sandy stone said:
Well, I guess I'm confused, too. I have a book by Leonard Susskind on non-relativistic QM, and he explicitly represents kets with column vectors, bras with row vectors, and linear operators with matrices. He does make it clear they are dependent on a particular choice of basis.
This is another sin in physics didactics! A vector is an abstract object, and it is represented as columns or rows of number only when referring to a basis and writing the linear decomposition of a vector in terms of this basis, putting the corresponding components of the vector in these handy schemes to perform calculations in terms of matrix-vector products.
In quantum theory you have abstract Hilbert-space vectors, written in the bra-ket notation as ##|\psi \rangle##. Then you have bases, which are usually determined as eigenbases of self-adjoint operators on Hilbert space that represent observables. Let's take the quantum theory of a single particle, moving in only one spatial dimension as an example. For its Hilbert space you can take the energy eigenstates of an harmonic oscillator as the basis. This is very convenient, because it's a discrete basis, and all basis vectors are true normalizable Hilbert-space vector. They are called ##|n \rangle##, where ##n \in \mathbb{N}_0=\{0,1,2,\ldots \}##. The energy eigenvalues of the harmonic oscillator in the usual convention are ##E_n=\hbar \omega (n+1/2)##.
What's more important in our context is the fact that these vectors ##|n \rangle## are a complete orthonormal set in the Hilbert space of our particles, i.e., you can decompose each vector in terms of a linear combination of these vectors:
$$|\psi \rangle =\sum_{n=0}^{\infty} |n \rangle \langle n|\psi \rangle=\sum_{n=0}^{\infty} \psi_n |n \rangle.$$
One can show that for any two vectors ##|\psi \rangle## and ##|\phi \rangle##
$$\langle \phi|\psi \rangle=\sum_{n=0}^{\infty} \phi_n^* \psi.$$
So you have mapping from the abstract Hilbert space to one specific realization, namely the Hilbert space of square summable sequences,
$$|\psi \rangle \mapsto (\psi_n).$$
The mapping is one-to-one, i.e., for any given sequence you can define also the vector according to it, using the above introduced basis.
Now, in analogy, to finite-dimensional unitary vector spaces, you can write the vector components as columns with the ##\psi_n## as entries and the co-vectors (represented by the bras in the Dirac notation) as rows,
$$\langle \phi | \mapsto (\phi_0^*,\phi_1^*,\ldots)$$
to have the usual matrix-vector notation.
You can also represent the operators for observables within this formalism. You just need completeness relations:
$$\hat{O}=\sum_{n_1,n_2=0}^{\infty} |n_1 \rangle \langle n_1|\hat{O}|n_2 \rangle \langle n_2| = \sum_{n_1,n_2=0}^{\infty} |n_1 \rangle \langle n_2 O_{n_1n_2}.$$
Then you have for any vector ##|\psi \rangle##
$$\hat{O} |\psi \rangle =\sum_{n_1,n_2=0}^{\infty} |n_1 \rangle O_{n_1 n_2} \langle n_2|\psi \rangle=\sum_{n_1,n_2=0}^{\infty} |n_1 \rangle O_{n_1n_2} \psi_{n_2},$$
i.e., the operation of ##\hat{O}## on ##|\psi \rangle## is maped to the usual matrix-vector product
$$(\hat{O} \psi)_{n_1}=\sum_{n_2=0}^{\infty} O_{n_1 n_2} \psi_{n_2}.$$
So you can write the operators as matrices with infinitely many rows and columns to be "applied" to the column-vector representation in terms of the vector components.
This specific formalism is also known as "Heisenberg's matrix mechanics".
In the same way you come to "Schrödinger's wave mechanics" by not using a discrete basis but a generalized basis of continuous eigenvalues, e.g., using the generalized position eigenstates.