n}e^{-i\lambda\tau_{3}}A^{\dagger}+$

$+iV_{\mu}\tau_{3}U^{-n}e^{-i\lambda\tau_{3}}A^{\dagger}+i\left({\partial_{\mu}\lambda}\right)\tau_{3}U^{-n}e^{-i\lambda\tau_{3}}A^{\dagger}$
,

$=\left[{\left({\partial_{\mu}+iV_{\mu}\tau_{3}}\right)U^{-n}}\right]e^{-\lambda\left(x\right)\tau_{3}}A^{\dagger}$
,

$\bf{=\left({D_{\mu}U^{-n}}\right)e^{-i\lambda\left(x\right)\tau_{3}}A^{\dagger}};$
(4.6)

$\left({D^{\mu}U^{n}}\right)^{\prime}=A\left[{\partial^{\mu}-i\left({V^{\mu}+\partial^{\mu}\lambda}\right)\tau_{3}}\right]\left({U^{n}e^{i\lambda\left(x\right)\tau_{3}}}\right)$,

$=A\left({\partial^{\mu}U^{n}}\right)e^{i\lambda\tau_{3}}+iA\left({\partial^{\mu}\lambda}\right)\tau_{3}U^{n}e^{i\lambda\tau_{3}}-iAV^{\mu}\tau_{3}U^{n}e^{i\lambda\tau_{3}}-iA\left({\partial^{\mu}\lambda}\right)\tau_{3}U^{n}e^{i\lambda\tau_{3}}$
,

$=A\left({\partial^{\mu}U^{n}}\right)e^{i\lambda\tau_{3}}-iAV^{\mu}\tau_{3}U^{n}e^{i\lambda\tau_{3}}=A\left[{\left({\partial^{\mu}-iV^{\mu}\tau_{3}}\right)U^{n}}\right]e^{i\lambda\tau_{3}}$
,

$\bf{=A\left({D^{\mu}U^{n}}\right)e^{i\lambda\tau_{3}}};$ (4.7)

$\Rightarrow\left({D_{\mu}U^{-n}}\right)^{\prime}\left({D^{\mu}U^{n}}\right)^{\prime}=\left({D_{\mu}U^{-n}}\right)e^{-i\lambda\left(x\right)\tau_{3}}A^{\dagger}Ae^{i\lambda\tau_{3}}\left({D^{\mu}U^{n}}\right)$
,

$=\left({D_{\mu}U^{-n}}\right)\left({D^{\mu}U^{n}}\right).$ (4.8) $\Rightarrow
Tr\left[{\left({D_{\mu}U^{-n}}\right)^{\prime}\left({D^{\mu}U^{n}}\right)^{\prime}}\right]=Tr\left[{\left({D_{\mu}U^{-n}}\right)\left({D^{\mu}U^{n}}\right)}\right].$
(4.9)

The second term of Lagrangian is proved similarly.

The gauge field $V_{\mu}\left(r\right)$ is defined

$V_{\mu}=-\frac{i}{{2n}}Tr\left({U^{-n}\partial_{\mu}U^{n}\tau_{3}}\right).$
(4.10)

We parameterize the $SU\left(2\right)$ matrix $U^{n}\left(r\right)$ in terms
of the complex scalars $A_{i}$

$U^{n}\left(r\right)=\left({\begin{array}[]{*{20}c}{A_{1}}&{-A_{2}^{*}}\\\
{A_{2}}&{A_{1}^{*}}\\\ \end{array}}\right),$ (4.11)

with the unitary constraint

$U^{\dagger}U=1\Rightarrow\bar{A}^{i}A_{i}=A_{1}^{*}A_{1}+A_{2}^{*}A_{2}=1.$
(4.12)

Now, (4.2) is

$D_{\mu}\left({\begin{array}[]{*{20}c}{A_{1}}&{-A_{2}^{*}}\\\
{A_{2}}&{A_{1}^{*}}\\\
\end{array}}\right)=\partial_{\mu}\left({\begin{array}[]{*{20}c}{A_{1}}&{-A_{2}^{*}}\\\
{A_{2}}&{A_{1}^{*}}\\\
\end{array}}\right)-iV_{\mu}\left({\begin{array}[]{*{20}c}{A_{1}}&{A_{2}^{*}}\\\
{A_{2}}&{-A_{1}^{*}}\\\ \end{array}}\right)$ (4.13)

or

$D_{\mu}A_{i}=\left({\partial_{\mu}-iV_{\mu}}\right)A_{i},$ (4.14)
$D_{\mu}\bar{A}_{i}=\left({\partial_{\mu}+iV_{\mu}}\right)\bar{A}_{i}.$ (4.15)

The form of gauge field is then

$V_{\mu}=-\frac{i}{{2n}}Tr\left({U^{-n}\partial_{\mu}U^{n}\tau_{3}}\right)$ ,

$=-\frac{i}{{2n}}Tr\left[{\left({\begin{array}[]{*{20}c}{A_{1}^{*}}&{A_{2}^{*}}\\\
{-A_{2}}&{A_{1}}\\\
\end{array}}\right)\partial_{\mu}\left({\begin{array}[]{*{20}c}{A_{1}}&{-A_{2}^{*}}\\\
{A_{2}}&{A_{1}^{*}}\\\ \end{array}}\right)\left({\begin{array}[]{*{20}c}1&0\\\
0&{-1}\\\ \end{array}}\right)}\right]$,

