Sharing Thoughts and Knowledge

Simple idea on Irreducible Representation

Posted on February 27, 2011 by ciksalma

We present here a good example of an irreducible representations. Studying group representations can give information about the group itself. In particular if one had an irreducible representation of a group, one can simple generate all the groups element from it.

The spin-up and -down states of the hydrogen ground state form an irreducible representation of SU(2). The element of SU(2) is no other than the Pauli matrices. It can easily be proved that the Pauli matrices satisfy the following relation

$\sigma_{i}\sigma_{j}=i\sigma_{k}$ ,

for

$\sigma_{1}=\begin{pmatrix} 0&1\\ 1&0\end{pmatrix}, \qquad \sigma_{1}=\begin{pmatrix} 0&-i\\ i&0\end{pmatrix}, \qquad \sigma_{1}=\begin{pmatrix} 1&0\\ 0&-1\end{pmatrix}.$

In words, this shows that each of $\sigma_{i}\in SU(2)$ can be rotated with all other $\sigma_{j}\in SU(2)$ and transformed into all other elements of SU(2).

Mathematically, matrix representation is usually states as a group representation that has no nontrivial invariant subspaces. It means, for the element of the group, say V the subgroup $W\in V$ can either be zero (i.e. trivial) or the group itself ( $W=0$ and W=V). Note that, most of the textbook usually used the term nontrivial for $W\neq 0$ and $W\neq V$ .

Remarks: We have discuss a bit on the group representation theory here.
Reference: 1. Weber and Arfken, Essential Mathematical Methods for Physicists 2. B.C. Hall, Lie Groups, Lie Algebras & Representations

Filed under: Group Theory, Mathematics | Leave a Comment »

Classical and Quantum Objects

Posted on February 17, 2011 by ciksalma

We present here a well-accepted table object on classical (CM) and quantum mechanics (QM).

The main objects are categorized in states, observables, dynamics and transformations.

States
CM- points $(p,q)$ of the phase space $\mathbb{R}^{2n}=\mathbb{R}^{n}_{q}\oplus\mathbb{R}^{n}_{p}$
QM- elements of Hilbert space $\mathcal{H}$ , $\psi\in\mathcal{L}^{2}(\mathbb{R}^{n})$
Observables
CM- Functions $f(q,p)$ on the phase space
QM- linear operators, $\hat{f}:\mathcal{H}\rightarrow \mathcal{H}$ (usually self-adjoint (hermitian) and unbounded)
Dynamics
CM- Hamiltonian, H system for states: $\dot{p}=-H_{q}$ , $\dot{q}=H_{p}$ . We have Liouville equation for observables $\dot{f}=\{H,f\}$
QM- Governed by Schrodinger equation for states: $i\hbar\psi=\hat{H}\psi$ . We have Heisenberg equation for observables $i\hbar\frac{d\hat{f}}{dt}=-[\hat{H},\hat{f}]$ . $\hat{H}$ is the energy operator
Transformations
CM- canonical transformation of phase space, $g: \mathbb{R}^{2n}_{q,p}\rightarrow\mathbb{R}^{2n}_{q,p}$ . $(p,q)\mapsto g(q,p)$ , $f(p,q)\mapsto (g*f)(p,q)\equiv f(g(p,q))$
QM- unitary transformation, $U:\mathcal{H}\rightarrow \mathcal{H}$ , $\psi \mapsto U\psi$ , $\hat{f}\mapsto U\hat{f}U^{-1}$

Remarks: $q$ and $p$ represent position and momentum respectively.

Reference: Quantization methods in differential equations / V.E. Nazaikinskii, B.-W. Schulze, B. Yu. Sternin

Filed under: Classical Mechanics, Quantum Mechanics | Leave a Comment »

Invariant Measure

Posted on June 2, 2010 by ciksalma

All Lie group, $g\in G$ possess the left- or right-invariant measure. In this post, we show how we compute the $d\mu(g)$ which indicates the left Haar measure and $d\mu_{r}(g)$ which indicates the right Haar measure. We refer to here for review on measures.

(Measure as the name in principle, tell us something that we evaluate which give a quantity of certain measurement as the output, e.g. length, volume ect.)

We take the example of the group from the previous post that is the affine group. For which the action of the group of $g\in G_{+}$ is of the form $gx=ax+b$ , together with the matrix representation

$g=\left(\begin{array}{cc}a&b\\ 0&1\end{array}\right)$ .

To compute if there is an invariant measure in this group we do the following.

From the multiplication of two elements of $G_{+}$ , we get
$g_{1}g_{2}=(a_{1}b_{2}+b_{1}, a_{1}a_{2})$ .
We parametrized as follows $a=a_{1}a_{2}$ and $b=a_{1}b_{2}+b_{1}$ .
Next we find the Jacobian matrix such as $d\mu(g)=\frac{\partial(a,b)}{\partial(a_{2},b_{2})}$
For the right-invariant $d\mu_{r}(g)=\frac{\partial(a,b)}{\partial(a_{1},b_{1})}$ .
Final step we can show that $d\mu(g)\neq d\mu_{r}(g)$ .

