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Let $H_1$ and $H_2$ denote arbitrary Hermitian operators (finite dimensional) and let $\left\|\ldots\right\|$ denote the usual operator norm. I conjecture that $$ \left\|\ln\left(e^{iH_1}e^{iH_2}\right)\right\|\leq\left\|H_1\right\|+\left\|H_2\right\|\ , $$ but have no idea how to prove this. If $H_1$ and $H_2$ commute, it reduces to the triangle inequality. Cauchy-Schwarz and Golden-Thompson come to mind, but do not seem to help.

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Since $\ln$ is not single valued when the domain is not restricted to positive definite matrices, I am confused about the meaning of $\ln(e^{iH_1}e^{iH_2})$. For example, let $H_1(t)=tI$ and $H_2=0$, where $I$ is the identity matrix. Then $\ln(e^{iH_1(t)}e^{iH_2})=e^{it}I$ is of periodicity $2\pi$, so I think it cannot be simply defined as $itI$. Or maybe you only focus on $H_1$ and $H_2$ in some neighborhood of $0$? –  23rd Nov 4 '12 at 18:01
    
Yes, what I had in mind was $\left\|H_1\right\|$ and $\left\|H_2\right\|$ being sufficiently small such that you have no ambiguities with the logarithm. So let the negative real axis be the branch cut of the complex logarithm and let $$\left\|H_1\right\|, \left\|H_2\right\|<\frac{\pi}{2}\ .$$ –  user47814 Nov 5 '12 at 8:51
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Given a unitary operator $U$ on $\mathbb{C}^n$, there exists an orthonormal basis $\{v_1,\dots,v_n\}$ and $\theta_1,\dots,\theta_n\in(-\pi,\pi]$, such that $Uv_k=e^{i\theta_k}v_k$, $k=1,\dots,n$. Let me define $\ln U$ be the unique operator which satisfies that $\ln U v_k=i\theta_kv_k$, $k=1,\dots,n$. With this definition, the conclusion is true. Note that $-i\ln U$ is Hermitian, $\|\ln U\|=\max_{1\le k\le n}|\theta_k|\le \pi$, and $\|\ln U\|\le\|H\|$ for every Hermitian operator $H$ with $U=e^{iH}$.

Let $\{e_1,\dots,e_n\}$ be a natural basis of $\mathbb{C}^n$ and let $\langle,\rangle$ be the standard inner product on $\mathbb{C}^n$, i.e. $\{e_1,\dots,e_n\}$ is an orthonormal basis of $\mathbb{C}^n$ w.r.t. $\langle,\rangle$. Since $\{e_1,\dots,e_n,ie_1,\dots,ie_n\}$ are $\mathbb{R}$-linearly independent, it gives a natural $\mathbb{R}$-linear isomorphism $$f:\mathbb{C}^n\to\mathbb{R}^{2n},\quad z\mapsto (\mathrm{Re}z,\mathrm{Im}z)$$ Moreover, $\langle,\rangle$ induces an inner product $\langle,\rangle_{\mathbb{R}}$ on $\mathbb{R}^{2n}$ in the following way: given $z,w\in\mathbb{C}^n$, $\langle f(z),f(w)\rangle_{\mathbb{R}}:=\mathrm{Re}\langle z,w\rangle$. Then given two unit vectors $z,w\in\mathbb{C}^n$, we can define the angle between them as $$\arccos\mathrm{Re}\langle z,w\rangle:=\angle(z,w)\in[0,\pi].$$

It is easy to show that for $\theta_k=\angle(z_k,w)$, $k=1,2$, $\angle(z_1,z_2)\le \theta_1+\theta_2$. To see this, firstly, since $\angle(z_1,z_2)\le\pi$, we may assume that $\theta_1+\theta_2\le\pi$. Secondly, up to a rotation, we may assume $w=e_1$. Then for $k=1,2$, $z_k=\cos\theta_ke_1+\sin\theta_kv_k$, where $\mathrm{Re}\langle e_1,v_k\rangle=0$ and $\|v_k\|=\sin\theta_k$. Therefore, Cauchy's inequality gives $$\cos\angle(z_1,z_2)\ge \cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2=\cos(\theta_1+\theta_2).$$ Therefore, $\angle(z_1,z_2)\le\theta_1+\theta_2$.

We claim that for any unitary operator $U$,

$$\|\ln U\|=\sup_{\|z\|=1}\angle(z,Uz).$$ To see this, let $\theta_1,\dots,\theta_n\in(-\pi,\pi]$ be all the eigenvalues of $-i\ln U$. Then $\|\ln U\|=\max_{1\le k\le n}|\theta_k|$. Let $\{v_1,\dots,v_n\}$ be an orthonormal basis of $\mathbb{C}^n$, such that $Uv_k=e^{i\theta_k}v_k$,$k=1,\dots,n$. Now for any unit vector $z=\sum_{k=1}^na_kv_k$, $$\mathrm{Re}\langle z,Uz\rangle=\mathrm{Re}\langle \sum_{k=1}^n a_kv_k,\sum_{k=1}^n a_ke^{i\theta_k}v_k\rangle=\sum_{k=1}^n|a_k|^2\cos\theta_k\ge\cos\|\ln U\|.$$ The equality can be realized by $z=v_k$ with $\theta_k=\|\ln U\|$, which completes the proof of the claim .

Now given Herimtian operators $H_1$ and $H_2$, denote $U_k=e^{iH_k}$, $k=1,2$ and $U=U_1U_2$. Since $U_1$ and $U_2$ are unitary, $U$ is unitary. Then $$\angle(z,Uz)=\angle(z,U_1U_2z)\le\angle(U_2z,U_1U_2z)+\angle(z,U_2z)\le \|\ln U_1\|+\|\ln U_2\|\le \|H_1\|+\|H_2\|.$$ Taking supremum on both sides, the conclusion follows.

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What can I say... thanks! –  user47814 Nov 5 '12 at 16:19
    
@AdrianHutter: You are welcome. Your question is very interesting to me.:) –  23rd Nov 5 '12 at 16:23
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