0
$\begingroup$

My understanding is that any wavefunction can be decomposed into a linear sum of basis vectors, which for momentum are something like sine waves and for position are delta functions. And then eigenstates are what you observe when you attempt to measure a property of the wavefunction...but I've also heard that you observe a basis vector...but eigenstates must be square integrable while basis vectors don't? I really just don't understand the difference between these two things and what actually happens to the wavefunction in real life when you make a measurement.

$\endgroup$
  • $\begingroup$ Since you already have several answers, I'll just leave this note as a comment. The states of the system aren't actually vectors (kets) but are instead rays. That is, the infinity of kets $e^{i\phi}|\psi\rangle$ (which are the same ket up to a phase) correspond to the same state since they belong to the same ray. $\endgroup$ – Alfred Centauri Aug 20 at 21:50
2
$\begingroup$

You can understand the distinction between a basis and a set of eigenstates by noting that the same concepts apply in a more simple setting, such as ordinary vectors and matrices in 3 dimensions.

A basis in 3 dimensions is any set of 3 linearly independent vectors. For convenience we would ordinarily choose them to be also mutually orthogonal and of unit size.

An eigenvector of a 3 x 3 matrix is any vector such that the matrix acting on the vector gives a multiple of that vector. A 3x3 matrix will ordinarily have this action for 3 vectors, and if the matrix is Hermitian then the vectors will be mutually orthogonal if their eigenvalues are distinct. Thus the set of eigenvectors can be used to form a basis. Equally, given any basis one can construct a matrix whose eigenvectors are the members of that basis.

All the above carries over directly to quantum mechanics. The matrix is one way of representing an operator; the vectors now may have any number of components and may have complex values.

When learning these conceptual ideas, an added level of difficulty comes in when dealing with eigenstates of either momentum or position, because then you get strictly unphysical cases such as states with infinite or zero spatial extent. It is not physically possible for a system to move from whatever state it is in to one with either infinite extent or zero extent. Equally, it is not possible to perform a measurement of either momentum or position with perfect precision. But we often invoke the notion of a delta function as a convenient mathematical tool, and physically possible states (with finite spatial extent) can conveniently be Fourier analysed as a sum of sin waves. In such cases the set of eigenstates form a continuum, and so does any basis. The wavefunction can be thought of as a vector with an infinite number of components.

$\endgroup$
  • $\begingroup$ Thank you, you actually answered another question that I just asked at the same time. Why do we learn that things collapse to eigenstates when that's not actually a physical phenomenon. That seems really unnecessarily confusing. $\endgroup$ – Jeff Bass Aug 20 at 19:16
  • $\begingroup$ @JeffBass, in general, the state vector immediately after a measurement is just one of the 'ingredients' of the state vector immediately before the measurement. Is this what is confusing? $\endgroup$ – Alfred Centauri Aug 20 at 21:56
  • $\begingroup$ @AlfredCentauri It's starting to become more clear. The biggest confusion was definitely that eigenstates that are either infinitely spread out in space or perfectly localized are not actually physically realizable. $\endgroup$ – Jeff Bass Aug 20 at 22:04
0
$\begingroup$

A function is really just a vector in infinite dimensions, and so an eigenstate is a basis function. In case you didn't know, the eigenstates/eigenvectors of a Hermitian operator (all physical observables correspond to Hermitian operators) form a complete basis. When we are working in infinite dimensions, our inner products use integrals: $$ \langle f|g\rangle = \int_{-\infty}^\infty f^*(x) g(x) dx. $$ So being square-integrable really just means that $\langle f|f \rangle$ is well-defined and finite.

When you measure the wave function, it collapses into one of the eigenstates of the operator representing the physical quantity you are measuring.

$\endgroup$
0
$\begingroup$

"Eigenstates" are properties of specific operators/observables. Different operators will have different Eigenstates. Eigenstates of the Hamiltonian are a natural set of basis "vectors" on which you can decompose your wave function. They are not the only set of possible basis "vectors" though. You can choose to decompose your wave function in any bases, but the calculations are generally easier if you use the Eigenbasis of the Hamiltonian. When you observe a variable, you observe one of the eigenstates of that variable.

The thing about square integrable is a separate issue and way more subtle which leads to the introduction of Hilbert spaces.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.