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Group Representation

  1. Jul 25, 2014 #1
    Good morning I'me french so excuse my bad language : so in this course : Cours 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 gonna 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 gonna 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 gonna 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:biggrin:.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 25, 2014 #2
    I'll wright all of this on the course because it's in french. It's to avoid to you to translate and to presents the context.

    Thank you in advance and have a nice afternoon:biggrin:.
     
  4. Jul 25, 2014 #3

    strangerep

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    FYI, the English word is spelled "write", not "right" nor "wright". :smile:

    I think you meant: $$\psi_{[a} \phi_{b]} = \epsilon_{ab}\psi_{m}\phi^{m} ~~~~~~~~ (1)$$
    Well, you've just shown that the expression (1) is invariant. And there's only 1 degree of freedom (i.e., ##\psi_m\phi^m##). So... which representation has elements that don't change under any rotation?

    I guess you meant: ##\psi_{(a} \phi_{b)}## .

    You've just got to count how many of the ##\psi_{(a} \phi_{b)}## products are independent. The ##a,b## indices take values in the set ##\{1,2\}##, so try writing all possible combinations of "a" and "b" with those values, and don't worry about the order. How many different combinations do you get?
     
  5. Jul 26, 2014 #4
    Good morning strangerep and many thanks,
    Euh there's 2 degree I think, this depend of [tex]\epsilon[/tex]. Indeed : [tex]\epsilon_{ab} = -\epsilon_{ba}[/tex].

    And it's not because : [tex]\psi_{a}\phi_{b} = \frac{1}{2}\psi_{[a} \phi_{b]}(= \epsilon_{ab}\psi_{m}\phi^{m}) + \frac{1}{2}\psi_{(a} \phi_{b)}[/tex] and that there's 3 possibility for [tex]\psi_{(a} \phi_{b)}[/tex] that you can said that there's 2 parts : one in the [tex]1[/tex] representation and the 3 others in the [tex]3[/tex] representation.

    All I know for the moment is that ther's 4 possibility of [tex]\psi_{a}\phi_{b}[/tex] which are the 4 componants of the tensor [tex]|\psi\rangle \otimes |\phi\rangle[/tex] in the base [tex](|+\rangle \otimes |+\rangle, |-\rangle \otimes |-\rangle, |+\rangle \otimes |-\rangle, |-\rangle \otimes |+\rangle)[/tex].

    Could you be more clear please?

    Maybe we should to develope [tex]|\psi\rangle \otimes |\phi\rangle[/tex] in the base form by [tex](|1,-1\rangle, |1,0\rangle, |1,1\rangle, |0,0\rangle)[/tex] who's is described page 25 of this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf.

    Thank you in advance and have a nice morning:biggrin:.
     
  6. Jul 26, 2014 #5

    strangerep

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    ?? What does that mean?

    I should have said "... independent degrees of freedom". (If a degree of freedom is a constant multiple of another, then there's really only 1 independent degree of freedom, not 2.)

    Since my French is much worse than your English, maybe I'll never be clear enough for you. But let us try one more thing...

    Write out an arbitrary 2x2 matrix. Then write out the symmetric and antisymmetric parts of the matrix separately. How many independent degrees of freedom are there in each part?
     
  7. Jul 26, 2014 #6
    I get 3 combinations(because of the symettry.). but it's not beacause I see 3 that those 3 combinations are a part of the components of ##|\psi\rangle \otimes |\phi\rangle ## in the parts of the base which is ##(|1,-1\rangle, |1,0\rangle, |1,1\rangle) ##.

    It's the [tex]1[/tex] representation(which is associated to the spin [tex]0[/tex].). Indeed in this representation a rotation is : [tex]e^{0} = 1[/tex]. It's the identity fonction so if you aply it you'll find the same components. The elements d'ont change under rotation.

    But there's 3 posibility for : ##\psi_{[a} \phi_{b]} = \epsilon_{ab}\psi_{m}\phi^{m}##.

    If ##a=b## it's ##0##, and ##\psi_{[a} \phi_{b]} = \psi_{[b} \phi_{a]}##.

