• 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.


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


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,054 questions , 2,207 unanswered
5,347 answers , 22,728 comments
1,470 users with positive rep
818 active unimported users
More ...

  A Question about Path-Integral Measure

+ 2 like - 0 dislike

I want to do the following path integral.

$$\mathcal{Z}=\int\mathcal{D}x e^{iS[\dot{x}]}$$

The action only denpends on $\dot{x}$. For some reason, I want to replace the integral measure $\mathcal{D}x$ by $\mathcal{D}\dot{x}$.

So I have

$$\mathcal{Z}=\int\mathcal{D}\dot{x}\mathrm{Det}\left(\frac{\delta x}{\delta\dot{x}}\right)e^{iS[\dot{x}]}.$$

The variable $x$ is related with $\dot{x}$ via the linear transformation


which implies 

$$\mathrm{Det}\left(\frac{\delta x}{\delta\dot{x}}\right)\equiv 1.$$

Am I correct in the above derivation?

asked Feb 13, 2019 in Theoretical Physics by Libertarian Feudalist Bot (270 points) [ no revision ]


@arnold neumaier Can you elaborate? 

1 Answer

+ 1 like - 0 dislike

''which implies'' does not follow. Rewrite the derivative as an operator applied to $x$, and perform the functional differentiation by working out the virtual displacement to get the correct answer.

answered Feb 13, 2019 by Arnold Neumaier (15,787 points) [ no revision ]

Is $\det(\frac{\delta x}{\delta\dot{x}})=\frac{1}{\det(\frac{d}{dt})}$ the correct answer?

Formally yes, but the result is not well-defined without regularization since the determinant of an operator with continuous spectrum makes no sense. On the other hand, the regularized determinant is a constant, hence can be absorbed in the normalization of the functional integral. 

Thank you very much. Would you please look at another question I posted here?

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:
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