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I Calculating the number of real parameters of a unitary matri

  1. Mar 2, 2016 #1
    So using the property
    UU=1
    I was able to get
    ∑uikujk*=δij
    where the sum is over k=1,...,n and u is a complex number.
    from this you get n equations when i=j. If i≠j I get n(n-1) equations but in the notes I'm reading from it should be n(n-1)/2. I'm struggling to see where the half comes from, any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 2, 2016 #2

    RUber

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    They are halved because the matrix should be symmetric.
     
  4. Mar 2, 2016 #3
    One way to think of this is that you start off with n2 complex matrix elements, which comprise 2 n2 real parameters.

    Then, each real equation — assuming they are independent — decreases the degrees of freedom by 1.

    Now, your equations, one for each i and each j, form n2 complex equations. But by taking a) the real part of both sides of a complex equation, and b) taking the imaginary part of each side, you end up with twice that number of real equations.

    So you have 2 n2 real equations that way. If the equations were independent, this would imply

    2 n2 - 2 n2 = 0​

    degrees of freedom. Wait — that's impossible. So maybe the equations are not all independent? Do you really have a separate equation for each pair (i,j) ?

    (Hint: Notice what happens if in your equation you interchange the i and the j.)

    P.S. I now see RUber's comment, but it isn't typically true that a unitary matrix is "symmetric", which would mean that the (i,j) entry is equal to the (j,i) entry for all i and j.
     
  5. Mar 2, 2016 #4

    RUber

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    Sorry, I didn't mean the matrix was symmetric, but the products have some symmetry. Especially when set equal to zero.
     
  6. Mar 2, 2016 #5
    $$\sum_{k}u_{ik}^*u_{kj} = \delta_{ij} $$
    I would like to point out that these equations are quadratic in the components of ##U##. The logic that applies to systems of linear equations does not necessarily apply to systems of other kinds of equations. One can carry out the assessment of the number of real parameters this way but there are other ways.

    The number of real parameters of a Lie group is going to be the dimension of the tangent space. This is just like the number of tangent directions on a surface is equal to the dimension of the surface. To get a nice simple result we calculate a tangent at the identity. With ##\epsilon \in \mathbb{R}##,
    $$V = I + \epsilon K $$
    Now we must restrict ##K## so that we move in a direction that makes V also unitary to first order.
    $$V^\dagger V = (I^\dagger +\epsilon K^\dagger)(I + \epsilon K) = I + \epsilon(K+K^\dagger) + \dots$$
    $$K+K^\dagger = 0 $$
    K has ##n^2## real parameters - n pure imaginary numbers on the diagonal, and then n(n-1)/2 arbitrary complex numbers for the upper diagonal.
    $$n(n-1) + n = n^2$$
    So the number of real parameters of K is ##n^2##, which is also the number of real parameters of ##U(n)##.
     
    Last edited: Mar 2, 2016
  7. Mar 2, 2016 #6
    Whether you are dealing with linear equations, or other kinds, in either case it's necessary to determine the number of independent equations if you want to know the dimension of the solution space.

    The general way to do this is to use the implicit function theorem, which requires linearizing the equations at a given point. In the case of linear equations, this is always the same at each point, since they are their own linearization. For nonlinear equations, you could potentially get differing results at different points.

    But each equation of the form

    Fj(x) = 0​

    for a function

    Fj : RnR

    that is of maximal rank on the inverse image Fj-1(0) of 0 will define a smooth hypersurface

    Sj = Fj-1(0).​

    If we assume all the equations

    Fj(x) = 0​

    to be true, then the points of Rn where all the equations hold are precisely the intersection of all the hypersurfaces

    Fj-1(0) = j Sj

    where the intersection is taken over all j.

    Then if the equations are all independent, each hypersurface lowers the dimension of the intersection by one.
     
  8. Mar 9, 2016 #7
    Correction: That last equation above should not say that Fj-1(0) = j Sj. It should just state the tautology:

    j Fj-1(0) = j Sj.​
     
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