0
$\begingroup$

Background

I have been reading about Symmetry operators and had a straight forward question. When we have a normal operator like momentum or Hamiltonian we know it acts over a wavefunction as $\hat{H}\psi$ or $\hat p\psi$. But when we talk about Symmetry Operators, everywhere I see it being operated along with its inverse like $U\psi U^{-1}$ where U is any of Symmetry operator.

Questions

Q What is the meaning of being operated with along with its inverse?

Q They also talk about the sense of even-ness and oddness whenever they talk about symmetry operator acting in this way. Is the same even-ness odd-ness concept applicable to say normal operator of Hamiltonian?

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ Which reference do you read? Please give a full reference (title, author, chapter, page and equation numbers), and also cite the definition of "symmetry operator" you are referring to. As of now, the question is totally unclear. For example, with the usual meaning and interpretation of symbols, the expression $U\psi U^{-1}$ is not well-defined. $\endgroup$
    – Tobias Fünke
    Commented yesterday
  • 1
    $\begingroup$ Recall vectors like ψ transform like Uψ, but operators/matrices like $\hat H$ transform like $U\hat H U^{-1}$. Is this what is confusing you? $\endgroup$
    – Cosmas Zachos
    Commented yesterday
  • $\begingroup$ @CosmasZachos Yes that is what's my confusion is $\endgroup$
    – Harshdeep Chhabra
    Commented yesterday
  • $\begingroup$ Rewrite your question then, emphasizing you understand this iron rule, so $U\hat H U^{-1} U \psi= U(\hat H \psi)$, for example, and that there is no such thing like the $U\psi U^{-1}$ you wrote, and which no serious text would write. $\endgroup$
    – Cosmas Zachos
    Commented yesterday

0

Reset to default