All Questions
17
questions
1
vote
2
answers
265
views
What are Test Functions in QFT (physics context)?
In terms of mode function solutions of the KG equation, the field operator can be written as
$$ \hat{\phi}(x) = \int_{R^3} d^3\textbf{p}\space \left(u_\textbf{p} \hat{a}_\textbf{p}+ u^*_\textbf{p}\hat{...
- 136
2
votes
0
answers
163
views
Can quantum fields be smeared in space (rather than spacetime)?
I am interested in what is known about the possibility of smearing interacting quantum fields on a Cauchy slice. This is easy to do for free fields and their conjugate momentum, and indeed this is ...
5
votes
2
answers
1k
views
What is an operator-valued distribution?
I am a mathematician who is trying to understand Wightman axioms.
I do not understand what an operator-valued distribution is, because in this context people say that operators can be unbounded, and ...
- 263
1
vote
1
answer
228
views
Does normal ordering (not) conflict with canonical commutators?
Normal ordering is pretty useful to stop expressions from diverging in quantum field theory and works out perfectly fine regarding this, but there is this little problem: Consider for example an ...
- 2,126
1
vote
1
answer
280
views
Why does QFT require operator-valued distributions? [duplicate]
I am new to QFT, and so far I have only gone through the basics up to defining what a quantum field is, which is an operator valued distribution. I have been struggling so far understanding why ...
- 3,350
0
votes
1
answer
351
views
Expressing the four-momentum operator in terms of field operators
There are a series of problems in chapter 3 of the book Quantum Field Theory of Point Particles and Strings by Hatfield that lead to a proof of Lorentz invariance in the canonical formulation of ...
- 785
8
votes
1
answer
566
views
Where does it become apparent in real scalar QFT that the field has to be an operator-valued distribution, as opposed to an operator-valued function?
It its very often stated that in QFT, we don't actually deal with operator valued functions (assign a field operator to each point in space time), but instead with operator valued distributions (in ...
- 6,763
0
votes
0
answers
214
views
Distributions in QFT
If field operators are really distributions then surely objects like the commutator, any correlation functions or even the free theory lagrangians are all ill-defined since products of distributions ...
- 766
0
votes
1
answer
150
views
"Trace" of a distribution?
In quantum field theory we have to sometimes take the "trace" of a distribution $M(x,y)$, $\text{tr}M\sim\int dx M(x,x)$. This happens for instance when we try to expand the determinant of a ...
- 3,514
-1
votes
1
answer
126
views
Dealing with $\delta^{(3)}(0)$ using normal ordering
On page 109 of David Tong's lecture notes on QFT, equations (5.11) and (5.12) read:
$$ H = \int \frac{d^{3}p}{(2\pi)^{3}} E_{\vec{p}}[(b_{\vec{p}}^{s})^{\dagger}b_{\vec{p}}^{s}-c_{\vec{p}}^{s}(c_{\vec{...
- 1,131
5
votes
2
answers
256
views
Confusion about quantum field in AQFT
As far as I known, quantum field is defined by operator-valued distribution mathematically. If I understand correctly, in AQFT, we use self-adjoint elements of $C$* algebra to describe algebra of ...
- 525
0
votes
0
answers
126
views
Proof of commutator in Tong's notes on QFT
I am following David Tong's notes on QFT: http://www.damtp.cam.ac.uk/user/tong/qft.html . In equation 2.21, he tries to prove $$[\phi(\vec{x}),\pi(\vec{y})] = i\delta^{(3)}(\vec{x}-\vec{y}).$$ Here, $\...
- 1,222
0
votes
1
answer
239
views
Do Equal Time Commutation / Anticommutation relations automatically also apply for spatial derivatives?
The question is basically in the title.
My naive thought was that when a commutation relation holds for all field operators $\Psi(\vec{x})$ (by "all" I mean "at all positions $\vec{x}$") on a fixed ...
- 6,763
0
votes
1
answer
366
views
Commutation relation in derivative of creation operator?
I am basically trying to get the BMS hard charge for the subleading soft theorem.
I have the usual commutation relation
$[a(\omega',z',\overline{z'}),a^\dagger(\omega,z,\overline{z})]=\delta(\omega'-\...
- 161
7
votes
2
answers
537
views
Eigenstates of a Hermitian field operator
Consider a Hermitian field operator $\phi(x)$ with eigenstates satisfying
$$
\phi(x) |\alpha\rangle = \alpha(x) | \alpha \rangle
$$
I'm trying to determine the inner product between the eigenstates. ...
- 26.6k