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Def. of unitary

  1. Apr 12, 2004 #1

    turin

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    I have read two definitions of unitary.
    A is unitary if:
    #1: det(A) = 1
    or
    #2: AadjointA = I

    Are these definitions equivalent?
     
  2. jcsd
  3. Apr 12, 2004 #2

    matt grime

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    Absolutely not. And you want to be careful with adjoint, better to specify that you mean A* the conjugate trasnpose.

    1 is sufficient for A to be SL

    2 says, with adjoint understood as above it is unitary.

    there are of course non-unitary matrices of det 1. exercise find a simple upper triangular real (integer) matrix to show this.
     
  4. Apr 12, 2004 #3

    turin

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    What is the difference between "adjoint" and "conjugate transpose?"




    What is "SL?"




    What is an "upper triangular" matrix? If it is what I think it is, then the determinant is just the product of the trace elements. This will not be a different value for the transpose of the matrix.
     
    Last edited: Apr 12, 2004
  5. Apr 12, 2004 #4

    matt grime

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    Adjoint is context dependent. if the matrices are real then adjoint is just transpose - part of the definition for unitary is that it is complex, and preserves an hermitian form (which is often how you'll see it defined). Just writing adjoint without specifying we mean complex at some point is omitting some information. One could argue that A^{adjoint}A =I defines the orthogonal (real) matrices.

    SL(F) is the special linear group - the group of matrices with determinant 1 (with entries in the field F)

    upper triangular means all entries below the main diagonal are zero.

    You can do the exerices with a diagonal matrix too, though obviously not one with integer coefficients.
     
  6. Apr 12, 2004 #5

    turin

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    matt grime,
    I did not understand the point you were trying to make about "adjoint." If the matrices are real, then the adjoint and transpose are the same thing, I get that. I don't see how this makes "adjoint" and "transpose conjugate" different (since the conjugate of a real number is itself).
     
    Last edited: Apr 12, 2004
  7. Apr 12, 2004 #6

    matt grime

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    If you just say that the matrix A satisfies A^{adjoint}A=I, we do not know if you mean orthogonal or unitary, at least indicate that you mean complex matrices.
     
  8. Apr 12, 2004 #7

    turin

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    This just seems a little strange to me. If the components of A are completely real-valued, then even though the condition were to specify that A is orthogonal, it would also be unitary. If the components were not all completely real-valued, are you saying that "adjoint" can strictly mean "transpose" in this case without complex conjugation?

    Please don't think I'm trying to be difficult. I was just taught that "adjoint" always means "transpose" and "complex conjugate" and I would like to be aware of any variation in usage for the sake of future conversations with the math people.
     
    Last edited: Apr 12, 2004
  9. Apr 13, 2004 #8

    matt grime

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    That is waht it means if the matirx is complex. Don't take this the wrong way, but as you asked if unitary meant det is 1 I assume you've not done much maths, thus I want to see that you understand that unitarity is a property of compelx matrices, and just saying adjoint doesn't do that.

    Let C be some space (what kind we won't say), with a bilinear map (?,?) on the elements of that space. THe (right) adjoint of a map T:C-->C is a map T* satisfying (TX,Y)=(X,T*Y)

    it occurs in many places, in the case of a complex vector space it means transpose conjugate and the bilinear map is the natural hermitian inner product.
     
  10. Apr 13, 2004 #9

    turin

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    No problem. And, you are correct.




    I don't follow this. What is a "bilinear map?" I'm assuming that X and Y are like vectors and that T is like an operator?

    I don't understand the following notation

    T:C-->C
     
  11. Apr 13, 2004 #10

    matt grime

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    I was trying not to have to specify what any of the things is, but C is some "space" X and Y are some elements of the space (it needn't be a vector space with vectors just a couple of sets with elements) and, T is a map from X to Y - that is what T:C--->C means.

    (?,?) is a just some way of paring up elements of C to get something in some other set, such as the real or complex numbers. (I didn't mean to say bilinear, just binary ie two inputs).

    *The* model for this is, yes, vector spaces (over R or C, this C isn't the other C in the post unfortunately) and vectors and linear maps and the inner product (or hermitian product), but there are other places where adjoint gets used.

    The main thing is if you ask:

    What is the group of matrices described by the law A^{adjoint}A=I=AA^{adjoint}?

    Then the answer is either the unitary group OR the orthogonal group depending on whether the matrices are real or complex. One is a subgroup of the other. It is nitpicking but if you don't make precise the terms you use then maths becomes hard.
     
    Last edited: Apr 13, 2004
  12. Apr 13, 2004 #11
    Or if the matrices are quaternionic, then we have the symplectic group
     
  13. Apr 13, 2004 #12
    You might be thinking of the group SU(n). the "S" in SU(n) means special linear, which means det(A) = 1. the "U" in SU(n) means unitary, which means A^adjoint*A=1 (where here, "adjoint" mean transpose conjugate)

    so a matrix in SU(n) satisfies both of your conditions. a matrix that satisfies only condition #1 (but not necessarily condition #2) is in SL(n,C), SL means special linear.

    if it satisfies condition #2 (but not necessarily condition #1) then the matrix is in U(n), for unitary.

    then the intersection of SL(n,C) and U(n) is SU(n). matrices that satisfy both condition #1 and #2
     
  14. Apr 13, 2004 #13

    matt grime

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    That's just evil, but as the quaternions aren't a field I don't feel too bad for not addressing them.
     
