The textbook reading session of our group meeting this year is dedicated to Zee’s group theory book. When coming to Part III where the physical applications are discussed, I finally realize the power of all the beautiful mathematical theorems in representation theory discussed in Part II. The point is that symmetry of a system can help break the original problem into little pieces, making it much easier to deal with. I will make this point clear in this study note. To understand the following discussion, you need to know a matrix can be thought of some action on a vector to transform it (for a concrete example, recall rotation matrices in 2D or 3D space acting on position vector). You also need to know eigenvalue problems in linear algebra. It would be better if you are familiar with harmonic motion discussed in your mechanics class, but it is not necessary. Knowledge on quantum mechanics is certainly not necessary at all, although I have to mention the true power of symmetry and group theory lies in quantum mechanics.
So What is the Problem?
Let’s start. The physical problem I want to discuss is several point particles with equal mass connected by a spring. We follow Zee’s book and discuss two specific cases (see Figure 1 and Figure 2 below also):
- Two particles connected by a spring moving in 1 dimension
- Three particles connected with each other by three springs moving in 2 dimension
First case is the simplest, the purpose is to get us familiar with the techniques. After getting comfortable with the game, we will play with the second one to see the power of group theory!
If you are a master in classical mechanics, you can solve this problem directly without any fancy group theory knowledge, calculating all the eigenvectors (or modes) and the corresponding eigenvalues. However, it is not the point here. What we want to know is that if all we know is the symmetry properties of the system, what can we say about its dynamics? You might be very surprised if I tell you although you cannot know the numeric values of eigenvalues, you can calculate some eigenvectors or the subspace several eigenvectors span! You can know the degeneracy structure of this problem. See how much symmetry tells you!
Two Key Theorems
Group theory is the math used to study symmetries. To extract information from symmetry properties, we need to introduce two theorems in group representation theory. However, I should tell you what the words like symmetry, group and representation mean exactly.
Symmetry means if you perform some transformation to the system, it remains the same. For two particles in Figure 1, swapping particle $a$ and $b$ is such symmetry transformation, which is also the simplest permutation $a \rightarrow b \rightarrow a$. We denote it mathematically as $(ab)$. All the possible transformations of a given symmetry form a group, $G$. The word group suggests the most important features of group in mathematics: if you do two consecutive symmetry transformations, the system still remains unchanged. We can define this consecutive transformations as product of two original transformations so the previous sentence means that product of two elements in a group remains in the group. Put it mathematically, take any $g_1, g_2 \in G$, we have $g_1 g_2 \in G$. As a result, we call this feature closure under multiplication. Other two features of a group are 1) the existence of identity element $e$, which means doing nothing to the system, and 2) the existence of inverse for any element in the group, which means the ability to undo a transformation. For our simplest example of two particles, the group contains two elements, doing nothing and swapping of two particles. $G = \{ I, (ab)\} := S_2$ where we use $I$ to denote identity element and we give the group a name $S_2$
Group and elements in the group are an abstract way of thinking about symmetry transformations, no numbers involved yet. We surely want to use numbers to describe the symmetry transformations. The first thing comes to our mind is matrices, since we know rotations can be well described using matrices. Let’s examine our two particles system again. The dynamic variable is indicated in Figure 1, the displacement of two particles. We put two together to form a column vector $(x_a, x_b)^T$. After the element $(ab)$ acts on the system, we have $(x_b, x_a)$. As a result, we see immediately that we can use the follow matrix to describe, or represent the element $(ab)$,
\begin{align} \label{eq:ab}
D((ab)) = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}.
\end{align}
We say the two displacements of the two particles furnish a two-dimensional representation for group $S_2$. The identity element is of course represented by a two-dimensional identity matrix.
Now we are ready for the two key theorems. The first is Schur’s lemma. The statement is that for an irreducible representation $D^{(r)}(g)$, if there is some matrix $H$ such that $H D^{(r)}(g) = D^{(r)}(g) H$ for all $H$, then $H = \lambda I$ for some constant $\lambda$. To put it in words, if a matrix $H$ commutes with representations matrix for all element in group $G$, this matrix is proportional to ideneity matrix.
