bohrpiphi
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I want to find a matrix such that it takes a spin z ket in the z basis,
| \; S_z + >_z
and operates on it, giving me a spin z ket in the x basis,
U \; | \; S_z + >_z = | \; S_z + >_x
I would have thought that I could find this transformation operator matrix simply by using the following argument:
U \; | \; S_z >_z = | \; S_z >_x
_z< S_z \; | U \; | \; S_z >_z = _z< S_z \; | \; S_z >_x
Therefore, elements of U are given by the inner product _z< S_z \; | \; S_z >_x
However, to compute the inner product _z< S_z \; | \; S_z >_x , I need to know | \; S_z >_x, which is exactly what I am trying to find.
Where is my misunderstanding?
I have shown that the matrix given by the inner products _z< S_z \; | \; S_x >_z gives the matrix:
\begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{matrix},
which cannot be correct, since _z< S_z \; | \; S_x >_z \neq _z< S_z \; | \; S_z >_x.
However, starting in the x basis and calculating | \; S_z >_x shows that the above matrix works. I imagine this is not a coincidence, but it seems to be implying that | \; S_z >_x = | \; S_x + >_z.
This is not a homework question.
| \; S_z + >_z
and operates on it, giving me a spin z ket in the x basis,
U \; | \; S_z + >_z = | \; S_z + >_x
I would have thought that I could find this transformation operator matrix simply by using the following argument:
U \; | \; S_z >_z = | \; S_z >_x
_z< S_z \; | U \; | \; S_z >_z = _z< S_z \; | \; S_z >_x
Therefore, elements of U are given by the inner product _z< S_z \; | \; S_z >_x
However, to compute the inner product _z< S_z \; | \; S_z >_x , I need to know | \; S_z >_x, which is exactly what I am trying to find.
Where is my misunderstanding?
I have shown that the matrix given by the inner products _z< S_z \; | \; S_x >_z gives the matrix:
\begin{matrix} 1/\sqrt{2} & 1/\sqrt{2} \\ 1/\sqrt{2} & -1/\sqrt{2} \end{matrix},
which cannot be correct, since _z< S_z \; | \; S_x >_z \neq _z< S_z \; | \; S_z >_x.
However, starting in the x basis and calculating | \; S_z >_x shows that the above matrix works. I imagine this is not a coincidence, but it seems to be implying that | \; S_z >_x = | \; S_x + >_z.
This is not a homework question.