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I Question(s) about Dirac notation

  1. Apr 3, 2017 #1
    I promise that anytime I have question about Dirac notation I will ask it in this thread.

    I do not know how to parse the following Dirac notation.

    [itex]|\Psi'\rangle = |u\rangle |U\rangle[/itex]

    Can someone please convert the Dirac notation to matrix notation?
  2. jcsd
  3. Apr 3, 2017 #2


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    It is most likely to be interpreted as the tensor product of the two states. How you would write it as a matrix depends on what the original Hilbert spaces are and on the tensor product basis you pick.
  4. Apr 3, 2017 #3
    There's a lot of freedom in how you write it. If we take u to mean spin up of particle 1 and U to mean spin up of particle 2, then you could perhaps write it like
    ##\Psi' = \begin{pmatrix}
    1 \\
    1 \\
    0\end{pmatrix}_2 =

    1 \\
    0 \\
    0 \\
    uU \\
    dU \\
    uD \\

    The uU dU uD dD are just labels for the basis states in the column vector and aren't really part of the notation.
  5. Apr 3, 2017 #4


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    Forget about matrix notation. It's almost always more confusing than helpful, except in cases where it's a clever calculational tool.
  6. Apr 3, 2017 #5
    I will forget about it but first I have to get comfortable with Dirac notation, which, to me, is confusing.

    I looked up a tensor product of two vectors and found in linear algeba that is the outer product of two vectors.( I use to call that the cartesian product)

    Here is what I found. "In linear algebra, the outer product is the tensor product of two vectors" For example the outer product of two vectors is \begin{equation}u\otimes v = uv^T=\begin{pmatrix} \\ u_1\\ u_2 \\ u_3 \\ . \\ . \\ . \end{pmatrix}\begin{pmatrix}v_1 & v_2\end{pmatrix}=\begin{pmatrix}u_1v_1 & u_1v_2 & u_1v_3 & \dots \\ u_2v_1 & u_2v_2 & u_2v_3 & \dots \\ u_3v_1 & u_3v_2 & u_3v_3 & \dots\\ \dots & \dots & \dots &\ddots\end{pmatrix}\end{equation}
    The tensor product (outer product) of two vectors enumerates all the possible combinations of the elements of each vector.

    However, in Dirac notation the outer product is given by
    $$A = |u\rangle\langle U|$$
    which is not the same as this
    $$|A\rangle = |u\rangle |U\rangle$$
    Last edited: Apr 3, 2017
  7. Apr 3, 2017 #6
    You can't just throw in a transpose like that. You should enumerate the possible combinations and then represent each combination as a term in a column vector, which is what I did in post 3. Matrices are kind of clumsy for treating tensors since you have to play around with the dimensions to get things right.
  8. Apr 3, 2017 #7
    Thanks. I understand what you just said. But let me ask you this, is the matrix I show in my previous post considered a tensor? In other words, does the outer product of two vectors result in a tensor data type?
    Last edited: Apr 3, 2017
  9. Apr 3, 2017 #8


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    No, it is a matrix representation of a tensor.

    A tensor is not a data type, it is a mathematical object.
  10. Apr 3, 2017 #9


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    ##|U\rangle## is a ket, and all kets are vectors in some vector space, so of course it is a vector. And of course ##|u\rangle|U\rangle## is also a ket, albeit one from a different vector space than either ##|U\rangle## or ##|u\rangle##, namely the product space of the two spaces (which may be the same) to which ##|U\rangle## and ##|u\rangle## belong. The following notations are all roughly equivalent, except in their capacity to annoy people who are annoyed by sloppy notation:

    Do remember that the symbols inside the ket are just labels, chosen for convenience in whatever problem you're working with. If I write something like ##A|a\rangle=a|a\rangle##, the only significance of the ##a## inside the ##|a\rangle## is that I've decided that for whatever I happen to be working on at the moment the letter "a" will be a really convenient label to use for "that ket which is an eigenket of operator ##A## with eigenvalue ##a##".
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