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B Projection operator

  1. Sep 27, 2016 #1
    What mathematical operation does this expression represent"

    |i><i|

    I know the i's are unit vectors, and I know <i|i> is the dot product, but what operation is the ket-bra?
     
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  3. Sep 27, 2016 #2

    vanhees71

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    The magic of Dirac's notation is that it is intuitive. Here you have an operator
    $$\hat{P}=|i \rangle \langle i|.$$
    To see what it does to a vector ##|\psi \rangle## just write it down:
    $$\hat{P}=|i \rangle \langle i|\psi \rangle,$$
    et voila, that's the right definition of the projector. It projects the vector ##|\psi \rangle## into the direction of ##|i \rangle##. This tells you that ##\psi_i=\langle i | \psi \rangle## is the component of ##|\psi \rangle## in direction of ##|i \rangle##. So it's precisely what a projection means!

    If you have a complete set of orthonormal vectors ##|j \rangle##, you have
    $$\sum_{j=1}^{\infty} |j \rangle \langle j |=\hat{1}.$$
    This means the decomposition of an arbitrary vector is given by (again just write it down; Dirac's magic gives the correct expression!):
    $$|\psi \rangle=\hat{1} |\psi \rangle = \sum_{j=1}^{\infty} |j \rangle \langle j|\psi \rangle.$$
     
  4. Sep 27, 2016 #3

    DrClaude

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  5. Sep 27, 2016 #4
    Thank you Vanhees, is there an expansion for the expression? e.g. a1i + a2i + a3i .....
     
  6. Sep 27, 2016 #5
    In matrix form the projector operation is the matrix resulting from column vector . row vector, where . is matrix multiplication. A ket is a column vector, a bra is a row vector, as you know if you do row vector . column vector you get a number, if you do column vector . row vector you get a matrix.
     
  7. Sep 28, 2016 #6

    vanhees71

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    A bra vector is not a column vector but a vector in an abstract Hilbert space. Usually you organize the components of a vector with respect to a basis as a column vector and the matrix elements of an operator as a matrix. Of course, it's much more convenient to use the Dirac formalism since then you don't need to think about this organization much. For the operation of an operator on a vector you have
    $$\hat{A} |\psi \rangle=\sum_{j,k=1}^{\infty} |j \rangle \langle j|\hat{A}|k \rangle \langle k|\psi \rangle=\sum_{j,k=1}^{\infty} |j \rangle A_{jk} \psi_k.$$
    Now you organize the vector components as a column and the matrix elements as a matrix,
    $$\tilde{A}=(A_{jk})=\begin{pmatrix} A_{11} & A_{12} &\ldots \\ A_{21} & A_{22} & \ldots \\ \vdots & \vdots & \vdots \end{pmatrix},$$
    $$\tilde{\psi}=(\psi_k)=\begin{pmatrix} \psi_1 \\ \psi_2 \\ \vdots \end{pmatrix}.$$
    Then you can have for the components of ##|\phi \rangle=\hat{A} |\psi \rangle##
    $$\tilde{\phi}=\tilde{A} \tilde{\psi}$$
    in the sense of the usual matrix-vector product. Note, however, that you have vectors and matrices with infinitely many components.
     
  8. Sep 28, 2016 #7
    I was referring to the isomorphism between ##L^2(\infty)## and ##l^2(\infty)##.
     
  9. Sep 28, 2016 #8

    vanhees71

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    It is very important to distinguish between the abstract vectors and various realizations. The Dirac bras and kets are in the abstract separable Hilbert space ##\mathcal{H}##. It can be realized as the function space ##L^2## of square integrable functions or the space ##\ell^2## of square-summable squences. These realizations are formally connected with the abstract Hilbert space of the bra-ket formalism by using (generalized) orthonormal bases. In the case of "wave mechanics" you use generalized position eigenstates as a generalized basis (note that these do not belong to the Hilbert space but to the dual space of the domain of the self-adjoint position operator). This leads to the one-to-one mapping between the abstract vector ##|\psi \rangle## to the wave function ##\psi(x)=\langle x|\psi \rangle##. If you use a discrete proper basis, e.g., the energy eigenfunctions of the harmonic oscillator. The map is to the sequence ##\psi_n=\langle n|\psi \rangle##.
     
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