Quantum states: only vectors?

[nomadreid]

TL;DR

The state is supposed to describe everything about a particle, but are vectors always sufficient to do this?

Elementary question: Is there ever a case where the solutions for a wave equation turn out not to be a vector (in Hilbert space of infinite complex-valued dimensions, or a restriction to a subspace thereof) , but something else -- say, (higher-order) tensors or bivectors, or some such?
My apologies that the question lacks a specific example; if I had one, I would be answering my own question.
I suspect the answer is "no", but I would like to make sure. Thanks.

 

[DarMM]

Density matrices are a more general type of state than a vector in the Hilbert space of the system.

Of course the space of density matrices is still a vector space.
EDIT: Correction as @jambaugh said it's the set of operators (which density matrices are a subset of) which is a vector space

Even in geometry the space of Tensors or bivectors is still a vector space thus they are still abstract vectors in a sense. "Tensor" refers more to transformation properties and in QM for states this isn't relevant in the particular sense you mean here.

 

[nomadreid]

Thank you, DarMM.

 

[jambaugh]

Actually, to be a bit pedantic, the density matrices reside in a vector space, but don't form one, not even projectively.

 

[nomadreid]

Thanks, jambaugh. What precisely does "reside" here mean?
From the Wiki article https://en.wikipedia.org/wiki/Density_matrix
"Describing a quantum state by its density matrix is a fully general alternative formalism to describing a quantum state by its state vector (its "ket") or by a statistical ensemble of kets. However, in practice, it is often most convenient to use density matrices for calculations involving mixed states, and to use kets for calculations involving only pure states."
This seems to imply that one could describe any state (pure or mixed) by a statistical ensemble of state vectors (with the number of vectors in the ensemble being 1 for pure states), which is equivalent to a density matrix. So the answer to my question, if I had included ensembles, is that everything can be done in terms of vectors, but doesn't have to be. Do I have that right?

 

[jambaugh]

"Reside = are elements of" The set of valid density matrices is a subset of the space of hermitian operators in the space of operators over the Hilbert space. But it doesn't close under the process of taking linear combinations which is the principle criterion for being a vector space. E.g. \rho_1 -3i\rho_2 is not a valid density operator (assuming \rho_1 and \rho_2 are.)

The actual role played by density operators is in their use (with the trace operation) as dual vectors or co-operators mapping operators representing observables to expectation values. [**]
\langle X\rangle_\rho = trace(\rho X)
Boolean observables (projection operators) then give us the probabilities. By knowing all the probabilities and expectation values for any observable you have a "complete" description of the system. (Complete in the sense of knowing everything the system description can tell us, not in traditional meaning of a given description maximally specifying the system.)

However it is done, the operational meaning of a system description (mode, state, whatever) is its prediction of what we will or might observe including the chances of so observing. How it's done selects the type of theory.

Classical descriptions presuppose a probability distribution over a manifold of objective states (typically phase space). As such it necessarily satisfies Bell type inequalities which are ways of expressing the fact that the probability distribution is an additive measure on the sets of states.

Quantum descriptions relax this supposition and directly express the empirical predictions of the system's behavior when observed. You can use density operators in classical mechanics too but all observables commute thus all density operators are simultaneously diagonalizable and we only consider these diagonal terms (the probabilities of each classical state).

EDIT: ** Footnote: see Riesz representation theorem. The trace of a product forms an inner product on the space of hermitian operators (as a real space).

 

[nomadreid]

Thank you very much, jambaugh. That gives me a lot to chew on.