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Unitary Matrices

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Find a third column so that U is unitary. How much freedom in column 3?

    [ 1/√3 i/√2 ]
    [1/√3 0 ] = U
    [i/√3 1/√2 ]


    2. Relevant equations

    UHU=I
    UH=U-1

    3. The attempt at a solution

    Obviously in order for the matrix to be unitary the inner product of their columns must be equal to zero.

    I've constructed to equations by multiplying the hermitian of each of the first two columns times by the third column I designated,
    [a]

    [c].
    These equations are a/√3+b/√3+ci/√3=0 and -ia/√2+c/√2=0.

    However, I have two equations and three unknowns.

    Is there another way to attempt the problem? Suggestions?
     
  2. jcsd
  3. Nov 6, 2011 #2

    I like Serena

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    Hi tatianaiistb! :smile:

    The length of the 3rd vector is free, so you're supposed to be left with 1 unknown.
    Fix any of the 3 unknowns to an arbitrary constant value and calculate the other 2.
    (If you can't solve it, pick another unknown to fix.)
     
  4. Nov 6, 2011 #3

    Deveno

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    the third column is completely determined, up to a multiple of a complex number lying on some circle.

    you should be able to solve for either a and b in terms of c, a and c in terms of b, or b and c, in terms of a.

    note that although we will still have one unknown, (a,b, or c), it's norm is known, and can be used to yield a vector in C3 with the desired properties (pick the positive real number with the proper norm, for example).

    i solved for b and c in terms of a, that worked out well.
     
  5. Nov 6, 2011 #4
    I'm still stuck with this problem. After doing the Hermitian inner products of columns 1 and 2 with my unknown column [a b c], I got the two equations I previously stated:

    a/sqrt3 + b/sqrtb - ci/sqrt3 = 0
    -ia/sqrt2 + c/sqrt2 = 0

    For my third equation, I used the length formula for complex numbers that:
    absolut value of a^2 + absolute value of b ^2 + absolute value c^2 = 1

    I then solved for a, b, and c, and got that a is 1/2, b is -1, and c is i/2.

    When I checked, this column is orthogonal and of lenght 1. However, when I checked with U^H*U=I, I'm not getting the identity matrix... almost, but not quite. So I know something is wrong.

    The answer should be (1/sqrt6, -2/sqrt6, i/sqrt6).

    Any ideas?
     
  6. Nov 6, 2011 #5

    I like Serena

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    What's the determinant?
    Is it perhaps -1?
    It should be +1.
     
  7. Nov 6, 2011 #6
    So how do you suggest I go about solving the problem?
     
  8. Nov 6, 2011 #7

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    If the determinant is -1, invert the sign of the 3rd vector.
     
  9. Nov 6, 2011 #8
    But I still cannot get U^H*U=I because as of right now, I'm getting the multiplication to equal
    [1 0 0
    0 1 0
    0 0 (3/2)]
    The (3/2) should be a 1...

    Any more suggestions? Thanks!
     
  10. Nov 6, 2011 #9

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    I'm just looking at your matrix again and noticed that your first 2 columns aren't orthogonal.
    Did you drop a minus sign?
     
  11. Nov 6, 2011 #10
    The first two columns, which are given by the problem, are orthogonal. Since we're dealing with complex numbers, if you take the inner product of X^H*y it does equal to 0.
     
  12. Nov 6, 2011 #11

    Deveno

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    first of all, what we have is aa*+ bb* + cc* =

    |a|2 + |b|2 + |c|2 = 1.

    this is the complex norm or modulus, not "absolute value"

    if you actually multiply:

    [tex]\begin{bmatrix}1/\sqrt{3}&i/\sqrt{2}&a\\1/\sqrt{3}&0&b\\i/\sqrt{3}&1/\sqrt{2}&c\end{bmatrix} \begin{bmatrix}1/\sqrt{3}&1/\sqrt{3}&-i/\sqrt{3}\\-i/\sqrt{2}&0&1/\sqrt{2}\\ \overline{a}&\overline{b}&\overline{c} \end{bmatrix}[/tex]

    and set equal to the identity matrix, you get:

    |a| = 1/6
    |b| = 2/3
    |c| = 1/6

    the vector (1/2, -1, i/2) is "the right direction" for your 3rd column, you just need to adjust by a scaling factor, z. note that:

    (1/2)2 + (-1)2 + (i/2)(-i/2) = 1/4 + 1 + 1/4 = 3/2, not 1.
     
  13. Nov 6, 2011 #12

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    Hmm.

    [tex]\begin{pmatrix} 1/√3 \\
    1/√3 \\
    i/√3 \\
    \end{pmatrix} \cdot \begin{pmatrix}
    i/√2 \\
    0 \\
    1/√2 \end{pmatrix} = i/√6 + 0 + i/√6 = 2i/√6[/tex]

    Am I missing something? :confused:
     
  14. Nov 6, 2011 #13

    Deveno

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    conjugate linearity?
     
  15. Nov 6, 2011 #14

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    Ah, okay!
     
  16. Nov 6, 2011 #15
    [1/sqrt3 1/sqrt3 -i/sqrt3] [i/sqrt2 0 1/sqrt2]^T = 0 the first column is the hermitian, meaning the transpose of the conjugate

    So are my first two equations correct? How are you solving for a,b,c using the U^H*U=I? I see what you're saying about the norm... The professor kept calling it "absolute," and that's part of my confusion.
     
  17. Nov 6, 2011 #16

    Deveno

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    read post #11 again, you obtained a solution, but you need to normalize your vector, it's not a unit vector.
     
  18. Nov 6, 2011 #17

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    As Deveno pointed out, it is not length 1.
    ||(1/2, -1, i/2)||2 = (1/2, -1, i/2)H (1/2, -1, i/2) = 3/2

    You can also write:
    ||(1/2, -1, i/2)||2 = |1/2|2 + |-1|2 + |i/2|2 = 3/2
     
  19. Nov 6, 2011 #18
    After normalizing the vector, I'm still not getting the right answer...

    a=(1/2)/sqrt(3/2)=sqrt6 / 6
    b=-1/sqrt(3/2)= - sqrt6 / 3
    c=(i/2)/sqrt(3/2)= (sqrt6 / 6)*i

    The answer is supposed to be (1/sqrt6) * (1,-2,i)

    Where did I err again? Thanks!
     
  20. Nov 6, 2011 #19

    Deveno

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    √6/6 = √6/(√6)2 = 1/√6

    √6/3 = (√2)(√3)/(√3)2 = √2/√3 = (√2)2/(√2)(√3) = 2/√6

    trust the force, luke.
     
  21. Nov 6, 2011 #20

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    No error, this is the right answer.
    [tex]{\sqrt 6 \over 6}={1 \over \sqrt 6}[/tex]
     
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