$=-\frac{i}{{2n}}Tr\left[{\left({\begin{array}[]{*{20}c}{A_{1}^{*}}&{A_{2}^{*}}\\\
{-A_{2}}&{A_{1}}\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{\partial_{\mu}A_{1}}&{\partial_{\mu}A_{2}^{*}}\\\
{\partial_{\mu}A_{2}}&{-\partial_{\mu}A_{1}^{*}}\\\
\end{array}}\right)}\right]$,

$=-\frac{i}{{2n}}Tr\left[{\begin{array}[]{*{20}c}{A_{1}^{*}\partial_{\mu}A_{1}+A_{2}^{*}\partial_{\mu}A_{2}}&{}\hfil\\\
{}\hfil&{-A_{2}\partial_{\mu}A_{2}^{*}-A_{1}\partial_{\mu}A_{1}^{*}}\\\
\end{array}}\right]$ ,

$=-\frac{i}{{2n}}\left[{\bar{A}^{i}\partial_{\mu}A_{i}-\left({\partial_{\mu}\bar{A}^{i}}\right)A_{i}}\right]$
,

$\bf\Rightarrow{V_{\mu}\left(r\right)=-\frac{i}{{2n}}\bar{A}^{i}\mathord{\buildrel{\lower
3.0pt\hbox{$\scriptscriptstyle\leftrightarrow$}}\over{\partial}}A_{i}}.$
(4.16)

With matrix $U^{n}\left(r\right)$ is defined in (4.11) and the gauge field is
defined in (4.16), the supersymmetric Lagrangian density (4.3) is then

${\cal
L}_{n}=-\frac{{f_{\pi}^{2}}}{{8n^{2}}}\bar{D}_{\mu}\bar{A}D^{\mu}A-\frac{1}{{16e^{2}n^{2}}}F_{\mu\nu}^{2}\left(V\right),$
(4.17)

with

$F_{\mu\nu}\left(V\right)=\partial_{\mu}V_{\nu}-\partial_{\nu}V_{\mu}.$ (4.18)

Really, we have \+ The first term

${\cal
L}_{n}^{1}=-\frac{1}{{16n^{2}}}f_{\pi}^{2}Tr\left[{D^{\mu}\left({\begin{array}[]{*{20}c}{A_{1}^{*}}&{A_{2}^{*}}\\\
{-A_{2}}&{A_{1}}\\\
\end{array}}\right)D_{\mu}\left({\begin{array}[]{*{20}c}{A_{1}}&{-A_{2}^{*}}\\\
{A_{2}}&{A_{1}^{*}}\\\ \end{array}}\right)}\right]$,

$=-\frac{1}{{16n^{2}}}f_{\pi}^{2}Tr\left({\begin{array}[]{*{20}c}{D^{\mu}A_{1}^{*}D_{\mu}A_{1}+D^{\mu}A_{2}^{*}D_{\mu}A_{2}^{*}}&{}\hfil\\\
{}\hfil&{D^{\mu}A_{2}D_{\mu}A_{2}^{*}+D^{\mu}A_{1}D_{\mu}A_{1}^{*}}\\\
\end{array}}\right)$ ,

$\Rightarrow{\bf{\cal
L}_{n}^{1}=-\frac{1}{{8n^{2}}}f_{\pi}^{2}D^{\mu}\bar{A}^{i}D_{\mu}A_{i}}.$
(4.19)

\+ The second term

$U^{-n}D_{\mu}U^{n}U^{-n}D_{\nu}U^{n}-U^{-n}D_{\nu}U^{n}U^{-n}D_{\mu}U^{n}$,

$=U^{-n}\left[{\left({\partial_{\mu}-iV_{\mu}\tau_{3}}\right)U^{n}}\right]U^{-n}\left[{\left({\partial_{\nu}-iV_{\nu}\tau_{3}}\right)U^{n}}\right]-$

$-U^{-n}\left[{\left({\partial_{\nu}-iV_{\nu}\tau_{3}}\right)U^{n}}\right]U^{-n}\left[{\left({\partial_{\mu}-iV_{\mu}\tau_{3}}\right)U^{n}}\right]$,

$=\left({U^{-n}\partial_{\mu}U^{n}-iU^{-n}V_{\mu}\tau_{3}U^{n}}\right)\left({U^{-n}\partial_{\nu}U^{n}-iU^{-n}V_{\nu}\tau_{3}U^{n}}\right)-$

$-\left({U^{-n}\partial_{\nu}U^{n}-iU^{-n}V_{\nu}\tau_{3}U^{n}}\right)\left({U^{-n}\partial_{\mu}U^{n}-iU^{-n}V_{\mu}\tau_{3}U^{n}}\right)$,

$=U^{-n}\partial_{\mu}U^{n}U^{-n}\partial_{\nu}U^{n}-U^{-n}\partial_{\nu}U^{n}U^{-n}\partial_{\mu}U^{n}.$
(4.20)

Using

$U^{-n}U^{n}=1\Rightarrow\partial_{\mu}U^{-n}U^{n}+U^{-n}\partial_{\mu}U^{n}=0,$
(4.21)

(4.20) is now

$\left({4.20}\right)=-\partial_{\mu}U^{-n}\partial_{\nu}U^{n}+\partial_{\nu}U^{-n}\partial_{\mu}U^{n}.$
(4.22)

