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Showing Complex Vectors are Orthonormal

  1. Apr 15, 2017 #1
    1. The problem statement, all variables and given/known data
    let [tex] \epsilon_1 [/tex] and [tex] \epsilon_2 [/tex] be unit vectors in R3. Define two complex unit vectors as follows:
    [tex] \epsilon_{\pm} = \frac{1}{\sqrt{2}}(\epsilon_1 \pm i \epsilon_2)[/tex]

    verify that epsilon plus minus constitutes a set of complex orthonormal unit vectors. That is, show that [tex] (\epsilon_\pm)^* \cdotp \epsilon_\mp = 0 [/tex]


    2. Relevant equations
    Dot Product.

    3. The attempt at a solution
    So... I don't know what I can possibly be missing. I do the dot product say of [tex] (\epsilon_+)^* \cdotp i \epsilon_- [/tex]
    and I'm ending up with, [tex] \frac{1}{2} [1-i \epsilon_2 \cdotp \epsilon_1-i \epsilon_2 \cdotp \epsilon_1-1] [/tex]
    So the complex parts don't go away? I'd appreciate any help... Thanks.
     
  2. jcsd
  3. Apr 15, 2017 #2

    PeroK

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    The complex inner product is not linear in both arguments.
     
  4. Apr 15, 2017 #3
    I'm sorry, I don't follow exactly.
     
  5. Apr 15, 2017 #4

    PeroK

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  6. Apr 15, 2017 #5
    Hrm, ok I see. When you take the dot product one of the vectors must be changed to it's complex conjugate? I still must be doing something wrong as I'm not getting the advertised result.
     
  7. Apr 15, 2017 #6

    PeroK

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    In a complex vector space, you have:

    ##\langle \alpha u, v \rangle = \alpha^* \langle u, v \rangle##

    ##\langle u, \alpha v \rangle = \alpha \langle u, v \rangle##

    This is the physicist's version. In pure mathematics, it's the other way round (like in the link I gave you). You need to know which approach is being used in your book or course.
     
  8. Apr 15, 2017 #7
    Oh no... I'm sorry I have a typo. I think that is what could be some of my confusion.

    I'm trying [tex] \epsilon_+^* \cdotp \epsilon_- [/tex] not [tex] \epsilon_+^* \cdotp i \epsilon_- [/tex]

    Sorry if that has wasted your time. I've definitely learned something about complex dot products but I still haven't been able to resolve this unfortunately.

    That first dot product is supposed to be zero...
     
  9. Apr 15, 2017 #8

    PeroK

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    You may be trying to take too many shortcuts. Write it out in full, starting with:

    ##\langle \epsilon_+, \epsilon_- \rangle = \langle \frac{1}{\sqrt{2}}(\epsilon_1 + i \epsilon_2), \frac{1}{\sqrt{2}}(\epsilon_1 - i \epsilon_2) \rangle##
     
  10. Apr 15, 2017 #9
    yes I have that, except that the epsilon plus is complex conjugate of itself. so [tex] \langle \epsilon_+^*, \epsilon_- \rangle = \langle \frac{1}{\sqrt{2}}(\epsilon_1 - i \epsilon_2), \frac{1}{\sqrt{2}}(\epsilon_1 - i \epsilon_2) \rangle [/tex]

    Is that not correct? Then I get the cross terms don't cancel? and if the i for the epsilon - is positive I get 1? I may just not be getting how to do a complex dot product. Funny we didn't go over this, perhaps we did it in our math methods last year... but I don't remember doing them at all if we did.
     
  11. Apr 15, 2017 #10

    PeroK

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    You don't take the complex conjugate in that way. The inner product of two vectors is:

    ##\langle u, v \rangle##

    What you are doing is:

    ##\langle u^*, v \rangle##

    Which is the inner product of ##u^*## and ##v##.

    You may be thinking of the inner product expressed as a product of coefficients:

    ##\langle u, v \rangle = u^*_1v_1 + u^*_2v_2 + u^*_3 v_3##
     
  12. Apr 15, 2017 #11
    The problem says to show that the complex conjugate of epsilon plus dotted with epsilon minus is 0 though. I'm just not seeing any possible way the real and complex can disappear... I've tried this so many different times trying different things...
     
  13. Apr 15, 2017 #12

    PeroK

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    This is wrong. You need to show that ##\epsilon_+ \cdotp \epsilon_- = 0## and ##\epsilon_\pm \cdotp \epsilon_\pm = 1##

    Perhaps whoever set the question got confused.

    Sorry, I didn't notice that this was part of the question, not your approach.
     
  14. Apr 15, 2017 #13
    Ok, I'll try that after some sleep. I was beginning to think there was an issue or typo perhaps! Thanks!
     
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