Search Results
| 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 operators
Search options questions only
not deleted
user 288281
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 $\times$, $+$!
0
votes
0
answers
126
views
Calculating the contraction of a field operator and a creation operator
In Folland's Quantum Field Theory (section 6.4) he considers a field:
$$\phi_\pi = \sum_\tau \int f(\textbf{q})\big[u(\textbf{q}, \tau, \pi)a(\textbf{q}, \tau, \pi) e^{-iq_\mu x^\mu} + v(\textbf{q}, …
- 3,888
1
vote
1
answer
301
views
Why does QFT require operator-valued distributions? [duplicate]
For example, the loss of determinism going from classical mechanics to quantum mechanics is (at least from what I understand) the motivation for upgrading observables from scalar-valued maps to operators … What is the analogue for upgrading operators to operator-valued distributions? …
- 3,888
1
vote
2
answers
146
views
Motivation behind introducing creation/annihilation operators into the Dirac equation
When studying the Klein-Gordon equation, the introduction of creation/annihilation operators was justified by recognizing a harmonic-oscillator-like equation which we know how to quantize. … Is there a similar justification when introducing these operators for the Dirac equation? Most of the resources I have looked at simply state them and move on. …
- 3,888
4
votes
2
answers
1k
views
Why do we care about the canonical commutation relations?
Suppose $\hat{x}$ and $\hat{p}$ are the position and momentum operators, it can be shown that
$$[\hat{x}, \hat{p}] = i\hbar\mathbb{I}.$$
The Stone-von Neumann theorem tells us that that the above is unique … My current interpretation of commutators is, informally speaking, that they measure the extent to which two operators commute. …
- 3,888
10
votes
3
answers
3k
views
How do we know that all quantum fields are Fourier transforms of creation and annihilation o...
I must have missed it while reading the text, but how do we know that all quantum fields
Involve only creation and annihilation operators and not any other kind of operator. … Are Fourier integrals of creation and annihilation operators. Where does this come from? (Why do we take the Fourier transform?) …
- 3,888
1
vote
1
answer
316
views
Why must we take the orthogonal projection to determine the probability distribution for mea...
This is inspired from Brian Hall's "Quantum Theory for Mathematicians", in which he says (page 125):
Suppose $A$ is a self-adjoint operator. Given a Borel set $E$ of $\mathbb{R}$, let $V_E$ be the cl …
- 3,888
0
votes
1
answer
284
views
As written, how is the spectral theorem useful?
For simplicity, I will only be considering bounded operators. … Let $H$ be an infinite dimensional separable Hilbert space, and let $B(H)$ denote the set of all bounded linear operators from $H$ to $H$. …
- 3,888
1
vote
1
answer
248
views
Creation and annihilation operator applied to non-basis vector
The creation and annihilation operators, $A^\dagger$ and $A$, increase/decrease the total number of particles by one. … rangle = | n_1, n_2, \ldots n_k-1,\ldots\rangle\\A^\dagger(e_k)| n_1, n_2, \ldots n_k,\ldots\rangle = | n_1, n_2, \ldots n_k+1,\ldots\rangle$$
In other words, when applied to the $k$th basis vector these operators …
- 3,888
5
votes
1
answer
342
views
What does it mean to apply a creation or annihilation operator to a free field, e.g. $\langl...
So far, I have studied free fields and some basic computations involving them, such as creation and annihilation operators. … Hence if $a$, $a^\dagger$ respectively, represent the annihilation and creation operators, then $a\varphi$ lowers the number of particles in $\varphi$ by 1 and similarly for the creation operator. …
- 3,888
0
votes
1
answer
516
views
How does the Klein-Gordon equation represent a classical field? [duplicate]
This is what I know about the Klein-Gordon equation so far. Suppose we are working with natural units such that $c = 1$. Then we may obtain the Klein-Gordon equation by considering any 4-vector $p^\mu …
- 3,888
0
votes
1
answer
509
views
Why is the wave function of a particle with definite momentum $p$ given as $e^{ipx/\hbar}$?
In many textbooks it is stated that, in position space, the wave function of a particle with definite momentum $p$ is given by $e^{ipx/\hbar}$. I know that the $\hbar$ comes from the de Broglie hypoth …
- 3,888
0
votes
0
answers
69
views
How does rotation appear in the components of the angular momentum operator?
.$$
Promoting the right hand side to operators:
$$X_i \rightarrow x_i\\
P_i \rightarrow -i\hbar\frac{\partial}{\partial x_i}$$
I would expect $\hat{J}_3$ to be
$$(\hat{J}_1\psi)(x) = -i\hbar\Big(x_1\frac …
- 3,888
1
vote
0
answers
129
views
Defining Wick/normal ordering beyond rearranging the order of annihilation and creation oper... [duplicate]
Most introductory quantum field theory books define Wick ordering as rearranging a product of creation and annihilation operators such that all the creation operators appear to the left of any annihilation … operators. …
- 3,888
0
votes
2
answers
312
views
How do we know that the operator $U(t) = e^{-itH/\hbar}$ corresponds to time?
By Stone's theorem on one-parameter unitary groups we know that there is a one-to-one correspondence between strongly continuous one-parameter unitary groups and self-adjoint operators. …
- 3,888
2
votes
1
answer
100
views
Equivalent definitions of Wick ordering
Let $\phi$ denote a field consisting of creation and annihilation operators. … before annihilation operators. …
- 3,888