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Hermitean operators?

  1. Feb 5, 2014 #1
    OK I'm not sure if this should go in the math or quantum forum, but as I'm learning these in introductory QM I post the questions here. Please move the thread if the section is inappropriate.

    Anyway, some questions:

    * What is an inner product space?

    * What is a hilbert space?

    * What are hermitian operators and what do they do? From what I can see, hermitian operators are supposed to kill the complex output when used on an eigenfunction. However, how does "being hermitian" do that?

    * In my lecture notes it is written an operator ##\hat{F}## is hermitian if ##\int (\hat{F} \Psi_1)^*\Psi_2 d\tau=\int (\Psi_1)^*\hat{F}\Psi_2 d\tau##, where the * represents complex conjugation. Uhm, why does that make something hermitian?
     
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  3. Feb 5, 2014 #2

    bhobba

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    It's a vector space with a defined inner product. If that's goobly gook then you need to study some linear algebra.

    In the context of QM its a complex inner product space with a countable basis. There is a bit more that can be said, but as far as QM is concerned that's about it. Again if it's gobbly gook you need to study some linear algebra.

    Hermitian operators are operators that equal their conjugate. Its a very simple theorem to show their eigenvalues must be real. It's so easy to prove once you know a bit of linear algebra the answer again is really the same - you need to study a bit of linear algebra.

    Just to show how easy the proof is here it is, if u is an eigenvector of A, A hermitian <u|A|u> = y <u|u> = conjugate <u|A|a> = conjugate (y) <u|u> ie y is real.

    Thanks
    Bill
     
  4. Feb 5, 2014 #3
    OK, so all possible dot products of the vectors in the vector-space make up the inner product space?
    With "complex", you mean the vectors can have complex components, right? But what does "countable basis" mean?

    I don't think I'm familiar with the notation (especially the "|"s and "<u|A|u>"). Could you explain it step by step? And yes, at this stage it's obvious
    , but I would appreciate it if you could at least give me some pointers on what to study. It's been a year since my course in linear algebra so I just need some derusting to get the mojo back.
     
  5. Feb 5, 2014 #4

    bhobba

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    Mate, this is like asking someone to explain calculus.

    There is really no choice but the slog of studying it.

    There are many free linear algebra books eg:
    http://linear.ups.edu/html/fcla.html

    Spend a bit of time going through it - its well worth it.

    What I gave is the so called Bra-Ket notation.

    Once you are familiar with Linear Algebra then myself or someone else can explain it, or give a link eg:
    http://www.mathpages.com/home/kmath638/kmath638.htm

    A basis means every vector can be formed as a linear combination of those vectors, and this is the minimum number that such can be applied to. That number is called the dimension of the space. There is a nifty theorem that says any vector space has a basis:
    http://www.proofwiki.org/wiki/Vector_Space_has_Basis

    Countable means, while it may be infinite, it is in one to one correspondence with the natural numbers.

    All this is basic math stuff and explaining it so you understand the terms really is what studying linear algebra is about.

    Thanks
    Bill
     
    Last edited: Feb 5, 2014
  6. Feb 5, 2014 #5
    I am familiar with linear algebra because, as I said, I took a course in proof-based linear algebra a year ago. Thing is I haven't used it since june 2013, so I am a bit rusty on the definitions, and some of the notation is foreign (like bra-kets: which seems to be specific to quantum mechanics only). However I can easily remember (and understand) the suff when it's properly explained. So no, it's not at all like explaining all of calculus.

    Back to your proof:

    Could you explain your notation, or better yet use latex? "A hermitian"? You mean A is hermitian? At any rate, I am with you until you say "y<u|u> = conjugate <u|A|a>". Why is that the case? where did the small "a" come from and what does it mean?

    PS: Thanks for the link explaining bra-ket notation.

    If these questions are too basic, perhaps this thread should be moved to a homework forum?
     
    Last edited: Feb 5, 2014
  7. Feb 5, 2014 #6

    bhobba

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    Vectors are written as |a> called kets. Given a vector space you have the set of linear functions mapping to complex numbers defined on those vectors (called functionals). This mapping, basically, means its a complex vector space. Formally they also form a vector space called its dual. They are called bras and are written as <b|. Since bras are linear functions defined on kets you have <b|a>, which is the linear function <b| acting on |a>, called a Bra-Ket (British for bracket :biggrin:).

    Now there is this nifty theorem, called the Riesz Representation Theorem, that shows the bras and kets can be put into one to one correspondence such that <a|b> = complex conjugate <b|a>, and this correspondence is generally assumed:
    http://en.wikipedia.org/wiki/Riesz_representation_theorem

    If A is a linear operator mapping kets to kets A|a> is also a ket. But since A|a> is a ket it can be acted on by a bra to give <b|A|a>. But this can be viewed as a linear functional defined on |a>. Hence <b|A is a bra and defines how linear operators act on bras. Thus by definition (<u|A)|v> = <u|(A|v>). The ket corresponding to the bra <a|A defines a linear operator on |a> called its Hermitian Conjugate A* so this ket is A*|a>. So we have <a|A|b> = conjugate <b|A*|a>. Operators such that A=A* are called Hermitian and by definition observables in QM are Hermitian.

    Hopefully that helps.

    Thanks
    Bill
     
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