0
$\begingroup$

Take a Hilbert space comprising of all square-integrable functions.

If we demand that each permissible function must have a decomposition in the Hamiltonian eigenbasis, does this ban some functions that were previously allowed in this Hilbert space?

I'm asking this because the Hamiltonian eigenbasis is usually discrete. I'm not convinced that every square-integrable function is producible by combining the eigenvectors linearly. It's like we have "too few" eigenvectors.

Take another situation where we discretise the eigenvectors. If we demand that each wave-function must have a discrete Fourier series expansion, then we ban non-periodic wave-functions.

I was also thinking the same about other operators like Spin, Angular momentum, etc. For quantum mechanics to work, we demand that each allowed wave-function must be expressible as a linear combination of the eigenvectors of each of these operators.

For instance, for particles with spin, is every square-integrable wave-function of the form $\psi (x,y,z,s)$ allowed?

Improve this question
$\endgroup$
6
  • $\begingroup$ In QM we assume that our Hilbert space is separable which by definition means there is a countable dense subset (indeed one can prove in general that $L^p (R^n)$ are separable with the standard Lebesgue measure). To be convinced that every square integrable function can be expanded in a discrete series you have to look at a functional analysis book: It is a theorem of functional analysis that for eg. the trigonometric and polynomial functions "span" $L^2[a,b]$ (in the sense of $L^2$ norm not pointwise). $\endgroup$
    – Leonid
    Feb 21, 2022 at 15:39
  • $\begingroup$ "I'm not convinced that every square-integrable function is producible by combining the eigenvectors linearly." Why not? That's exactly what the spectral theorem says (for sufficiently well-behaved operators). $\endgroup$
    – ACuriousMind ♦
    Feb 21, 2022 at 16:55
  • $\begingroup$ Although I agree with the answers and comments here this question made me wonder about the infinite square well: how do its eigenvectors cover the whole space? Do we have to consider eigenvectors at infinite energy for that? $\endgroup$
    – Lucas Baldo
    Feb 21, 2022 at 17:20
  • $\begingroup$ @LucasBaldo I am not sure I have understood your comment. The eigenvectors of the square well Hamiltonian are a basis for the Hilbert space of the square integrable functions on the interval $[0,L]$. It is not the same Hilbert space of the square integrable functions on $ \mathbb R$ $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Feb 21, 2022 at 18:09
  • $\begingroup$ @GiorgioP Hm, but what if we consider the infinite square well as the limit case of the finite square well when the outside potential $V_0 \rightarrow \infty$. Then for all finite values of $V_0$ the Hilbert space is the integrable functions on $\mathbb{R}$, but in the limit the Hilbert space suddenly changes? $\endgroup$
    – Lucas Baldo
    Feb 21, 2022 at 21:13

2 Answers 2

Reset to default
3
$\begingroup$

If we demand that each permissible function must have a decomposition in the Hamiltonian eigenbasis, does this ban some functions that were previously allowed in this Hilbert space?

It does not imply any restriction on the admissible vectors in the Hilbert space. Every vector in a Hilbert space is a (generally infinite) superposition of eigenvectors of a selfadjoint operator with pure point spectrum. If a Hamiltonian, or every other operator, like a component of the angular momentum, has pure point spectrum this is the case.

For instance, for particles with spin, is every square-integrable wave-function of the form $\psi(x,y,z,s)$ allowed?

Yes, it is allowed. Actually you need $2S+1$ such wavefunctions if the spin is $S$ and the integer $s$ varies from $-S$ to $S$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ hmm $2s+1$ wavefunctions? What's that about? So far in my study, there's only one wave function. Or maybe 2 if you split up the spin up and spin down wavefunctions. $\endgroup$
    – Ryder Rude
    Feb 21, 2022 at 18:05
  • 1
    $\begingroup$ The quantum state has the form, in that case $\sum_{j=-s}^s |j\rangle \otimes \psi_j$, where the $|j\rangle$s form a basis of $2s+1$ elements in the space of spin. That is equivalent to give a set of $2s+1$ wavefunctions (spin 0) $\psi_j$. For $s=1/2$ you have $2= 2/2+1$ wavefunctions, indeed. Spin up/spin down. $\endgroup$
    – Valter Moretti
    Feb 21, 2022 at 18:16
2
$\begingroup$

No, there is no such restriction because we don't demand that every state-vector must be expressed as a linear combination of the eigenstates of a Hermitian operator. Rather, it's a property of Hermitian operators that their orthonormal eigenstates span the whole Hilbert space. See, the spectral theorems.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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