Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Hermitian and not Hermitian

  1. Jul 7, 2004 #1
    Let's say you're wondering around P(oo) (which I'll use to represent the space of polymomials of any degree on the interval [-1,1]) and you decide to calculate the matrix X representing the position operator x. Let's say you do this in the basis: 1, t, t^2, ..., t^n,... you'll find that the matrix X is not hermitian even though x is hermitian on P(oo). Then you decide to calculate the matrix again, but now you do it with the orthonormal basis generated from the above basis via the Gram-schmidt method (i.e. the properly normalized Legendre polynomials) Now you find that X is hermitian. What gives? should the hermiticity of an operator be independent of its representation in a particular basis? (maybe not and this is super obvious, feel free in that case to slap me).

    Kevin
     
  2. jcsd
  3. Jul 7, 2004 #2

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    SLAP SLAP SLAP :-)

    The hermiticity of a matrix is not depending on its basis *as long as the basis is orthonormal* of course, but your previous basis of 1, t, t^2 ... wasn't....

    cheers,
    Patrick.
     
  4. Jul 8, 2004 #3
    Please slap me some more :smile: because I didn't know the above. Is it the same with properties like unitary-ism, symmetry, etc?

    Kevin
     
  5. Jul 8, 2004 #4
    Wait a second?!

    A linear operator A is symmetric iff (g,Af) = (Ag,f) for all f,g in Domain(A).

    A linear operator A is self-adjoint iff: (i) A is symmetric, (ii) the adjoint At exists, and (iii) Domain(A) = Domain(At).

    (Your uses of the term "Hermitian" coincide with my use of "self-adjoint" in the above [note: the definition I have given is incomplete because the term "adjoint" in (ii) has not been defined].)

    Now, let A be self-adjoint, and let fi be an arbitrary basis. Define the "matrix" of A in this basis by

    Aij = (fi,Afj) .

    Then,

    Aij = (fi,Afj) = (Afi,fj) = (fj,Afi)* = Aji* .

    So, why does the basis have to be orthonormal?

    ************************************
    * SLAP SLAP SLAP :-) ... on me
    *
    * These "matrix" elements only make sense for an orthonormal basis!
    *************************************

    Homology, how are you defining the "matrix" of an operator in the tn-basis?

    *************************************
    * Oh, I see ...
    *
    * since f(x) = Sigma_n { fnxn }, define the
    * "vectors" as the "infinituples" {fn} and the "matrices"
    * as the maps Amn of the "infinituple" components
    * according to
    *
    * Sigma_n Amnfn .
    *
    * ... In any case, the definition of "self-adjoint" which I gave above
    * is the one that is meant in QM and it is quite independent of
    * any choice of basis.
    *
    * Using a definition like this one down here, is equivalent to
    * changing the "inner product" associated with the Hilbert space
    * in the definition I gave above.
    *************************************
     
    Last edited: Jul 8, 2004
  6. Jul 8, 2004 #5
    to Eye_in_the_Sky

    For an inner product (v,w) the matrix form is Transpose(v)*A*w where A is the matrix of the bilinear form defining the inner product. So if we have an operator M such that (Mv,w)=(v,Mw) then the matrix form for this is
    Transpose(M*v)*A*w=Transpose(v)*Tranpose(M)*A*w and this must be equal to Tranpose(v)*A*(M*v) so in order for the matrix version of the operator M to be self-adjoint Transpose(M)*A=A*M. Now it may be the case that this implies symmetry in M, but it doesn't seem so.

    Some options that occur to me now are: we could try to find a basis for which both A and M are diagonal and then they would both be symmetric and the above requirement for self-adjoint-ness would be met. Or if the matrix A can be reduced to the identity by an orthogonal change of basis then that would imply that the transformed version of M would symmetric.

    However I think that it is clear from above that a matrix can be self-adjoint but not be symmetric, it really depends on the inner product.

