Symmetries

Molecular Vibrations

Some additional groups

We already mentioned the orthogonal group $\OO(n)$ of all linear isometries (also called orthogonal transformations) of the euclidean space $\R^n$; it can be identified with a subgroup of the group $\Gl(n,\R)$ - the set of all invertible $n\times n$-matrices, more precisely $\OO(n)=\{U\in\Gl(n): U^*U=1\}$. The special orthogonal group $\SO(n)$ is the subgroup of $\OO(n)$ comprising all matrices $U\in\OO(n)$ satisfying $\det U=+1$. The complex analogue of $\OO(n)$ is the unitary group $\UU(n)$, it's the set of all linear isometries (also called unitary transformations) of the complex euclidean space $\C^n$ and can in turn be identified with the set of matrices $U\in\Gl(n,\C)$ such that $U^*U=1$. Remember: if $A=(a_{jk})$ is in $\Ma(n,\C)$, then $A^*$ is given by $(\bar a_{kj})$, i.e. $A^*$ is the complex conjugate of the transposed of $A$!

$\OO(n)$ is a subgroup of $\UU(n)$

Prove that for all $U\in\UU(n)$: $|\det U|=1$.

Prove that $\OO(n)$ is isomorphic to $\Z_2\times\SO(n)$.

$\SU(n)\colon=\{U\in\UU(n):\det U=+1\}$ is called the special unitary group . $\UU(1)$ is the unit circle $S^1\colon=\{z\in\C:|z|=1\}$ (the group operation is multiplication) which is isomorphic to the torus $\TT\colon=\R/2\pi\Z$, i.e. addition in $\R$ $\modul\ 2\pi$. The space of functions $f:\TT\rar\C$ may thus be identified with the space of $2\pi$-periodic functions $f:\R\rar\C$: if $f:\R\rar\C$ is $2\pi$-periodic then there is exactly one function $F:S^1\rar\C$ such that for all $x\in\R$: $F(e^{ix})=f(x)$. $F$ is continuous, smooth, etc. if and only if $f$ is continuous, smooth, etc.

Suppose $E$ is an $n$-dimensional complex euclidean space and $\UU(E)$ the group of all linear isometries $U:E\rar E$. Prove that $\UU(E)$ is isomorphic to $\UU(n)$.

Let $e_1,\ldots,e_n$ be the standard ONB of $\C^n$ and $b_1,\ldots,b_n$ any ONB of $E$. Define an isomorphism $J:\C^n\rar E$ by $Je_j=b_j$, then $U\mapsto JUJ^{-1}$ is an isomorphism of $\UU(n)$ onto $\UU(E)$.

The site representation of a molecule

The symmetry group of a molecule simply permutes atoms of the same type. This gives rise to what is called the site representation of the molecule in some space $\C^n$. It is a particular case of a unitary representation, i.e. a homomorphism $\Psi:G\rar\UU(n)$ into the group of unitary matrices. Of course, if the group of symmetries is generated by $g_1,\ldots,g_n$, then it suffices to determine $\Psi(g_1),\ldots,\Psi(g_n)$. In case $\Psi$ is injective the representation is called faithful . The site representation of a molecule need not be faithful: the reflection $\s_2$ in the symmetry group of $\chem{H_2O}$ (cf. subsection) leaves all three atoms invariant, i.e. $\Psi(\s_2)=1$. If $\Psi$ is a faithful representation, then the subgroup $\Psi(G)$ of $\UU(n)$ is isomorphic to $G$.

$\Psi:G\rar\UU(n)$ is faithful iff $\ker\Psi\colon=\{g\in G:\Psi(g)=1\}$ contains only the neutral element of $G$.

For example, any $g\in T_d$ permutes the hydrogen atoms of methane $\chem{CH_4}$ and leaves the carbon atom invariant, cf. the group $T_d$. Thus we obtain the following representation $\Psi$ of $T_d$ in $\C^4$ - we disregard the central carbon atom: $$ \Psi(C_3)=\left(\begin{array}{cccc} 1&0&0&0\\ 0&0&0&1\\ 0&1&0&0\\ 0&0&1&0 \end{array}\right), \Psi(C_2)=\left(\begin{array}{cccc} 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0 \end{array}\right), \Psi(S_4)=\left(\begin{array}{cccc} 0&0&0&1\\ 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0 \end{array}\right), \Psi(\s)=\left(\begin{array}{cccc} 0&1&0&0\\ 1&0&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right), $$

Check that the site representation of methane is faithful.

Determine the site representation of benzene $\chem{C_6H_6}$, i.e. the regular hexagon.

In math groups are somehow the abstract mathematical structures that can describe transformations without referring to any objects out there and a representation of a group is the action of the group on a particular set of objects. In the previous example the object is a model of the methane molecule.

The standard representation, $\det$ and the trivial representation

Since any symmetry group $G$ is a subgroup of $\OO(3)$, the canonical inclusion map $G\rar\OO(3)\rar\UU(3)$ is a unitary representation, it's called the standard representation - it's clearly faithful. Also $g\mapsto\det(g)$ is a representation and it's only 'inequivalent' to the trivial representation $g\mapsto1$ if $G$ containts both proper and improper rotations - both are in general not faithful.

The standard representation of $S(n)$

Assigning each permutation $\pi$ the matrix $P(\pi)=(p_{jk})$, where $p_{\pi(k)k}=1$ and otherwise $p_{jk}=0$, i.e. $P(\pi)e_k=e_{\pi(k)}$, we get a unitary representation $P$ of $S(n)$ in $\C^n$ (or an orthogonal representation of $S(n)$ in $\R^n$). Indeed for all $k$: $P(\s)P(\pi)e_k=P(\s)e_{\pi(k)}=e_{\s(\pi(k))}=e_{\s\pi(k)}=P(\s\pi)e_k$.

Vibrations

Let us assume $n$-atoms comprise a molecule and the potential energy of the molecule attains a (local) minimum when the atoms are located at positions $s_1,\ldots,s_n$ - that means the molecule is stable. The $j$-th atom we assign a $3$-dimensional euclidean space $E_j(=\R^3)$ and put $E\colon=E_1\oplus\cdots\oplus E_n(=\R^{3n})$. Let $x_1\in E_1,\ldots,x_n\in E_n$ be small pertubations of the atoms, so that the $j$-th atoms is located at $s_j+x_j$, then putting $$ X\colon=(x_1,\ldots,x_n) =(x_{11},x_{12},x_{13},x_{21},\ldots,x_{n3}) =(X_1,\ldots,X_{3n})\in\R^{3n}: $$ we get by Taylor's Theorem for sufficiently small pertubations $x_j$: $$ V(X)=V(x_1,\ldots,x_n) \sim V_0+\tfrac12\Hess V(0,\ldots,0)(X,X) =V_0+\tfrac12\la X,HX\ra $$ where $V_0=V(0,\ldots,0)$, $H$ is a positive linear operator on $E$ and $\la.,.\ra$ the canonical euclidean product on $E=\R^{3n}$, i.e. $H$ is self-adjoint and for all $X\in E$: $\la HX,X\ra\geq0$. The latter property follows from the fact that $V$ attains a local minimum at $X=(0,\ldots,0)$. Elements $X\in E$ essentially represent the positions of the atoms and are therefore called configurations, making $E$ the configuration space .

