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Quantum Mechanics: Matrix Representation
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[QUOTE="barefeet, post: 4995716, member: 490981"] 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) [/QUOTE]
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