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Determine condition on invariants under transformation

  1. Mar 26, 2014 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider a ##j=1, SU(2)## representation (or fundamental ##S0(3)## representation). Suppose that ##a_i, b_i## and ##c_i## (i=1,2,3) are vectors transforming under this representation i.e ##a_i' = [\rho_1 (x)]_{ij} a_j = \rho_{ij} a_j## and similarly for b and c. Consider $$I_1 (\vec{a}, \vec{b}) = \delta_{ij} a_i b_j\,\,\,\,\,\,\,\,\,\,I_2(\vec{a}, \vec{b}, \vec{c}) = \epsilon_{ijk} a_i b_j c_k$$

    1)Deduce the conditions on ##\delta_{ij}## and ##\epsilon_{ijk}## for which ##I_1## and ##I_2## are invariants under the transformation.

    2. Relevant equations
    Want to show that ##I_1 (\vec{a}, \vec{b}) = I_1 (\vec{a'}, \vec{b'}) ## and similarly for ##I_2##. and ##\rho_1## is an element of the ##SO(3)## group.

    3. The attempt at a solution
    So need to show that ##I_1 (\vec{a}, \vec{b}) = I_1(\rho\vec{a}, \rho \vec{b})## under the transformation ##\rho##. I also know that $$\rho_j (x)_{mm'} = \langle j,m | e^{ix^a T_a}| j,m' \rangle$$ where ##T_a## are the Lie algebra elements.

    To be invariant, under the transformation ##\delta_{ij}' a_i' b_j' = \delta_{ij}a_i b_j##. I am looking for a hint in the right direction to perhaps use what I have written down here. Thanks.
     
    Last edited: Mar 26, 2014
  2. jcsd
  3. Mar 26, 2014 #2

    strangerep

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    You seem to be jumping around between SU(2) and SO(3), but most of your question seems to be in terms of SO(3), so I'll go with that...

    Starting with the simple representation of SO(3) by a 3x3 matrices, let ##R## be a 3x3 matrix. What then is the condition ##R## must satisfy for it to be an element of (this representation of) SO(3) ?

    Hint: Look up SO(3) on Wikipedia.
     
    Last edited: Mar 26, 2014
  4. Mar 26, 2014 #3
    Not OP but also working on this problem and having a similar problem in getting started so thought I'd post in here rather than start my own thread.

    Anyway, from what I've seen on the Wikipedia page the condition is that [itex]RR^{T} = I[/itex]? Which goes along with [itex]det(R) = 1[/itex] in defining [itex]SO(3)[/itex].
     
  5. Mar 26, 2014 #4

    strangerep

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    Yes, those are the conditions. Now re-express them in subscript notation, and relate to the original question.
     
  6. Mar 26, 2014 #5
    Well for the first I get[itex]RR^{T} = r_{ij}r{ji} = \delta_{ij}[/itex] is this the condition? Then we have [itex]\epsilon_{imr}det(R) = \epsilon_{jns}r_{ij}r_{mn}r_{rs} [/itex] with [itex]det(R) = 1[/itex] leaving the equation invariant.

    However the second part of the problem asks:

    Show that these conditions are related to the canonical definititon of [itex]SO(3), SO(3)=\{O\in GL(3, \mathbb{R}): O^{T}O = I, det(O) = 1\} [/itex]

    Which to me seems what the conditions above already describe this, which makes me think the conditions above aren't what we are looking for in the first part.
     
  7. Mar 26, 2014 #6

    strangerep

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    Yeah, I find the question a bit weird too. Sometimes extremely basic questions are like that.
    But let's persevere...

    Progress further that part of the question involving ##\delta_{ij}' a_i' b_j' = \delta_{ij}a_i b_j##. Write it out with ##\rho##'s and see what it implies about ##\delta_{ij}## (i.e., act as if you don't already know what ##\delta_{ij}## is).
     
  8. Mar 26, 2014 #7
    Ok, so with [itex]a'_{i} = \rho_{ij}a_{j}[/itex] I find that [itex]\delta'_{ij}a'_{i}b'_{j} = \delta'_{ij}\rho_{ik}a_{k}\rho_{jl}b_{l}[/itex] which then I'm not sure if I can then just equate the delta dashed with the rho's to the non transformed delta I assume not due to the summation?
     