The second term is now

$\Rightarrow{\cal
L}_{n}^{2}=\frac{1}{{32e^{2}n^{4}}}Tr\left[{\left({\partial_{\mu}U^{-n}\partial_{\nu}U^{n}-\partial_{\nu}U^{-n}\partial_{\mu}U^{n}}\right)^{2}}\right].$
(4.23)

We define

$F_{\mu\nu}\left(V\right)\equiv\partial_{\mu}V_{\nu}-\partial_{\nu}V_{\mu},$
(4.24)
$=-\frac{i}{{2n}}Tr\left[{\partial_{\mu}\left({U^{-n}\partial_{\nu}U^{n}\tau_{3}}\right)-\partial_{\nu}\left({U^{-n}\partial_{\mu}U^{n}\tau_{3}}\right)}\right],$
(4.25)
$=-\frac{i}{{2n}}Tr\left[{\partial_{\mu}U^{-n}\partial_{\nu}U^{n}\tau_{3}-\partial_{\nu}U^{-n}\partial_{\mu}U^{n}\tau_{3}}\right].$
(4.26) $\Rightarrow
F_{\mu\nu}^{2}=\frac{1}{{2n^{2}}}Tr\left[{\left({\partial_{\mu}U^{-n}\partial_{\nu}U^{n}-\partial_{\nu}U^{-n}\partial_{\mu}U^{n}}\right)^{2}}\right].$
(4.27)

From (4.23) and (4.27), we have

${\bf{\cal L}_{n}^{2}=\frac{1}{{16e^{2}n^{2}}}F_{\mu\nu}^{2}}.$ (4.28)

To supersymmetrise this model, we extend $A_{i}$ to chiral scalar multiples
$\left({A_{i},\psi_{\alpha i},F_{i}}\right)$ $\left({i,\alpha=1,2}\right)$ and
the vector $V_{\mu}\left(r\right)$ to real vector multiples
$\left({V_{\mu},\lambda_{\alpha},D}\right)$. Here, the fields $F_{i}$ are
complex scalars, $D$ is real scalar, $\psi_{\alpha i}$, $\lambda_{\alpha}$ are
Majorana two-component spinors. $\psi_{\alpha i}$ corresponds to a left-handed
chiral spinor, $\bar{\psi}^{\alpha i}=\left({\psi_{i}^{\alpha}}\right)^{*}$
corresponds to a right-handed one. The supersymmetric Lagrangian density is
given by

${\cal
L}_{susy}=\frac{{f_{\pi}^{2}}}{{8n^{2}}}\left[{-D^{\mu}\bar{A}^{i}D_{\mu}A_{i}-\frac{1}{2}i\bar{\psi}^{\dot{\alpha}i}\left({\sigma_{\mu}}\right)_{\alpha\dot{\alpha}}\mathord{\buildrel{\lower
3.0pt\hbox{$\scriptscriptstyle\leftrightarrow$}}\over{D}}^{\mu}\psi_{i}^{\alpha}+\bar{F}^{i}F_{i}-}\right.$
(4.29) $\left.{-i\bar{A}^{i}\lambda^{\alpha}\psi_{\alpha
i}+iA_{i}\bar{\lambda}^{\dot{\alpha}}\bar{\psi}_{\dot{\alpha}}^{i}+D\left({\bar{A}^{i}A_{i}-1}\right)}\right]+$
(4.30)
$+\frac{1}{{8e^{2}n^{2}}}\left[{-\frac{1}{2}F_{\mu\nu}^{2}-i\bar{\lambda}^{\dot{\alpha}}\left({\sigma^{\mu}}\right)_{\dot{\alpha}}^{\alpha}\partial_{\mu}\lambda_{\alpha}+D^{2}}\right],$
(4.31)

is invariant under supersymmetric transformations

$\delta A_{i}=-\varepsilon^{\alpha}\psi_{\alpha i},$ (4.32)
$\delta\psi_{\alpha
i}=-i\bar{\varepsilon}^{\dot{\alpha}}\left({\sigma^{\mu}}\right)_{\alpha\dot{\alpha}}D_{\mu}A_{i}+\varepsilon_{\alpha}F_{i},$
(4.33) $\delta
F_{i}=-i\bar{\varepsilon}^{\dot{\alpha}}\left({\sigma^{\mu}}\right)_{\dot{\alpha}}^{\alpha}D_{\mu}\psi_{\alpha
i}-i\bar{\varepsilon}^{\dot{\alpha}}A_{i}\bar{\lambda}_{\dot{\alpha}},$ (4.34)
$\delta
V_{\mu}=-\frac{1}{2}i\left({\sigma_{\mu}}\right)^{\alpha\dot{\alpha}}\left({\bar{\varepsilon}_{\dot{\alpha}}\lambda_{\alpha}+\varepsilon_{\alpha}\bar{\lambda}_{\dot{\alpha}}}\right),$
(4.35)
$\delta\lambda_{\alpha}=\varepsilon^{\beta}\left({\sigma^{\mu\nu}}\right)_{\beta\alpha}F_{\mu\nu}+i\varepsilon_{\alpha}D,$
(4.36) $\delta
D=\frac{1}{2}\left({\sigma^{\mu}}\right)_{\alpha\dot{\alpha}}\partial_{\mu}\left({\bar{\varepsilon}^{\dot{\alpha}}\lambda^{\alpha}-\varepsilon^{\alpha}\bar{\lambda}^{\dot{\alpha}}}\right).$
(4.37)

The field equation and their supersymmetric transformations lead to the
following constraints with

