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Bra-Ket notation confusion

  1. Oct 13, 2015 #1
    I'm new to bra-ket notation and am slightly confused; given an infinite square well with eigenvectors:

    [tex] \phi = \sqrt{2/a}\sin( (n\pi x)/a) [/tex]

    And we assume the form: H |φ> = E_n |φ>

    How would you then represent φ in terms of a column matrix, because that what I thought |φ> represents. Given some operator H (matrix) how would I operate with that matrix on |φ>?

    Thanks
     
  2. jcsd
  3. Oct 13, 2015 #2

    Orodruin

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    States are not column matrices and operators are not matrices. For separable spaces, it is possible to represent the state with a (generally infinite) column vector and the operators with a (generally infinite) matrix. What they will look like depends on the basis you use.
     
  4. Oct 16, 2015 #3
    Well, I guess I'm confused, too. I have a book by Leonard Susskind on non-relativistic QM, and he explicitly represents kets with column vectors, bras with row vectors, and linear operators with matrices. He does make it clear they are dependent on a particular choice of basis.
     
  5. Oct 16, 2015 #4
    And this is actually a good approach to learning QM. In full QM the bras, kets, and operators are mathematical objects which are more general than that, but basically they still have the same operational properties as vectors and matrices.
     
  6. Oct 16, 2015 #5

    Orodruin

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    In no way does this contradict what I said. Matrices and row vectors are perfectly fine ways of representing linear operators and kets in finite dimensional Hilbert spaces. The problem comes when you try to do this in infinite dimensional ones (where you would get infinite matrices and vectors) or, even worse, non-separable ones.
     
  7. Oct 16, 2015 #6
    OK, I understand now. Thanks.
     
  8. Oct 17, 2015 #7

    vanhees71

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    This is another sin in physics didactics! A vector is an abstract object, and it is represented as columns or rows of number only when referring to a basis and writing the linear decomposition of a vector in terms of this basis, putting the corresponding components of the vector in these handy schemes to perform calculations in terms of matrix-vector products.

    In quantum theory you have abstract Hilbert-space vectors, written in the bra-ket notation as ##|\psi \rangle##. Then you have bases, which are usually determined as eigenbases of self-adjoint operators on Hilbert space that represent observables. Let's take the quantum theory of a single particle, moving in only one spatial dimension as an example. For its Hilbert space you can take the energy eigenstates of an harmonic oscillator as the basis. This is very convenient, because it's a descrete basis, and all basis vectors are true normalizable Hilbert-space vector. They are called ##|n \rangle##, where ##n \in \mathbb{N}_0=\{0,1,2,\ldots \}##. The energy eigenvalues of the harmonic oscillator in the usual convention are ##E_n=\hbar \omega (n+1/2)##.

    What's more important in our context is the fact that these vectors ##|n \rangle## are a complete orthonormal set in the Hilbert space of our particles, i.e., you can decompose each vector in terms of a linear combination of these vectors:
    $$|\psi \rangle =\sum_{n=0}^{\infty} |n \rangle \langle n|\psi \rangle=\sum_{n=0}^{\infty} \psi_n |n \rangle.$$
    One can show that for any two vectors ##|\psi \rangle## and ##|\phi \rangle##
    $$\langle \phi|\psi \rangle=\sum_{n=0}^{\infty} \phi_n^* \psi.$$
    So you have mapping from the abstract Hilbert space to one specific realization, namely the Hilbert space of square summable sequences,
    $$|\psi \rangle \mapsto (\psi_n).$$
    The mapping is one-to-one, i.e., for any given sequence you can define also the vector according to it, using the above introduced basis.

    Now, in analogy, to finite-dimensional unitary vector spaces, you can write the vector components as columns with the ##\psi_n## as entries and the co-vectors (represented by the bras in the Dirac notation) as rows,
    $$\langle \phi | \mapsto (\phi_0^*,\phi_1^*,\ldots)$$
    to have the usual matrix-vector notation.

    You can also represent the operators for observables within this formalism. You just need completeness relations:
    $$\hat{O}=\sum_{n_1,n_2=0}^{\infty} |n_1 \rangle \langle n_1|\hat{O}|n_2 \rangle \langle n_2| = \sum_{n_1,n_2=0}^{\infty} |n_1 \rangle \langle n_2 O_{n_1n_2}.$$
    Then you have for any vector ##|\psi \rangle##
    $$\hat{O} |\psi \rangle =\sum_{n_1,n_2=0}^{\infty} |n_1 \rangle O_{n_1 n_2} \langle n_2|\psi \rangle=\sum_{n_1,n_2=0}^{\infty} |n_1 \rangle O_{n_1n_2} \psi_{n_2},$$
    i.e., the operation of ##\hat{O}## on ##|\psi \rangle## is maped to the usual matrix-vector product
    $$(\hat{O} \psi)_{n_1}=\sum_{n_2=0}^{\infty} O_{n_1 n_2} \psi_{n_2}.$$
    So you can write the operators as matrices with infinitely many rows and columns to be "applied" to the column-vector representation in terms of the vector components.

    This specific formalism is also known as "Heisenberg's matrix mechanics".

    In the same way you come to "Schrödinger's wave mechanics" by not using a discrete basis but a generalized basis of continuous eigenvalues, e.g., using the generalized position eigenstates.
     
  9. Oct 17, 2015 #8
    Thanks. I think Susskind was trying to keep it simple for us beginners by concentrating on cases with a finite number of dimensions in Hilbert space - in this case, spin. Again, he made it clear that he was working with a specific choice of basis. Speaking as a beginner, it's nice to start off with something more or less concrete, and relatable to familiar math formulations, before moving on to more abstract ideas. For instance, I personally find this treatment of the Dirac equation to be easier to follow than many, perhaps only because the notation is a little easier for me: http://www.mathpages.com/home/kmath654/kmath654.htm
     
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