Hamiltonian for a Klein-Gordon field can be written as - H=∫d3p(2π)3ω→p[a†→pa→p+12(2π)3δ(3)(0).] In one of my lecture notes on QFT, It is written that - In absence of gravity, we can neglect the second term in above equation which will lead us to- H=∫d3p(2π)3ω→pa†→pa→p. It is obvious that second term in the first equation will make vacuum energy infinite. But Neglecting this term is the best we can do? Something is infinite and just for our convenience, we are putting it to zero. How is this logically and mathematically justifiable? Secondly, What is the role of gravity in neglecting this term? Why can we not neglect this term if gravity is present?
Subscribe to:
Post Comments (Atom)
Understanding Stagnation point in pitot fluid
What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...
-
I have an hydrogenic atom, knowing that its ground-state wavefunction has the standard form ψ=Ae−βr with $A = \frac{\bet...
-
At room temperature, play-dough is solid(ish). But if you make a thin strip it cannot just stand up on it's own, so is it still solid? O...
-
Sometimes I am born in silence, Other times, no. I am unseen, But I make my presence known. In time, I fade without a trace. I harm no one, ...
-
I want to know what happens to the space a black hole crosses over as our galaxy travels through space.
-
Small vessels generally lean into a turn, whereas big vessels lean out. Why do ships lean to the outside, but boats lean to the inside of a ...
-
I'm sitting in a room next to some totally unopened cans of carbonated soft drinks (if it matters — the two affected cans are Coke Zero...
-
What exactly are the spikes, or peaks and valleys, caused by in pictures such as these Wikipedia states that "From the point of view of...
No comments:
Post a Comment