$\bar{A}^{i}A_{i}=0,$ (4.38) $\bar{A}^{i}\psi_{\alpha i}=0,$ (4.39)
$\bar{A}^{i}F_{i}=0$ (4.40)

and following algebraic expressions for

$V_{\mu}=-\frac{1}{2}\left\\{{i\bar{A}^{i}\mathord{\buildrel{\lower
3.0pt\hbox{$\scriptscriptstyle\leftrightarrow$}}\over{\partial}}_{\mu}A_{i}+\left({\sigma_{\mu}}\right)^{\alpha\dot{\alpha}}\bar{\psi}_{\dot{\alpha}}^{i}\psi_{\alpha
i}}\right\\},$ (4.41) $\lambda_{\alpha}=-i\left\\{{\bar{F}^{i}\psi_{\alpha
i}+i\left({\sigma^{\mu}}\right)_{\alpha\dot{\alpha}}\left({D_{\mu}A_{i}}\right)\bar{\psi}^{\dot{\alpha}i}}\right\\},$
(4.42)
$D=D^{\mu}\bar{A}^{i}D_{\mu}A_{i}+\frac{1}{2}i\bar{\psi}^{\dot{\alpha}i}\left({\sigma^{\mu}}\right)_{\alpha\dot{\alpha}}\left({\mathord{\buildrel{\lower
3.0pt\hbox{$\scriptscriptstyle\leftrightarrow$}}\over{D}}_{\mu}\psi_{i}^{\alpha}}\right)-\bar{F}^{i}F_{i}.$
(4.43)

Setting $\psi_{\alpha}=F_{i}=0$, then

${\cal
L}_{susy}=-\frac{{f_{\pi}^{2}}}{{8n^{2}}}\bar{D}^{\mu}\bar{A}D_{\mu}A+\frac{1}{{8e^{2}n^{2}}}\left[{-\frac{1}{2}F_{\mu\nu}^{2}+\left({\bar{D}^{\mu}\bar{A}D_{\mu}A}\right)^{2}}\right].$
(4.44)

The second term is fourth-order in derivatives. However, there is other
possible fourth-order term

$\square\bar{A}\square A-\left({\bar{D}^{\mu}\bar{A}D_{\mu}A}\right)^{2},$
(4.45)

where $\square\equiv D^{\mu}D_{\mu}$ is the gauge covariant d’Alembertian.
Thus, we may let it to the Lagrangian density

${\cal
L}_{susy}=-\frac{{f_{\pi}^{2}}}{{8n^{2}}}\bar{D}^{\mu}\bar{A}D_{\mu}A+\frac{1}{{8e^{2}n^{2}}}\left[{\alpha\left\\{{-\frac{1}{2}F_{\mu\nu}^{2}+\left({\bar{D}^{\mu}\bar{A}D_{\mu}A}\right)^{2}}\right\\}+}\right.$
(4.46) $\left.{+\beta\left\\{{\square\bar{A}\square
A-\left({\bar{D}^{\mu}\bar{A}D_{\mu}A}\right)^{2}}\right\\}}\right].$ (4.47)

Now, we recall the hedgehog ansatz

$U^{n}\left(r\right)=\cos nf\left(r\right)+i\vec{\tau}\frac{{\vec{r}}}{r}\sin
nf\left(r\right).$ (4.48)

From (4.11), we have

$A_{1}=\cos nf\left(r\right)+i\cos\theta\sin nf\left(r\right),$ (4.49)
$A_{2}=ie^{i\varphi}\sin\theta\sin nf\left(r\right).$ (4.50)

Inserting the hedgehog ansatz into the supersymmetric Lagrangian density, we
obtain the supersymmetric static energy

$E=4\pi\frac{{f_{\pi}}}{e}\int{dxx^{2}}\left\\{{\frac{1}{{12}}\left({f^{\prime
2}+\frac{{2\sin^{2}nf}}{{n^{2}x^{2}}}}\right)+\left({\frac{{\alpha+\beta}}{{15}}}\right)\left({f^{\prime
2}-\frac{{\sin^{2}nf}}{{n^{2}x^{2}}}}\right)+}\right.$ (4.51)
$\left.{+\frac{\beta}{{12}}\left({f^{\prime\prime}+\frac{{2f^{\prime}}}{{nx}}-\frac{{\sin
2nf}}{{n^{2}x^{2}}}}\right)}\right\\}.$ (4.52)

The Euler-Lagrange equation takes the field equation

$-x^{2}f^{\prime\prime}-2xf^{\prime}+\frac{{\sin
2nf}}{n}+\frac{{4\left({\alpha+\beta}\right)}}{5}\left[{\frac{{2f^{\prime\prime}\sin^{2}nf}}{{n^{2}}}-6x^{2}f^{\prime
2}f^{\prime\prime}-4xf^{\prime 3}+}\right.$ (4.53) $\left.{+\frac{{f^{\prime
2}\sin 2nf}}{n}+\frac{{\sin^{2}nf\sin
2nf}}{{n^{3}x^{2}}}}\right]+\beta\left[{x^{2}f^{\left(4\right)}+\frac{{4xf^{\left(3\right)}}}{n}-\frac{{4f^{\prime\prime}\cos
2nf}}{n}+}\right.$ (4.54) $\left.{+\frac{{4f^{\prime 2}\sin
2nf}}{{n^{2}}}-\frac{{4\sin^{2}nf\sin 2nf}}{{n^{3}x^{2}}}}\right]=0,$ (4.55)

with $x=ef_{\pi}r$ is the dimensionless variable and boundary conditions

$f\left(0\right)=\pi,$ (4.56) $f\left(\infty\right)=0.$ (4.57)

Figure 4.1: Plots of supersymmetric numerical solution of $f\left(x\right)$
for a case of $\alpha=1,\beta=0,n=2,3$ [26].