Therefore the group posses non-invariant measure. Since the left- and right-invariant measure are not equal. This will take us to the term of quasi-invariant measure.

Filed under: Measure Theory | Leave a Comment »

Homogeneous Spaces

Posted on May 31, 2010 by ciksalma

Let $G$ be locally compact (i.e. metrizable) group; usually we denote $G$ to be the Lie groups. We call $X$ being the transformation of space of $G$ . Then the following is true.

For all $x,y\in X$ , then there is $g\in G$ so that the equation

$y=gx$ is solved.

Lets try to understand this further. Take, in terms of the affine group, $G_{+}$ which has the following properties

$G_{+}$ has the transformation of element such as $x\mapsto ax+b$ .
the group element is written as $g=(b,a)$ for $g\in G_{+}$ .
the group multiplication takes as follows
$g_{1}*g_{2}=(b_{1},a_{1})(b_{2},a_{2})=(b_{1}+a_{1}b_{2},a_{1}a_{2})$ .
the matrix group representation is $g=\left(\begin{array}{cc}a&b\\ 0&1\end{array}\right)$ .

Next we do the following, take the subgroup $H$ with element $h=(0,a)\in H$ . Then one can easily shows

$(b,a)=(b,1)(0,a)$ for $b\in\mathbb{R}$ . For which the quotient $G_{+}/H\cong\mathbb{R}$ .

Also if for $x\in\mathbb{R}$ , then

$(b,a)(x,1)=\left(\begin{array}{cc}a&ax+b\\ 0&1\end{array}\right)$ . Which gives the left coset of g,

$gx=ax+b$ .

This is non other than the homogeneous space for which the space of the reals, $\mathbb{R}$ being the transformation space of the affine group $G_{+}$ . To summarize

For $x\in\mathbb{R}$ , there is $g=(b,a)\in G_{+}$ such that $gx=ax+b$ is solved.

Filed under: Group Theory | Leave a Comment »

Beamer Package

Posted on May 18, 2010 by ciksalma

For those who using Windows operating system.

When installing your MikTeX compiler in you machine, do the following so that the Beamer packages are also install.

During the setting set your preferences, choose yes for the install missing packages on-the-fly.

Then you can compile your tex file of Beamer.

When I failed to execute my Beamer file last time, I assumed that my current editor which I used (i.e. TeXnicCenter) is not compatible with the MikTeX compiler. This is because when I’m using WinEdt at my office, the file I wanted to execute running perfectly well.

Then I thought perhaps the current version of MiKTeX that I installed in my net-book which is 2.8 is causing the problem. Since at my office they still used the 2.7 version. But I don’t thing this is really necessary. So that is why I reinstalling my MiKTeX to see what I’ve missed without paying attention to the versions.

Some users have suggested that you can update your MiKTeX packages using either package manager or just update. I did try this but nothing happened. So I just came up with the very unprofessional step i.e reinstalling.

Good Luck

Filed under: General | Leave a Comment »

Why they called inner product

Posted on April 7, 2010 by ciksalma

When I studied physics during my degree level, one usually use the term of dot product/scalar product instead of inner product. As a physicist it is well known that this two terms give a similar meaning. In addition many authors used the terms interchangeably.

However, in mathematics, there is a reason why they used the inner product. It is because in maths, one will also find an outer product.

If an inner product of two vectors $f,g\in V$ vector space, is given by

$\langle f\mid g\rangle$ ,

consequently the outer product of two vectors will take the form

$\mid f\rangle\langle g\mid$ .

If the inner product give the norm of any two vectors in a vector space, the outer product is used to construct the projection operator, P.

We know that both product rules played an important roles in quantum mechanics.

Filed under: Mathematics, Quantum Mechanics | Leave a Comment »

Eigenvalues must be real

Posted on March 10, 2010 by ciksalma

Given a symmetric (or hermitian or self-adjoint) operator as the following, $\textbf{A}$ on $\mathcal{H}$ , Hilbert space with the inner product of two arbitrary vectors in $\mathcal{H}$

$<\phi|\textbf{A}\psi>=<\textbf{A}\phi|\psi>$ . For $\phi,\psi \in \mathcal{H}$

Then the eigenvalues of a hermitian operator are reals.

Proof:

If $A\phi=\lambda \phi$ and $\phi \neq 0$ .

Then

$\lambda<\phi|\phi>=<\phi|\lambda\phi>=<\phi|\textbf{A}\phi>=<\textbf{A}\phi|\phi>=<\lambda\phi|\phi>$
$=\bar{\lambda}<\phi|\phi>$ .

