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Matrix or differential operator?

  1. May 20, 2010 #1
    Hi,
    I've been reading a couple of different books on quantum mechanics and have come a mathematical difficulty. I understand that the Hamiltonian is an operator but in some books, it represented as a matrix and in others as a differential operator? How can they both be equivalent approaches?

    I understand that they have some of the same properties, eg. are non-commutative. But how can both give the same results? I think (although I may be wrong) that I am asking for the relations between a function and a vector.

    Thanks
     
  2. jcsd
  3. May 20, 2010 #2

    Fredrik

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    See this post (or any book on linear algebra) for the relationship between matrices and linear operators between finite-dimensional vector spaces.
     
  4. May 20, 2010 #3
    Thanks for the speedy and clear response. Does that suggest that any linear operator A can be written as a matrix and if so is there a general method?
     
  5. May 20, 2010 #4

    Fredrik

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    Yes it does, when the vector spaces are finite-dimensional. The stuff I wrote about there is the method. I decided to edit my post, but you replied while I was typing so I'm posting the edited version here instead:


    See the quote below (or any book on linear algebra) for the relationship between matrices and linear operators between finite-dimensional vector spaces.
    Also, consider the special case U=V, with an orthonormal basis [itex]\{u_i\}[/itex]. (Set [itex]v_i=u_i[/itex]).

    [tex]\langle u_i,Au_j\rangle=\langle u_i,(Au_j)_k u_k\rangle=(Au_j)_k\langle u_i,u_k\rangle=(Au_j)_k\delta_{ik}=(Au_j)_i=A_{ij}[/tex]
     
    Last edited: May 20, 2010
  6. May 20, 2010 #5
    Using discrete approximations may help seeing what's going on.

    Suppose you are interested in some functions [itex]f:[0,1]\to\mathbb{C}[/itex].

    You can approximate the interval [itex][0,1][/itex] as a discrete set

    [tex]
    \big\{0,\frac{1}{N},\frac{2}{N},\ldots,\frac{N-1}{N}\big\}
    [/tex]

    with some large [itex]N[/itex]. Then the functions [itex]f:[0,1]\to\mathbb{C}[/itex] can approximated as functions

    [tex]
    f:\big\{0,\frac{1}{N},\frac{2}{N},\ldots,\frac{N-1}{N}\big\}\to\mathbb{C}.
    [/tex]

    But now these new functions are merely vectors in N-dimensional vector space. Like

    [tex]
    (v_1,v_2,\ldots, v_N) \approx \Big(f(0), f\big(\frac{1}{N}\big), f\big(\frac{2}{N}\big),\ldots, f\big(\frac{N-1}{N}\big)\Big)
    [/tex]

    Any operator that was form [itex]A:D(A)\to \mathbb{C}^{[0,1]}[/itex] with some domain [itex]D(A)\subset \mathbb{C}^{[0,1]}[/itex], can now be approximated as a [itex]N\times N[/itex]-matrix.

    This approach gives some insight into what it means that function spaces are infinite dimensional vector spaces.
     
  7. May 20, 2010 #6
    So roughly speaking are we saying that each value of a function can be thought of as one of infinitely many components of a vector?
     
  8. May 20, 2010 #7
    Yes the argument becomes the index and the value of the function at that argument becomes the component of that index.
     
  9. May 21, 2010 #8
    Last edited by a moderator: May 4, 2017
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