    Last edited: Mar 26, 2014
  9. Mar 27, 2014 #8

    strangerep

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    You can do that here because the a's and b's are variables. The equation $$\delta'_{ij}\rho_{ik}a_{k}\rho_{jl}b_{l} ~=~ \delta_{ij} a_i b_j$$must hold for all values of the a's and b's. (But you'll need to rename some dummy summation indices for it to make sense.)
     
  10. Mar 27, 2014 #9

    CAF123

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    Before coming back to this thread, I had done what has been suggested but ferret123, what you wrote in #5 is my concern too. I'm not sure what they mean by conditions on the ##\delta_{ij}## and ##\epsilon_{ijk}##. It seems to me provided we suppose they transform like isotopic tensors between any two frames, then it works out. Could this be what they want? Then again, as you said, this leaves very little to do for part 2.
     
  11. Mar 27, 2014 #10

    CAF123

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    Does the equation not make sense as it is? All indices on LHS are summed over and likewise on RHS.
    Given ##\delta_{ij}' \rho_{ik} a_k \rho_{jl}b_l = \delta_{ij}a_i b_j##, if I suppose ##\delta_{ij}' = \delta_{ij}## then the rest follows. So my question is, am I allowed to make this supposition and then later conclude that this is the condition for ##\delta_{ij}##? Exact same reasoning for ##\epsilon_{ijk}##.

    Alternatively, do some relabeling of the indices on the ##\delta_{ij}'## term, use the tensor transformation law (on the assumption that we know ##\delta_{ij}## is at least a tensor) and then the rest follows without any of the above suppositions.

    I have some problems with ##\epsilon_{ijk}## though. If I assume that this is a 3rd rank tensor then, again using the transformation law, I get $$\epsilon_{ijk}' a_i' b_j' c_k' = a_{\alpha} b_{\beta} c_{\gamma} \epsilon_{\alpha \beta \gamma}$$ which is exactly what I want (i.e from this infer ##\epsilon_{ijk}' = \epsilon_{ijk}## since the indices are all dummy.) The problem is, this is not how ##\epsilon_{ijk}## transforms. (I know that it transforms like a pseudotensor and thus by using the correct transformation law, I get the above but with a factor of ##\det \rho##. Any ideas? Of course, using the supposition method, I do not run into such problems.
     
    Last edited: Mar 27, 2014
  12. Mar 27, 2014 #11
    Surely by the supposition method you have to make the assumption that [itex]\rho\rho^{T} = I\ \&\ det(\rho) = 1[/itex] to gain the equalities for the transformed tensors?
     
  13. Mar 27, 2014 #12

    CAF123

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    Do you think that the condition that is needed is that the two tensors are invariant under any transformation? But yes, I agree, even by supposition you still have to invoke the conditions you wrote. I guess for part 2) it is just a case of noting that these conditions are the ones the matrices satisfy in SO(3). I think I'll go with the transformation law approach, and then for part 2), just say that these are the conditions that elements in SO(3) obey. What do you think?

    I'll see what comments strangerep has with regard to what I wrote in #10.
     
    Last edited: Mar 27, 2014
  14. Mar 27, 2014 #13

    CAF123

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    There is also a final part of this question about the analogous expressions of the invariants ##\delta_{ij}a_ib_j## and ##\epsilon_{ijk}a_ib_jc_k## for the case ##j=1/2## SU(2) representation.

    A ##j=1/2## SU(2) representation has the generator matrices of size 2x2, and to make the matrix product meaningful necessarily vectors a, b and c are two dimensional. So how to generalize the invariant in SO(3) ##\epsilon_{ijk}a_ib_jc_k##? (since this contains a cross product, of which there is no notion in 2D).

    I am also unsure of how to incorporate the ##U^{\dagger}U=1## condition.
    Thanks.
     
  15. Mar 27, 2014 #14

    strangerep

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    Yes, but that's not what I meant. See below.

