- #1
Calabi
- 140
- 2
Good morning I'me french so excuse my bad language : so in this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf take a look at page 16.
They say that all rotation auround a unitary vector [tex]\vec{u}[/tex] of angle [tex]\theta[/tex] in the conventionnal space
could be right like this with the matrix : [tex]e^{\theta(u_{x}A_{x} + u_{y}A_{y} + u_{z}A_{z})}[/tex], the 3 matrix [tex]A[/tex] are given in the relation I-47 page 16.
So now take a look at those 3 matrix, we're going to right it [tex]X[/tex]. They check the falowing commutation relation : [tex][X_{a}, X_{b}] = \epsilon_{abc}X_{c}(1)[/tex], the [tex]\epsilon[/tex] are the Levi Civita symbol.
Now we're page 17 near relation I-55. We're going to associate at the matrix [tex]X_{x}, X_{y}, X_{z}[/tex] 3 nxn matrix that we right [tex]A_{x}, A_{y}, A_{z}[/tex] or like this [tex]A_{1}, A_{2}, A_{3}[/tex]. Those 3 matrx nxn check the same commutation relation (1).
Actually we try to represent [tex]SO(3)[/tex] and his Lie algebra [tex]so(3)[/tex] in a [tex]\mathbb{C}[/tex] vector space of dimension [tex]n[/tex]. We call this space [tex]V[/tex].
Each matrix [tex]e^{\theta^{a}X_{a}}[/tex] of [tex]SO(3)[/tex] is represant by the unitary automorphism [tex]e^{\theta^{a}A_{a}}[/tex] which proceed in [tex]V[/tex].
Now let's interrset us to the [tex]X_{z}[/tex] matrox. She's represented in [tex]V[/tex] by [tex]A_{z}[/tex]. Now let's define [tex]H_{a} = iA_{a}[/tex]. For exemple we have [tex]H_{3} = iA_{3}[/tex].
The matrix [tex]H_{3}[/tex] could be considerate as an automorphism in [tex]V[/tex]. Let's suppose there's a base in [tex]V[/tex] where [tex]H_{3}[/tex] is diagonalisable. Let's right the vector of this base(the eigenvectors of [tex]H_{3}[/tex].). like this : [tex]|m\rangle[/tex]. We have [tex]H_{3}|m\rangle = m|m\rangle[/tex].
And now we define [tex]H_{+} = H_{1} + iH_{2}[/tex] and [tex]H_{-} = H_{1} - iH_{2}[/tex].
We demonstrate that [tex]H_{+}|m\rangle = \beta_{m}|m+1\rangle[/tex]. And [tex]H_{-}|m\rangle = \alpha_{m}|m-1\rangle[/tex].
The [tex]\alpha[/tex] and [tex]\beta[/tex] are to determine.
Now we can suppose that it exist in the base(form by the eigenvectors of [tex]H_{3}[/tex].). a eigenvector [tex]|j\rangle[/tex] which is associated to a maximal eigenvalue [tex]j[/tex]. There's also a minimal eigenvalue(which we demosntrate after is equats to [tex]-j[/tex].).
As you saw if you apply [tex]H_{+}[/tex] on a eigenvector you obtein another eigenvector whose the eigenvalue was increase of 1.
In the same way if you apply [tex]H_{-}[/tex] on a eigenvector you obtein another eigenvector whose the eigenvalue was decrease of 1.
Moreover you know that all the base of [tex]V[/tex] contain n vectors.
So if you apply [tex]H_{+}[/tex] on [tex]|j\rangle[/tex] you obtein [tex]\vec{0}[/tex]. In the same way if you apply [tex]H_{-}[/tex] on [tex]|-j\rangle[/tex] you obtein [tex]\vec{0}[/tex].
It's a sort of security.
So if you know [tex]|j\rangle[/tex] you can obtein all the vectors of the base(form by the eigenvectors of [tex]H_{3}[/tex].). by applying [tex]n-1[/tex] time. on [tex]|j\rangle[/tex] until you obtein [tex]|-j\rangle[/tex].
NOw we could demonstrate that [tex]j = \frac{n-1}{2}(2)[/tex] and we also demosntrate that : [tex]\beta_{m} = \sqrt{j(j+1) - m(m+1)}[/tex] and
[tex]\alpha_{m} = \sqrt{j(j+1) - m(m-1)}[/tex].
So now we're going to wright the vector of this base(the eigenvectors of [tex]H_{3}[/tex].). like this : [tex]|j, m\rangle[/tex]. We have [tex]H_{3}|j, m\rangle = m|j, m\rangle[/tex].
Now with (2) you could see that : the eigenvalue are half integer(en français le mot est demi entière comme 1.5 3.5 4.5 5.5 enfin tu vois quoi.). if [tex]n[/tex] is a peer number(2,4,4,8.). or whole number(1,2,3,4,5.). if [tex]n[/tex] a odd number (1,3,5,7,9.).
