• 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
5,064 questions , 2,215 unanswered
5,347 answers , 22,734 comments
1,470 users with positive rep
818 active unimported users
More ...

  Line element in relativistic quantum mechanics?

+ 1 like - 0 dislike


I noticed the following. Let us take a look at the massless Klien Gordon equation (in relativistic quantum mechanics) along a paticular axis  $x$:

$$  - \hbar^2\partial_t^2 \psi =  - \hbar^2 c^2 \partial_x^2 \psi$$

Writing in terms of the translation operator $\hat T$ and the unitary operator $\hat U$ and using the notation that $\delta k$ is a small shift in $k$ and cancelling factors:

$$\frac {2I - U(\delta t) - U(-\delta t)}{c^2 \delta t^2} | \psi \rangle =  \frac {2I - T(\delta x) - T(-\delta x)}{\delta x^2} | \psi \rangle $$

Taking the inner product and taking a ratio:

$$ \frac{\langle 2I - U(\delta t) - U(-\delta t) \rangle}{\langle 2I - T(\delta x) - T(-\delta x) \rangle } = \frac{c^2 \delta t^2}{\delta x^2} $$

Substracting $1$ both sides:

$$ \frac{ \langle - U(\delta t) - U(-\delta t) + T(\delta x) + T(-\delta x)  \rangle}{\langle 2I - T(\delta x) - T(-\delta x) \rangle } = \frac{c^2 \delta t^2 - \delta x^2}{\delta x^2} $$

Multplying $\delta x^2$ both sides:

$$ 0 = {c^2 \delta t^2 - \delta x^2} - {\delta x^2} \frac{ \langle - U(\delta t) - U(-\delta t) + T(\delta x) + T(-\delta x)  \rangle}{\langle 2I - T(\delta x) - T(-\delta x) \rangle } $$

This can only classically become:

$$ \delta s^2 = {c^2 \delta t^2 - \delta x^2} = 0$$

if one postulates:

$$\frac{\delta x}{\delta t} = c$$


Is there a way to prove

$$\frac{\delta x}{\delta t} = c$$

 this without postulating it?

asked Apr 26, 2021 in Theoretical Physics by Asaint (90 points) [ no revision ]

1 Answer

+ 1 like - 0 dislike

If you have the equation

$$(\partial_0^2-\partial_1^2)\Psi=0$$ then for any twice differentiable function $f$ a solution is $\Psi(x^0,x^1):=f(x^0-x^1)$. This is a constant shape (given by $f$) moving along the $x^1$-axis at constant speed $c$. So if I consider the same solution at a different time, $x^0+\delta x^0$, this implies $\delta x^1=\delta x^0$ for looking at the same part of the moving shape.

Introducing the operators $U$ and $T$ you write a discretisation of the partial derivatives of the KG equation. The "$=$"-sign between the l.h.s. and the r.h.s. of this discretisation implies a relation between $c\delta t$ and $\delta x$, or $\delta x^0$ and $\delta x^1$ in my above notation, which quantities are classical throughout your entire argument. As $|\Psi\rangle$ is to correspond to a solution of the KG equation, this relation is the one mentioned above, $\delta x^1=\delta x^0$.

Put somewhat differently, with the above relation between $\Psi$ and $f$ we have $$\Psi(x^0,x^1+\delta)=f(x^0-(x^1+\delta))=f((x^0-\delta)-x^1)=\Psi(x^0-\delta,x^1)$$ A shift from $x^0-\delta$ to $x^0$ is accompanied by a shift from $x^1$ to $x^1+\delta$.

answered Apr 26, 2021 by Flamma (90 points) [ no revision ]

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