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I have a Hamiltonian

$$H = a(t) \cdot (S_z) ^2$$

where $H$ and $S_z$ are operators, $S_z$ being the $z$-component of the spin operator.

I am trying to show that the commutators for the spin operators with each other and with the Hamiltonian are the same in both Heisenberg and Schrödinger representations.

For some operator $A,$ denote the Heisenberg representation by subscript $h$ and the Schrödinger representation by subscript $s$

Then $$A_h = U^\dagger A_s U$$

where $U$ is the unitary evolution operator.

I'm currently working on the commutator $[S_{x_h},~ S_{y_h}]$ which has boiled down to

$$[S_{x_h} ,~ S_{y_h}] = i \cdot h \cdot U^\dagger S_{z_s} U$$

I'm a bit confused about how to proceed.

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    $\begingroup$ U involves the exponential of ... H (which is the same in both pictures--why?) and which is proportional to the square of $S_z$, so it commutes with it, no? So you found $S_{z,s}= S_{z, h}$. Now, then... $\endgroup$ – Cosmas Zachos Sep 20 '16 at 23:49
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Don't you have the property proven which you already want to prove? By the definition of Heisenberg operators which you've provided, the equation you wrote down directly implies $ \left [ S_{x_h}, S_{y_h} \right ] = i \hbar S_{z_h} $, which is what you want.

Since I'm probably mis-understanding your work, here is mine. It is a general property of operators that commutation relations that hold in the Schrodinger Picture also hold in the Heisenberg Picture.
Proof: If $ A_s $ and $ B_s $ are any two operators in the Schrodinger picture (in your case a Spin operator or the Hamiltonian) such that $ \left [ A_s, B_s \right ] = C_s $, then $ \left [ A_h, B_h \right ] = C_h $. To see this:

\begin{align} \left [ A_h, B_h \right ] & = A_h B_h - B_h A_h \\ & = U^{\dagger} A_s U U^{\dagger} B_s U - U^{\dagger} B_s U U^{\dagger} A_s U \\ & = U^{\dagger} A_s B_s U - U^{\dagger} B_s A_s U \\ & = U^{\dagger} \left [ A_s, B_s \right ] U \\ & = U^{\dagger} C_s U = C_h \end{align}

where I used the property $ U $ is unitary and therefore $ U^{\dagger} U = U^{-1} U = \mathbb{1} $.
This proof also works in reverse to show the converse. Take whatever operators you're dealing with and insert them into this proof and you've got it.

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