##  Chapter 5 Appendix

###  5.1 Some notations for Chapter 2 and 3

There are some notations to calculations in chapter 2 and 3

$\frac{\partial}{{\partial
x^{\mu}}}\equiv\partial_{\mu}=\left({\partial_{0},\partial_{i}}\right)=\left({\frac{\partial}{{\partial
t}},\vec{\nabla}}\right);$ (5.1) $\frac{\partial}{{\partial
x_{\mu}}}\equiv\partial^{\mu}=\left({\partial^{0},\partial^{i}}\right)=g^{\mu\nu}\partial_{\nu}=diag\left({1,-1,-1,-1}\right)\partial_{\nu}=\left({\frac{\partial}{{\partial
t}},-\vec{\nabla}}\right);$ (5.2) $\partial_{i}F\left(r\right)=\frac{{\partial
F\left(r\right)}}{{\partial x^{i}}}=\frac{{\partial F}}{{\partial
r}}\frac{{\partial r}}{{\partial x^{i}}}=F^{\prime}\frac{{x^{i}}}{r}.$ (5.3)
$U_{o}^{n}\left(r\right)=\exp\left[{i\tau.\hat{r}\left\\{{nF\left(r\right)}\right\\}}\right]=\cos\left\\{{nF\left(r\right)}\right\\}+i\vec{\tau}.\hat{r}\sin\left\\{{nF\left(r\right)}\right\\};$
(5.4)
$U_{o}^{-n}\left(r\right)=\exp\left[{-i\tau.\hat{r}\left\\{{nF\left(r\right)}\right\\}}\right]=\cos\left\\{{nF\left(r\right)}\right\\}-i\vec{\tau}.\hat{r}\sin\left\\{{nF\left(r\right)}\right\\};$
(5.5)

Some identities of Pauli matrices are

$\tau_{i}\tau_{j}=i\varepsilon_{ijk}\tau_{k};$ (5.6)
$\left({\tau_{i}}\right)^{2}=1;$ (5.7) $\tau_{i}\tau_{j}=-\tau_{j}\tau_{i};$
(5.8) $Tr\left({\tau_{i}^{2}}\right)=2;$ (5.9) $Tr\left({\tau_{i}}\right)=0;$
(5.10)

We will define

$\partial_{\mu}U_{o}^{n}\left(r\right)=\partial_{\mu}\left({\cos
nF}\right)+i\vec{\tau}\partial_{\mu}\left({\hat{r}\sin nF}\right)$ ,

$=-n\partial_{i}F\sin nF+\frac{{i\vec{\tau}}}{r}\partial_{i}\vec{r}\sin
nF+i\vec{\tau}\vec{r}\partial_{i}\left({\frac{1}{r}}\right)\sin
nF+in\vec{\tau}\frac{{\vec{r}}}{r}\partial_{i}F\cos nF$ ,

$=-n\frac{{F^{\prime}x^{i}}}{r}\sin nF+\frac{{i\tau_{i}\vec{e}^{i}}}{r}\sin
nF-\frac{{i\vec{\tau}\vec{r}x^{i}}}{{r^{3}}}\sin
nF+in\frac{{\vec{\tau}\vec{r}F^{\prime}x^{i}}}{{r^{2}}}\cos nF;$ (5.11)

$\partial_{\mu}U_{o}^{-n}\left(r\right)=\partial_{\mu}\left({\cos
nF}\right)-i\vec{\tau}\partial_{\mu}\left({\hat{r}\sin nF}\right)$ ,

$=-n\frac{{F^{\prime}x^{i}}}{r}\sin nF-\frac{{i\tau_{i}\vec{e}^{i}}}{r}\sin
nF+\frac{{i\vec{\tau}\vec{r}x^{i}}}{{r^{3}}}\sin nF-
in\frac{{\vec{\tau}\vec{r}F^{\prime}x^{i}}}{{r^{2}}}\cos nF;$ (5.12)

$\partial^{\mu}U_{o}^{n}\left(r\right)=\partial^{\mu}\left({\cos
nF}\right)+i\vec{\tau}\partial^{\mu}\left({\hat{r}\sin nF}\right)$ ,

$=n\frac{{F^{\prime}x_{i}}}{r}\sin nF-\frac{{i\tau_{i}\vec{e}_{i}}}{r}\sin
nF+\frac{{i\vec{\tau}\vec{r}x_{i}}}{{r^{3}}}\sin nF-
in\frac{{\vec{\tau}\vec{r}F^{\prime}x_{i}}}{{r^{2}}}\cos nF;$ (5.13)

The first term of the Lagrangian density (2.2) is defined

$\Rightarrow{\cal
L}_{n}^{1}=Tr\left({\partial_{\mu}U_{o}^{-n}\partial^{\mu}U_{o}^{n}}\right)$

$=Tr\left\\{{\left({-n\frac{{F^{\prime}x^{i}}}{r}\sin
nF-\frac{{i\tau_{i}\vec{e}^{i}}}{r}\sin
nF+\frac{{i\vec{\tau}\vec{r}x^{i}}}{{r^{3}}}\sin nF-
in\frac{{\vec{\tau}\vec{r}F^{\prime}x_{i}}}{{r^{2}}}\cos
nF}\right)\times}\right.$