I won’t give a formal definition for irreducible representation here, just remember they are basic building blocks for a larger reducible representation $D(g)$, so any reducible representation can be written as direct sum of irreducible representations. Concretely, this means in suitable basis,
\begin{align} \label{eq:Ddiag}
D(g) = \begin{pmatrix}
\ldots & 0 & 0 & 0 \\
0 & D^{(r)}(g) & 0 & 0 \\
0 & 0 & D^{(s)}(g) & 0 \\
0 & 0 & 0 & \ldots
\end{pmatrix}.
\end{align}
In this case, if the matrix $H$ satisfies $H D(g) = D(g) H$, Schur’s lemma ensures $H$ also has block-diagonal form and each block is proportional to identity matrix,
\begin{align}
H = \begin{pmatrix}
\ldots & 0 & 0 & 0 \\
0 & \lambda^{(r)} I & 0 & 0 \\
0 & 0 & \lambda^{(s)} I & 0 \\
0 & 0 & 0 & \ldots
\end{pmatrix}.
\end{align}
So much for Schur’s lemma. The second is the most important theorem in representation theory: the Great Orthogonality Theorem. The statement is that, suppose we have a group $G$ with $N(G)$ group elements, and two of its irreducible representations $D^{(r)}, D^{(s)}$ whose dimension is $d_r, d_s$, then we have the following orthogonality relation,
\begin{align} \label{def:GOT}
\sum_{g} D^{(r)\dagger}(g) ^i_{\, j} D^{(s)}(g)^k_{\,l} = \frac{N(G)}{d_r} \delta^{rs} \delta^i_{\, l} \delta^k_{\, j}.
\end{align}
This is a very powerful theorem which poses a strong restriction to irreducible representations. Now it’s time to see what these two mathematical theorem tells us about the above two physical systems!
The Baby Problem: Two Particles in 1D
Let’s look at our baby problem. It has $S_2$ symmetry.
Since the potential energy of the system depends only on the difference of two dynamic variables $V = f(x_b – x_a)$, if we exchange two dynamic variables, the potential energy of the system is invariant. To put it in another form, the energy is in a quadratic form (why quadratic form? Remember the dynamic variable is deviation from equilibrium configuration, and the equilibrium configuration is by definition a local minimum of potential energy! ), if we put two dynamic variables together to form a vector $\mathbf{x} = (x_a, x_b)^T$,
\begin{align} \label{def:potEnergy}
V = \frac{1}{2}\mathbf{x}^T H \mathbf{x},
\end{align}
with some 2 by 2 matrix $H$. The invariance of energy under exchange of two particles indicates
\begin{align}
\mathbf{x}^T H \mathbf{x} \rightarrow \mathbf{x}^T D^T((ab)) H D((ab))\mathbf{x} = \mathbf{x}^T H \mathbf{x},
\end{align}
which means $D^T((ab)) H D((ab)) = H$ , where $D((ab))$ is given in eq. $\eqref{eq:ab}$. Notice this is an orthogonal matrix (it turns out to be a general result for representation matrices: we can always make them orthogonal matrix by some similar transformations), so we have $ H D((ab)) = D((ab)) H $. The other element in $S_2$ is represented by a 2 by 2 identity matrix which trivially commutes with matrix $H$. Now we are about to apply Schur’s lemma! However, is this two-dimensional representation of $S_2$ irreducible or not? It turns out for a group as simple as $S_2$, the irreducible representations can just be one dimensional (see Zee’s book for detailed explanation, for our purpose, just remember once a symmetry group is given, its irreducible representation can be known from purely group theoretical analysis). It has two 1-d irreducible representations and the so-called character table is shown in Figure 4 above. Character of a representation for a group element is defined as the trace of its representation matrix.
