Interpreting the energy density of QFT vacuum states

  • #1
math-physicist
5
0
I am reading Peskin-Schroeder's QFT text, and there on pg. 98 Equation (4.56) they derive the expression for the vacuum energy density (relative to the zero of energy set by ##H_0|0\rangle = 0##):
$$ \frac{E_0}{\rm{Volume}} = \frac{i\,\sum\text{(all disconnected pieces)}}{(2\pi)^4\,\delta^4(0)}\ . $$
My questions are:
  • What do they mean by "relative to the zero of energy set by ##H_0|0\rangle = 0##" ?
  • The right side of the above expression has a delta function evaluated at 0. Doesn't that imply that the right side is essentially negligible/zero? Then how does one interpret this equation? The vacuum has zero energy?
  • How does one derive Equation (4.56) starting from Equation (4.55) which states:
    $$ e^{\sum V_i}\propto e^{-2iE_0T}\ ? $$
 
Last edited:
Physics news on Phys.org
  • #2
math-physicist said:
I am reading Peskin-Schroeder's QFT text, and there on pg. 98 Equation (4.56) they derive the expression for the vacuum energy density (relative to the zero of energy set by ##H_0|0\rangle = 0##):
$$ \frac{E_0}{\rm{Volume}} = \frac{i\,\sum\text{(all disconnected pieces)}}{(2\pi)^4\,\delta^4(0)}\ . $$
My questions are:
  • What do they mean by "relative to the zero of energy set by ##H_0|0\rangle = 0##" ?
A Hamiltonian is defined only up to a constant energy shift. The condition fixes this constant by requiring that that the ground state (vacuum) has energy zero.
math-physicist said:
  • The right side of the above expression has a delta function evaluated at 0. Doesn't that imply that the right side is essentially negligible/zero? Then how does one interpret this equation? The vacuum has zero energy?
No. The delta function is in the denominator! This is sloppy notation, What is meant is that after multiplication with the denominator, reinterpreting the formula in a distributional sense, and integration over a test function, the equation is valid in perturbation theory.
math-physicist said:
  • How does one derive Equation (4.56) starting from Equation (4.55) which states:
    $$ e^{\sum V_i}\propto e^{-2iE_0T}\ ? $$
This is a well-known combinatorial results for generating functionals, relating the moment generating function and the cumulant generating function. See https://en.wikipedia.org/wiki/Moment-generating_function#Relation_to_other_functions
 
  • Like
Likes math-physicist
  • #3
@A. Neumaier , Thank you for your answer. Your answer to my first two questions makes sense. but in answer to my third question, you said:
A. Neumaier said:
This is a well-known combinatorial results for generating functionals, relating the moment generating function and the cumulant generating function. See https://en.wikipedia.org/wiki/Moment-generating_function#Relation_to_other_functions
I think you misinterpreted what I was asking. I meant to ask why does that proportionality imply the energy density relation? I.e.,
$$ e^{\sum V_i}\propto e^{-2iE_0T} \stackrel{\implies}{\rm{(Why??)}} \frac{E_0}{\rm{Volume}} = \frac{i\,\sum\text{(all disconnected pieces)}}{(2\pi)^4\,\delta^4(0)}\ . $$
Could you please at least sketch the steps of this derivation? TIA.
 
  • #4
math-physicist said:
@A. Neumaier , Thank you for your answer. Your answer to my first two questions makes sense. but in answer to my third question, you said:

I think you misinterpreted what I was asking. I meant to ask why does that proportionality imply the energy density relation? I.e.,
$$ e^{\sum V_i}\propto e^{-2iE_0T} \stackrel{\implies}{\rm{(Why??)}} \frac{E_0}{\rm{Volume}} = \frac{i\,\sum\text{(all disconnected pieces)}}{(2\pi)^4\,\delta^4(0)}\ . $$
Could you please at least sketch the steps of this derivation? TIA.
One takes the logarithm in the left relation and applies the combinatorial result I stated to the expansion of the left hand side, which is the moment generating function.
 

Similar threads

Replies
1
Views
869
Replies
8
Views
1K
Replies
7
Views
963
  • Quantum Physics
Replies
8
Views
2K
  • Quantum Physics
Replies
1
Views
947
Replies
3
Views
917
  • Quantum Physics
Replies
4
Views
1K
  • Quantum Physics
Replies
19
Views
3K
  • Quantum Physics
Replies
13
Views
863
  • Science and Math Textbooks
Replies
7
Views
623
Back
Top