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The original question was to prove

$$e^{i\sigma_z\phi} \sigma_y e^{-i\sigma_z\phi}=e^{2i\sigma_z\phi} \sigma_y.$$

Since left hand side is equal to $\sigma_y$, I thought in order for right hand side to be equal to $\sigma_y$,I have to show that $$e^{2i\sigma_z\phi}=1$$ For this,I tried to expand: $$e^{2i\sigma_z\phi}=\cos(2\sigma_z\phi)+i\sin(2\sigma_z\phi)$$ And use, $$\cos(2\sigma_z\phi)=1-\frac{(2\sigma_z\phi)^2}{2!}+\frac{(2\sigma_z\phi)^4}{4!}....$$ $$\sin(2\sigma_z\phi)=2\sigma_z\phi-\frac{(2\sigma_z\phi)^3}{3!}+\frac{(\sigma_2\phi)^5}{5!}...$$ where $\sigma_z$ is a Pauli matrix ($z$-component)

I have tried to expand the exponential in the similar way as above, but could not verify the above result. Don't know, whether my question is wrong or there are any other methods to solve this problem.

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  • $\begingroup$ How can $e^{2i\sigma_z\phi}=1$ be true for any $\phi$? At most it holds for $\phi=0$ or $\pi$ mod $2\pi$. $\endgroup$ – Meng Cheng Jul 9 '15 at 3:20
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    $\begingroup$ As someone already mentioned, your identity only holds for special values of $\phi$. To find out which values of $\phi$ it holds for, note that $\sigma_z^2 = 1$. Combined with the identities you already provided, you can then show that $\exp(2i\sigma_z\phi) = \cos(2\phi) + i\sigma_z \sin(2\phi)$, and check when this equals one. $\endgroup$ – jabirali Jul 9 '15 at 3:33
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    $\begingroup$ There is a $\sigma_y$ in between, and these are non-commuting operators. You can not just switch the order of multiplication. $\endgroup$ – Meng Cheng Jul 9 '15 at 3:42
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    $\begingroup$ @RoshanShrestha you're dealing with operators (matrices) here, so the order matters. In general, $A B C \neq A C B$ for three operators. $\endgroup$ – Kyle Arean-Raines Jul 9 '15 at 3:43
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    $\begingroup$ You are just one step away. Using $\sigma_z^2=1$, it is fairly easy to see that $e^{i\phi\sigma_z}=\cos\phi+i\sigma_z\sin \phi$. Then plug this expression into the left-hand side, expand it and simply. $\endgroup$ – Meng Cheng Jul 9 '15 at 4:03
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LHS: $(\cos\phi+i\sigma_z\sin\phi)\sigma_y(\cos\phi-i\sigma_z\sin\phi)=\cos^2\phi\sigma_y-i\cos\phi\sin\phi[\sigma_y,\sigma_z]+\sin^2\phi \sigma_z\sigma_y\sigma_z=(\cos^2\phi-\sin^2\phi)\sigma_y+2\cos\phi\sin\phi\sigma_x=\cos 2\phi \sigma_y+\sigma_x\sin2\phi=(\cos 2\phi+i\sin 2\phi\sigma_z)\sigma_y=e^{2i\phi\sigma_z}\sigma_y$

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  • $\begingroup$ +1 I can't believe my more complicated answer is getting more votes. I've a good mind to delete it, even though it isn't wrong. I guess it shows people some interesting other things, but your answer's clearly the better one. $\endgroup$ – Selene Routley Jul 9 '15 at 6:38
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Use fact that the matrices $i\,\sigma$ form the Lie algebra $\mathfrak{su}(2)$ and then the adjoint representation "braiding relationship" that:

$$\exp(\phi\,\mathrm{ad}(Z))\,Y = \mathrm{Ad}(e^{\phi\,Z})\,Y\tag{1}$$

and here $Z$ and $Y$ are $3\times 1$ column matrices that stand for superpositions of basis matrices in the Lie algebra in question. $e^{i\sigma_z\phi}.\sigma_y.e^{-i\sigma_z\phi}$ is the linear map $\mathrm{Ad}(e^{i\,\phi\,\sigma_z})$ acting on the vector that stands for $i\,\sigma_y$.

