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Let $H$ be a self-adjoint $n \times n$ matrix with complex entries. $H$ gives rise to a continuous 1-parameter group of unitaries $t \mapsto U_t = \exp(itH) : \mathbb{R} \to U(n)$.

Let $A$ be some other $n \times n$ matrix. If $A$ commutes with $H$, then it commutes with $itH$ for all $t$ and hence with $U_t$ for all $t$ (by considering the series expansion). Thus we have $$ U_t A U_t^* = AU_tU_t^* = A \ \ \ \text{for all }t \ \ \ (*)$$ I suspect that the converse is true. That is, if $(*)$ holds, then $A$ must commute with $H$, but I'm not sure how to prove this and would appreciate some help.

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up vote 6 down vote accepted

Differentiate $AU_t=U_tA$ and evaluate at $t=0$. (Relevant: matrix calculus.)

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I like it! Thanks. – Mike F Aug 19 '12 at 20:49

This is probably another way to view anon's answer above. Notice that because $H$ is self - adjoint, we have that

$$\begin{eqnarray*} U_tAU_t^\ast &=& e^{itH}A(e^{itH})^\ast\\ &=& e^{itH}Ae^{{(itH)}^{\ast}} \\ &=&e^{itH}Ae^{-itH^\ast}\\ &=&e^{itH}Ae^{-itH}.\end{eqnarray*}$$ Now if you take the derivative of this at $t =0$ you will get that

$$\begin{eqnarray*} 0 &=& \frac{d}{dt}A\bigg|_{t=0} \\ &=& \frac{d}{dt}(U_tAU_t^\ast) \bigg|_{t = 0} \\ &=& [H,A] \\ &=& HA - AH\end{eqnarray*}$$

from which it follows that $A$ commutes with $H$.

Edit: The trick of differentiating on both sides is very useful in proving things like given two matrices $A,B$ in the Lie algebra $\mathfrak{g}$ of some matrix Lie group, the commutator $[A,B]$ is also in $\mathfrak{g}$.

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Thanks for the extra perspective and for re-tagging. – Mike F Aug 22 '12 at 19:42

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