Skip to main content
Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Code code:"if (foo != bar)"
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Saves in:saves
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with
Search options not deleted user 250811

In physics, an operator is almost always either a square matrix or a linear mapping between two function spaces (defined on, say, $\mathbb R^n$). Operators serve as observables and as time evolution operators in Quantum Mechanics. This tag will most often find valid use in quantum mechanics; don't use this tag just because your equations contain "everyday operations" like $\times$, $+$!

10 votes

Commutator and Taylor series in quantum mechanics

This is simply an identity of Taylor expansion and has nothing to do with the fact that you have operators around, if $$f(x)=\sum_n \frac{f^{(n)}(0)}{n!} … The reason for that is that analytic functions of operators are defined by their Taylor series. …
Yarden Sheffer's user avatar
0 votes

Using Schwarz's Inequality to show an expectation value relationship of a particle

For the Cauchy-Schwartz inequality you need to have some inner product that you can use. The inner product you used $$\left\langle f,g\right\rangle=\int f(x)g(x)dx$$ is valid, but the definition $$\le …
Yarden Sheffer's user avatar
0 votes
Accepted

Examples of the physical significance and importance of matrix diagonalization and eigenvalu...

This is not a physics application per se, but I remember that as a first year undergrad I was convinced in the usefulness of matrix diagonalization by the application to obtain a closed expression for …
Yarden Sheffer's user avatar
1 vote

How should I solve the Schrödinger equation by diagonalization in QFT?

Assuming that I understand the question correctly, the answer is not specifically about the AIM, but in general about second-quantized operators. … The crucial thing to understand is that the ladder operators $c_{k\sigma}$ do not act on the Hilbert space $L^2(\mathbb{R}^n)$, but rather on the Fock space, whose basis is given by the occupation numbers …
Yarden Sheffer's user avatar