An important things on that representation with [tex]H_{3}[/tex] : we represent [tex]so(3)[/tex] and in the same time [tex]su(2)[/tex]. Indeed those 2 Lie algebra are isomorph.
The representation I talk about from the beginning is irreducible. We wright it [tex]n[/tex].
For exemple [tex]2[/tex] is the representation associated to the space whose on of base is [tex](|\frac{1}{2},\frac{-1}{2}\rangle, |\frac{1}{2},\frac{1}{2}\rangle)[/tex]. In this base [tex]H_{3}[/tex] is diagonalisable. And [tex]H_{3} = \frac{1}{2}\sigma_{z} = \begin{pmatrix} \frac{1}{2} 0 \\ 0 \frac{-1}{2} \end{pmatrix}[/tex].
And finally : we demonstrate that the representation [tex]2 \otimes 2 = 3 \oplus 1[/tex] and the representation [tex]3 \otimes 2 = 4 \oplus 2[/tex].
We have an illustration of the Clebsh Gordon theorem : a tensorial representation could be reduce at a addition of irreducible representation.
If you want I can explain more about this point. Just ask for it.
the quantity ##\psi^a\psi_a## invariant on a [tex]SU(2)[/tex] transformation, Indeede I represant the [tex]SU(2)[/tex] group whose de Lie algebra is isomorph to the Lie algebra of [tex]SO(3)[/tex]. At one element(rotation.). of [tex]SO(3)[/tex] is associatted 2 element of [tex]SU(2)[/tex]. It come frome the fact that a spinor have to be turn in [tex]4\pi[/tex] to be equal as himself.
Whate ever : here is the probleme 30 translate in english :
"First check that the antysymetric part [tex]\psi_{[a \phi b]}[/tex] could be wright like this : [tex]\psi_{[a \phi b]} = \epsilon_{ab}\psi_{m}\phi^{m}[/tex]"
I've made it. It was easy.
Then they said : "show the antysymetric part is invariant on a [tex]SU(2)[/tex] action"
I've made it to. Thene the course said : "In which representation is the antysyetric part" le cour dit : à quelle représentation la partie antysymetrique apartient t elle.
Then there's : "Show that the symetrics parts [tex]\psi_{(a \phi b)}[/tex] has 3 components and comments".
I stump on those 2 question.
Could you help me to resolve it please?
Thank you very much and have a nice afternoon.
They say that all rotation auround a unitary vector [tex]\vec{u}[/tex] of angle [tex]\theta[/tex] in the conventionnal space
could be right like this with the matrix : [tex]e^{\theta(u_{x}A_{x} + u_{y}A_{y} + u_{z}A_{z})}[/tex], the 3 matrix [tex]A[/tex] are given in the relation I-47 page 16.
So now take a look at those 3 matrix, we're going to right it [tex]X[/tex]. They check the falowing commutation relation : [tex][X_{a}, X_{b}] = \epsilon_{abc}X_{c}(1)[/tex], the [tex]\epsilon[/tex] are the Levi Civita symbol.
Now we're page 17 near relation I-55. We're going to associate at the matrix [tex]X_{x}, X_{y}, X_{z}[/tex] 3 nxn matrix that we right [tex]A_{x}, A_{y}, A_{z}[/tex] or like this [tex]A_{1}, A_{2}, A_{3}[/tex]. Those 3 matrx nxn check the same commutation relation (1).
Actually we try to represent [tex]SO(3)[/tex] and his Lie algebra [tex]so(3)[/tex] in a [tex]\mathbb{C}[/tex] vector space of dimension [tex]n[/tex]. We call this space [tex]V[/tex].
Each matrix [tex]e^{\theta^{a}X_{a}}[/tex] of [tex]SO(3)[/tex] is represant by the unitary automorphism [tex]e^{\theta^{a}A_{a}}[/tex] which proceed in [tex]V[/tex].
Now let's interrset us to the [tex]X_{z}[/tex] matrox. She's represented in [tex]V[/tex] by [tex]A_{z}[/tex]. Now let's define [tex]H_{a} = iA_{a}[/tex]. For exemple we have [tex]H_{3} = iA_{3}[/tex].
The matrix [tex]H_{3}[/tex] could be considerate as an automorphism in [tex]V[/tex]. Let's suppose there's a base in [tex]V[/tex] where [tex]H_{3}[/tex] is diagonalisable. Let's right the vector of this base(the eigenvectors of [tex]H_{3}[/tex].). like this : [tex]|m\rangle[/tex]. We have [tex]H_{3}|m\rangle = m|m\rangle[/tex].
And now we define [tex]H_{+} = H_{1} + iH_{2}[/tex] and [tex]H_{-} = H_{1} - iH_{2}[/tex].
