# Does the dimensions of the wavefunction vector in Hilbert space depend on the number of eigenfunctions it is a superposition of?

I've seen people say that wavefunctions represented as vectors in a Hilbert space can (but don't have to) have infinite dimensions. So if a state vector requires X number of basis eigenfunctions linearly combined to represent it the state vector itself must have X components (dimensions)?

So in the case of momentum space, the total wavefunction is some linear combination of plane wave form eigenfunctions, if there are infinite eigenfunctions this is an infinite dimensional Hilbert space? Or if the wavefunction state vector required only 100 eigenfunctions to describe it, this state vector exists in a 100 dimensional Hilbert space?

Maybe I'm misunderstanding something fundamental, I'm still very new to this stuff. Thank you in advance for any answers.

• It is usually customary to speak about the dimension of a (topological) vector space, not about the dimension of a vector (element in that space). Dec 13, 2017 at 17:17
• Is that just a convention? Am I right then in thinking that when you expand the wave function as a linear combination of e^(ikx) plane wave eigenfunctions, the number of eigenfunctions needed to construct the wavefunction is the same as the dimensionality of the Hilbert space it exists in? Dec 13, 2017 at 17:36

The dimensionality of a vector space $V$ is determined by how many vectors are present in a basis of $V$. It's relatively simple to demonstrate that all basis sets of a vector space are the same size.
There is no corresponding notion of dimensionality for an individual (non-zero) vector $v\in V$. The most obvious way to see this is to simply choose a basis of which $v$ is a member, in which case the number of basis vectors necessary to "build" $v$ is trivially 1.
In other words, the number of basis vectors required to construct some non-zero vector $v\in V$ depends on which basis set you choose to work with.
• Anytime you write down a column vector, you are (perhaps implicitly) choosing a basis in which to work. The column vector $\pmatrix{a \\b}$ is shorthand for "$a$ times the first basis vector plus $b$ times the second basis vector." Dec 14, 2017 at 0:54