Linked Questions

4
votes
1answer
2k views

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 ...
3
votes
1answer
2k views

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?
0
votes
0answers
68 views

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 ...
12
votes
2answers
2k views

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 ...
13
votes
4answers
2k views

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 ...
21
votes
1answer
2k views

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 ...
14
votes
2answers
1k views

How should Dirac notation be understood?

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{...
13
votes
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 ...
10
votes
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 ...
5
votes
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})\...
2
votes
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 ...
3
votes
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?
10
votes
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 ...
1
vote
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 ...
0
votes
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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