Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Where does a fermionic coherent state live (which Hilbert space)?

+ 7 like - 0 dislike
1387 views

There have been a couple of questions on fermionic coherent states, but I didn't find any that covered the following question:

If I define a coherent fermionic state in the 2-level-system spanned by $|0\rangle$ and $|1\rangle$, I will write it as \begin{align} |\gamma\rangle=e^{a^\dagger\gamma-\overline{\gamma}a}|0\rangle\,, \end{align} where $\gamma$ is a Grassmann variable and $a^\dagger$ and $a$ are fermionic creation and annihilation operators. Such a state has the property that the expectation value of $a^\dagger$ is given by $\gamma$ and the one of $a$ by $\overline{\gamma}$. How can a regular operator have an expectation value that is not a complex number?

To me it seems that we formally extend our Hilbert space to something where vectors cannot just have complex numbers as coefficients, but also polynomials of Grassmann variables. What's the best way to think of this? Can I use such a state to describe a concrete physical state? What happens if the number operator has an expectation value containing Grassmann variables?

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user LFH
asked Jul 24, 2016 in Theoretical Physics by LFH (40 points) [ no revision ]
Related: physics.stackexchange.com/q/40746/2451

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user Qmechanic
I want to point out one thing. Physical quantities will be bilinear with two $a$ or $a\dagger$s. So they will be Grassman even objects. So I don't think there should be a problem in it being a physical state. Number operator too will have two of them. But I'm with you on a regular operator having a non complex number expectation value, I think that part needs a better explanation.

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user BoundaryGraviton
I agree that this is how one makes physically sense out of it. I don't recall the details for Grassmann even operators - is there a natural identification of Grassmann even objects with real or complex numbers?

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user LFH
BTW: The reference that Qmechanic added cleared up a lot of things. I assume that one needs to extend the Hilbert space to a super Hilbert space or something like that - where coefficients can be Grassmann-valued.

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user LFH
For this, you can see the question link shared by QMechanic.

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user BoundaryGraviton
I think this hri.res.in/~cmschool/coldatom/talks/school_rajdeep.pdf lecture slides explain Fermionic coherent states well and address the questions that you ask. See page 49 and onward.

This post imported from StackExchange Physics at 2017-09-17 13:03 (UTC), posted by SE-user BoundaryGraviton

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysic$\varnothing$Overflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...