Undergrad Finding ##\partial^\mu\phi## for a squeezed state in QFT

Click For Summary
The discussion focuses on calculating the derivative ##\partial^\mu\phi## for a massless squeezed state in quantum field theory, defined as $$\phi=\sum_k\left(a_kf_k+a^\dagger_kf^*_k\right)$$ with specific mode functions. The challenge arises from handling the sum over k and the ladder operators during differentiation. It is clarified that ##\partial_\mu## can be applied term by term, treating the ladder operators as constants if they are independent of ##x##. Additionally, a different dummy summation index should be used to avoid conflicts with the free index on ##\partial_\mu##. The discussion emphasizes that the expression represents a field operator rather than a state in Fock space.
Sciencemaster
Messages
129
Reaction score
20
TL;DR
I'm trying to apply an operator to a massless and minimally coupled squeezed state, I'm having trouble calculating ##\partial^\mu\phi## but due to a sum over k and the ladder operators.
I'm trying to apply an operator to a massless and minimally coupled squeezed state. I have defined my state as $$\phi=\sum_k\left(a_kf_k+a^\dagger_kf^*_k\right)$$, where the ak operators are ladder operators and fk is the mode function $$f_k=\frac{1}{\sqrt{2L^3\omega}}e^{ik_\mu x^\mu}$$ (assuming periodic boundary condition in a three-dimensional box of side L where k is the wave number).
However, I'm having trouble calculating ##\partial^\mu\phi## due to the sum over k and the ladder operators. I would very much appreciate it if someone could help me through the math of this step!
 
Physics news on Phys.org
Well, ##\partial_\mu## is a linear operator, so you can apply it term by term in the sum. As for the ladder operators, if they're independent of ##x## you can just treat them like constants during the partial differentiation.

Btw, you'll need to use a different dummy summation index in the exponent so as not to conflict with the free index ##\mu## on ##\partial_\mu##. E.g., change ##k_\mu x^\mu## to ##k_\alpha x^\alpha##.
 
  • Like
Likes topsquark and vanhees71
One should also note that this doesn't describe a state but a field operator in terms of free-field energy eigenmodes or a neutral scalar field. The ##\hat{a}_k## are annihilation and ##\hat{a}_k^{\dagger}## in Fock space.
 
I am slowly going through the book 'What Is a Quantum Field Theory?' by Michel Talagrand. I came across the following quote: One does not" prove” the basic principles of Quantum Mechanics. The ultimate test for a model is the agreement of its predictions with experiments. Although it may seem trite, it does fit in with my modelling view of QM. The more I think about it, the more I believe it could be saying something quite profound. For example, precisely what is the justification of...

Similar threads

  • · Replies 1 ·
Replies
1
Views
556
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K