So we have $\lambda=\bar{\lambda}$ . Thus $\lambda$ mus be real. $\Box$

In quantum mechanics this last equation is the well-known eigenequation with eigenfunction $\phi$ and the eigenvalue $\lambda$ . This real eigenvalues is with respect to the energy if one computes the Hamiltonian given by $\hat{H}\psi=E \psi$

Filed under: Mathematics, Quantum Mechanics | Leave a Comment »

Operator Q,P both act on infinite-dim spaces

Posted on March 2, 2010 by ciksalma

The canonical commutator relation CCR is given by

$[Q,P]=QP-PQ=\imath\hbar 1$ .

For which $Q$ is the position operator and $P$ is the momentum operator.

Then it has been observed that the CCR was impossible if both $Q$ and $P$ acting on a finite-dimensional spaces.

We prove this.

W apply the trace of matrix for both side of the equation. It is given by the summation of the diagonal of matrix.

We work first for the LHS

$Tr[Qp-PQ]=Tr(QP)-Tr(PQ)=0$ using the trace of matrix properties.

Now for the RHS

$Tr[\imath\hbar 1]=\imath\hbar Tr(1_{n})=\imath\hbar n$ for $n$ being reals

where $1_{n}$ is $n\times n$ identity matrix.

It shows that both sides is not satisfied unless $\hbar$ vanishes.

This agrees with the fact that physical observables in quantum mechanics are represented mathematically by linear operators on Hilbert spaces.

Filed under: Mathematics, Quantum Mechanics | Leave a Comment »

Exercise #10: QM

Posted on February 9, 2010 by ciksalma

If $A$ and $B$ are bounded operators on the Hilbert space $\mathcal{H}$ , show that

$(AB)=B^{*}A^{*}$ .
If $A=A^{*}$ and $B=B^{*}$ , then $(AB)=B^{*}A^{*}$ iff $[A,B]=0$ .

Answer:

Let $f,g \in \mathcal{H}$ . Then $AB-BA=0$ . Also true is $A^{*}B{*}-B^{*}A^{*}=0$ . This implies

$<B^{*}A^{*}f|g>=<A^{*}f|Bg>=<f|ABg>=<(AB)^{*}f|g> \Box$ .
This is a two ways prove (i.e, iff condition).

$\Leftarrow$

Let $[A,b]=0$ . Then $AB-BA=0$ .
$A^{*}B{*}=B^{*}A^{*}=(AB)^{*}$ . Since $A=A^{*}$ and $B=B^{*}$ , then $AB=(AB)^{*}$ .

$\Rightarrow$

Let $(AB)^{*}=AB$ . Then, it is self-adjoint. So using the self-adjoint property we know $B^{*}A^{*}=(AB)^{*}$ .
Since $A=A^{*}$ and $B=B^{*}$ , then
$B^{*}A^{*}=AB=BA$
$AB-BA=0$
$[A,B]=0 \Box$

Filed under: Quantum Mechanics | Leave a Comment »

Momentum Operator Being Unbounded

Posted on February 4, 2010 by ciksalma

We do this exercise yesterday during the lecture, where we learned mathematics of quantum mechanics.

Look at this plane wave. We first ignore the normalization constant and also take it without the time parameter. So we have the following

$\psi(x)=e^{i\frac{k}{\hbar}x}$ .

We would like to see the corresponding eigenvector, eigenvalue and thus the eigenequation of a plane wave after we act a momentum operator $P=-i\frac{d}{dx}$ on it. We have the following

$P\psi(x)=-i\frac{d}{dx}e^{i\frac{k}{\hbar}x}$ ;
$=i\frac{k}{\hbar}e^{i\frac{k}{\hbar}x}$ ;
$=k\psi(x)$ . (taking $\hbar=1$ )

What we have is an eigenvector of $\psi(x)=e^{i\frac{k}{\hbar}x}$ , and also the eigenvalue of $k$ . Well everything seems perfect. And physicist must agree but not for a mathematician.

However there is one problem arise. The eigenvector is somehow doesn’t exist. We say this if we try to calculate the norm of the function $\psi(x)=e^{i\frac{k}{\hbar}x}$ . We solve using the formula for norm as in $L^{2}$ space, $L^{2}(\Re,dx)$ with respect to Lebesgue measure dx as the following

$<\psi(x)|\psi(x)>= \int_{\Re}|e^{i\frac{k}{\hbar}x}|^{2}dx$ ;
$=\int_{\Re}\bar{\psi(x)}\psi(x)dx$ ;
$=\int_{\Re}1dx=0$ .

This means that the eigenvector is not exist in the $L^{2}$ space. Eventhough this $L^{2}$ space is already an infinte dimension of space which one can work on with.

This crucial example shows simply that the momentum operator is an unbounded operator. Similar happens to the case of the position operator $Q=x$ .

Now, can we find any bounded operator but have no eigenvectors defined on any space? Another exercise.

Filed under: Quantum Mechanics | Leave a Comment »

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Simple idea on Irreducible Representation

Classical and Quantum Objects

Invariant Measure

Homogeneous Spaces

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Why they called inner product

Eigenvalues must be real

Operator Q,P both act on infinite-dim spaces

Exercise #10: QM

Momentum Operator Being Unbounded

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