    If I understand the true intent of the question, you are not allowed to make this supposition here. Rather, you must derive it. I.e., you have
    $$\delta_{ij}' \rho_{ik} a_k \rho_{jl}b_l = \delta_{ij}a_i b_j$$ but nothing else at this point. I.e., you don't yet know that ##\delta_{ij}## is the Kronecker delta. (It might have been better if the question had used some other symbol.)

    Anyway, for the next step, you need to re-arrange this equation into the form ##X_{ij} a_i b_j = 0##. That's where my hint about subscript relabelling comes in.

    From there, the next step is obviously to deduce rigorously that ##X_{ij} = 0##, but I'm not sure how much forward-hinting I should give you, so I'll leave it at that for now.

    Finish the ##\delta_{ij}## case first. Then it should be clearer how to generalize the method to find properties of ##\epsilon_{ijk}##.

    That's because you're assuming what you're being asked to prove (if I understand the question correctly).

    Oh, regarding the SU(2) case, you'll have to post the entire question, exactly as given. It's too hard for me to guess in the dark.

    However, I'll mention that there is an antisymmetric matrix in 2D.
     
    Last edited: Mar 27, 2014
  16. Mar 27, 2014 #15
    So we start with [itex]\delta'_{ij}\rho_{ik}\rho_{jl}a_{k}b_{l} = \delta_{ij}a_{i}b_{j}[/itex] I feel like I then want subract the RHS to give me an expression equal to 0 from which I can then factor out the [itex]a[/itex] and [itex]b[/itex] terms letting whats left become [itex]X_{ij}[/itex] although I'm not particularly confident this is allowed or the subscript relabeling that would be required?
     
  17. Mar 27, 2014 #16

    strangerep

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    Yes.

    Remember that summed indices are dummy indices. E.g., ##Z_i Z_i = Z_k Z_k##.
    Just rename the indices in one of your terms.

    (Probably better if you stop 2nd-guessing. Just do it and post it. That way, I can see more clearly what mental blocks or knowledge gaps you might have.)
     
  18. Mar 27, 2014 #17
    Ok, [itex]\delta'_{ij}\rho_{ik}\rho_{jl}a_{k}b_{l} = \delta_{ij}a_{i}b_{j}[/itex] then to me factoring out the [itex]a[/itex] and [itex]b[/itex] terms should require all terms being summed over together so we have [itex]\delta'_{ij}\rho_{ik}\rho_{jl}a_{k}b_{l} - \delta_{ij}a_{i}b_{j} = 0 [/itex] the relabeling [itex]i=k,\ j=l[/itex] I get [itex](\delta'_{ij}\rho_{ik}\rho_{jl} - \delta_{kl})a_{k}b_{l}= 0[/itex] which would then give [itex]X_{ij} = \delta'_{ij}\rho_{ik}\rho_{jl} - \delta_{kl}[/itex] is this the right logic?
     
  19. Mar 27, 2014 #18

    strangerep

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    Yes, that's (almost) what I intended -- except that your lhs should be ##X_{kl}##, since ##k,l## are the free (unsummed) indices here.

    But... thinking about it... I'm now even less sure that I understand the intent of the question. In which case, I might be leading you down the wrong path. :frown:

    Are you sure that the original post contains the question exactly as given? No typos? Also, you said there was a 2nd part to the question, but that doesn't make much sense to me in the context of the 1st part.

    So... maybe you should re-post the entire question? Also, where did the question come from? If this is part of a course, are there associated lecture notes which I could examine online to check the background and emphasis?
     
  20. Mar 27, 2014 #19
    Yeah I'll repost the entire question should be able to access it here http://www2.ph.ed.ac.uk/~rzwicky2/SoQM/Ast_SoQM_14_3.pdf [Broken]
    with the lecture notes here http://www2.ph.ed.ac.uk/~rzwicky2/SoQM/notes.pdf [Broken] I've also had doubts about what the first part of this question means hopefully some context to the course and the whole problem will give some insight, thanks for attempting to get your head around this.
     
    Last edited by a moderator: May 6, 2017
  21. Mar 27, 2014 #20

    strangerep

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    Umm,.... is this for credit towards your final result in the course? If so, does the lecturer allow you to get outside expert help in doing it?
     
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