    Is this clear to you? Also I'm not sure what you mean by
    Do you mean how to come about it? well one way is to see what the effect of x is on the basis vectors 1, t, t^2 etc. This will define the rows of x. Besides that its the result of an excercise in Griffith's QM book in which he says: "If this is a Hermitian operator (and it is), how come the matrix is not equal to its transpose conjugate?"

    Try it, you'll find that the matrix X for the position operator in the 1, t, t^2, ... is not hermitian/self-adjoint however it is in the orthonormal basis generated by this basis.

    Kevin
     
  7. Jul 9, 2004 #6

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    First of all, in finite-dimensional vector spaces, hermitic, self-adjoint and symmetrical (conjugate-symmetrical) are the same.
    Second, you can only write that the matrix element of an operator A is
    a_i,j = (e_i, Ae_j) if {e_i} form an orthonormal set of course, such that
    (e_i, e_j) = delta_{i,j}.

    Finally, if you transform a matrix A into another base through a basis transformation S, you have B = S A S^(-1).

    The adjoint of B, B' = S^(-1)' A' S'.

    If we assume the matrix A to be self-adjoint, then A = A', but in order for B to be equal to B', we need that S^(-1) = S', which is only true for a unitary matrix S ; meaning that the new basis is orthonormal.

    cheers,
    Patrick.
     
  8. Jul 9, 2004 #7
    Again, this is not true as I have already pointed out. here is an example. Let's use the following syntax for matrices B={{1-1,1-2},{2-1,2-2}} where the first number indicates row and the second indicates column so the entry 2-1 in B indicates that this entry is in the second row first column.

    Okay, let A={{4,1},{1,1}} define an inner product on R^2. A is positive definite and symmetric so it does the job. What this means is that for two vectors v,w we can evaluate (v,w)=Transpose(v)*A*w (keeping in mind here that this inner product is defined by A). Now let M={{1,2},{10,3}}. I claim that this nonsymmetric matrix is self-adjoint.
    (Mv,w)=Transpose(Mv)*A*w=Tranpose(v)*Tranpose(M)*A*w.

    Now, Tranpose(M)*A={{14,11},{11,5}}=A*M so we can conclude that

    Transpose(v)*Transpose(M)*A*w=Transpose(v)*A*M*w = (v,Mw)

    I haven't used this anywhere, I'm not sure why you bring it up.

    This I agree with certainly. Certainly if a matrix is symmetric then it remains so under unitary/orthogonal changes of basis, by just your argument. However, I remain fixed on the fact that self-adjoint only implies symmetric for orthonormal bases whether finite dimensional (as considered above in this post) or infinite-dimensional as considered in my original post.

    Thanks again for your feedback, these conversations really help flesh out ideas.

    Kevin
     
  9. Jul 9, 2004 #8

    vanesch

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Right. This is what I would call putting the wagon before the horses :-)
    To me, the in-product is the canonical one (by the matrix A = delta_ij), and orthogonal bases are defined with respect to this base.
    What you do with your definition of the inner product with matrix A is to work (in my language) in a non-orthogonal basis, and you conclude that the matrix is not symmetrical. I reformulate this by saying that (assuming the canonical in product) a hermitean operator in a non-orthogonal basis is not symmetrical. But that's saying the same thing !!

    Defining a general inproduct just comes down to saying that you use the canonocal inproduct, but in a non orthonormal basis. As any symmetrical matrix can be diagonalised, the relationship is straightforward...

    cheers,
    patrick.
     
  10. Jul 9, 2004 #9
    Ahh, this explains our misunderstanding, since my notion of an inner product is just something induced by a bilinear form with certain properties.

    Ha ha, I think we understand one another and I've figured out my original problem with your help! It was originally such a surprise to me that such things could happen since I have been raised on inner products that use the identity. However I see now the importance of the orthonormal basis in the context of these words: hermitian/self-adjoint and the implications to their representative matrices.

    Thanks again for your help!