Suppose $E$ is real euclidean and $B:E\times E\rar\R$ a bi-linear symmetric mapping, i.e. for all $x,y\in E$ the mappings $u\mapsto B(x,u)$ and $v\mapsto B(v,y)$ are linear and $B(x,y)=B(y,x)$. Then there is exactly one self-adjoint operator $H\in\Hom(E)$ such that for all $x,y\in E$: $B(x,y)=\la Hx,y\ra$.

Compute the operator $H$ for $V(x_1,\ldots,x_n)=-(1+\sum x_j^2)^{-2}$.

We try to sidestep partial derivatives, which are a last resort! Since the Taylor expansion of $(1+x^2)^{-2}$ up to order $2$ is given by $1-2x^2$, we get $V(X)=-1+2\sum X_j^2+\cdots$ and thus $HX=4X$.

Compute the operator $H$ for $V(x_1,\ldots,x_n)=1-\exp(-\sum jx_j^2)$.

Compute the operator $H$ for $V(x_1,x_2,x_3)=\tfrac12\sum_{j < k}\norm{x_k-x_j}^2$.

A motion of the molecule is just a smooth curve $X:\R\rar E$ describing the positions of the atoms over time. For simplicity's sake let us assume that all atoms have the same mass, then the classical equation of motion (cf. example) gives us the following linear differential equation: $X^\dprime(t)=-HX(t)$, which has the unique solution: $$ X(t)=\cos(t\sqrt H)X(0)+\frac{\sin(t\sqrt H)}{\sqrt H}X^\prime(0)~. $$ where $\sqrt H$ is the unique positive linear operator $A$ that comes up to $A^2=H$. The operator $\sin(t\sqrt H)/\sqrt H$ is defined by power series expansion of the analytic function $\sin(tz)/z$ replacing $z$ with $\sqrt H$. In our context the uniqueness is not that significant, it's more important to know how to caclculate $\sqrt H$:

If $H=UDU^*$ for some $U\in\OO(n)$ and some diagonal matrix $D=diag\{a_1,\ldots,a_n\}$ with non negative entries $a_j$, then $\sqrt H=U\sqrt D U^*$, where $\sqrt D=diag\{\sqrt{a_1},\ldots,\sqrt{a_n}\}$.

Check by series expansion that for $A\in\Hom(E)$ (or $A\in\Ma(n,\C)$) and $x\in E$ ($x\in\C^n$) the derivative of the curve $c(t)\colon=\cos(tA)x$ is given by $c^\prime(t)=-\sin(tA)Ax$.

The roots of the eigen-values of $H$ are called the eigen-frequencies of the molecule, cf wikipedia. An important tool for computing the eigen-spaces of $H$ (or modes, as a physicist would put it) is the group of symmetries $G$ of the molecule: Recall, a symmetry $g\in G$ just permutes the atoms, thus we may think of it as a permutation $j\mapsto\pi(j)$ assigning each displacement vector $x_j$ of the $j$-th atom the displacement vector $g(x_j)$ of the $\pi(j)$-th atom. Now, putting $$ \Prn_k(\Psi(g)X) \colon=\d_{k\pi(j)}g(\Prn_j(X)) $$ we get an isometry $\Psi(g)$ of $E$ and it's pretty obvious that $\Psi(g)\in\OO(E)$ and for all $g,h\in G$: $\Psi(gh)=\Psi(g)\Psi(h)$, to state it differently: $\Psi$ is a homomorphism of the group $G$ into the group $\OO(E)$ of isometries of $E$, i.e. $\Psi$ is an orthogonal representation of $G$ in $E$ and a fortiori a unitary representation in $\C^{3n}$ - we will explicate this in more detail when discussing vibrations of $\chem{H_2O}$ and $\chem{CH_4}$ below. Since $G$ is the group of symmetries of the molecule, we hypothesize that the potential energy doesn't change under $\Psi(g)$, i.e.: $V(\Psi(g)X)=V(X)$. This is of course not a statement that can be proved, it's rather an assumption that can be made plausible. In our context it's equivalent (cf. lemma) to the statement: \begin{equation}\label{mvi1}\tag{MVI1} \forall g\in G:\qquad H\Psi(g)=\Psi(g)H \end{equation} i.e. the operator $H$ commutes with all isometries $\Psi(g)$, $g\in G$.

Approximate stable states

Remarkably a variety of problems in QM comes down to the same mathematical problem: suppose $n$ electrons, which only interact electrically, are bound by some molecule. What are the "stable states" of the $n$ electrons in the vicinity of positions $s_1,\ldots,s_n\in\R^3$ with minimal electric potential (we assume the atoms are pinned down, thus in this problem $x_1,\ldots,x_n$ are dislocations of the electrons, not of the atoms): approximating the electric potential by its second order Taylor polynomial we have to solve Schrödinger's eigen-equation on $E=\R^{3n}$: \begin{equation}\label{mvi2}\tag{MVI2} -\D\psi(X)+\tfrac12\la X,HX\ra\psi(X)=(E-V_0)\psi(X), \end{equation} where $V_0\colon=V(0,\ldots,0)$ and $E$ is the energy of the state - mathematically speaking: $\psi$ is an eigen-function and $E-V_0$ the corresponding eigen-value. To solve it, we only need to find the eigen-vectors and eigen-values of $H$; once we've got them, the eigen-functions $\psi$ are well known: Hermite polynomials multiplied by a Gaussian. If $\psi\in L_2(\R^{3n})$ is a normalized solution of \eqref{mvi2}, i.e. $\int|\psi|^2=1$, then the function $X\mapsto|\psi(X)|^2$ is a probability density and $|\psi(x_1,x_2,\ldots,x_n)|^2$ is the probability density for finding one electron in position $s_1+x_1$, another in position $s_2+x_2$ and so on! The stable state with the lowest energy $E$ is called the ground state; mathematically it's the eigen-function with lowest eigen-value!

Prove that the functions $e^{-ax^2}$ and $xe^{-ax^2}$ are eigen-functions of the operator $-\psi^\dprime(x)+4a^2x^2\psi(x)$. What are the corresponding eigen-values?

This just needs two lines in maxima: enter

diff(-exp(-a*x^2),x,2)+4*a^2*x^2*exp(-a*x^2);
diff(-x*exp(-a*x^2),x,2)+4*a^2*x^3*exp(-a*x^2);

and you will see that these two functions are indeed eigen-functions with eigen-values $2a$ and $6a$, respectively.

