How do the two sides of the Schrodinger equation match? This is the form of the Schrodinger equation given by Wikipedia.
$$i\hbar\frac{\partial}{\partial{t}}|\Psi(t)\rangle=\hat{H}|\Psi(t)\rangle$$
My question is a simple one, but difficult for me to understand the answer to. To what space does each side of the equation belong? The left hand side is a constant scalar factor times the time derivative of a column vector, so it belongs to the same space as the vector. The right hand side is the Hamiltonian operator applied to a column vector, which yields energy, a scalar. I must be misunderstanding one of these two expressions since, in my interpretation, they belong to different spaces and are completely incomparable.
 A: $|\Psi\rangle$ is not a vector in the usual sense. It is a vector if viewed as a solution in the vector space of all solutions, or in physicists' words, a quantum state.
For an anology, we can say all polynomials $\mathbb R[x]$ over $\mathbb R$ is a vector space, with basis $1,x,x^2,\cdots$. Then any polynomial $p(x)$ is a vector in the sense you view it as an element in $\mathbb R[x]$. On the other hand, you can still put it into an equation (assuming $x\in\mathbb{R}$)
$$\frac{dp}{dx}=2x$$
in which case $p(x)$ is just a real variable depending on another real variable $x$.
So in the Schrodinger's equation, instead of thinking of $|\Psi\rangle$ as a vector, you should think of it as a complex variable depending on time and the space coordinates, so both sides are simply complex variables.
p.s. even so the RHS is not the energy.

Reply to the comment (too long to put in the comment):
I am not saying it is not a vector. I am saying it is simultaneously a vector and a complex variable. Let me start from the beginning.
WHAT IS THIS "VECTOR" WE ARE TALKING ABOUT
First, when you mention vector, you need to be sure of what vector space you are referring to. If you have seen some examples of solving Schrodinger's equation, say, the infinite well problem, you'll know by separating the space and time variables $\Psi(x,y,z,t)=\Phi(x,y,z)\psi(t)$one can derive from the time-dependent equation a time-independent one:
$$\hat H\Phi=E\Phi$$
where $E$ is a constant. Usually the equation has a system of time-independent solutions $\Phi_n(r)$ ($r=(x,y,z)$), called the eigenfunctions (also eigenvectors, but I'd like to avoid the word "vector" for now), which satisfy
$$\hat H\Phi_n=E_n\Phi_n$$
where $E_n$ are called the eigenvalues. All other solutions has the form
$$\Psi(r,t)=\sum_n c_ne^{-iE_nt/\hbar}\Phi_n(r)$$
To make a solution normalizable, one must impose the condition
$$\int_{\rm all\ space}\bar\Psi\Psi dV<\infty\iff\sum_n|c_n|<\infty$$
Now comes the crucial part: what is the vector space we keep talking about? It is the vector space of all solutions! In the case above, it is
$$S=\left\{\sum_n c_ne^{-iE_nt/\hbar}\Phi_n(r):\ \sum_n|c_n|<\infty\right\}$$
One can verify that this is a vector space, and has an inner product
$$<f,g>=\int(\bar fg)dV$$
on it, under which $e^{-iE_nt/\hbar}\Phi_n$ is a system of orthogonal basis of $S$.
SUMMARY:
So there's no problem saying that $\Psi$ is a vector, as long as you see it as one. But you have to distinguish between the two explanations below:
(1) $\Psi(r,t)$ as a function, assumes value in a vector space ($\mathbb C^n$ or $\mathbb C^\infty$) over $\mathbb{C}$.
(2) $\Psi(r,t)$ assumes value in $\mathbb C$, and as a function itself (i.e. a mapping, a rule of correspondence), belongs to a set of functions, which constitute a vector space.
I guess you were thinking (1), but I am trying to explain (2) to you.
I'd like to say more but too tired now. I suggest reading some textbooks instead of Wikipedia:
SUGGESTED READING MATERIALS
[1] Sect. 1.2-1.5 Mathematical Concepts of Quantum Mechanics by Stephen J. Gustafson and Israel Michael Sigal
[2] Sect. 8.1-8.2 Quantum Mechanics by Franz Schwabl
p.s. I am not sure whether there's a name for $\hat H\Psi$.
A: If $\Psi$ is an eigenvector of $H$ then $H\Psi = E\Psi$, a vector.
If $\Psi$ is not an eigenvector of $H$ then $H\Psi$ is a vector but not a constant multiple of $\Psi$.