$\times\left.{\left({n\frac{{F^{\prime}x_{i}}}{r}\sin
nF-\frac{{i\tau_{i}\vec{e}_{i}}}{r}\sin
nF+\frac{{i\vec{\tau}\vec{r}x_{i}}}{{r^{3}}}\sin nF-
in\frac{{\vec{\tau}\vec{r}F^{\prime}x_{i}}}{{r^{2}}}\cos nF}\right)}\right\\}$

$=Tr\left\\{{-n^{2}F^{\prime
2}\sin^{2}nF+inF^{\prime}\frac{{\vec{\tau}\vec{r}}}{{r^{2}}}\sin^{2}nF-
inF^{\prime}\frac{{\vec{\tau}\vec{r}}}{{r^{2}}}\sin^{2}nF+}\right.$

$+in^{2}F^{\prime 2}\frac{{\vec{\tau}\vec{r}}}{{2r}}\sin 2nF-
inF^{\prime}\frac{{\vec{\tau}\vec{r}}}{{r^{2}}}\sin^{2}nF-3\frac{{\sin^{2}nF}}{{r^{2}}}+\frac{{\sin^{2}nF}}{{r^{2}}}-$

$-n\frac{{F^{\prime}}}{{2r}}\sin
2nF+inF^{\prime}\frac{{\vec{\tau}\vec{r}}}{{r^{2}}}\sin^{2}nF+\frac{{\sin^{2}nF}}{{r^{2}}}-\frac{{\sin^{2}nF}}{{r^{2}}}+n\frac{{F^{\prime}}}{{2r}}\sin
2nF-$

$\left.{-in^{2}F^{\prime 2}\frac{{\vec{\tau}\vec{r}}}{{2r}}\sin
2nF-n\frac{{F^{\prime}}}{{2r}}\sin 2nF+n\frac{{F^{\prime}}}{{2r}}\sin
2nF-n^{2}F^{\prime 2}\cos^{2}nF}\right\\}$

$=Tr\left({-n^{2}F^{\prime 2}-2\frac{{\sin^{2}nF}}{{r^{2}}}}\right)$

$\Rightarrow{\cal
L}_{n}^{1}=-2n^{2}\left({\frac{{dF}}{{dr}}}\right)^{2}-4\frac{{\sin^{2}nF}}{{r^{2}}}.$
(5.14)

To defined the second term of the Lagrangian density (2.2), we will go to the
intermediary step using the unitary condition of $U_{o}^{n}$

$U_{o}^{-n}U_{o}^{n}=1\Rightarrow\partial_{\mu}U_{o}^{-n}U_{o}^{n}+U_{o}^{-n}\partial_{\mu}U_{o}^{n}=0.$
(5.15)

The second term of Lagrangian density (2.2) is now

$\Rightarrow{\cal
L}_{n}^{2}=Tr\left[{\left({\partial_{\mu}U_{o}^{-n}\partial_{\nu}U_{o}^{n}-\partial_{\nu}U_{o}^{-n}\partial_{\mu}U_{o}^{n}}\right)^{2}}\right].$
(5.16)

Note (5.11), (5.12) we can define

${\cal
L}_{n}^{2}=32n^{2}\frac{{\sin^{2}nF}}{{r^{2}}}\left({\frac{{dF}}{{dr}}}\right)^{2}+16\frac{{\sin^{4}nF}}{{r^{4}}}.$
(5.17)

###  5.2 Some notations for Chapter 4

Dotted and undotted spinor indices run from 1 to 2 and are denoted by the
early letters of the English alphabet. Spinor indices are raised and lowered
by the $\varepsilon$ tensors

$\varepsilon^{ab}=\left({\begin{array}[]{*{20}c}0&1\\\ {-1}&0\\\
\end{array}}\right),$ (5.18)
$\varepsilon_{ab}=\left({\begin{array}[]{*{20}c}0&{-1}\\\ 1&0\\\
\end{array}}\right).$ (5.19)

For example

$\psi^{a}=\varepsilon^{ab}\psi_{b},$ (5.20)
$\chi^{a}=\varepsilon_{ab}\chi^{b}.$ (5.21)

The product $\psi\chi$ is

$\psi\chi=\psi^{a}\chi_{a}=-\psi_{a}\chi^{a}=\chi^{a}\psi_{a}=\chi\psi.$
(5.22)

In the special case

$\psi^{a}\psi^{b}=-\frac{1}{2}\psi\psi\varepsilon^{ab}.$ (5.23)

The hermitian conjugates of the spinors $\psi$ and $\chi$ are $\bar{\psi}$ and
$\bar{\chi}$

$\bar{\psi}^{\dot{a}}=\left({\psi^{a}}\right)^{\dagger},$ (5.24)
$\bar{\chi}=\left({\chi_{a}}\right)^{\dagger}.$ (5.25)

Their dotted indices are raised and lowered by the tensors
$\varepsilon^{\dot{a}\dot{b}}$ and $\varepsilon_{\dot{a}\dot{b}}$ which are
equal to their undotted counterparts $\varepsilon^{ab}$ and
$\varepsilon_{ab}$. The product of them is

$\bar{\psi}\bar{\chi}=\bar{\psi}_{\dot{a}}\bar{\chi}^{\dot{a}}=-\bar{\psi}^{\dot{a}}\bar{\chi}_{\dot{a}}=\bar{\chi}_{\dot{a}}\bar{\psi}^{\dot{a}}=\bar{\chi}\bar{\psi}.$
(5.26)

