# Linked Questions

1answer
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### Discrete vs Continuous spectra of operators [duplicate]

Why is it that if an operator $Q$ has a discrete spectra, that the eigenfunctions are all in Hilbert space? Why is it that if the spectrum is continuous we automatically know that the eigenfunctions ...
1answer
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### Continuous spectrum (quantum mechanics) [duplicate]

Does a continuous spectrum of an observable always imply that the corresponding eigenvectors will not be normalizable? If yes, how to prove it?
0answers
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### What lives in the Hilbert Space? [duplicate]

Consider the eigenvalue equation: $$\hat{Q}\Psi = q\Psi$$ where $q$ and $\Psi$ are eigenvalues and eigenfunctions of the hermitian operator $\hat{Q}$. If the spectrum of the hermitian operator is ...
2answers
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### Are all scattering states un-normalizable?

I am an undergraduate studying quantum physics with the book of Griffiths. in 1-D problems, it said a free particle has un-normalizable states but normalizable states can be obtained by sum up the ...
4answers
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### Why are the inner products of the eigenfunctions of an operator with a discrete eigenvalue spectrum guaranteed to exist?

I was reading through a textbook, and the statement was made that the inner products are guaranteed to exist if the eigenvalue spectrum of the operator is discrete. I have come across no support for ...
1answer
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### Is the existence of a sole particle in an hypothetical infinite empty space explicitly forbidden by QM?

Suppose the universe is completely empty with one sole particle trapped in it. To simplify, I will only be looking at the one dimensional case. However, all arguments are applicable for three ...
2answers
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If vectors $|\vec{r}⟩$ and $|\vec{p}⟩$ are defined as $$\hat{\vec{r}} |\vec{r}⟩ = \vec{r} |\vec{r}⟩ \\ \hat{\vec{p}} |\vec{p}⟩ = \vec{p} |\vec{p}⟩$$ then one can see that products like $$⟨\vec{... 2answers 1k views ### Must bounded operators have normalisable eigenfunctions and discrete eigenvalues? When we have bound states, to my knowledge, we have states that are normalisable and a discrete energy spectrum. However, in the case of scattering states that have a continuous energy spectrum, the ... 3answers 1k views ### What fundamental reasons imply quantization? In classical wave mechanics, quantization can occur simply from a finite potential well. In quantum mechanics, the quantization is obtained from the Schrödinger equation, which is, to my knowledge, a ... 2answers 2k views ### When is the spectrum of the Hamiltonian (purely) continuous? Given a quantum hamiltonian H = \frac{1}{2m}\vec{p}^2 +V(\vec{x}) in n-dimensions, the general rule-of-thumb is that the energy will be discrete for energies E for which \{ \vec{x} | V(\vec{x})\... 1answer 978 views ### Is the energy always discrete? In the von Neumann axioms for quantum mechanics, the first postulate states that a quantum state is a vector in a separable Hilbert space. It means it is assumed the Hilbert space has a basis with at ... 1answer 724 views ### Eigenvalues of Infinite Dimensional Matrix [duplicate] If I take a infinite-dimensional square matrix, what can I say about its eigenvalue spectrum? Will they have a discrete infinity of eigenvalues or continuous infinity of them? 1answer 516 views ### Is the Hilbert space spanned by both bound and continuous hydrogen atom eigenfunctions? As e.g. Griffiths says (p. 103, Introduction to Quantum Mechanics, 2nd ed.), if a spectrum of a linear operator is continuous, the eigenfunctions are not normalizable, therefore it has no ... 2answers 443 views ### About the orthogonality of the Hamiltonian eigenstates for the the continuous energy spectrum I would like first to describe a strange case that I encountered.  \ \ -  I solved the Schrodinger equation with a potential barrier (a potential well limited by a finite height wall which decrease ... 1answer 235 views ### Is it possible to decompose into eigenstates of Dirac Hamiltonian? If we have the Hilbert space \mathcal H = L^2(\mathbb R^3, \mathbb C^4) and a Hamiltonian:$$H=\gamma^i p_i + m \gamma^0 where $\gamma^i$ are matrices and $\{\gamma^i,\gamma^j\}=\delta^{ij}$. A ...

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