# Prove that $e^{\frac{i\lambda}{\hbar} S_x}S_ze^{-\frac{i\lambda}{\hbar}S_x}=S_z\cos(\lambda)+S_y\sin(\lambda)$ [closed]

Prove using Hadamard's lemma that $$e^{\frac{i\lambda}{\hbar} S_x}S_ze^{-\frac{i\lambda}{\hbar}S_x}=S_z\cos(\lambda)+S_y\sin(\lambda)$$ where $\lambda$ is a complex number.

I get: \begin{align} e^{\frac{i\lambda}{\hbar} S_x}S_ze^{-\frac{i\lambda}{\hbar}S_x} &= S_z+[S_x,S_z]\frac{i\lambda}{\hbar}+\frac{[S_x,[S_x,S_z]]}{2!}\biggl(\frac{i\lambda}{\hbar}\biggr)^2+\frac{[S_x,[S_x,[S_x,S_z]]]}{3!}\biggl(\frac{i\lambda}{\hbar}\biggr)^3+\cdots \\ &=S_z+\lambda S_y-\frac{1}{2!}\lambda^2S_z-\frac{1}{3!}\lambda^3S_y+\cdots \end{align}

where i get $[S_x,S_z]=-i\hbar S_y$, $[S_x,[S_x,S_z]]=\hbar^2S_z$, $[S_x,[S_x,[S_x,S_z]]]=-i\hbar^3 S_y$

But I don't know how to bring it into the form required by the question. How do I do that?

## closed as off-topic by Jim, ACuriousMind♦, Emilio Pisanty, John Rennie, JamalSApr 3 '15 at 8:26

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• You already have it in the form if you know the Taylor expansion of cos and sin! Just put the Sz terms together and Sy terms as well. – Luboš Motl Apr 2 '15 at 14:06
• Oh! Ok so the even factorials will sum up to form cos and the odd to form sine? – Paradox 101 Apr 2 '15 at 14:08
• There is a shorter way. Both sides satisfy the differential equation $d^2f/d\lambda^2 = -f(\lambda)$ and they satisfy the same initial conditions $f(0)=S_z$ and $df/d\lambda|_0 = S_y$, so they coincide. – Valter Moretti Apr 2 '15 at 16:15
• For the record, my edit should not be interpreted to mean that I think the question is on topic and fits with the homework policy. Maybe it is, maybe not, but I removed the parts that are definitely not permitted by the policy. – David Z Apr 2 '15 at 20:57
• @Paradox101 - it is often useful (but not necessary, as Valter shows) to consider power series for the functions. But if you use them for the exponential, you should probably know what these series (and the factorials and signs in them) look like for the sine and cosine, too. – Luboš Motl Apr 3 '15 at 5:55