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0 votes
1 answer
107 views

Mathematical meaning of a position eigenbra $\langle x_0 |$

Let $|x_0\rangle$ be an position eigenket. The physical picture I have for $|x_0\rangle$ is a particle located at $x_0$. Thus it should be represented by a delta function $\delta(x-x_0)$. For $f\in L^...
CBBAM's user avatar
  • 3,350
3 votes
1 answer
290 views

Mathematical definition of annihilation and creation operators

I am self-studying quantum field theory and have gotten to creation and annihilation operators, respectively denoted $A^\dagger$ and $A$. Conceptually I understand what these objects are, at least on ...
CBBAM's user avatar
  • 3,350
0 votes
1 answer
364 views

Radial position operator

While trying to find the expectation value of the radial distance $r$ of an electron in hydrogen atom in ground state the expression is: $$\begin{aligned}\langle r\rangle &=\langle n \ell m|r| n \...
Kashmiri's user avatar
  • 1,270
0 votes
0 answers
54 views

Inner product evaluation in QM

On wikipedia on the page for inner product it states that for any two $x,y$ in a vector space $V$ the inner product $(\cdot , \cdot)$ satisfies $(ax, y) = a(x,y)$ where $a\in\mathbb{C}$. The inner ...
Chan's user avatar
  • 73
4 votes
2 answers
500 views

Completeness of Norm in Hilbert Space

I am not sure what it really means for the norm to be complete in a Hilbert Space. Can you provide me a proper definition? I am aware of the formula $||\Psi|| = <\Psi|\Psi>^{1/2}$. What are ...
Raj's user avatar
  • 43
3 votes
2 answers
432 views

Precise definition of the Hilbert space in QM?

In QM books (at least those I have read) the definition of the Hilbert space used is somewhat blurred (the "space of square integrable functions" is not enough to define it precisely : which kind of ...
huurd's user avatar
  • 131
4 votes
3 answers
597 views

Why does Griffiths define the complex inner product differently? [closed]

I have just now noticed that Griffiths (in his book Introduction to Quantum Mechanics) defines the complex inner product as $$\big<z,w\big>~=~\sum_{i=1}^n\overline{z}_iw_i.$$ In all ...
B. Pasternak's user avatar