Jump to content
Main menu
Main menu
move to sidebar
hide
Navigation
Main page
Recent changes
Random page
freem
Search
Search
Appearance
Create account
Log in
Personal tools
Create account
Log in
Pages for logged out editors
learn more
Contributions
Talk
Editing
Openai/69174844-9774-8012-8b69-32262ca5e35a
(section)
Add languages
Page
Discussion
English
Read
Edit
Edit source
View history
Tools
Tools
move to sidebar
hide
Actions
Read
Edit
Edit source
View history
General
What links here
Related changes
Special pages
Page information
Appearance
move to sidebar
hide
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
===== 现在我们要求相位变换允许在时空中“变化”: ===== ψ(x)→eiα(x)ψ(x)\psi(x) \rightarrow e^{i\alpha(x)}\psi(x)ψ(x)→eiα(x)ψ(x) 这是局域 U(1) 对称性(local gauge symmetry)。 问题:原来的导数项不再不变,因为 ∂μψ→eiα(x)[∂μ+i(∂μα)]ψ\partial_\mu \psi \rightarrow e^{i\alpha(x)}\big[\partial_\mu + i(\partial_\mu \alpha)\big]\psi∂μψ→eiα(x)[∂μ+i(∂μα)]ψ 多出了 ∂μα\partial_\mu\alpha∂μα 项。 为了恢复不变性,我们必须引入一个新的场 Aμ(x)A_\mu(x)Aμ(x),定义 Dμ≡∂μ+ieAμD_\mu \equiv \partial_\mu + i e A_\muDμ≡∂μ+ieAμ 并让 AμA_\muAμ 在局域变换下相应地变化: Aμ(x)→Aμ(x)−1e∂μα(x)A_\mu(x) \rightarrow A_\mu(x) - \frac{1}{e}\partial_\mu \alpha(x)Aμ(x)→Aμ(x)−e1∂μα(x) 于是整个拉格朗日: L=ψˉ(iγμDμ−m)ψ\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psiL=ψˉ(iγμDμ−m)ψ 在局域 U(1) 对称下保持不变。 👉 这个引入的新场 AμA_\muAμ,就是电磁势,电磁相互作用的载体。
Summary:
Please note that all contributions to freem are considered to be released under the Creative Commons Attribution-ShareAlike 4.0 (see
Freem:Copyrights
for details). If you do not want your writing to be edited mercilessly and redistributed at will, then do not submit it here.
You are also promising us that you wrote this yourself, or copied it from a public domain or similar free resource.
Do not submit copyrighted work without permission!
Cancel
Editing help
(opens in new window)