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Homework Help: Quantum Mechanics: Matrix Representation

  1. Feb 1, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the matrix representation of ##\mathbb{\hat J}_z## using the states ##|+y\rangle## and ##|-y\rangle## as a basis?

    2. Relevant equations

    ##|\pm y\rangle =\frac{1}{\sqrt{2}}|+z\rangle \pm \frac{i}{\sqrt{2}}|-z\rangle##

    3. The attempt at a solution

    A solution was given:

    ##\mathbb{\hat J}_z =\left[{\begin{array}{cc} \langle +y|+z\rangle & \langle +y|-z\rangle \\ \langle -y|+z\rangle & \langle -y|-z\rangle \\\end{array}}\right]\frac{\hbar}{2}\left[{\begin{array}{cc} 1 & 0 \\ 0 & -1 \\\end{array}}\right]
    \left[{\begin{array}{cc} \langle +z|+y\rangle & \langle +z|-y\rangle \\ \langle -z|+y\rangle & \langle -z|-y\rangle \\\end{array}}\right] = \frac{\hbar}{2}\left[{\begin{array}{cc} 0 & 1 \\ 1 & 0 \\\end{array}}\right],##

    but I am confused on what is going on? Can anyone explain exactly what is going on, please?
    Last edited: Feb 1, 2015
  2. jcsd
  3. Feb 1, 2015 #2
    You have the operator [itex] \mathbb{\hat J}_z [/itex]. Normally you represent that in the z-basis, so in matrix notation the 4 matrix elements that you already know are [itex] \langle \pm z \mid \mathbb{\hat J}_z \mid \pm z \rangle [/itex]. If you want to represent [itex] \mathbb{\hat J}_z [/itex] in another basis like y you have to know the matrix elements [itex] \langle \pm y \mid \mathbb{\hat J}_z \mid \pm y \rangle [/itex] . One way to do this is to insert a unit operator/identity matrix on both sides as this has no effect:
    [tex] \langle \pm z \mid \mathbb{I} \mathbb{\hat J}_z \mathbb{I} \mid \pm z \rangle [/tex]

    Now if you know that you can write the identity matrix as [itex] \sum \mid z \rangle \langle z \mid [/itex] using the completeness relation you are practically done:

    [tex] \langle \pm y \mid \mathbb{I} \mathbb{\hat J}_z \mathbb{I} \mid \pm y \rangle = \langle \pm y \mid z \rangle \langle z \mid \mathbb{\hat J}_z \mid z \rangle \langle z \mid \pm y \rangle [/tex]

    So the middle part [itex] \langle z \mid \mathbb{\hat J}_z \mid z \rangle [/itex] is the matrix in the already known basis
    [itex] \frac{\hbar}{2}\left[{\begin{array}{cc} 1 & 0 \\ 0 & -1 \\\end{array}}\right] [/itex] And the outer parts are the matrices to transform it into an y basis.

    (Note: I mix up the notation for matrix elements and the matrix itself. I probably miss a summation here and there, but I hope you understand the difference)
  4. Feb 1, 2015 #3
    I think I understand now. Thank you very much!
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