Prove that the functions $\psi_0(x)\colon=e^{-\sum a_jx_j^2}$ on $\R^n$ is an eigen-functions of the operator $-\D\psi(x)+4(\sum a_j^2x_j^2)\psi(x)$. What is the corresponding eigen-value? The function $\psi_0$ is actually the ground state!

Vibrations of $\chem{H_2O}$

We assume that the molecule lies in the $yz$-plane (cf. subsection).

Fixing the central oxygen atom our state space reduces to $\R^6$. Compute the matrices of the above representation (cf. subsection) of $C_{2v}$ in this state space.

Both the $C_2$ rotation and the reflection $\s_1$ interchange the hydrogen atoms and thus the matrix of the site-representation of these symmetries is $$ \left(\begin{array}{cc} 0&1\\ 1&0 \end{array}\right) $$ To get the matrices of $\Psi(C_2)$ and $\Psi(\s_1)$ we just have to substitute the matrices of $C_2$ and $\s_1$, respectively, for $1$ in the above matrix; this gives us two block matrices: $$ \left(\begin{array}{cc} 0 &C_2\\ C_2& 0 \end{array}\right), \left(\begin{array}{cc} 0&\s_1\\ \s_1&0 \end{array}\right), $$ which are tensor products - cf. section; in unblocked form they are given by $$ \left(\begin{array}{cccccc} 0 & 0&0&-1& 0&0\\ 0 & 0&0& 0&-1&0\\ 0 & 0&0& 0& 0&1\\ -1& 0&0& 0& 0&0\\ 0 &-1&0& 0& 0&0\\ 0 & 0&1& 0& 0&0 \end{array}\right), \left(\begin{array}{cccccc} 0& 0&0&1& 0&0\\ 0& 0&0&0&-1&0\\ 0& 0&0&0& 0&1\\ 1& 0&0&0& 0&0\\ 0&-1&0&0& 0&0\\ 0& 0&1&0& 0&0 \end{array}\right) $$ The reflection $\s_2$ leaves both hydrogen atoms invariant and therefore the matrix of $\Psi(\s_2)$ is given by $$ \left(\begin{array}{cc} \s_2&0\\ 0&\s_2 \end{array}\right) =\left(\begin{array}{cccccc} -1& 0&0& 0& 0&0\\ 0& 1&0& 0& 0&0\\ 0& 0&1& 0& 0&0\\ 0& 0&0&-1& 0&0\\ 0& 0&0& 0& 1&0\\ 0& 0&0& 0& 0&1 \end{array}\right) $$

Vibrations of tetrahedral molecules like methane $\chem{CH_4}$

We retain the notation from subsection. Fixing the central $C$-atom and labeling the surrounding $H$ atoms $1,2,3,4$ we see that $g\in\{E,C_3,C_2,S_4,\s\}$ permutes the hydrogen atoms as follows: $$ E:(1234),\quad C_3:(1342),\quad C_2:(3412),\quad S_4:(2341),\quad \s:(2134)~. $$ These give rise to associated permutation matrices in $\OO(4)$ - the site representation of $\chem{CH_4}$ (disregarding the central $C$ atom): $$ \left(\begin{array}{cccc} 1 & 0& 0& 0\\ 0 & 0& 0& 1\\ 0 & 1& 0& 0\\ 0 & 0& 1& 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0& 1& 0\\ 0 & 0& 0&1\\ 1& 0& 0& 0\\ 0 &1& 0& 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0& 0&1\\ 1& 0& 0& 0\\ 0 &1& 0& 0\\ 0 & 0&1& 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 1& 0& 0\\ 1 & 0& 0& 0\\ 0 & 0& 1& 0\\ 0 & 0& 0& 1 \end{array}\right)~. $$ Thus we get an orthogonal representation of $T_d$ in $\R^{12}$ and a fortiori a unitary representation of $T_d$ in $\C^{12}$. We write out the matrices of $\Psi(C_3)$, $\Psi(C_2)$, $\Psi(S_4)$ and $\Psi(\s)$ in block form: $$ \left(\begin{array}{cccc} C_3& 0& 0& 0\\ 0 & 0& 0&C_3\\ 0 &C_3& 0& 0\\ 0 & 0&C_3& 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0&C_2& 0\\ 0 & 0& 0&C_2\\ C_2& 0& 0& 0\\ 0 &C_2& 0& 0 \end{array}\right), \left(\begin{array}{cccc} 0 & 0& 0&S_4\\ S_4& 0& 0& 0\\ 0 &S_4& 0& 0\\ 0 & 0&S_4& 0 \end{array}\right), \left(\begin{array}{cccc} 0 & \s& 0& 0\\ \s & 0& 0& 0\\ 0 & 0& \s& 0\\ 0 & 0& 0& \s \end{array}\right)~. $$ Again, these are just the tensor products of the permutation matrices of the site representation and the symmetries itself, cf. section.

Compute all eigen-values and eigen-spaces of $\Psi(\s)\in\UU(12)$.

The following series of maxima commands will essentially do it:

s: matrix([0,0,-1],[0,1,0],[-1,0,0]);
n: zeromatrix(3,3);
A: matrix([n,s,n,n],[s,n,n,n],[n,n,s,n],[n,n,n,s]);
a: mat_unblocker(A);
eigenvectors(a);

or in sage:

s=matrix([[0,0,-1],[0,1,0],[-1,0,0]])
a=block_matrix([[0,s,0,0],[s,0,0,0],[0,0,s,0],[0,0,0,s]])
a.eigenvalues()
a.eigenspaces_right()

instead of the last two commands you may use:

a.eigenmatrix_right()

or by means of the NumPy library in sage:

import numpy
E,V=numpy.linalg.eig(a)
print(E);print(V)

In any case: the eigen-values are $-1$ ($5$-fold) and $+1$ ($7$-fold) and the eigen-vectors are the columns of the following matrix: $$ \left(\begin{array}{cccccccccccc} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & -1 & 0 \end{array}\right) $$

Compute all eigen-values and eigen-spaces of $\Psi(C_3)\in\UU(12)$.

Determine the block matrices $\Psi(C_6)$ and $\Psi(\s_1)$ of the representation $\Psi:C_{6v}\rar\UU(18)$ for the vibrations of the benzene molecule $\chem{C_6H_6}$. $C_6$ and $\s_1$ denote the two generators of the dihedral group $C_{6v}$, cf. subsection.

Determine the block matrices $\Psi(C_4)$ and $\Psi(S_6)$ of the representation $\Psi:O_h\rar\UU(18)$ for the vibrations of sulfur hexafluoride $\chem{SF_6}$ (cf. wikipedia). $C_4$ and $S_6$ denote two generators of the octahedral group $O_h$, cf. subsection.