    And even if they were one possibility in the same way, it's not beacause I see 1 that this 1 is a part of the components of ##|\psi\rangle \otimes |\phi\rangle ## in the parts of the base which is ##|0,0\rangle ##. It's on this point I'm lock.

    Indeed we try to made the Clebsh Gordon theorem but with the compnentns and not with the base this time(as we have made in the problem 25 page 25 of this course : cours.).

    Do you see what I mean please?

    Thank you in advance and have a nice afternoon:biggrin:.
     
  8. Jul 26, 2014 #7

    strangerep

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    Yes, they are.

    Yes.

    Yes, I see exactly where you are confused.

    I also see that you did not make any attempt to follow my suggestion at the end of my post #5.
    If you do not at least try to work through my suggestions, then it's difficult for me to help you.
    I will try one (but only one) more time...

    Can you work out the following outer product, and then decompose the resulting matrix into symmetric and antisymmetric parts?
    $$
    \psi \phi^T ~=~
    \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} ~
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ \cdots ~?
    $$
     
    Last edited: Jul 26, 2014
  9. Jul 28, 2014 #8
    I get :
    $$

    \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ \begin{pmatrix}\psi_1 \phi_1 & \psi_1 \phi_2 \\ \psi_2 \phi_1 & \psi_2 \phi_2\end{pmatrix} = \frac{1}{2}(\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
    \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} + \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} ) = \frac{S+A}{2}
    $$

    Where $$S = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
    \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix}$$ is the symetric part and $$A = \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix}$$ is the antysymetric part.

    Good afternoon:biggrin:.
     
  10. Jul 28, 2014 #9
    More : As we prevously show $$A = \begin{pmatrix} 0 & \psi_{1} \phi_{2} - \phi_{1} \psi_{2} \\ -(\psi_{1} \phi_{2} - \phi_{1} \psi_{2}) & 0\end{pmatrix} = \begin{pmatrix} 0 & \epsilon_{12}\psi_{m}\phi^{m} \\ \epsilon_{21}\psi_{m}\phi^{m} & 0\end{pmatrix}$$, since $$\epsilon_{12} = -\epsilon_{21}$$ this matrix could be considerate as to be in a 1 dimension matrix space. And we can also said that $$\psi_{m}\phi^{m} (1)$$ is the coponents of a vector in the base $$|0,0\rangle$$ which is in the irreducble 1 representation(which is a sub space of $$2 \otimes 2$$.). beacause (1) is invariant under a [tex]SU(2)[/tex] action.

    Then we can say that $$S = \begin{pmatrix} \psi_{1} \phi_{1} + \phi_{1} \psi_{1} & \psi_{1} \phi_{2} + \phi_{1} \psi_{2} \\ \phi_{1} \psi_{2} + \psi_{1} \phi_{2} & \psi_{2} \phi_{2} + \phi_{2} \psi_{2} \end{pmatrix} $$. I see 3 different comonents.

    But I always can't see how to develope [tex]|\psi\rangle \otimes |\phi\rangle[/tex] in the base form by [tex](|1,-1\rangle, |1,0\rangle, |1,1\rangle, |0,0\rangle)[/tex] who's is described page 25 of this course : http://lapth.cnrs.fr/pg-nomin/salati/TQC_UJF_13.pdf.

    What can I do please?

    Thank you in advance and have a nice afternoon:biggrin:.
     
  11. Jul 28, 2014 #10
    THis matrix space is isomorph to the 1 representation.

    Good afternoon:biggrin:.
     
  12. Jul 28, 2014 #11

    strangerep

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    OK. Now what happens if you perform a rotation: ##\psi \to \psi' = R\psi## (and similarly for ##\phi##). What happens on the right hand side of the 1st equation?
     
  13. Jul 29, 2014 #12
    Good morning,
    I transforme the components and the value of : [tex]\psi_{m}\phi^{m}[/tex] is invariant under this rotation.

    Whcih equation please?

    If you talk about : [tex]\frac{S + A}{2}[/tex] I know that [tex]A[/tex] is invariant but I don't know what hapend to [tex]S[/tex].

    What can I do please?