In the special case

$\bar{\psi}^{\dot{a}}\bar{\psi}^{\dot{b}}=\frac{1}{2}\bar{\psi}\bar{\psi}\varepsilon^{\dot{a}\dot{b}}.$
(5.27)

The hermitian conjugates of the product $\chi\psi$ is

$\left({\chi\psi}\right)^{\dagger}=\left({\chi^{a}\psi_{a}}\right)^{\dagger}=\bar{\psi}_{\dot{a}}\bar{\chi}^{\dot{a}}=\bar{\psi}\bar{\chi}=\bar{\chi}\bar{\psi}.$
(5.28)

The Pauli matrices $\sigma_{a\dot{b}}^{m}$ are

$\sigma^{0}=\left({\begin{array}[]{*{20}c}{1}&0\\\ 0&{1}\\\
\end{array}}\right),\sigma^{1}=\left({\begin{array}[]{*{20}c}0&1\\\ 1&0\\\
\end{array}}\right)$ ,

$\sigma^{2}=\left({\begin{array}[]{*{20}c}0&{-i}\\\ i&0\\\
\end{array}}\right),\sigma^{3}=\left({\begin{array}[]{*{20}c}1&0\\\ 0&{-1}\\\
\end{array}}\right).$ (5.29)

Other useful identities are

$\chi\sigma^{n}\bar{\psi}=-\bar{\psi}\bar{\sigma}^{n}\chi,$ (5.30)
$\chi\sigma^{m}\bar{\sigma}^{n}\psi=\psi\sigma^{n}\bar{\sigma}^{m}\chi,$
(5.31)
$\left({\chi\sigma^{m}\bar{\psi}}\right)^{\dagger}=\psi\sigma^{m}\bar{\chi},$
(5.32)
$\left({\chi\sigma^{m}\bar{\sigma}^{n}\psi}\right)^{\dagger}=\bar{\psi}\bar{\sigma}^{n}\sigma^{m}\bar{\chi}.$
(5.33)

Using these above notations, we can define (from…)

$\delta\bar{A}_{i}=-\bar{\psi}_{\dot{\alpha}i}\bar{\varepsilon}^{\dot{\alpha}},$
(5.34)
$\delta\bar{\psi}_{\dot{\alpha}i}=i\left({\sigma^{\mu}}\right)_{\alpha\dot{\alpha}}\varepsilon^{\alpha}D_{\mu}\bar{A}_{i}+\bar{\varepsilon}_{\dot{\alpha}}\bar{F}_{i},$
(5.35)
$\delta\bar{F}_{i}=iD_{\mu}\bar{\psi}_{\dot{\alpha}i}\left({\sigma^{\mu}}\right)_{\dot{\alpha}}^{\alpha}\varepsilon^{\alpha}+i\bar{A}_{i}\lambda_{\alpha}\varepsilon^{\alpha},$
(5.36)
$\delta\bar{V}_{\mu}=\frac{1}{2}i\left({\bar{\lambda}_{\dot{\alpha}}\varepsilon_{\alpha}+\lambda_{\alpha}\bar{\varepsilon}_{\dot{\alpha}}}\right)\left({\sigma_{\mu}}\right)^{\alpha\dot{\alpha}},$
(5.37)
$\delta\bar{\lambda}_{\dot{\alpha}}=\left({\sigma^{\mu\nu}}\right)_{\beta\alpha}\bar{\varepsilon}^{\dot{\beta}}F_{\mu\nu}-i\bar{\varepsilon}_{\dot{\alpha}}\bar{D},$
(5.38)
$\delta\bar{D}=\frac{1}{2}\partial_{\mu}\left({\bar{\lambda}^{\dot{\alpha}}\varepsilon^{\alpha}-\lambda^{\alpha}\bar{\varepsilon}^{\dot{\alpha}}}\right)\left({\sigma^{\mu}}\right)_{\alpha\dot{\alpha}}.$
(5.39)

Since $\delta$ and $D_{\mu}$ can permute together, $\delta{\cal L}_{susy}$ is
defined

$\delta{\cal
L}_{susy}=\frac{{f_{\pi}^{2}}}{{8n^{2}}}\left[{-D^{\mu}\left({\delta\bar{A}^{i}}\right)D_{\mu}A_{i}-D^{\mu}\bar{A}^{i}D_{\mu}\left({\delta
A_{i}}\right)-\frac{1}{2}i\left({\delta\bar{\psi}^{\dot{\alpha}i}}\right)\left({\sigma_{\mu}}\right)_{\alpha\dot{\alpha}}\mathord{\buildrel{\lower
3.0pt\hbox{$\scriptscriptstyle\leftrightarrow$}}\over{D}}^{\mu}\psi_{i}^{\alpha}-}\right.$

$-\frac{1}{2}i\bar{\psi}^{\dot{\alpha}i}\left({\sigma_{\mu}}\right)_{\alpha\dot{\alpha}}\mathord{\buildrel{\lower
3.0pt\hbox{$\scriptscriptstyle\leftrightarrow$}}\over{D}}^{\mu}\left({\delta\psi_{i}^{\alpha}}\right)+\delta\bar{F}^{i}F_{i}+\bar{F}^{i}\delta
F_{i}-i\left({\delta\bar{A}^{i}}\right)\lambda^{\alpha}\psi_{\alpha
i}-i\bar{A}^{i}\left({\delta\lambda^{\alpha}}\right)\psi_{\alpha
i}-i\bar{A}^{i}\lambda^{\alpha}\delta\psi_{\alpha i}+$

