Dual of the TDSE Quite a quick and hopefully simple question. The TDSE takes the form $$i\hbar\frac{\partial\lvert\psi\rangle}{\partial t}=H\lvert\psi\rangle$$ and so if we take the dual of this to find the time evolution of a bra, we find $$-i\hbar\frac{\partial\langle\psi\lvert}{\partial t}=\langle\psi\lvert H$$ which is all pretty obvious to me, apart from the time derivative. If we want to take the dual of $\frac{\partial\lvert\psi\rangle}{\partial t}$ why are we simply allowed to  bypass the derivative operator and take the dual of the ket. On the right hand side we had to reverse the order of operator and ket and then take their respective duals (H is hermitian). Why doesn't that apply here, so we get $\langle\psi\lvert\frac{\partial}{\partial t}$ instead?
 A: 
Why doesn't that apply here, so we get $<\psi|\frac{\partial}{\partial t}$ instead?

For one thing, the expression:
$$
<\psi|\frac{\partial}{\partial t}
$$
makes no sense because the derivative is not operating on anything...
To explain further: The quantity 
$$
\frac{\partial|\psi\rangle}{\partial t}\;,
$$
is itself a ket (you could call it, e.g., "$|\chi\rangle$") because it is the limit of the difference of two kets
$$
\frac{\partial|\psi\rangle}{\partial t}\equiv\lim_{\delta\to0}\frac{|\psi(t+\delta)\rangle-|\psi(t)\rangle}{\delta}\equiv|\chi\rangle\;.
$$
And so, the dual is
$$
\langle\chi|\equiv\lim_{\delta\to0}\frac{
\langle\psi(t+\delta)|-\langle\psi(t)|}{\delta}\equiv\frac{\partial\langle\psi|}{\partial t}\;.
$$
A: Just to add to what has been said above.
Your notation is a little bit confusing as in the final line you have the Hamiltonian not acting on anything (convention that operators act from the left). I wouldn't swap the order of operators in Dirac notation but would represent the Hermitian conjugate Hamiltonian by $H^\dagger$. So you would go from
$$H\vert\psi\rangle=i\hbar\frac{\partial\vert\psi\rangle}{\partial t}$$
to
$$\left(H\vert\psi\rangle\right)^\dagger=\left(i\hbar\frac{\partial\vert\psi\rangle}{\partial t}\right)^\dagger$$ when you take the conjugate.
But this is equal to
$$H^\dagger\langle\psi\vert=-i\hbar\left(\frac{\partial\vert\psi\rangle}{\partial t}\right)^\dagger$$
But since we have $H=H^\dagger$ as the Hamiltonian is a Hermitian operator and $\left(\frac{\partial\vert\psi\rangle}{\partial t}\right)^\dagger=\frac{\partial\langle\psi\vert}{\partial t}$ as shown above you end up with $$H\langle\psi\vert=-i\hbar\frac{\partial\langle\psi\vert}{\partial t}$$
A: 
If we want to take the dual of ∂∣ψ⟩∂t why are we simply allowed to
  bypass the derivative operator and take the dual of the ket.

If I'm not mistaken, $|\psi(t)\rangle$ is a ket valued function of time; a function that associates a ket with each value of the parameter $t$.
A simple example is
$$|\psi(t)\rangle = |\psi_n\rangle e^{-i\frac{E_n}{\hbar}t}$$
A more complicated example is
$$|\psi(t)\rangle = \sum_n c_n(t)\;|\psi_n\rangle $$
When taking the time derivative, the function is taken to a new ket valued function but the time derivative operator does not operate on kets.
For example:
$$\frac{\partial}{\partial t}|\psi(t)\rangle = \sum_n \frac{d}{dt}(c_n(t))\;
|\psi_n\rangle $$
Similarly, $\langle\psi(t)|$ is bra valued function of time and it follows that
$$\frac{\partial}{\partial t}\langle\psi(t)| = \sum_n \frac{d}{dt}(c^*_n(t))\;
\langle\psi_n|$$
A: In non relativistic qm time is simply a parameter and is not represented by a hermitian operator, therefore the time derivative does not have to be moved.
