Sharing Thoughts and Knowledge

Simple idea on Irreducible Representation

ciksalma — Sun, 27 Feb 2011 16:30:06 +0000

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

for

In words, this shows that each of can be rotated with all other 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 can either be zero (i.e. trivial) or the group itself ( and W=V). Note that, most of the textbook usually used the term nontrivial for and .

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

Classical and Quantum Objects

ciksalma — Thu, 17 Feb 2011 17:32:09 +0000

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 of the phase space
QM- elements of Hilbert space ,
Observables
CM- Functions on the phase space
QM- linear operators, (usually self-adjoint (hermitian) and unbounded)
Dynamics
CM- Hamiltonian, H system for states: , . We have Liouville equation for observables
QM- Governed by Schrodinger equation for states: . We have Heisenberg equation for observables . is the energy operator
Transformations
CM- canonical transformation of phase space, . ,
QM- unitary transformation, , ,

Remarks: and represent position and momentum respectively.

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

Invariant Measure

ciksalma — Wed, 02 Jun 2010 22:02:14 +0000

All Lie group, possess the left- or right-invariant measure. In this post, we show how we compute the which indicates the left Haar measure and 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 is of the form , together with the matrix representation

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

From the multiplication of two elements of , we get
.
We parametrized as follows and .
Next we find the Jacobian matrix such as
For the right-invariant .
Final step we can show that .

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.

Homogeneous Spaces

ciksalma — Mon, 31 May 2010 19:00:42 +0000

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

For all , then there is so that the equation

is solved.

Lets try to understand this further. Take, in terms of the affine group, which has the following properties

has the transformation of element such as .
the group element is written as for .
the group multiplication takes as follows
.
the matrix group representation is .

Next we do the following, take the subgroup with element . Then one can easily shows

for . For which the quotient .

Also if for , then

. Which gives the left coset of g,

This is non other than the homogeneous space for which the space of the reals, being the transformation space of the affine group . To summarize

For , there is such that is solved.

Beamer Package

ciksalma — Tue, 18 May 2010 16:17:45 +0000

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

Why they called inner product

ciksalma — Wed, 07 Apr 2010 21:42:24 +0000

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 vector space, is given by

consequently the outer product of two vectors will take the form

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.

Eigenvalues must be real

ciksalma — Wed, 10 Mar 2010 22:08:02 +0000

Given a symmetric (or hermitian or self-adjoint) operator as the following, on , Hilbert space with the inner product of two arbitrary vectors in

=<\textbf{A}\phi|\psi>' title='<\phi|\textbf{A}\psi>=<\textbf{A}\phi|\psi>' class='latex' />. For

Then the eigenvalues of a hermitian operator are reals.

Proof:

If and .

Then

=<\phi|\lambda\phi>=<\phi|\textbf{A}\phi>=<\textbf{A}\phi|\phi>=<\lambda\phi|\phi>' title='\lambda<\phi|\phi>=<\phi|\lambda\phi>=<\phi|\textbf{A}\phi>=<\textbf{A}\phi|\phi>=<\lambda\phi|\phi>' class='latex' />
' title='=\bar{\lambda}<\phi|\phi>' class='latex' />.

So we have . Thus mus be real.

In quantum mechanics this last equation is the well-known eigenequation with eigenfunction and the eigenvalue . This real eigenvalues is with respect to the energy if one computes the Hamiltonian given by

Operator Q,P both act on infinite-dim spaces

ciksalma — Tue, 02 Mar 2010 16:53:09 +0000

The canonical commutator relation CCR is given by

For which is the position operator and is the momentum operator.

Then it has been observed that the CCR was impossible if both and 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

using the trace of matrix properties.

Now for the RHS

for being reals

where is identity matrix.

It shows that both sides is not satisfied unless vanishes.

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

Exercise #10: QM

ciksalma — Tue, 09 Feb 2010 15:46:37 +0000

If and are bounded operators on the Hilbert space , show that

.
If and , then iff .

Answer:

Let . Then . Also true is . This implies

===<(AB)^{*}f|g> \Box' title='===<(AB)^{*}f|g> \Box' class='latex' />.
This is a two ways prove (i.e, iff condition).

Let . Then .
. Since and , then .

Let . Then, it is self-adjoint. So using the self-adjoint property we know .
Since and , then

Momentum Operator Being Unbounded

ciksalma — Thu, 04 Feb 2010 22:10:04 +0000

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

We would like to see the corresponding eigenvector, eigenvalue and thus the eigenequation of a plane wave after we act a momentum operator on it. We have the following

;
;
. (taking )

What we have is an eigenvector of , and also the eigenvalue of . 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 . We solve using the formula for norm as in space, with respect to Lebesgue measure dx as the following

= \int_{\Re}|e^{i\frac{k}{\hbar}x}|^{2}dx' title='<\psi(x)|\psi(x)>= \int_{\Re}|e^{i\frac{k}{\hbar}x}|^{2}dx' class='latex' />;
;
.

This means that the eigenvector is not exist in the space. Eventhough this 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 .

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