Linked Questions

11
votes
2answers
528 views

Ambiguity in number of basis vectors [duplicate]

The dimension of the Hilbert space is determined by the number of independent basis vectors. There is a infinite discrete energy eigenbasis $\{|n\rangle\}$ in the problem of particle in a box which ...
1
vote
1answer
413 views

Could two different bases of a Hilbert space have different cardinality? [duplicate]

Here is a quote from http://en.m.wikipedia.org/wiki/Hilbert_space#Hilbert_dimension (accessed: Nov. 22, 2013) : As a consequence of Zorn's lemma, every Hilbert space admits an orthonormal basis; ...
3
votes
0answers
546 views

Continuous and discrete basis of Hilbert Space [duplicate]

Any state in Hilbert space $|\phi\rangle$ can be expressed in terms of a complete basis $\{| v_i\rangle, i=1,2,...\}$ as $$|\phi\rangle=\sum|v_i\rangle \langle v_i |\phi\rangle . $$ Now, if I ...
1
vote
1answer
220 views

How can the energy eigenkets and position eigenkets of a quantum system both be complete? [duplicate]

A quantum system is often given as a wavefunction of position. This is a vector in a continuous, infinite-dimensional vector space, with uncountable dimension. However we also know that the energy ...
2
votes
1answer
51 views

On the dimensionality of the Hilbert space for a central potential [duplicate]

So, we have that for particles in a central potential, the wavefunction can be expressed as: $$\psi(\mathbf{x})=R(r)Y^m_l(\theta,\phi)$$ Explanation (not crucial in the question): My question is: ...
23
votes
3answers
23k views

Don't understand the integral over the square of the Dirac delta function

In Griffiths' Intro to QM [1] he gives the eigenfunctions of the Hermitian operator $\hat{x}=x$ as being $$g_{\lambda}\left(x\right)~=~B_{\lambda}\delta\left(x-\lambda\right)$$ (cf. last formula on ...
5
votes
3answers
2k views

Normalization of basis vectors with a continuous index?

I have an infinite basis which associates with each point, $x$, on the $x$-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ ...
7
votes
2answers
3k views

Basis in quantum mechanics

My quantum mechanics textbook (Primer of Quantum Mechanics, by Marvin Chester) says that both the momentum space and the position space are basis spaces. It also says that the momentum space is ...
6
votes
2answers
2k views

Normalization of the path integral

When one defines the path integral propagator, there is the need to normalize the propagator (since it would give you a probability density). There are two formulas which are used. 1) Original (v1+v2)...
3
votes
3answers
784 views

Open problem? Square of the wave function $\Psi(x)_{x_o} = \delta(x-x_0)$ of a particle localized at a point $x_0$?

Does anybody know the status of the problem to define the wave function (non-relativistic Quantum Mechanics) of a particle localized at a definite point? Landau-Lifshitz says in chapter 1 that this ...
11
votes
1answer
512 views

Conceptual difficulty in understanding Continuous Vector Space

I have an extremely ridiculous doubt that has been bothering me, since I started learning quantum mechanics. If we consider the finite dimensional vector space for the spin$\frac{1}{2}$ particles, ...
10
votes
1answer
1k views

Hilbert space of a free particle: Countable or Uncountable?

This is obviously a follow on question to the Phys.SE post Hilbert space of harmonic oscillator: Countable vs uncountable? So I thought that the Hilbert space of a bound electron is countable, but ...
5
votes
3answers
1k views

Can eigenstates of a Hilbert space be thought of as delta functions?

Say we have an observable that describes a Hilbert space and that observable acts on state kets. Lets take the position observable for example. Then $\langle y|x\rangle = \delta(y - x)$. But can the ...
4
votes
3answers
519 views

Implicit Postulate of Quantum Mechanics

Consider the following quantum system: a particle in a one dimensional box (= infinite potential well). The energy eigenstates wave functions all vanish outside the box. But the position eigenstates ...
1
vote
3answers
400 views

Momentum operator representation

If $\hat{p}$ acts on position eigenstate, it is $$\tag{1}\hat{p}\left|x\right\rangle=+i\hbar\frac{\partial }{\partial x}\left|x\right\rangle .$$ But in general $$\tag{2}\hat{p} = -i\hbar \frac{\...

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