Which Schrodinger equation is correct?

In the coordinate representation, in 1D, the wave function depends on space and time, $\Psi(x,t)$, accordingly the time dependent Schrodinger equation is

$$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$

In a representation-free notations we deal instead with the ket $|\Psi\rangle = |\Psi(t)\rangle$. Now how to write Schrodinger equation in this case? I find some books write

$$H|\Psi\rangle = i\hbar\frac{\partial}{\partial t}|\Psi\rangle$$

and others write

$$H|\Psi\rangle = i\hbar\frac{d}{dt}|\Psi\rangle$$

So which one of the last 2 equations is correct?

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$$H\Psi(x,t) = i\hbar\frac{\partial}{\partial t}\Psi(x,t)$$

says that we will differentiate wrt to time, keeping x constant.

$$H|\Psi\rangle = i\hbar\frac{d}{dt}|\Psi\rangle$$

says that we differentiate wrt to time, and moreover this is all the state vector $|\Psi\rangle$ depends.

$$H|\Psi\rangle = i\hbar\frac{\partial}{\partial t}|\Psi\rangle$$

has the same content as the second equation, its just that we don't need to be careful by writing a partial derivative since there isn't any other variables in $|\Psi\rangle$ besides time anyway, so either a partial or total derivative will do.

So in all 3 equations we are differentiating wrt to the same time dependance, its just that sometimes we have to be careful if there are other variables around or not.

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Your second and third equations are the same equation. They just use a different notation for the time derivative. Since in this "abstract" form $|\Psi \rangle$ only depends on time perhaps it is more correct to use the last one, but it is matter of taste.

In order to get your first equation (a wave equation), you must project on $\langle x|$: $$H(P=-i\hbar \partial _x , X=x)\Psi (x,t)=i\hbar \partial _t \Psi (x,t)$$

In this case $\partial _t$ (rather than $\frac{d}{dt}$) is a better notation because now $\Psi$ also depends on the coordinate $x$.

Regarding Nick Kidman's comment:

i) In the abstract SE the Hamiltonian $H$ is a function of "abstract" operators $H=H(P, X)$ (capital letters refer to operators).

ii) One has the canonical commutation relations $[X,P]=i\hbar$. A realization or representation of this commutation relation in a (certain) space of functions $f(x)$ (the lower case $x$ is a coordinate instead an operator) $X\,f(x)=x\,f(x)$ and $Pf(x)=-i\hbar\partial _x \,f(x)$ since $[x,-i\hbar\partial _x]f(x)=i\hbar f(x)$. (One can prove that this is the only representation of the commutation relation modulo unitarity equivalence, in a finite-dimensional system. In QFT one has truly different representations.) Therefore $P|x\rangle =-i\hbar\partial _x \,|x\rangle$ and $X|x\rangle =x \,|x\rangle$

iii) By definition $\Psi (x)\equiv \langle x|\Psi \rangle$.

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I would find it helpful to elaborate a little more on how one get's to the coordinate represenation, especially why after projecting with $\langle x|$, the $H$ operator is still there "outside of the function". –  NikolajK Aug 14 '12 at 20:38
Ok, I'll add it to my answer. –  drake Aug 14 '12 at 20:40
Okay, one could still improve on how the bra $\langle x |$, (now with the derivatives in between working to the right) gets through to the $|\Psi\rangle$ untill it can be replaced by the function $\Psi(x,t)$ with the derivatives acting on it, but it's quite okay. –  NikolajK Aug 14 '12 at 21:08

I) Let us reformulate OP's question(v1) as:

What time differentiation symbol should one use on the right-hand side of the time-dependent ket Schrödinger equation?

Answer: Whatever symbol that means $$\lim_{\Delta t \to 0} \frac{|\Psi(t+\Delta t )\rangle_S-|\Psi(t) \rangle_S }{\Delta t},$$

so apparently both OP's suggestions work. Here the subscript "$S$" (and "$H$") denotes the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively.

II) Let us mention for completeness that in the Heisenberg picture,

$$|\Psi \rangle_H\quad \text{does not evolve in time},$$

$${}_H\langle x,t |\quad \text{does not evolve in time},$$

$$\psi (x,t) ~=~ {}_H\langle x,t |\Psi \rangle_H,$$

$${}_H\langle x,t |\hat{x}(t)~=~ x ~{}_H\langle x,t |, \quad \Leftrightarrow \quad{}_H\langle x,t |\hat{x}(t)|\Psi \rangle_H ~=~ x \psi (x,t),$$

$${}_H\langle x,t |\hat{p}(t)~=~ \frac{\hbar}{i} \frac{\partial}{\partial x} ~{}_H\langle x,t |, \quad \Leftrightarrow \quad{}_H\langle x,t |\hat{p}(t)|\Psi \rangle_H ~=~ \frac{\hbar}{i} \frac{\partial}{\partial x} \psi (x,t),$$

$$i\hbar\frac{\partial}{\partial t} {}_H\langle x,t | ~=~{}_H\langle x,t |\hat{H}(t) \quad \Leftrightarrow \quad i\hbar\frac{\partial}{\partial t}\psi (x,t) ~=~{}_H\langle x,t |\hat{H}(t) |\Psi \rangle_H~=~(-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}+V(x))\psi (x,t) .$$

For a full explanation, see e.g. J.J. Sakurai, Modern Quantum Mechanics.

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Yeah I understand that. But my point is, if one uses $d/dt$ where the state is in the ket form, and one transforms that equation into the coordinate representation, then how $d/dt$ will turn into $\partial/\partial t$? –  Revo Aug 14 '12 at 21:01
@Revo: As there is nothing like a trajectory $x(t)$ (=$x$ as function of $t$, for which $\frac{\text d}{\text d t}$ is relevant) in the $x$-slot of the wave function (that is, the first argument in $\Psi(x,t)$), it doesn't matter. –  NikolajK Aug 14 '12 at 21:10
@NickKidman Sorry Nick, but that is not what I meant. It is obvious that time and space are independent, that goes without saying. –  Revo Aug 14 '12 at 21:58
@Revo: Mhm, if this is not what you ment, where is the difference between $\frac{\text d}{\text d t}$ and $\frac{\partial }{\partial t}$ for you then? –  NikolajK Aug 15 '12 at 7:50
@Revo: There is no "waiting" here, the state always depends on time. Take the functional analytic stance. Consider $p:\mathbb{R}\rightarrow \mathbb{R}, x\mapsto 3+5x-x^7$. That map is a set (subset of $\mathbb{R}\times\mathbb{R}$) and can for example be taken to be an element of the space of polynomials $\mathcal P$. In this space $p\in\mathcal P$ does not depend on $x$. All elements of $\mathcal P$ can be written using $x$, but it's only the image $p(x)\in\mathbb{R}$ which has $x$-dependence. Now $|\Psi\rangle$ actually depends on $t$ and $\Psi_t(x)$ is only its association with $|x\rangle$. –  NikolajK Aug 15 '12 at 9:34
$$\frac{df(t)}{dt}$$