  15. Apr 13, 2004 #14

    turin

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    Sorry. I'm probably the wrong guy for you to be helping, then.




    I think that it makes sense to me in this way. But I am still confused by T:C-->C. That is, it seems like T should be the identity in this case.




    I definitely don't mind the nitpicking. Thanks for the help.
     
  16. Apr 13, 2004 #15

    turin

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    lethe,
    What does the C mean in SL(n,C)?
     
  17. Apr 13, 2004 #16

    Janitor

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    Here is something that Zee says in his book on Q.F.T.

     
  18. Apr 13, 2004 #17

    matt grime

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    Lethe's C is the complex numbers. My C was supposed to suggest Category.

    No one will ever agree on notation it seems. In Hilbert space theory adjoint is often just written * because overline denotes complex conjugation (damn physicists mucking it up), and who on earth wants to just take the conjugate of a matrix?
     
    Last edited: Apr 13, 2004
  19. Apr 15, 2004 #18
    how about sesquilinear? i like that word.
     
  20. Apr 15, 2004 #19
    you're too discriminating! you'll never cut it as a physicist!

    anyway, in case anyone is interested see this page for an explanation of the following analogy:

     
  21. Apr 15, 2004 #20
    my favorite notation convention (one that is, i think, shared by many physicists) is for [itex]z[/itex] and [itex]\overline{z}[/itex] to be your independent complex variables, and restrict yourself to the 2d plane where [itex]z^*=\overline{z}[/itex]
     
  22. Apr 15, 2004 #21

    turin

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    Well, I am interested, and I followed the link, but I did not get an explanation. I understood -17% (I now feel dumber). What is a quaternion? I've heard of it, and I've even seen people bring it up, but i have never understood it. Apparently, it is an extension of the complex field. But I thought that the complex field was as high as one needed to go in order to accomplish ... something, I forgot what it was called, something like "algebraic closure," maybe. So, anyway, can you give an algebraic excercise or something that would help us (me) understand quaternions, there purpose, etc.?
     
  23. Apr 15, 2004 #22
    OK, yes, the complexes are the algebraic closure of the reals, so for many purposes, the complexes are the most general field you would ever need. furthermore, the quaternions are not commutative, so they aren't even really a field, under most peoples definitions of field. (some people allow skew fields, however, then the quaternions might be allowed)

    another link on John Baez' homepage is about the whole family of normed division algebras. he explains how the only normed division algebras are the reals, the complexes, the quaternions, and the octonions. this is related to Bott periodicity, and the fact that only the 0 sphere, the 1 sphere, the 3 sphere and the 7 sphere are parallalizable. so there is some really deep mathematics here.

    by the way, a division algebra is a set of numbers where you are allowed add, subtract, multiply, divide, and take a norm. it might not be a field because we don't require that it be commutative (ab=ba), or perhaps even associative (a(bc)=(ab)c). for example, the octonions are not associative.

    so what are the quaternions? well there are a couple of descriptions. the easiest one, i think, is to just take the complexes, and add a few numbers.

    let i2=j2=k2=-1 and ij=k, jk=i and ki=j

    so you have 3 square roots of -1, instead of just one. it is a 4 dimensional vector space over the reals, and with those last relations, you can figure out how to multiply, divide and take the norm of any quaternion like a+bi+cj+dk. that last set of equations should remind you of the vector cross product in R3. it's no coincidence!

    another construction of the quaternions is the Pauli sigma matrices. you have seen them before, right? well, they satisfy the equations i wrote above (with perhaps a factor of i thrown in for good luck), so the quaternions can also be thought of as those 2x2 complex matrices spanned by the Pauli spin matrices and the identity matrix.
     
  24. Apr 15, 2004 #23

    matt grime

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    It is a non-abelian extension of the complex field - multiplication is not commutative. They can be described in terms of pauli spin matrices which is partly why physicist start to dribble unbecomingly when they come up in conversation.

    They are the set of all things of the form a+bi+cj+dk where a,b,c,d are real numbers and i,j,k satisfy ij = k jk=i ki=j i^2=j^2=k^2=-1=ijk

    or they are they can be realized as a degree two extension of the complx numbers.

    C is algebraically closed is exactly what you should have thunk.

    the quaternions aren't a filed, they are a division algebra.
     
  25. Apr 15, 2004 #24

    turin

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    Could I say, with any sort of meaning, that the Pauli matrices (with the 2x2 identity) are a realization of the quaternions? I'm trying to learn this group theory stuff, and there are some suspicious similarities here.

    How would one have a matrix with quaternion elements, which are themselves matrices apparently? Can you give an example? Would it be a matrix of matrices? Would the vectors on which it operates also need matrix components?

    This is the first time I've seen the relationship to Pauli matrices (it has made things orders of magnitude more clear); I hope I'm not abusing the relationship.
     
  26. Apr 15, 2004 #25

    matt grime

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    I suppose you could say that they are a realiztion (of each other - they are isomorphic algebras).

    A matrix of quaternions is just that - each ij'th entry is a quaternion and you multiply them in the obvious manner. They operate on (though that isn't necessary, or useful, often, when thinking of matrices) the vectors whose entries are quaternions. In the same way you can write z x+iy for complex numbers, there's no reason not to write w = x+iy+jz+wk
     
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