Now, $\mathrm{ad}(Z)\,Y$ stands for the commutator bracket $[i\,\sigma_z,\,i\,\sigma_y] = -[\sigma_z,\,\sigma_y] = 2\,i\,\sigma_x$. Thus $\mathrm{ad}(Z)^2\,Y$ stands for the commutator bracket $[i\,\sigma_z,\,2\,i\,\sigma_x] = -2\,[\sigma_z,\,\sigma_x] = -4\,i\,\sigma_y$. Keep on repeating these commutator brackettings to find the LHS of (1), which is the superposition:

$$\begin{array}{lcl}\exp(\phi\,\mathrm{ad}(Z))\,Y &=& Y + 2 \,\phi\,X - \frac{(2\,\phi)^2}{2!}\,Y - \frac{(2\,\phi)^3}{3!}\,X + \frac{(2\,\phi)^4}{4!}\,Y + \frac{(2\,\phi)^5}{5!}\,X -\cdots\\\\&=& \cos(2\,\phi)\,Y + \sin(2\,\phi)\,X\end{array}\tag{2}$$

which Pauli matrix does this superposition stand for?


More First Principles Answer

Actually you use a generalized version of something very like this procedure to prove the braiding relationship above.

Differentiate the LHS wrt $\phi$ to find:

$$\mathrm{d}_\phi e^{i\sigma_z\phi}.\sigma_y.e^{-i\sigma_z\phi} = i\,e^{i\sigma_z\phi}.[\sigma_z,\,\sigma_y].e^{-i\sigma_z\phi} = i\,[\sigma_z,\,e^{i\sigma_z\phi}.\sigma_y.e^{-i\sigma_z\phi}]\tag{3}$$

Use the commutation relationships to prove that this function $f(\phi)$ of $\phi$ equals $2\,\sigma_x$ at $\phi=0$.

Do the same for the RHS to prove:

$$\mathrm{d}_\phi e^{2i\sigma_z\phi}.\sigma_y = 2\,i\,e^{2i\sigma_z\phi}.\sigma_z.\sigma_y\tag{4}$$

A function $g(\phi)$ of $\phi$ which also equals $2\,\sigma_x$ at $\phi=0$ (prove this by direct evaluation).

Now find a differential equation that describes $f$ in terms of its derivatives alone. Do likewise for $g$. Then prove that the two sets are the same Cauchy initial value problem, thus by uniqueness of solution (Picard-Lindelöf theorem) $f(\phi) = g(\phi)$.

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Here is a some different proof: \begin{align*} e^{i\sigma_{z}\phi}\sigma_ye^{-i\sigma_z\phi} &=e^{i\sigma_{z}\phi}(e^{-i\sigma_z\phi}\sigma_y+[\sigma_y,e^{-i\sigma_z\phi}])\\ &=\sigma_y+e^{i\sigma_{z}\phi}[\sigma_y,cos(\phi)-i sin(\sigma_z\phi)]\\ &=\sigma_y+e^{i\sigma_{z}\phi}[\sigma_y,-i\sigma_zsin(\phi)]\\ &=\sigma_y-i sin(\phi) e^{i\sigma_{z}\phi}[\sigma_y,\sigma_z]\\ &=\sigma_y+2isin(\phi) e^{i\sigma_z\phi}\sigma_z\sigma_y\\ &=e^{i\sigma_z\phi}(e^{-i\sigma_z\phi}\sigma_y+2i\sigma_zsin(\phi)\sigma_z\sigma_y)\\ &=e^{i\sigma_z\phi}(cos(\sigma_z\phi)+i\sigma_zsin(\phi))\sigma_y\\ &=e^{2i\sigma_z\phi}\sigma_y \end{align*}

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