    Kevin
     
  11. Jul 11, 2004 #10
    For Homology: a small point of clarification

    First off, I'd like to write

    Transpose(v)*A*w

    as

    t(v)*Bw .

    Now, it looks to me like the "v" and "w" in "(v,w)" are "abstract" vectors, whereas the "v" and "w" in "t(v)*Bw" are the components of those vectors - relative to some basis - arranged as column matrices. Is that what you mean?

    If so, then let's write "vectors" in bold font. Also, let's select a basis, and give it a name, say bi. Then, using the "summation convention":

    v = vibi , w = wibi .

    In this notation, we then have for the components

    t(v)*Bw <--> vi*Bijwj ,

    and from (a "consistent" definition of inner product)

    t(v*)Bw = (v,w) , for all v,w ,

    it follows that

    Bij = (bi,bj) .

    This "connects" the component representation (relative to basis bi) to the abstract vector representation.

    Next, let L be a linear operator in/on the vector space, and let L <--> Lij be its matrix representation relative to the basis bi; that is,

    L <--> L <--> Lij ,

    where Lij is defined by

    Lbj = Lijbi .

    You can check that this is the "correct" definition to give the correspondences

    Lv <--> Lv <--> Lijvj .

    Finally, if we define "symmetric" as

    (v,Lw) = (Lv,w) , for all v,w in Domain(L) ,

    then the equivalent condition in component form - relative to the basis bi - is given by

    t(L)*B = BL , on the domain of L ,

    where, to repeat,

    Bij = (bi,bj) .
     
  12. Jul 11, 2004 #11
    are you implying something?

    Perhaps we've mixed up terms. As I have learnt it, self-adjoint is the term for such linear operators L such that (Lv,w)=(v,Lw) v,w in a real vector space V. Hermitian is simply the complex version. If we represent a self-adjoint operator in a orthonormal basis then the resulting matrix representation will be symmetric, i.e. unchanged under transpose. While for the hermitian case, the resulting matrix is unaltered under transpose and conjugation.

    But in any event, we're not arguing are we? You can freely calculate the matrix entries of the position operator in the nonorthonormal basis 1, t, t^2, t^3, ... and in the orthonormal basis derived from them and you will see that in the nonorthonormal basis the X matrix is not hermitian even though x (as an abstract operator) certainly is and that in the orthonormal basis X is hermitian.

    Kevin
     
  13. Jul 12, 2004 #12
    Homology:

    My (admittedly poor) attempt at offering a small point of clarification has resulted in a larger point of "obscurification". What I wrote was originally intended to be a response to your reply to my earlier post in this thread. But by the time I got to writing it and posting it, I saw that you had already reached a resolution through your exchanges with Vanesch ... so, I changed it around posted it anyways, thinking it would serve as sort of clarification and summary of the main conclusions.

    All I really wanted to do was lay down the basic relationship between the "abstract" vector representation and the "component" representation in clear terms. Specifically, I wanted to say:

    The condition

    [1] (Lv,w) = (v,Lw) , for all v,w ,

    is equivalent to

    [2] t(L)*B = BL ,

    where

    [3] Bij = (bi,bj) .

    If a basis bi is not orthonormal with respect to an inner product ( , ), then by [3] the matrix B is different from the identity. In that case, the Hermicity condition [1], which is equivalent to [2], will not coincide with t(L)* = L (except in the trivial case L = 0).

    ... and that is all.
    ---------------------------------

    Regarding the definition of "symmetric" which I stated, ... well yeah ... just forget about it (... until you see it again somewhere else soon).
     
  14. Jul 12, 2004 #13
    You Betcha! :biggrin: no worries, and again: thanks.

    Kevin
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Hermitian and not Hermitian
  1. Hermitian operator (Replies: 2)

  2. Hermitian operators (Replies: 1)

  3. Hermitian Operator (Replies: 3)

  4. Hermitian Operators? (Replies: 34)

Loading...