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

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

146 submissions , 123 unreviewed
3,953 questions , 1,403 unanswered
4,889 answers , 20,762 comments
1,470 users with positive rep
507 active unimported users
More ...

Interpreting the Klein-Gordon Annihilation Operator Expression

+ 2 like - 0 dislike
223 views

I can derive $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ for a free real scalar Klein-Gordon field in three ways mathematically: the usual Fourier transform way in Peskin/Srednicki, an awesome direct $a = \frac{1}{2}(2a) = \dots$ way (exercise!), and as a by-product of a clever way of mode-expanding the Hamiltonian, but I can't qualitatively tell you why $$a(k) = \int d^3 x e^{ik_{\mu} x^{\mu}} (\omega_{\vec{k}} \psi + i \pi)$$ should be the answer we'll get before doing any mathematical derivations. Good books, like Zee, will sometimes give two or three derivations of the same thing after all: 

Thus, in chapter VI.1, instead of deriving Einstein’s field equation as a true Confucian scholar would, I try to get to it as quickly as possible by a method I dub “winging it southern California style.” Similarly, in chapter VI.2, I get to cosmology as quickly as possible.

Zee - Gravity, Intro.

How would you interpret $a(k)$, $\psi$ & $\pi$ in such a way so as to make the above expression, & similarly for the creation operator, in the real and complex case, obvious - without sweeping the problem under the rug by referring to the analogous expression for quantum harmonic oscillators which should also be explainable with such a description?

Attempt:

What I see so far is pretty interesting, it's not as elementary as I'd like but it's really nice nonetheless: There seems to be a deep link between the $U(1)$ phase invariance symmetry of the expected value of wave functions $$<g|f> \rightarrow <g'|f'> = <g|e^{-i\theta} e^{i\theta} |f> = <g|f>$$

and the fact that the Klein-Gordon action under a $U(1)$ Noether symmetry produces the Lorentz invariant conserved current $$j_{\mu} = \mathrm{i} (\psi^* \partial_{\mu} \psi - \psi \partial_{\mu} \psi^*)=\mathrm{i} \psi^* \overleftrightarrow{\partial_{\mu}} \psi$$ so that, with a hint of genius, we are somehow motivated to define $<g|f>$ explicitly for functions satisfying Klein-Gordon via
$$\langle g|f \rangle= \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \mathrm{i} g^* \overleftrightarrow{\partial_0} f$$  (seeing it in books before, it seemed like a trick with no reasoning) so that the projection of the wave function onto the $k$'th frequency gives the $a(k)$ coefficient.

Thus it seems that, for a function $f$ satisfying Klein-Gordon, since a wave of charge (time-component of current vector) $k_0$ is represented by $e^{-ik_{\mu} x^{\mu}}$, the amount of $f$ of charge $k$ would be found by analyzing $$<e^{-i k_{\mu} x^{\mu}}|f> = a(k).$$

This seems very haphazard, but indicative.

Could anybody give an answer that a) completely wings it to explain $a(k)$ easily, then b) cleans up my attempt?

asked Jul 12, 2016 in Theoretical Physics by bolbteppa (110 points) [ revision history ]
edited Jul 14, 2016 by bolbteppa

2 Answers

+ 3 like - 0 dislike

A free quantum field is usually defined as a distribution, a linear map from a space of test functions $\mathcal{F}$ into an algebra of operators $\mathcal{A}$. We can write $\hat\phi:\mathcal{F}\rightarrow\mathcal{A}; f\mapsto\hat\phi_f=a_{f^*}+a^\dagger_f$. We take the vacuum vector to be annihilated by $a_f$, $a_f|0\rangle=0$. Using $\hat\phi_f$ we can write down, for example, a one--particle vector state $\hat\phi_f|0\rangle=|f\rangle$. We could construct a free quantum field at a point as $\hat\phi(x)=\hat\phi_{\delta_x}$, except that this is improper insofar as the space of test functions is usually taken not to include distributions such as the Dirac delta function $\delta_x(z)$, or we can work with $\hat\phi(x)$ provided we keep clear in our minds that it is a distribution, not a function, with correspondingly different rules.

For a mass $m$ real Klein-Gordon quantum field operator $\hat\phi_f$, we can write the commutator as $[\hat\phi_f,\hat\phi_g]=(f^*,g)-(g^*,f)$, where the two--particle vacuum expectation value (VEV) $\langle 0|\hat\phi^\dagger_f\hat\phi_g|0\rangle=\langle 0|a_fa^\dagger_g|0\rangle=(f,g)$ is a positive semi--definite inner product $$(f,g)=\int \tilde f^*(k)2\pi\delta(k\cdot k-m^2)\theta(k_0)\tilde g(k)\frac{\mathrm{d}^4k}{(2\pi)^4}.$$

This is a manifestly Lorentz and translation invariant construction, which, loosely, projects the test functions $f$ and $g$ to specific, positive-frequency (because of the ${\small \theta(k_0)}$) solutions of the Klein--Gordon equation (because of the ${\small 2\pi\delta(k\cdot k-m^2)}$). Without the introduction of components (as for the electromagnetic field, say) $(f,g)$ is the only positive semi--definite inner product with the required properties that we can write down. The properties of the Klein--Gordon quantum field are fixed by this construction (together with common ideas about algebras and Hilbert spaces).

Whether you will like this construction better than what you will find in most textbooks, and whether you will find it helpful, will depend on how you like your mathematics. The problem with this construction, however, is that interacting quantum fields are constructed heuristically using $\hat\phi(x)$, they cannot be constructed using $\hat\phi_f$.

answered Jul 14, 2016 by Peter Morgan (1,125 points) [ no revision ]
+ 2 like - 1 dislike

The standard analogy of $\phi$ and $\pi$ with $q$ and $p$, using the corresponding facts for the harmonic oscillator, is perfectly adequate. It works since free fields are essentially given by a continuum of independent oscillators, one for each 3-momentum. Nothing is swept under the rug. One can do exactly the same on the field level directly, but it doesn't add any insight.

Alternatively, and perhaps the cleanest and most direct way to set up everything is to start in momentum space with creation and annihilation operators, then form $a(t,x)=\int dp e^{-iE(p)t+p\cdot x} a(p)$, show that its adjoint $a^*(x)$ creates a general 1-particle state form the vacuum, and verify that $\phi(x)=const(a(x)+a^*(x))$ satisfies the free field equations.

answered Jul 14, 2016 by Arnold Neumaier (12,570 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:
p$\hbar$y$\varnothing$icsOverflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...