Reading Ballentine's "Quantum Mechanics; A Modern Development" I got stuck on his really short proof of what I think is Stone's theorem. On page 65 (paperback, reprint of 2008) he writes about about a one-parameter group of unitary operators $U(s)$. To cite slightly extended formulas for hopefully better understanding:

The unitary condition requires that $$ U(s)U^\dagger(s) = I+s\left[\frac{dU}{ds}+\frac{dU^\dagger}{ds}\right]_{s=0} + O(s^2)$$ should simply be equal to $I$, independent of the value of $s$. Hence the coefficient of $s$ must vanish, and we may write $$\frac{dU}{ds}\Biggr\rvert_{s=0} = iK,\; \text{with}\; K=K^\dagger.$$

for some Hermitian operator $K$.

The angle brackets are just the chain rule where $U(0)=I=U^\dagger(0)$ is already left out. I understand "should simply be $I$" (by definition of 'unitary') and "hence the coefficient of $s$ must vanish". But why does this lead to the last line. I am obviously missing something obvious :-(


2 Answers 2


$\frac{dU}{ds}+\frac{dU^\dagger}{ds}=0$ implies that $\frac{dU}{ds}$ is anti-selfadjoint: set $A = \frac{dU}{ds}$, then what you have is $A + A^\dagger = 0$. Therefore by setting $A = iK$ you have that $K$ is a self-adjoint operator.

  • $\begingroup$ So $(\frac{dU}{ds})^\dagger = \frac{dU^\dagger}{ds}$ ? $\endgroup$
    – Harald
    Apr 6, 2015 at 11:45
  • $\begingroup$ @Harald, yes. You can verify that from the definition of derivative. $\endgroup$
    – Phoenix87
    Apr 6, 2015 at 11:47

This just means that any hermitian Operator $K$ will fullfill the requirement of unitarity for $U(s)$: $$s\left[\frac{dU}{ds}+\frac{dU^\dagger}{ds}\right]_{s=0}=0$$

This is because for any hermitian operator $K$ you will get $$s\left[\frac{dU}{ds}+\frac{dU^\dagger}{ds}\right]_{s=0}=s\left[iK+(iK)^\dagger \right]=0$$.


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