Computing eigenvalues exactly in sage

Usually the unitary matrices $U$ we will run across have finite order, i.e. there is some $n\in\N$ such that $U^n=1$. To compute its eigen-values exactly in sage you may possibly consider this matrices as matrices over the cyclotomic field $\Q(e^{2\pi i/N})$ - the field generated by $\Q$ and the complex number $e^{2\pi i/N}$, also known as the cyclotomic field $\Q(N)$ of order $N$, where $N$ is some suitable multiple of $n$:

Compute all eigen-values and eigen-spaces of $U\in\UU(5)$: $$ U=\left(\begin{array}{ccccc} 0&1&0&0&0\\ 0&0&0&0&1\\ 0&0&0&1&0\\ 0&0&1&0&0\\ 1&0&0&0&0 \end{array}\right) $$

The order of the matrix is $6$ and since all the entries of the matrix are certainly elements of $\Q(e^{2\pi i/6})$, we choose the cyclotomic field of order $6$:

F.<zeta>=CyclotomicField(6)
U=matrix(F,[[0,1,0,0,0],[0,0,0,0,1],[0,0,0,1,0],[0,0,1,0,0],[1,0,0,0,0]])
U.eigenspaces_right()

This will give you the eigen-values: $\z-1,-\z,-1,1,1$, where $\z=e^{2\pi i/6}$ and the corresponding eigen-vectors, which we omit.

Compute all eigen-values and eigen-spaces of $U\in\UU(3)$: $$ U=\left(\begin{array}{cc} 0&i&0\\ 0&0&1\\ -i&0&0 \end{array}\right) $$

The order of the matrix is $3$ but $i$ is not an elements of $\Q(e^{2\pi i/3})$, thus we choose the cyclotomic field of order $12$ because this field contains both $e^{2\pi i/3}$ and $\pm i$:

F.<zeta>=CyclotomicField(12)
U=matrix(F,[[0,I,0],[0,0,1],[-I,0,0]])
U.eigenspaces_right()

The eigen-values of $U$ are: $1,\z^2-1,-\z^2$ where $\z=e^{2\pi i/12}$. The eigen-values must be numbers in $\Q(3)$, indeed: $\z^2-1=e^{2\pi/3}$ and $-\z^2=e^{4\pi i/3}$.

Compute all eigen-values and eigen-spaces of $U\in\UU(5)$: $$ U=\left(\begin{array}{ccccc} 0&0&1&0&0\\ 0&0&0&0&1\\ 0&1&0&0&0\\ 1&0&0&0&0\\ 0&0&0&1&0 \end{array}\right) $$

Homogeneous and Harmonic Homogeneous Polynomials

Cf. students version in german

Homogeneous Polynomials

This section is not about representations per se but only about particular spaces, the groups may be represented in. For $l\in\N_0$ let ${\cal P}_l$ be the space of $l$-homogeneous complex-valued polynomial functions on $\R^3$. Then ${\cal P}_l$ is a complex vector-space and for $\a=(\a_1,\a_2,\a_3)\in\N_0^3$ satisfying $|\a|\colon=\a_1+\a_2+\a_3=l$ the polynomials $P_\a=x^\a\colon=x_1^{\a_1}x_2^{\a_2}x_3^{\a_3}$ form a basis of ${\cal P}_l$. Its dimension is $\dim{\cal P}_l=(l+2)(l+1)/2$: indeed write the sequence $\a_1,\a_2,\a_3$ as $1\ldots101\ldots101\ldots1$ with ones in the first $\a_1$ slots followed by zero, $\a_2$ ones in the following slots followed by zero and $\a_3$ ones in the remaining slots. Thus we have to fill in two zeros in $l+2$ slots!

What is the dimension of the space of $l$-homogeneous polynomial functions on $\R^n$?

Suppose $G$ is any subgroup of the orthogonal group $\OO(3)$; for $P\in{\cal P}_l$ we put \begin{equation}\label{hhpeq1}\tag{HHP1} \G(g)P(x)\colon=P(g^{-1}x) \end{equation} Then $\G$ is a representation of $G$ similar to the one we encountered in subsection. In order to turn it into a unitary representation we need to define a euclidean product on ${\cal P}_l$. So let us take the following product: \begin{equation}\label{hhpeq2}\tag{HHP2} \la P,Q\ra \colon=\frac1{c_l}\int_{\R^3}P(x)\cl{Q(x)}e^{-\Vert x\Vert^2/2}\,dx =\int_{S^2}P(x)\cl{Q(x)}\s(dx)~. \end{equation} where $c_l$ is a normalizing constant and $\s$ denotes the normalized surface measure on $S^2$, i.e. for all measurable subsets $A$ of $S^2$: $\s(A)\colon=\l((0,1]A)/\l(B)$, where $\l$ denotes Lebesgue measure in $\R^3$, $(0,1]A\colon=\{t\z:\,t\in(0,1],\z\in A\}$ and $B=(0,1]S^2$ the euclidean unit ball:

With respect to this euclidean product $\G$ is unitary because of rotational invariance of Lebesgue-measure.

The constant $c_l$ is given by $4\pi\G(l+3/2)2^{l+1/2}$, where $\G$ denotes gamma function (cf. wikipedia); in particular: $c_0=(2\pi)^{3/2}$ and $c_{l+1}=(2l+3)c_l$.

We do not check the last two statements. By integration in spherical coordinates: $r\in\R^+$, $\z\in S^2$, we have for all integrable functions $f:\R^3\rar\C$: $$ \int_{\R^3} f(x)\,dx=4\pi\int_0^\infty r^2\int_{S^2}f(r\z)\,\s(d\z)\,dr $$ For $P,Q\in{\cal P}_l$ we put $f(x)=P(x)\bar Q(x)e^{-\Vert x\Vert^2/2}$; the above formula yields: $$ \la P,Q\ra =\frac{4\pi}{c_l}\int_0^\infty r^{2+2l}e^{-r^2/2} \,dr\int_{S^2}P(\z)\bar Q(\z)\,\s(d\z) $$ and therefore $$ c_l=4\pi\int_0^\infty r^{2+2l}e^{-r^2/2}\,dr=4\pi\G(l+3/2)2^{l+1/2}~. $$

The polynomials $x^\a$, $|\a|=l$, form an orthogonal basis of ${\cal P}_l$ for $l=1$ but not for $l=2$. Solution by T. Speckhofer

The mapping $\Psi(z)f\colon=f(zw)$ is a unitary representation of $S^1=\UU(1)$ in the space $L_2(\C,\g)$ with the gaussian measure $\g(dw)=(2\pi)^{-1}e^{-|w|^2/2}\,dw$. 2. Verify by means of exam that for smooth functions $f$: $$ \ttdl t0\Psi(e^{it})f(w)=iw\pa_wf(w)-i\bar w\pa_{\bar w}f(w) $$ Solution by T. Speckhofer

Let $E$ be the space of all complex valued polynomials on $\R$ of degree less or equal $n$ and $P\in\Hom(E)$ the operator $Pf\colon=-if^\prime$. Compute $\tr P$ and $\det P$.

