Change of Basis For Pauli Matrix From Z Diagonal to X Diagonal Basis

  • Thread starter bohrpiphi
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Main Question or Discussion Point

I want to find a matrix such that it takes a spin z ket in the z basis,

[itex] | \; S_z + >_z [/itex]

and operates on it, giving me a spin z ket in the x basis,

[itex] U \; | \; S_z + >_z = | \; S_z + >_x [/itex]

I would have thought that I could find this transformation operator matrix simply by using the following argument:

[itex] U \; | \; S_z >_z = | \; S_z >_x [/itex]

[itex] _z< S_z \; | U \; | \; S_z >_z = _z< S_z \; | \; S_z >_x [/itex]

Therefore, elements of U are given by the inner product [itex] _z< S_z \; | \; S_z >_x [/itex]

However, to compute the inner product [itex] _z< S_z \; | \; S_z >_x [/itex] , I need to know [itex] | \; S_z >_x [/itex], which is exactly what I am trying to find.

Where is my misunderstanding?

I have shown that the matrix given by the inner products [itex] _z< S_z \; | \; S_x >_z [/itex] gives the matrix:

[itex] \begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{matrix} [/itex],

which cannot be correct, since [itex] _z< S_z \; | \; S_x >_z \neq _z< S_z \; | \; S_z >_x [/itex].

However, starting in the x basis and calculating [itex] | \; S_z >_x [/itex] shows that the above matrix works. I imagine this is not a coincidence, but it seems to be implying that [itex] | \; S_z >_x = | \; S_x + >_z [/itex].

This is not a homework question.
 

Answers and Replies

  • #2
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I also just realized that this should have been posted in the Advanced Homework section.
 

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