# Is it a total or an explicite time derivative in the Schrödinger equation?

I am always dubious when I need write Schrödinger equation: do I write $\partial / \partial t$ or $d/dt$ ?

I suppose it depends on the space in which it is considered. How?

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The most general Schrödinger equation has total derivatives $$i\hbar \frac{d}{dt}|\psi\rangle = \hat H |\psi\rangle$$ because the state vector $|\psi\rangle$ only depends on one variable, $t$. It's a complicated object that knows about the probability of anything in the given state, but this is hidden "inside" the state vector.
However, if you rewrite the state vector in a given representation, e.g. as $\psi(t,x,y,z,X,Y,Z)$ for the wave function of two particles, then the dependence on $x,y,z,X,Y,Z$, the coordinates of two particles, is put on equal footing with the $t$-dependence, and therefore the $t$-derivatives have to be written as partial ones, $\partial/\partial t$, to emphasize that $x,y,z,X,Y,Z$ are kept fixed during the differentiation. $$i\hbar \frac{\partial}{\partial t}\psi(t,x,y,z,X,Y,Z) = \hat H \psi(t,x,y,z,X,Y,Z)$$ where the Hamiltonian contains things like the kinetic energy of the first particle $$\hat H = \dots -\frac{\hbar^2}{2m} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \right)+\dots$$ and similarly the kinetic energy of the second particle $$\hat H = \dots -\frac{\hbar^2}{2M} \left( \frac{\partial^2}{\partial X^2} + \frac{\partial^2}{\partial Y^2} + \frac{\partial^2}{\partial Z^2} \right)+\dots$$ Note that there are partial derivatives everywhere because $\psi$ is now not a "general state vector" whose information is compactified; it is a complex-valued function of many variables.
Well, that is what I would have said. But both in my course and in Oxford quantum course $\partial$ is used instead of $d$ even in "the most general Schrödinger equation" ... So I am still not convinced. – Isaac Jan 14 '12 at 10:20