We demonstrate that [tex]H_{+}|m\rangle = \beta_{m}|m+1\rangle[/tex]. And [tex]H_{-}|m\rangle = \alpha_{m}|m-1\rangle[/tex].
The [tex]\alpha[/tex] and [tex]\beta[/tex] are to determine.
Now we can suppose that it exist in the base(form by the eigenvectors of [tex]H_{3}[/tex].). a eigenvector [tex]|j\rangle[/tex] which is associated to a maximal eigenvalue [tex]j[/tex]. There's also a minimal eigenvalue(which we demosntrate after is equats to [tex]-j[/tex].).
As you saw if you apply [tex]H_{+}[/tex] on a eigenvector you obtein another eigenvector whose the eigenvalue was increase of 1.
In the same way if you apply [tex]H_{-}[/tex] on a eigenvector you obtein another eigenvector whose the eigenvalue was decrease of 1.
Moreover you know that all the base of [tex]V[/tex] contain n vectors.
So if you apply [tex]H_{+}[/tex] on [tex]|j\rangle[/tex] you obtein [tex]\vec{0}[/tex]. In the same way if you apply [tex]H_{-}[/tex] on [tex]|-j\rangle[/tex] you obtein [tex]\vec{0}[/tex].
It's a sort of security.
So if you know [tex]|j\rangle[/tex] you can obtein all the vectors of the base(form by the eigenvectors of [tex]H_{3}[/tex].). by applying [tex]n-1[/tex] time. on [tex]|j\rangle[/tex] until you obtein [tex]|-j\rangle[/tex].
NOw we could demonstrate that [tex]j = \frac{n-1}{2}(2)[/tex] and we also demosntrate that : [tex]\beta_{m} = \sqrt{j(j+1) - m(m+1)}[/tex] and
[tex]\alpha_{m} = \sqrt{j(j+1) - m(m-1)}[/tex].
So now we're going to wright the vector of this base(the eigenvectors of [tex]H_{3}[/tex].). like this : [tex]|j, m\rangle[/tex]. We have [tex]H_{3}|j, m\rangle = m|j, m\rangle[/tex].
Now with (2) you could see that : the eigenvalue are half integer(en français le mot est demi entière comme 1.5 3.5 4.5 5.5 enfin tu vois quoi.). if [tex]n[/tex] is a peer number(2,4,4,8.). or whole number(1,2,3,4,5.). if [tex]n[/tex] a odd number (1,3,5,7,9.).
An important things on that representation with [tex]H_{3}[/tex] : we represent [tex]so(3)[/tex] and in the same time [tex]su(2)[/tex]. Indeed those 2 Lie algebra are isomorph.
The representation I talk about from the beginning is irreducible. We wright it [tex]n[/tex].
For exemple [tex]2[/tex] is the representation associated to the space whose on of base is [tex](|\frac{1}{2},\frac{-1}{2}\rangle, |\frac{1}{2},\frac{1}{2}\rangle)[/tex]. In this base [tex]H_{3}[/tex] is diagonalisable. And [tex]H_{3} = \frac{1}{2}\sigma_{z} = \begin{pmatrix} \frac{1}{2} 0 \\ 0 \frac{-1}{2} \end{pmatrix}[/tex].
And finally : we demonstrate that the representation [tex]2 \otimes 2 = 3 \oplus 1[/tex] and the representation [tex]3 \otimes 2 = 4 \oplus 2[/tex].
We have an illustration of the Clebsh Gordon theorem : a tensorial representation could be reduce at a addition of irreducible representation.
If you want I can explain more about this point. Just ask for it.
the quantity ##\psi^a\psi_a## invariant on a [tex]SU(2)[/tex] transformation, Indeede I represant the [tex]SU(2)[/tex] group whose de Lie algebra is isomorph to the Lie algebra of [tex]SO(3)[/tex]. At one element(rotation.). of [tex]SO(3)[/tex] is associatted 2 element of [tex]SU(2)[/tex]. It come frome the fact that a spinor have to be turn in [tex]4\pi[/tex] to be equal as himself.
Whate ever : here is the probleme 30 translate in english :
"First check that the antysymetric part [tex]\psi_{[a \phi b]}[/tex] could be wright like this : [tex]\psi_{[a \phi b]} = \epsilon_{ab}\psi_{m}\phi^{m}[/tex]"
I've made it. It was easy.
Then they said : "show the antysymetric part is invariant on a [tex]SU(2)[/tex] action"
I've made it to. Thene the course said : "In which representation is the antysyetric part" le cour dit : à quelle représentation la partie antysymetrique apartient t elle.
Then there's : "Show that the symetrics parts [tex]\psi_{(a \phi b)}[/tex] has 3 components and comments".
I stump on those 2 question.
Could you help me to resolve it please?
Thank you very much and have a nice afternoon.