The idea that the state of a system does not depend on the basis that we choose to represent it in, has always puzzled me. Physically I can imagine that the basis ought to just yield an equivalent representation of the system, but I cannot convince myself of this independence. I apologize for the naive character of this post.
For example, in most QM books one frequently comes across statements such as: "The state of the system is represented by a vector (or ket) $|\psi\rangle$ in the corresponding Hilbert space", or that a "vector is not the same thing as the set of its components in a basis". Physically this means that e.g. the momentum $\mathbf{p}$ of a system should not depend on how we choose to orient the axes of our basis.
It is a very abstract idea that the state is just a vector lying in the entire Hilbert space (spanned by the set of all the common eigenvectors of the observables describing the system). How can I write or speak of this $|\psi\rangle$ without having a basis to represent it in. As soon as I want to say anything about this $\psi$, e.g. its norm, I will need its components in some basis, to compute $\sqrt{\langle \psi|\psi \rangle}.$
- So what do I know about $|\psi\rangle$ without a chosen basis? How can I express that knowledge (of $\psi$) without a basis?
- Is this independence better illustrated when one considers the fact that the set of eigenvalues for any chosen observable of the system, are the same regardless of the chosen basis?
- Why it seems so difficult to imagine vector spaces, or vectors lying in abstract high-dimensional spaces ($|\psi\rangle \in \mathcal{H}$), without a basis? In what sense do we mean that a vector is more than just its components $(\langle v_1|\psi\rangle, \langle v_2|\psi\rangle, \dots)$?
- But as soon as I want to compute overlaps such as $\langle v_1|\psi\rangle$ or norms $||\psi||$ I need the components of $|\psi\rangle$ in some basis. So how can I convince myself that no matter what basis I choose, this abstract $|\psi\rangle \in \mathcal{H}$ will not depend on it?
- Finally, how should I interpret $|\psi\rangle \in \mathcal{H}$ without necessarily having an observable in mind? (i.e. the general statement that the state of the system lies in the Hilbert space).
Best Answer
The state is a vector in the Hilbert space of the Hamiltonian, which gives it a natural basis in terms of the eigenvectors; distinct eigenvalues then exist in distinct (orthogonal) subspaces - for degenerate values the subspaces are larger, but they are still distinct from all others. Clearly this situation gives many advantages in analysis.
However, this representation, though natural, and dependent only upon the spectral decomposition of the Hamiltonian, is not unique. The experimentalist will prefer to establish a basis which corresponds to his experimental setup! Chosing a different basis will change all of the coordinates, but does not change the states.
To make this clear recall that the state vectors of our Hilbert space can also be viewed a as rays, which are similar to the geometric rays of Euclidean geometry. Imagine the ray first, then superimpose a grid upon it - as you rotate the grid about the origin of the ray the intersections of the grid with the ray define the coordinates for that basis. The coordinates change, but the ray doesn't change. For ordinary geometry the relationships between two rays (angles, projections) don't change, so it is clear that some relationships between the coordinates are fixed - these are properties of a metric space.
Our Hilbert space is a normed space - distances don't mean anything, but a normalized state vector always has a length of 1, even when you change the basis; nor does the length change under unitary operators - hence their name.
All of this becomes clear in a good linear algebra course.