# Interpreting the energy density of QFT vacuum states

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• math-physicist
math-physicist
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:
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

math-physicist
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.

math-physicist 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.

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