Harmonic homogeneous polynomials

Let ${\cal H}_l$ be the subspace of ${\cal P}_l$, containing all harmonic polynomials, i.e. $P\in{\cal H}_l$ iff $P\in{\cal P}_l$ and $\D P=0$, cf wikipedia. Since for all $g\in\OO(3)$: $\D P\circ g=(\D P)\circ g$ - remember: all $g\in\OO(3)$ are symmetries of the Laplacian, the above representation $\G(g)P\colon=P\circ g^{-1}$ restricts to a representation of $G$ in the spaces ${\cal H}_l$, $l\in\N_0$.

The constant polynomial and the polynomials $x,y,z$ form orthogonal basis of the spaces ${\cal H}_0$ and ${\cal H}_1$ respectively. Moreover the polynomials $xy,yz,zx,x^2-y^2,2z^2-x^2-y^2$ form an orthogonal basis of ${\cal H}_2$. The corresponding normalized polynomials will be denoted by $s_1$, $p_1,p_2,p_3$ and $d_1,\ldots,d_5$ respectively. We have e.g. $s_1=1$, $p_1=\sqrt3x$, $d_1=\sqrt{15}xy$, $d_4=\sqrt{15/4}(x^2-y^2)$, $d_5=\sqrt{5/4}(2z^2-x^2-y^2)$.

We do not check orthogonality! By putting $Q=P\in{\cal P}_l$ in \eqref{hhpeq2} we obtain: $$ \int_{S^2}|P(\z)|^2\,\s(d\z) =\frac1{c_l}\int_{\R^3}|P(x)|^2e^{-\Vert x\Vert^2/2}\,dx~. $$ For $P(\z)=x^2$ the right hand side equals $3(2\pi)^{3/2}/c_2=1/5$ and for $P(\z)=xy$: $(2\pi)^{3/2}/c_2=1/15$, it follows that $$ \int_{S^2}x^4\,\s(d\z)=\frac1{5} \quad\mbox{and}\quad \int_{S^2}x^2y^2\,\s(d\z)=\frac1{15} $$ Our first goal is to determine the dimensions of the spaces ${\cal H}_l$. We'll employ the following simple result in linear algebra:

Suppose $E$ is a finite dimensional Hilbert-space and $A\in\Hom(E)$. Then $(\im A)^\perp=\ker A^*$ and $\im A^*=(\ker A)^\perp$. What can you say in the infinite dimensional case for bounded $A$?

For every $Q\in{\cal P}_l$ there exist uniquely determined polynomials $H\in{\cal H}_l$ and $P\in{\cal P}_{l-2}$, such that $Q=H+\Vert x\Vert^2 P$. The dimension of ${\cal H}_l$ thus equals $\dim{\cal P}_l-\dim{\cal P}_{l-2}=2l+1$.

$\proof$ We introduce a new complex euclidean product: Suppose $P=\sum_{|\a|=l}c_\a x^\a\in{\cal P}_l$ is a homogeneous polynomial of deree $l$; let $P(\pa)$ be the partial differential operator $\sum_{|\a|=l}c_\a\pa^\a$, then for $Q=\sum_{|\a|=l}d_\a\pa^\a$ we put $$ [P,Q]\colon=P(\pa)\bar Q =\sum_{|\a|=|\b|=l} c_\a \bar d_\b\pa^\a x^\b =\sum_{|\a|=l}\a!c_\a\bar d_\a $$ and get another euclidean product on ${\cal P}_l$. Now the kernel of $\D:{\cal P}_l\rar{\cal P}_{l-2}$ is ${\cal H}_l$ and for all $P\in{\cal P}_{l-2}$ and all $Q\in{\cal P}_l$ we have $$ [\D^*P,Q] =[P,\D Q] =P(\pa)\D\bar Q =\D P(\pa)\bar Q =[\Vert x\Vert^2P,Q] $$ and thus the adjoint $\D^*$ of $\D$ is $P\mapsto\Vert x\Vert^2P$. Since all spaces are finite dimensional, we conclude: $$ {\cal P}_l =\ker\D\oplus(\ker\D)^\perp ={\cal H}_l\oplus\im\D^*~. $$ $\eofproof$

The polynomials $x^\a/\sqrt{\a!}$, $|\a|=l$, form an orthonormal basis of $({\cal P}_l,[.,.])$.

The polynomials $xy,yz,zx,x^2-y^2,2z^2-x^2-y^2$ form an orthogonal basis of $({\cal H}_2,[.,.])$. Normalize these polynomials.

What is the dimension of the space of $l$-homogeneous harmonic polynomials on $\R^n$?

In dimension $n=2$ we have $\dim{\cal H}_l=2$. Show by means of exam that the polynomials $p_l(x,y)=\Re(x+iy)^l$ and $q_l(x,y)\colon=\Im(x+iy)^l$ form an orthogonal basis.

A homogeneous harmonic polynomial $H$ is certainly determined by its restriction $H|S^2$, which is called a spherical harmonic on $S^2$. Apparently, the previous lemma implies, that the restriction of every homogeneous polynomial to $S^2$ is a linear combination of spherical harmonics. But we are going to prove that much more is true: the spaces ${\cal H}_l$ form an orthogonal decomposition of the Hilbert-space $L_2(S^2)\colon=L_2(S^2,\s)$:

Suppose $f$ is a smooth function and $X$ a vector field on $\R^n$. Prove that $\divergence(fX)=\la X,\nabla f\ra+f\divergence(X)$.

The spaces ${\cal H}_l$ of spherical harmonics of dergee $l\in\N_0$ form an orthogonal decomposition of the Hilbert-space $L_2(S^2)$.

$\proof$ Assume $k\neq l$, $P\in{\cal H}_k$, $Q\in{\cal H}_l$ and let $N$ be the outer unit normalfield of $S^2=\pa B$. Then we get by homogeneity: $\la N,\nabla P\ra=kP$ and $\la N,\nabla Q\ra=lQ$. By the divergence theorem we therefore have \begin{eqnarray*} (k-l)\int_{S^2} P\bar Q\,d\s &=&\int_{S^2}\la N,\bar Q\nabla P-P\nabla\bar Q\ra\,d\s\\ &=&\frac1{4\pi}\int_{B}\divergence(\bar Q\nabla P-P\nabla\bar Q) =\frac1{4\pi}\int_{B}\bar Q\D P-P\D\bar Q=0~. \end{eqnarray*} By the previous remark and the Stone Weierstraß Theorem the sum $\sum{\cal H}_l$ is dense in $C(S^2)$ and thus dense in $L_2(S^2)$. $\eofproof$

Can you explain the last argument of the previous proof in detail?

