Inner product between a state and its derivative (quantum mechanics) I seem to have $\langle\varphi|\frac{d}{dt} \varphi\rangle=0$ for any ket $|\varphi\rangle$, which I doubt very much...
For any quantum state $|\varphi\rangle$, we know it's normalized and therefore $\langle\varphi|\varphi\rangle = 1$.
Now $$\frac{d}{dt} \langle\varphi|\varphi\rangle = 2\langle\varphi|\left(\frac{d}{dt} \varphi\rangle\right)$$
So $\langle\varphi|\left(\frac{d}{dt} \varphi\rangle\right) = \langle\varphi| \frac{d}{dt} \varphi\rangle = \frac{1}{2} \frac{d}{dt} 1 = 0$?
Is it actually true? If not, what mistake(s) am I making?

EDIT: I appreciate much the reference to Schrödinger Equation in the answers, which relates this question to physical intuitions. However, IMHO it needs to be noted: the state $|\varphi;t\rangle$ in the question doesn't have to be actually involved in any dynamics, i.e. it's time-dependence doesn't have to be related to a Hamiltonian. For example, $|\varphi;t\rangle$ can be defined as the ground state of a time-dependent Hamiltonian. Therefore, the Schrödinger Equation approach is not a necessary.
 A: The inner product in Quantum Mechanics is not symmetric if you interchange the terms, since $$\langle \phi | \psi \rangle = \langle \psi | \phi \rangle^*.$$
As a result, $$\frac{\text{d}}{\text{d}t}\langle \psi | \psi \rangle = \Big\langle \frac{\text{d}\psi}{\text{d}t}\Big|\psi\Big\rangle + \Big\langle \psi \Big| \frac{\text{d}\psi}{\text{d}t}\Big\rangle = 2 \,\text{Re}\left(\Big\langle \psi \Big| \frac{\text{d}\psi}{\text{d}t}\Big\rangle\right).$$
Thus, the only thing you can say using this argument is that $\Big\langle \psi \Big| \frac{\text{d}\psi}{\text{d}t}\Big\rangle$ is purely imaginary.Of course, you could arrive at this same conclusion from the Schrodinger Equation: $$i\hbar \frac{\text{d}}{\text{d}t}|\psi\rangle = \hat{H}|\psi\rangle \quad \quad \iff \quad\quad \Big\langle \psi \Big|\frac{\text{d}}{\text{d}t}\psi\Big\rangle = - \frac{i}{\hbar} \langle \psi | \hat{H} | \psi\rangle.$$ Since $\langle \psi | \hat{H} | \psi\rangle$ is the expectation value of the Hamiltonian, it must always be real, and so the right hand side is purely imaginary.
A: You did a mistake in evaluating the derivative.
$$\frac{d}{dt} \langle \psi \vert \psi \rangle = \left( \frac{d}{dt} \langle \psi \right)\vert\psi \rangle +  \langle \psi \vert \left( \frac{d}{dt} \psi \rangle \right)  \neq 2  \langle \psi \vert \left( \frac{d}{dt} \psi \rangle \right). $$
Recalling the Scrodinger equation $\vert \dot \psi \rangle= (- i \hat H) \vert \psi \rangle$, and $\langle \dot \psi \vert = \langle \psi \vert (i \hat H)$, you see how the sum of the two terms is equal to zero:
$$ \left( \frac{d}{dt} \langle \psi \right)\vert\psi \rangle +  \langle \psi \vert \left( \frac{d}{dt} \psi \rangle \right) = i \langle \psi \vert \hat H \vert \psi \rangle - i \langle \psi \vert \hat H \vert \psi \rangle = 0. $$
To be more accurate,
$$\frac{d}{dt} \langle \psi \vert \psi \rangle = 2   Re \biggl[\langle \psi \vert \left( \frac{d}{dt} \psi \rangle \right)\biggr],$$
which happens to be zero and makes or theory consistent $:)$ . This may have many consequences, but I' m not aware of any resulti towards this direction.
