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Basic Bra-Ket Question

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Consider a three-dimensional vector space spanned by an orthonormal basis [itex] |1\rangle, |2 \rangle, |3 \rangle [/itex]. Kets [itex] |\alpha \rangle, |\beta \rangle [/itex] are given by
    [tex] |\alpha \rangle = i|1\rangle -2|2 \rangle -i|3\rangle, \qquad |\beta \rangle = i|1\rangle +2 |3\rangle. [/tex]

    part a) Construct [itex] \langle \alpha| \text{ and } \langle \beta | [/itex] (in terms of the dual basis [itex] \langle 1|, \langle 2|, \langle 3| [/itex]).

    3. The attempt at a solution

    I just want to check that I understand this correctly. Is the Bra the row vector that is basically the complex conjugate of the Ket, leading to the inner product? In this case,
    [tex] \langle \alpha | = -i \langle 1 | -2 \langle 2| +i \langle 3| \qquad \langle \beta | = -i\langle 1| + 2 \langle 3| [/tex]
    Last edited: Feb 17, 2013
  2. jcsd
  3. Feb 18, 2013 #2


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    Staff: Mentor

  4. Feb 18, 2013 #3
    Thanks DrClaude. So if I have part a right,
    [tex] \langle \alpha | = (-i, \, -2, \, i) \quad \text{ and } \quad \langle \beta | = (-i, \, 0, \, 2)[/tex]

    My calculation in the second part should be correct:
    [tex]\langle \alpha | \beta \rangle = 1 + 2i \quad \text{ and } \quad \langle \beta | \alpha \rangle = 1 - 2i [/tex]

    Then the third part asks:
    Find all nine matrix elements of the operator [itex]\hat{A} = |\alpha\rangle \langle \beta| [/itex], in this basis, and construct the matrix A. Is it hermitian?

    How do I go about this? I don't see how to multiply [itex]|\alpha\rangle\langle \beta |[/itex] to get nine elements.
  5. Feb 18, 2013 #4
    If [itex]|\alpha\rangle [/itex] represents a column vector (3,1) and [itex] \langle \beta |[/itex] a row vector (1,3) in terms of matrix multiplication what should (3,1)x(1,3) give you?
  6. Feb 18, 2013 #5


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    In vector terms the product [itex]<\alpha||\beta>[/itex] is the "inner product"- after taking the complex conjugate, multiply corresponding terms and add. If [itex]|\alpha>= <a_1, a_2, a_3>[/itex] and [itex]|\beta>= <b_1, b_2, b_3>[/itex], then [itex]<\alpha||\beta>= a_1b_1^*+ a_2b_2^*+ a_3b_3^*[/itex]. ("*" is the complex conjugate.)

    For the "exterior product", you form the nine products of every member of [itex]|\alpha>[/itex] with every member of [itex]<\beta|[/itex] as a matrix:
    [tex]|\alpha><\beta|= \begin{bmatrix}a_1b_1^* & a_1b_2^* & a_1b_3^* \\ a_2b_1^* & a_2b_2^* & a_2b_3^* \\ a_3b_1^* & a_3b_2^* & a_3b_3^*\end{bmatrix}[/tex]
  7. Feb 18, 2013 #6


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    Kevin and Ivy beat me to it. I will just stress that kets should be seen as column vectors, not row vectors as you wrote, and bras as their Hermitian conjugate.
  8. Feb 18, 2013 #7
    Thanks for the replies, I wasn't thinking of the Bra as a row vector. After doing the multiplication, I get
    [tex] \hat{A} = | \alpha \rangle \langle \beta | =
    \begin{bmatrix} 1 &0 &2i\\ 2i &0 &-4\\ -1 &0 &-2i \end{bmatrix} [/tex]
    Which is not hermitian as [itex] \hat{A} \neq \hat{A}^\dagger [/itex]. Correct?
  9. Feb 19, 2013 #8


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    That looks correct.
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