Remark: Apparently the theorem holds in any dimension!
The significance of spherical harmonics in chemistry is due to the fact that they in a way describe the directional dependency of what is called an atomic orbital of a spherically symmetric nucleus. Atomic orbital is nothing but another name of a particular stable state: a stable state of an electron bound by a nucleus. The state space is the Hilbert-space $L_2(\R^3)$ and it's known that all stable states $\psi$ can be described by functions of the form $\psi(r\z)\colon=u(r)H(\z)$, $r>0$, $\z\in S^2$ - this is due to the fact that in spherical coordinates $(r,\z)$ the Hamiltonian of a spherically symmetric nucleus is a sum of tensor products (cf. section in particular exam). $u$ reflects the radial dependency and $H$ the directional dependency; the latter are exactly the spherical harmonics. Thus if $H:S^2\rar\C$ denotes a normalized spherical harmonic, then $|H(\z)|^2$ is interpreted as the probability density to come across the electron in state $\psi$ in the direction $\z\in S^2$: for any (measurable) subset $A$ of $S^2$ the integral $\int_A|H(\z)|^2\s(d\z)$ is the probability of finding the electron in the cone $\R^+A\colon=\{ra:r > 0,a\in A\}$ generated by $A$, cf. wikipedia.

Work out the representation \eqref{hhpeq1} for $C_{2v}$ in ${\cal H}_l$ for $l=0,1,2$.

We determine the matrices of this representations with respect to the orthonormal bases given in example. First of all we have $C_2^{-1}(x,y,z)=(-x,-y,z)$, $\s_1^{-1}(x,y,z)=(x,-y,z)$, $\s_2^{-1}(x,y,z)=(-x,y,z)$.
For $l=0$ we get the trivial representation: $\G(g)=1$.
For $l=1$ the matrices of $\G(C_2),\G(\s_1),\G(\s_2)$ are given by $$ \left(\begin{array}{ccc} -1&0&0\\ 0&-1&0\\ 0&0&1 \end{array}\right),\quad \left(\begin{array}{ccccc} 1&0&0\\ 0&-1&0\\ 0&0&1 \end{array}\right),\quad \left(\begin{array}{ccccc} -1&0&0\\ 0&1&0\\ 0&0&1 \end{array}\right) $$ For $l=2$ we have: $\G(C_2)d_1=d_1$, $\G(C_2)d_2=-d_2$, $\G(C_2)d_3=-d_3$, $\G(C_2)d_4=d_4$ and $\G(C_2)d_5=d_5$, $\G(\s_1)d_1=-d_1$, $\G(\s_1)d_2=-d_2$, $\G(\s_1)d_3=d_3$, $\G(\s_1)d_4=d_4$ and $\G(\s_1)d_5=d_5$, $\G(\s_2)d_1=-d_1$, $\G(\s_2)d_2=d_2$, $\G(\s_2)d_3=-d_3$, $\G(\s_2)d_4=d_4$ and $\G(\s_2)d_5=d_5$. Thus we get for the matrices of $\G(C_2),\G(\s_1),\G(\s_2)$: $$ \left(\begin{array}{ccccc} 1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{array}\right),\quad \left(\begin{array}{ccccc} -1&0&0&0&0\\ 0&-1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{array}\right),\quad \left(\begin{array}{ccccc} -1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&-1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1 \end{array}\right) $$

Atomic orbitals

The stable states of a hydrogen atom are usually expressed by a triple of natural numbers: $n,l$ and $m$ - the quantum numbers: 1. $n\in\N$ is called the first quantum number, it determines the energy $E_n$ of the stable state: $E_n=-\g/n^2$. 2. $l\in\N_0$ is called the angular momentum quantum number and, of course, indicates the angular momentum; mathematically it's just an index for the spherical harmonics of dergee $l$. Given $n$ the number $l$ must be one of the numbers $0,1,\ldots,n-1$. 3. $m\in\Z$ is called the magnetic quantum number : for given $l$ we have $m\in\{-l,-l+1,\ldots,0,1,\ldots,l\}$; these are exactly $2l+1$ numbers and thus they build a convenient index set for a basis of the space ${\cal H}_l$ of spherical harmonics of degree $l$.
The energy $E_n$ of a stable state is an eigen-value of an operator - the Hamiltonian of the hydrogen atom - and the state itself is a corresponding eigen-function. Let us say a few words on the multiplicity of the eigen-value $E_n$: For every $n$ there are exactly $n$ values of $l$, namely $0,\ldots,n-1$; now for each of these values $l$ there are $2l+1$ pairwise orthonormal eigen-functions - an orthonormal basis of ${\cal H}_l$. All these functions give rise to eigen-functions of the Hamiltonian of the hydrogen atom and all these eigen-functions form a basis for the eigen-spaces of this Hamiltonian. Therefore the multipliticity of $E_n$ is: $1+3+5+\ldots+(2n-1)=n^2$. Now there is an additional quantum number: the spin, it can only have two values, which we simply denote by $\pm1/2$. It follows that the multipliticity of $E_n$ is $2n^2$ - we see that each energy level $E_n$ can only be occupied by at most $2n^2$ electrons. This is one of the basic principles of QM: two electrons bound to a nucleus, which are described by the same four quantum numbers, are identical.
A note on notational inconveniences: Let us lable the stable states of an electron in an hydrogen atom: we need to have a lable for the energy - we take $n\in\N$, a lable for the angular momentum - we take $l\in\{0,1,\ldots,n-1\}$, a lable for an orthonormal basis of ${\cal H}_l$ - we take $m\in\{-l,\ldots,l\}$ and a lable for the spin - we take $s\in\pm1/2$. In math these are labels for some function, so we would come up with something like $\psi_{n,l,m,s}$, but physicists familiar with the bra-ket notation just use $|n,l,m,s\ra$ or $|n,l,m,\uar\ra$ or any combination of four symbols, which properly lables the stable states. Apropos of notation: in chemistry or physics electrons whose first quantum number is $n$ and whose angular momentum quantum number is $l=0$, $l=1$, $l=2$ and $l=3$ respectively are called $ns$-, $np$-, $nd$- and $nf$ electrons. Electrons with rotational invariant orbits, i.e. directionally independent orbits are $s$ electrons. Sample points of a typical $f$-orbital are animated below: high density of points indicates a larger probability of finding the electron in the corresponding direction (the distribution was generated by C code).