$\left.{+i\left({\delta
A_{i}}\right)\bar{\lambda}^{\dot{\alpha}}\bar{\psi}_{\dot{\alpha}}^{i}+iA_{i}\left({\delta\bar{\lambda}^{\dot{\alpha}}}\right)\bar{\psi}_{\dot{\alpha}}^{i}+iA_{i}\bar{\lambda}^{\dot{\alpha}}\left({\delta\bar{\psi}_{\dot{\alpha}}^{i}}\right)+\left({\delta
D}\right)\left({\bar{A}^{i}A_{i}-1}\right)+D\left({\delta\bar{A}^{i}}\right)A_{i}+D\bar{A}^{i}\left({\delta
A_{i}}\right)}\right]+$

$+\frac{1}{{8e^{2}n^{4}}}\left[{-F_{\mu\nu}\delta
F_{\mu\nu}-i\left({\delta\bar{\lambda}^{\dot{\alpha}}}\right)\left({\sigma^{\mu}}\right)_{\dot{\alpha}}^{\alpha}\partial_{\mu}\lambda_{\alpha}-i\bar{\lambda}^{\dot{\alpha}}\left({\sigma^{\mu}}\right)_{\dot{\alpha}}^{\alpha}\partial_{\mu}\left({\delta\lambda_{\alpha}}\right)+2D\delta
D}\right].$ (5.40)

## Bibliography

  * [1] T. H. R. Skyrme, Nucl. Phys., 31 (1962) 566.
  * [2] G. S. Adkins, C. R. Nappi and E. Witten, Nucl. Phys., B228  (1983) 552. G. S. Adkins, Nucl. Phys., B249  (1985) 507. G.S. Adkins, C. R. Nappi, Nucl. Phys., B233 (1984) 109. E. Witten, Nucl. Phys., B223 (1983) 422. E. Witten, Nucl. Phys., B223 (1983) 433.
  * [3] N. A. Viet, P. T. Tuyen, J. Phys., G15 (July 1989) 937.
  * [4] H. Y. Cheung, Feza Gursey, Modern Physics Letters A, Vol.5 No.21  (1990) 1685
  * [5] E. A. Bergshoeff, R. I. Nepomachie, H. J. Schnitzer, Nucl. Phys., B249  (1985) 93
  * [6] N. Shiiki, N. Sawado, S.Oryu, hep-th/0603069 (2006)
  * [7] A. Acus, hep-ph/9901240 (1999)
  * [8] C. Houghton, S. Magee, hep-th/0602227 (2006)
  * [9] J. Wess, B. Zumino, Nucl. Phys., B70 (1974) 39. J. Wess, B. Zumino, Nucl. Phys., B78 (1974) 1.
  * [10] A. Salam, J. Strathdee, Nucl. Phys., B76  (1974) 477. A. Salam, J. Strathdee, Nucl. Phys., B80  (1974) 499.
  * [11] M. F. Sohnius, Physics Report, 128, Nos 2 and 3, (1985), 39 - 204
  * [12] P. T. Tuyen, Quantum Mechanics, in Vietnamese, College of Science, VNU, (2004)
  * [13] P. T. Tuyen, Theory of Fundamental Particles, in Vietnamese, College of Science, VNU, (2006)
  * [14] P. T. Tuyen, Theory of Group, in Vietnamese, College of Science, VNU, (2006)
  * [15] L. Ryder, Quantum Field Theory, second edition, Cambridge University Press, (1996)
  * [16] S. Weinberg, The Quantum Theory of Fields, vol I, II, III, Cambridge University Press, (1995)
  * [17] M. E. Peskin, D. V. Shroeder, An Introduction to Quantum Field Theory, Westview Press, (1995)
  * [18] H. Georgi, Lie Algebras in Particle Physics, second edition, Perseus Books, (1999)
  * [19] W. Greiner, J. Reinhardt, Field Quantization, Springer, (1996)
  * [20] A. Bilal, hep-th/0101055 (2001)
  * [21] K. Cahill, hep-ph/9907295 (1999)
  * [22] D. Q. Tuan, P. T. Tuyen, J. Science. (Math - Phys), VNU, T.XXII, No 2AP (2006) 194
  * [23] D. Q. Tuan, P. T. Tuyen, Phenomenal models of Baryon, Talk in National Conference on Theoretical Physics 31, Cua Lo, Nghe An, 2006 (no-published)
  * [24] D. Q. Tuan, P. T. Tuyen, The composite Skyrme model with the mass of pion, (no-published) 2006
  * [25] D. Q. Tuan, P. T. Tuyen, The composite Skyrme model with N=4, (no-published) 2006
  * [26] D. Q. Tuan, P. T. Tuyen, The supersymmetric composite skyrmion, (no-published) 2007
  * [27] Nikolaev V.A. , Tkachev O.G. , hep-ph/0109192 (2001)
  * [28] O.V.Pavlovskii, hep-th/0312343 (2003)
  * [29] R. A. Battye, P. M. Sutcliffe, hep-th/0602220 (2006)
  * [30] D. Q. Tuan, P. T. Tuyen, The formalism of gravitational composite Skyrme model, Talk in National Conference on Theoretical Physics 32, Nha Trang, Khanh Hoa, 2007 (no-published)