- #1
geoduck
- 258
- 2
Is there a reason why we have to expand a field ψ about the true vacuum |Ω>? Can't we just do field theory about ψ=0 instead of about ψ=<Ω|ψ|Ω>?
Also, I'm a bit confused about other fields. For the E&M potential, under the true vacuum, wouldn't we need to expand about A=<Ω|A|Ω> instead of A=0?
Also, how do we find the true vacuum anyhow? The way that it seems to be done is to take the derivative of the potential V, and set it equal to zero. The potential V usually has a negative mass or something shifting the true vacuum away from ψ=0. Is this an approximation? Don't we have to take the derivative of the effective potential Veff instead of V?
And how can one show that two derivatives of Veff gives the mass generated? I thought to get the mass generated, you have to shift the fields, and identify the coefficients of terms proportional to the square of the field. But a paper I'm reading claims you can just take two derivatives of Veff to get the mass generated?
Also, I'm a bit confused about other fields. For the E&M potential, under the true vacuum, wouldn't we need to expand about A=<Ω|A|Ω> instead of A=0?
Also, how do we find the true vacuum anyhow? The way that it seems to be done is to take the derivative of the potential V, and set it equal to zero. The potential V usually has a negative mass or something shifting the true vacuum away from ψ=0. Is this an approximation? Don't we have to take the derivative of the effective potential Veff instead of V?
And how can one show that two derivatives of Veff gives the mass generated? I thought to get the mass generated, you have to shift the fields, and identify the coefficients of terms proportional to the square of the field. But a paper I'm reading claims you can just take two derivatives of Veff to get the mass generated?