Now, what's about more complex atoms like nitrogen $\chem{N}$: in a first approximation nitrogen will be viewed as a helium atom with $5$ additional protons. As a consequence the two electrons get bounded more tightly. This gives us a $5$-fold positively charged nucleus, whose potential is assumed to be rotational invariant. But this assumption in turn implies that the directional dependency of the eigen-functions is exactly the same as the directional dependency of the eigen-functions of hydrogen. We won't elaborate on this statement, we just take it for granted! It remains to adjoin $5$ electrons to this nucleus. For these electrons there are $2\cdot 2^2=8$ possible states of minimal energy. Now we further assume that all these $5$ electrons can be adjoint independently, i.e. without changing the potential essentially. Well that's obviously a crude approximation, nevertheless it works quite well. Cf. e.g. Feynman Lectures III, 19, Periodic Table. In summary, nitrogen in its base configuration is written as $\chem{He.2s^2.2p^3}$, meaning: it's perceived as helium $\chem{He}$ plus two electrons in a $2s$ orbital and three electrons in a $2p$ orbital. Here are the configurations of the first elements: $\chem{H}$: $\chem{1s^1}$, $\chem{He}$: $\chem{1s^2}$, $\chem{Li}$: $\chem{He.2s^1}$, $\chem{Be}$: $\chem{He.2s^2}$, $\chem{B}$: $\chem{He.2s^2.2p^1}$, $\chem{C}$: $\chem{He.2s^2.2p^2}$, $\chem{N}$: $\chem{He.2s^2.2p^3}$, $\chem{O}$: $\chem{He.2s^2.2p^4}$, $\chem{F}$: $\chem{He.2s^2.2p^5}$, $\chem{Ne}$: $\chem{He.2s^2.2p^6}$, $\chem{Na}$: $\chem{Ne.3s^1}$, etc.
There is one point we have to keep in mind when simplifying things this way: in our simplification of nitrogen all $8$ possible states have the same energy but except for $\chem{H}$ and $\chem{He}$ the energy of the states increases as their angular momentum number increases, i.e. a $2s$ electron has lower energy than a $2p$ electron. Consequently it may happen, that the energy of a $3d$-state is higher than the energy of a $4s$-state: for example potassium $\chem{K}$ has configuration: $\chem{Ne.3s^2.3p^6.4s^1}$ and not $\chem{Ne.3s^2.3p^6.3d^1}$, also calcium $\chem{Ca}$ has configuration: $\chem{Ne.3s^2.3p^6.4s^2}$ and not $\chem{Ne.3s^2.3p^6.3d^2}$. Also the next inert gas argon $\chem{Ar}$ has configuration: $\chem{Ne.3s^2.3p^6}$ and there are still $2(2\cdot2 +1)=10$ possible $3d$-states. This indicates that there is a big energy gap seperating $3d$ electrons from $3p$ electrons.

Molecular orbitals: $\chem{H_2O}$

The molecular orbital method, which we refer to (cf. wikipedia, ZLibrary), considers valence electrons as being distributed over the whole molecule. Instead of giving a detailed account on this theory, we highlight the method by a simple example: $\chem{H_2O}$. Again, we retain the notions from subsection and assume that the molecule lies in the $yz$-plane with the oxygen placed at the center! In its base configuration oxygen $\chem{O}$ is essentially helium $\chem{He}$ with $6$ additional electrons, two in a $2s$-orbital and four in a $2p$-orbital. For the state space $E_O$ of oxygen we therefore choose $E_O\colon={\cal H}_0\oplus{\cal H}_1$ - one $2s$-orbital and three $2p$-orbitals (notice that each of these orbitals can be occupied by two electrons with opposite spin). As orthonormal basis we choose $$ s_O=1,\quad p_1=\sqrt3x,\quad p_2=\sqrt3y\quad\mbox{and}\quad p_3=\sqrt3z~. $$ Each of the two hydrogen atoms has a $1s$ electron, thus we choose one dimensional state spaces $E_1=\C s_1$ and $E_2=\C s_2$. The symmetry group of water $\chem{H_2O}$ is $C_{2v}$; for each symmetry $g\in C_{2v}$ we define a unitary operator $\Psi(g)\in\UU(E_1\oplus E_2\oplus E_O)$ as follows: $g$ leaves the central oxygen atom invariant and for $Y\in E_O$ we set $\Psi(g)Y\colon=Y\circ g^{-1}$; $g$ either interchanges the hydrogen atoms or leaves them invariant; in the former case $\Psi(g)$ interchanges the spaces $E_1$ and $E_2$ thus: \begin{eqnarray*} &&\Psi(C_2)s_1=s_2, \Psi(C_2)s_2=s_1, \Psi(C_2)s_O=s_O, \Psi(C_2)p_1=-p_1, \Psi(C_2)p_2=-p_2, \Psi(C_2)p_3=p_3,\\ &&\Psi(\s_1)s_1=s_2, \Psi(\s_1)s_2=s_1, \Psi(\s_1)s_O=s_O, \Psi(\s_1)p_1=p_1, \Psi(\s_1)p_2=-p_2, \Psi(\s_1)p_3=p_3,\\ &&\Psi(\s_2)s_1=s_1, \Psi(\s_2)s_2=s_2, \Psi(\s_2)s_O=s_O, \Psi(\s_2)p_1=-p_1, \Psi(\s_2)p_2=p_2, \Psi(\s_2)p_3=p_3~. \end{eqnarray*} This representation is similar to the one discussed in subsection and in particular subsection. Eventually we get for the matrices of $\Psi(C_2),\Psi(\s_1),\Psi(\s_2)$: $$ \left(\begin{array}{cccccc} 0&1&0&0&0&0\\ 1&0&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&-1&0&0\\ 0&0&0&0&-1&0\\ 0&0&0&0&0&1 \end{array}\right), \left(\begin{array}{cccccc} 0&1&0&0&0&0\\ 1&0&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&-1&0\\ 0&0&0&0&0&1 \end{array}\right), \left(\begin{array}{cccccc} 1&0&0&0&0&0\\ 0&1&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&-1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{array}\right) $$ So far we've got only a representation of the symmetry group of water in a six dimensional space. The molecular orbitals we are looking for are linear combinations of the five chosen atomic orbitals - that's what molecular orbital theory postulates, but which linear combinations? Well, the exact answer involves the Hamiltonian of the molecule or at least some simplification of this animal and it's not that kind of thing we're after. We simply want to exploit symmetry to limit the possible variety of linear combinations without even knowing a smidgon about the Hamiltonian, beside its symmetries of course and that means: the Hamiltonian $H$ commutes with the representation $\Psi$, i.e. for all $g\in C_{2v}$: $[H,\Psi(g)]=0$. That's our only prerequisite (cf. section)!

Find the representation $\Psi:C_{3v}\rar\UU(E_N\oplus E_1\oplus E_2\oplus E_3)$ of the symmetry group of ammonia $\chem{NH_3}$, where $E_N={\cal H}_0\oplus{\cal H}_1$ and $E_1=E_2=E_3={\cal H}_0$.

