# 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). – DanielC Dec 13 '17 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? – Terry Dec 13 '17 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.

• So can the wavefunction be written as a column vector in the same way vectors in R^3 can? Are the theoretical components of the wavefunction column vector just the list of eigenfunctions multiplied by their coefficients? – Terry Dec 13 '17 at 17:38
• The numbers which are stacked in a column vector are the coefficients of the basis vectors. As long as your Hilbert space is finite dimensional, then you can represent its vectors in the same way. If the space is infinite dimensional, then the column vector would need to be infinitely long, but the same idea applies. – J. Murray Dec 13 '17 at 17:44
• Sorry just to be clear, the components of the hypothetical wave function in column vector form are only the coefficients? If this is the case where do the eigenfunctions exist? – Terry Dec 14 '17 at 0:31
• 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." – J. Murray Dec 14 '17 at 0:54