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I Angles between complex vectors

  1. Mar 31, 2016 #1
    So I was trying to learn how to find the angle between two complex 4-dimentional vectors. I came across this paper, http://arxiv.org/pdf/math/9904077.pdf which I found to be a little confusing and as a result not overly helpful. I was wondering if anyone could help at all?

    Many thanks in advance :)
     
  2. jcsd
  3. Mar 31, 2016 #2

    Ssnow

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    Assuming that you have a ##4## dimensional complex vector space ##V_{\mathbb{C}}## you have essentially two possibility, You can consider the complex space isometric to the real space ##\mathbb{R}^{8}## so you have the relation for the Euclidean angle ##\Theta##:

    [tex]\cos{\Theta(A,B)}=\frac{(A,B)}{|A||B|}[/tex]

    where ##(,)## is the product in ##\mathbb{R}^{8}## or you can consider your complex space isometric to ##\mathbb{C}^{4}## and have the relation for the Hermitian angle ##\Theta_{c}##:

    [tex]\cos{\Theta_{c}(A,B)}=\frac{(A,B)_{\mathbb{C}}}{|A||B|}[/tex]

    where now ##(,)_{\mathbb{C}}## is the hermitian product on ## \mathbb{C}^{4}##. Defining and almost complex structure you can have other kind of angles (Kahler) but depends what you need ... the last part of the article works in order to find relations between these kind of angles ...
     
  4. Mar 2, 2018 #3
    Just to expand on the Euclidean and Hermitian angles, since complex angles can be a bit confusing: if a Hermitian (complex) inner product is defined on ##\mathbb{C}^{4}##, then the complex angle between two complex vectors ##v## and ##w## is defined as

    $$\cos\theta_{c}\equiv\frac{\left\langle v,w\right\rangle }{\left\Vert v\right\Vert \left\Vert w\right\Vert }.$$

    Both the angle and its cosine are in general complex. The Euclidean angle is defined by taking the real part of the cosine, to get a real angle:

    $$\cos\theta_{E}\equiv\frac{\mathrm{Re}\left(\left\langle v,w\right\rangle \right)}{\left\Vert v\right\Vert \left\Vert w\right\Vert }.$$

    If we take an orthonormal basis in ##\mathbb{C}^{4}## based on the Hermitian inner product, this gives an orthonormal basis of ##\mathbb{R}^{8}## via decomplexification (removing the possibility of complex multiplication of scalars), which in turn defines a real inner product on ##\mathbb{R}^{8}##. The Euclidean angle is equal to the angle between the vectors under this decomplexification. Note that a Euclidean angle of ##\pi/2## does not ensure a vanishing Hermitian inner product.

    If we instead take the modulus (absolute value) of the cosine, this defines the Hermitian angle, which is again real:

    $$\cos\theta_{H}\equiv\frac{\left|\left\langle v,w\right\rangle \right|}{\left\Vert v\right\Vert \left\Vert w\right\Vert }.$$

    This angle, just like the angle in Euclidean space, is the ratio of the orthogonal projection of ##v## onto ##w## over the norm of ##v## (or the reverse). However, it's important to remember that the orthogonal projection here uses the Hermitian inner product, so that parallel vectors in ##\mathbb{C}^{4}## may be orthogonal using the corresponding real inner product in ##\mathbb{R}^{8}##.

    More details here: https://www.mathphysicsbook.com/mathematics/abstract-algebra/generalizing-vectors/norms-of-vectors/
     
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