# Wigner function of $|n\rangle\langle m|$

For $$n=m$$, the Wigner function is given by, $$W_n(\alpha) = \frac{2}{\pi} (-1)^n \exp(-2 |\alpha|^2) L_n(4 |\alpha|^2),$$

And for $$n \neq m$$, it is, $$f_{mn}=\sqrt{\frac{m!}{n!}} e^{i(m-n) \arctan\left( p/x\right) } \frac{\left( -1\right)^{m}}{\pi\hbar}\left ( \frac{x^{2}+p^{2}}{\hbar/2} \right ) ^ {\left( n-m\right) /2} L_{m}^{n-m}\left( \frac{x^{2}+p^{2}}{\hbar/2}\right) e^{-\left( x^{2}+p^{2}\right) /\hbar}$$ and in this case, there will be always a $$e^{i(m-n) \arctan\left( p/x\right)}$$ term. But the Wigner function is a real valued function. How to reconcile this?

• What is $f_{mn}$?
– Alex
Oct 27, 2023 at 13:53
• Linked. Note this operator is not Hermitian. Oct 27, 2023 at 16:03

In general, Wigner transforms of Hermitian operators are real, $$g(x,p)= \hbar \int \!\! dy ~ e^{-iyp} \langle x+\hbar y/2| \hat G |x-\hbar y /2 \rangle , \implies \\ g(x,p)^*= \hbar \int \!\! dy ~ e^{iyp} \langle x+\hbar y/2| \hat G |x-\hbar y /2 \rangle ^* \\ =\hbar \int \!\! dy ~ e^{iyp} \langle x-\hbar y/2| \hat G ^\dagger |x+\hbar y /2 \rangle ,$$ so $$g=g^*$$ for $$\hat G= \hat G^\dagger$$, like WF diagonal elements, otherwise not. $$|m\rangle \langle n|$$ is not hermitian.
The text passage provides a detailed explanation of the Wigner function for different cases and includes a question about reconciling the real-valued nature of the Wigner function with the presence of a complex exponential term.
single_question
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Wigner function of $|n\rangle\langle m|$

For $$n=m$$, the Wigner function is given by, $$W_n(\alpha) = \frac{2}{\pi} (-1)^n \exp(-2 |\alpha|^2) L_n(4 |\alpha|^2),$$

And for $$n \neq m$$, it is, $$f_{mn}=\sqrt{\frac{m!}{n!}} e^{i(m-n) \arctan\left( p/x\right) } \frac{\left( -1\right)^{m}}{\pi\hbar}\left ( \frac{x^{2}+p^{2}}{\hbar/2} \right ) ^ {\left( n-m\right) /2} L_{m}^{n-m}\left( \frac{x^{2}+p^{2}}{\hbar/2}\right) e^{-\left( x^{2}+p^{2}\right) /\hbar}$$ and in this case, there will be always a $$e^{i(m-n) \arctan\left( p/x\right)}$$ term. But the Wigner function is a real valued function. How to reconcile this?

• What is $f_{mn}$?
– Alex
Oct 27, 2023 at 13:53
• Linked. Note this operator is not Hermitian. Oct 27, 2023 at 16:03

In general, Wigner transforms of Hermitian operators are real, $$g(x,p)= \hbar \int \!\! dy ~ e^{-iyp} \langle x+\hbar y/2| \hat G |x-\hbar y /2 \rangle , \implies \\ g(x,p)^*= \hbar \int \!\! dy ~ e^{iyp} \langle x+\hbar y/2| \hat G |x-\hbar y /2 \rangle ^* \\ =\hbar \int \!\! dy ~ e^{iyp} \langle x-\hbar y/2| \hat G ^\dagger |x+\hbar y /2 \rangle ,$$ so $$g=g^*$$ for $$\hat G= \hat G^\dagger$$, like WF diagonal elements, otherwise not. $$|m\rangle \langle n|$$ is not hermitian.
user
The text passage provides a detailed explanation of the Wigner function for different cases and includes a question about reconciling the real-valued nature of the Wigner function with the presence of a complex exponential term.

TYPE: SINGLE_QUESTION