Just focus on last three columns. We can regard one column in character table as a column vector, where the first element denotes the character for group element I and the second for $(ab)$. Notice one irreducible representation is called $1$, where both group elements are represented by number 1. This representation is also known as trivial representation for obvious reasons. The other irreducible representation is called $\bar{1}$, where the odd permutation $(ab)$ is represented by $-1$ instead of $1$. If we denote our 2-dimensional representation as $2$, the corresponding column in character table should be $(2,0)^T$ (just trace over a 2-d identity matrix and the matrix in eq. $\eqref{eq:ab}$). Notice $(2,0)^T = (1,1)^T + (1,-1)^T$, so our 2-dimensional representation will break into one $1$ and one $\bar{1}$, or write abstractly as $2 = 1 \oplus \bar{1}$, or if we go to suitable basis,
\begin{align} \label{eq:D}
D((g)) = \begin{pmatrix}
D^{(1)}(g) & 0 \\
0 & D^{(\bar{1})}(g)
\end{pmatrix},
\end{align}
for $g = I, (ab)$. Schur’s lemma then ensure the $H$ matrix is in block-diagonal form in this basis.
\begin{align} \label{eq:H}
H = \begin{pmatrix}
\lambda^{(1)} & 0 \\
0 & \lambda^{(\bar{1})}
\end{pmatrix}.
\end{align}
You may think it is a trivial example since all of this are as expected. This is because $S_2$ group only has 1-dimensional representation so there is no degeneracy in $H$. We will see more interesting degeneracy structures in our three-particle problem, which is by no means trivial.
However, this is not the end of our baby problem. What if I tell you we can calculate the eigenvectors for $H$ without knowing any details of $H$ except its symmetry properties? You may laugh and give me the answer: one “zero mode” where two particles moving in the same direction $\mathbf{x} \sim (1,1)$, the other “breath mode” where two moves in opposite directions $\mathbf{x} \sim (1,-1)$. That’s right! But our intuition only works for the very simple example like this one, what I want to show is a systematic analysis of symmetry using group theory, which can be generalized to more complicated problems.
Projection Operator into desired Representation Mode
Notice in eq. $\eqref{eq:D}$ and $\eqref{eq:H}$, the eigenvalues of $H$ are labeled by two irreducible representations the 2-dimensional representation breaks into. This means the eigenmodes are also labeled by two irreducible representations. Next, we want to construct the projection operator into design mode, given the representation $D(g)$ determined naturally from our physical problem (here we mean the 2-dimensional representation furnished by two displacements). To this end, we will need to use the Great Orthogonality Theorem in eq. $\eqref{def:GOT}$. We set $i=j$ and sum over this index, let’s denote the character $\chi^{(r)}(g) = \sum_i D^{(r)}(g) ^i_{\, i}$, we have
\begin{align} \label{eq:chiD}
\sum_{g} \chi^{(r)*}(g) D^{(s)}(g)^k_{\,l} = \frac{N(G)}{d_r} \delta^{rs} \delta^k_{\, l}.
\end{align}
With this result in mind, we define the projection operator into the mode corresponding to $r$ representation as
\begin{align}
P_{(r)} =\frac{d_r}{N(G)} \sum_{g} \chi^{(r)*}(g) D(g).
\end{align}
Remember $d_r, N(G)$ and $\chi^{(r)}(g)$ are properties of the underlying symmetry group, while $D(g)$ is determined by the physical system in hand. To understand why the projection operator is defined in such way, just stare at eq. $\eqref{eq:Ddiag}$ and $\eqref{eq:chiD}$, I will repeat eq. $\eqref{eq:Ddiag}$ for your convenience here,
\begin{align}
D(g) = \begin{pmatrix}
\ldots & 0 & 0 & 0 \\
0 & D^{(r)}(g) & 0 & 0 \\
0 & 0 & D^{(s)}(g) & 0 \\
0 & 0 & 0 & \ldots
\end{pmatrix}. \nonumber
\end{align}
Notice when we sum over all $D(g)$ weighted by character $\chi^{(r)*}(g)$, all blocks in $D(g)$ except $D^{(r)}(\cdot)$ vanish, the end result is in the desired projection operator in its diagonal basis. Let’s try our brand-new projection operator on the baby problem. Remember for our baby problem,
\begin{align} \label{eq:Dbaby}
D(I) = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix},
D((ab)) = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},
\end{align}
while the character table of $S_2$ is given in Figure 4. The projection operator into trivial representation is
\begin{align}
P_{(1)} &=\frac{1}{2} \left( \chi^{(1)*}(I) D(I) + \chi^{(1)*}(ab) D(ab)\right) \nonumber\\
&= \frac{1}{2} \left(
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
+ \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \right) \nonumber\\
&=\frac{1}{2}\begin{pmatrix}
1 & 1 \\
1 & 1
\end{pmatrix}.