    Thank you in advance and have a nice morning:biggrin:.
     
  14. Jul 29, 2014 #13

    strangerep

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    Well, work it out. How did you figure out that A is invariant? Do a similar calculation for S and see what you get. (And if you still can't do it, show me how you figured out that A is invariant.)
     
  15. Jul 30, 2014 #14
    Good morning,
    Well we've already said that [tex]A = \begin{pmatrix} 0 & \psi_{1} \phi_{2} - \phi_{1} \psi_{2} \\ -(\psi_{1} \phi_{2} - \phi_{1} \psi_{2}) & 0\end{pmatrix} = \begin{pmatrix} 0 & \epsilon_{12}\psi_{m}\phi^{m} \\ \epsilon_{21}\psi_{m}\phi^{m} & 0\end{pmatrix}[/tex] so it's in a one dimension matrix space, so it's isomorph to [tex]1[/tex]. In [tex]1[/tex] the components of the matrix are represented by [tex]\psi_{m}\phi^{m}[/tex] which is invariant under a [tex]SU(2)[/tex] action.

    So we can conclude that [tex]A[/tex] is invariant under a [tex]SU(2)[/tex] action.

    Good afternoon:biggrin:.
     
  16. Jul 30, 2014 #15
    So I'm gonna start from the beginning. We have : $$\psi_{a}\phi_{b} = \frac{1}{2}\psi_{[a} \phi_{b]}(antisymetric) + \frac{1}{2}\psi_{(a} \phi_{b)}(symetric)$$.

    First I've show that : $$\psi_{[a \phi b]} = \epsilon_{ab}\psi_{m}\phi^{m}$$.

    More : we can say in matrix language that $$\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} ~=~ \begin{pmatrix}\psi_1 \phi_1 & \psi_1 \phi_2 \\ \psi_2 \phi_1 & \psi_2 \phi_2\end{pmatrix} = \frac{1}{2}(\begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} +
    \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} + \begin{pmatrix} \psi_1 \\ \psi_2 \end{pmatrix} \otimes
    \begin{pmatrix} \phi_1 & \phi_2 \end{pmatrix} - \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} \otimes \begin{pmatrix} \psi_1 & \psi_2 \end{pmatrix} ) = \frac{S+A}{2}$$

    And $$A = A = \begin{pmatrix} 0 & \psi_{m}\phi^{m} \\ -\psi_{m}\phi^{m} & 0\end{pmatrix}$$, so this matrix is in a 1 dimension matrix space with the components $$\psi_{m}\phi^{m}$$ in the base $$\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}$$.
    So since we know that $$\psi_{m}\phi^{m}$$ is invariant under a $$SU(2)$$ rotation
    we can say that this space matrix is a $$1$$ representation.

    So the antysymetric part is a $$1$$ representation. The spin $$0$$. Indeed all rotation of $$SU(2)$$ are represented by the unity $$\mathbb{I}_{1} = 1$$ in the $$1$$ representation.The components are invariant under a $$SU(2)$$ rotation.

    Now because of the symetry, there's 3 possibility(components.). for the symetric part : $$S = \begin{pmatrix} \psi_{1} \phi_{1} + \phi_{1} \psi_{1} & \psi_{1} \phi_{2} + \phi_{1} \psi_{2} \\ \phi_{1} \psi_{2} + \psi_{1} \phi_{2} & \psi_{2} \phi_{2} + \phi_{2} \psi_{2} \end{pmatrix}$$

    So we can say I've resolve the problem. But they say in the course to comment.

    What do you'll say please?

    Thank you in advance and have a nice afternoon:biggrin:.
     
  17. Jul 31, 2014 #16

    strangerep

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    Consider just the symmetric part. I'll call it $$\chi_{ab} ~:=~ \psi_{(a} \phi_{b)} ~.$$Now perform the rotation transformation on ##\psi## and ##\phi## that I mentioned in post #11. What is the explicit expression for the transformed ##\chi## ?

    Then decompose the transformed ##\chi## into symmetric and antisymmetric parts. What do you find? And can you now conclude a bit more than you have so far?
     
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