$C_3$ permutes the hydrogen atoms cyclically: $1\to2\to3\to1$; thus $\Psi(C_3)$ maps the $s$-orbital $s$ of nitrogen onto itself and the $s$-orbitals $s_1,s_2,s_3$ of the hydrogen atoms are permuted: $\Psi(C_3)s_1=s_2$, $\Psi(C_3)s_2=s_3$ and $\Psi(C_3)s_3=s_1$. As for the remaining three $p$-orbitals $p_1=\sqrt3x,p_2=\sqrt3y,p_3=\sqrt3z$ of nitrogen we get: \begin{eqnarray*} \Psi(C_3)p_1(x,y,z) &=&p_1\circ C_3^{-1}(x,y,z)\\ &=&p_1(-\tfrac12x+\tfrac{\sqrt3}2y,-\tfrac{\sqrt3}2x-\tfrac12y,z)\\ &=&\sqrt3(-\tfrac12x+\tfrac{\sqrt3}2y) =-\tfrac12p_1+\tfrac{\sqrt3}2p_2 \end{eqnarray*} and $\Psi(C_3)p_2=-\tfrac{\sqrt3}2p_1-\tfrac12p_2$, $\Psi(C_3)p_3=p_3$. Similarly, $\s_1$ permutes the hydrogen atoms as follows: $1\to1,2\to3,3\to2$. $\Psi(\s_1)s_1=s_1$, $\Psi(\s_1)s_2=s_3$ and $\Psi(\s_1)s_3=s_2$. $$ \Psi(\s_1)p_1(x,y,z) =p_1\circ\s_1^{-1}(x,y,z) =p_1(x,-y,z) =\sqrt3x =p_1 $$ and $\Psi(\s_1)p_2=-p_2$, $\Psi(\s_1)p_3=p_3$. This gives us the matrices: $$ \Psi(C_3)= \left(\begin{array}{ccccccc} 1&0&0&0&0&0&0\\ 0&-\tfrac12&-\tfrac{\sqrt3}2&0&0&0&0\\ 0&\tfrac{\sqrt3}2&-\tfrac12&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&0&0&1\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&1&0 \end{array}\right),\quad \Psi(\s_1)=\left(\begin{array}{ccccccc} 1&0&0&0&0&0&0\\ 0&1&0&0&0&0&0\\ 0&0&-1&0&0&0&0\\ 0&0&0&1&0&0&0\\ 0&0&0&0&1&0&0\\ 0&0&0&0&0&0&1\\ 0&0&0&0&0&1&0 \end{array}\right)~. $$

Find the representation $\Psi:T_d\rar\UU(E_C\oplus E_1\oplus E_2\oplus E_3\oplus E_4)$ of the symmetry group of methane $\chem{CH_4}$, where $E_C={\cal H}_0\oplus{\cal H}_1$ and $E_1=E_2=E_3=E_4={\cal H}_0$.

Spin $l/2$ particles

A spin $l/2$ particle, $l\in\N_0$, is a particular quantum mechanical system: a complex Hilbert space $E$ of dimension $l+1$ - the state space of the particle - and an irreducible (cf. section) unitary representation $\Psi:\SU(2)\rar\UU(E)$ of $\SU(2)$ in $E$. The unitary representation is just the way the angular momentum operators act on the particle (cf. section). Analogously, irreducible unitary representations of $\SU(3)$ play a prominent role in e.g. quantum chromodynamics, cf. wikipedia. In the simplest non trivial case of a spin $1/2$ particle (also called qubit) the representation is the standard representation of $\SU(2)$ and the angular momentum operators (in the direction of an orthonormal basis of $\R^3$) are given by the Pauli spin matrices: $$ \s_1\colon= \left(\begin{array}{cc} 0&1\\ 1&0 \end{array}\right),\quad \s_2\colon= \left(\begin{array}{cc} 0&-i\\ i&0 \end{array}\right),\quad \s_3\colon= \left(\begin{array}{cc} 1&0\\ 0&-1 \end{array}\right) $$ multiplied by $1/2$.

Verify that $[\s_1,\s_2]=2\s_3$, $[\s_2,\s_3]=2\s_1$ and $[\s_3,\s_1]=2\s_2$

The matrices $\s_j$ are unitary, but that's not the point, it's rather the fact that the matrices $\s_j/2i$ generate the Lie-algebra of $\SU(2)$: Suppose $a=(a_1,a_2,a_3)$ is a normalized vector in $\R^3$ and put $$ S_a\colon=\tfrac12(a_1\s_1+a_2\s_2+a_3\s_3)~. $$ In QM the operator $S_a$ is an observable (i.e. it's self-adjoint on $\C^2$) and it's called the angular momentum operator of a spin $1/2$ particle in the direction of $a$, so $\s_j/2$ is the angular momentum operator in the direction of the $j$-th axis.

Prove that for all $t\in\R$: $\exp(-itS_a)\in\SU(2)$. Suggested solution.

Remember: for any $x\in\C^2$ the function $\psi:t\mapsto\exp(-itS_a)x$ is the solution to the time dependent Schrödinger equation $\psi^\prime(t)=-iS_a\psi(t)$, $\psi(0)=x$. In math the operators $\s_1/2i,\s_2/2i,\s_3/2i$ are called generators of the Lie-algebra $\su(2)$ of $\SU(2)$, cf. chapter. For the time being this means that $\exp:\su(2)\rar\SU(2)$ and every element of $\su(2)$ can be written as a (unique) linear combination of these operators and since $$ -itS_a=t\sum_j a_j\tfrac1{2i}\s_j $$ the operators $iS_a$ are elements of $\su(2)$.

Compute the eigen-values and eigen-vectors of $S_a$ for $a=(\cos\vp\sin\theta,\sin\vp\sin\theta,\cos\theta)\in S^2$. Suggested solution.

For a spin $l/2$ particle we will see that the angular momentum operator in any direction admits the eigen-values: $l/2,l/2-1,l/2-2,\ldots,-l/2$ (cf. exam, exam and the subsequent remarks). We'll take this up again in the chapter on matrix groups!

Further references

M. Robinson: Symmetry and the Standard Model, Mathematics and Particle Physics, A. Flournoy, Particle Physics.

What should you be acquainted with?

Acknowledgments

Symmetries

Subgroups of $\OO(3)$

Symmetries of a molecule

Matrices of symmetries

The group $C_{nv}$

The group $C_{nh}$

$C_n$: the cyclic group of order $n$

$T_d$: the group of symmetries of the tetrahedron

$O_h$: the group of symmetries of the cube and the octahedron

$I_h$: the group of symmetries of the icosahedron

Molecular Vibrations

Some additional groups

The site representation of a molecule

The standard representation, $\det$ and the trivial representation

The standard representation of $S(n)$

Vibrations

Approximate stable states

Vibrations of $\chem{H_2O}$

Vibrations of tetrahedral molecules like methane $\chem{CH_4}$

Computing eigenvalues exactly in sage

Hamiltonians

Diagonalizable operators

Weights and weight vectors

An example

Observables

The bra-ket notation in QM

Emmy Noethers Theorem on Symmetries and Conservations

Coordinate transforms

A glance on Emmy Noethers theorem: equation of motion

A glance on Emmy Noethers theorem: field equations