\end{align}
We read off the eigenmode corresponding to trivial representation as the first column of the above projection operator $\sim (1,1)^T$. If we try the $\bar{1}$ mode, we will have
\begin{align}
P_{(\bar{1})} &=\frac{1}{2} \left( \chi^{(\bar{1})*}(I) D(I) + \chi^{(\bar{1})*}(ab) D(ab)\right) \nonumber\\
&= \frac{1}{2} \left(
\begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}
– \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} \right) \nonumber\\
&=\frac{1}{2}\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}.
\end{align}
So we read off the eigenmode $\sim (1,-1)^T$. Remember all we use is the symmetry properties of our physical problem and two powerful theorems from group representation theory, and the mathematical operations we use are only multiplications and additions. We find the two eigenmodes successfully! Power of group theory!
Explicit Connection to the Old Way of Solving the Problem in Classical Mechanics
Up to this point, I assume the eigenvectors and eigenvalues of matrix $H$ in eq. $\eqref{def:potEnergy}$, that defines the potential energy are the eigenmodes and energy of the corresponding harmonic motion problem. In case it is not apparent for some of you, let’s solve this harmonic problem for a concrete problem in good old fashion. Assume the spring constant $k=1 $ so the potential energy is simply $V = 1/2 (x_b – x_a)^2 $, and it can be put into the following form,
\begin{align} \label{def: V}
V = \frac{1}{2} (x_a, x_b)
\begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}
\begin{pmatrix}
x_a \\
x_b
\end{pmatrix} = \frac{1}{2} \mathbf{x}^T H \mathbf{x},
\end{align}
where we denote $\mathbf{x} = (x_a, x_b)^T$, and from which we can read off the matrix $H $ as
\begin{align} \label{eq:Hsym}
H = \begin{pmatrix}
1 & -1 \\
-1 & 1
\end{pmatrix}.
\end{align}
Notice it is necessary to write the matrix $H $ in the symmetric form in order to make an explicit connection with our above discussion. The Lagrangian of the system is
\begin{align} \label{def:Lagr}
L = \frac{1}{2} \mathbf{x}^2 – \frac{1}{2} \mathbf{x}^T H \mathbf{x}.
\end{align}
Apply Euler-Lagrangian equation, the equation of motion of the system is
\begin{align} \label{eq:EOM}
\ddot{\mathbf{x}} + H \mathbf{x} = 0.
\end{align}
To find the modes of the system, we need to use the wave ansatz $\mathbf{x}(t) = \mathbf{u} e^{i\omega t}$, where the vector nature and time-dependent nature of the dynamic varible $\mathbf{x}$ are separated as factor $\mathbf{u}$ and $e^{i \omega t}$ respectively. The vibration factor $e^{i \omega t}$ is the reason why we call it a wave ansatz. Plug it into eq. \eqref{eq:EOM}, two time derivatives will bring down a factor $-\omega^2$, after cancelling the vibration factor, we get an eigenvalue problem
\begin{align}
H\mathbf{x} = \omega^2 \mathbf{x}.
\end{align}
So we have shown explicitly that the eigenvectors and eigenvalues of matrix $H $ is the eigenmodes and frequencies for the original harmonic problem.
That’s all for this post. Will leave the three particles in 2D problem for my next post, maybe in next month! There you will appreciate even more power of group theory! Stay tuned.
