I am confused about the definition of a derivative that appears in quantum mechanics. A quantum state is usually denoted by $\lvert \Psi(t) \rangle$, and the Schrödinger equation is $$ i\hbar \frac{d}{dt}\lvert \Psi(t) \rangle = \hat{H}\lvert \Psi(t) \rangle. $$ This would mean that $d/dt$ is operating on a vector in a Hilbert space, which I have never heard of.

However, this state $\lvert \Psi (t)\rangle$ could more conveniently be written as $f(t)$ where $f:\mathbb{R}\to\mathcal{H}$, where $\mathcal{H}$ is a hilbert space and then I guess it might make sense to definie an analouge of the derivative on functions like $f$. In that case we could write the equation like $$ i\hbar\left(\frac{d}{dt}f\right)(t) = \hat{H}f(t). $$ I tried to make a definition like $$ \left(\frac{d}{dt}f\right)(t_0) = \lim_{t\to t_0} \frac{f(t)-f(t_0)}{t-t_0}, $$ and on the right hand side I would then use the definition of the limit for a sequence of vectors and then maybe use that $\mathcal{H}$ is complete in order to ensure that the limit always exists.

Does this seem legit? What keywords should I use to find texts on functions such as $f:\mathbb{R}\to \mathcal{H}$?

  • $\begingroup$ The state depends on $t$ and so the derivative can be calculated. $\endgroup$
    – Wuestenfux
    Sep 18, 2018 at 13:49
  • $\begingroup$ By what definition of the derivative? $\lvert \Psi(t) \rangle$ is a vector in some Hilbert space. $\endgroup$ Sep 18, 2018 at 13:52
  • $\begingroup$ "... is complete in order to ensure that the limit always exists." False. The existence of the limit has no relation with the completeness of the space. $\endgroup$ Sep 18, 2018 at 14:17

1 Answer 1


The usual definition of derivative works for the case $$f:\Bbb R\longrightarrow E$$ with $E$ normed space: $$f'(t_0) = \lim_{t\to t_0}\frac{f(t) - f(t_0)}{t - t_0}.$$ (check yourself that the expression makes sense)


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