Peskin Schroeder Enigmatic Equation

[IRobot]

Hi,

I am learning QFT in the Peskin/Schroeder book and I found 4.56 on page 98 really weird, it is:
\rho_{vaccum\: energy\: density} = \frac{i\sum_{all\: disconnected\: diagramms}amplitude}{(2\pi)^4\delta^{(4)}(0)}

The authors do not comment really this result, but could someone tell me at least if this is finite, calculable, ...Thank you.

 

[QuantumClue]

Hi,

I am learning QFT in the Peskin/Schroeder book and I found 4.56 on page 98 really weird, it is:
\rho_{vaccum\: energy\: density} = \frac{i\sum_{all\: disconnected\: diagramms}amplitude}{(2\pi)^4\delta^{(4)}(0)}

The authors do not comment really this result, but could someone tell me at least if this is finite, calculable, ...


Thank you.

I recognize the 2\pi^4\delta^{4}(0)} term. This term comes from a quantum field amplitude equation, usually seen in this form : 2\pi^4\delta^{4}(p-p)}. It's maybe not surprising as the energy density of the vacuum is also measured similarly.

 

[byzheng]

He is equating the exponents on 4.55, and using what he discusses in 4.49

 

[RedX]

Hi,

I am learning QFT in the Peskin/Schroeder book and I found 4.56 on page 98 really weird, it is:
\rho_{vaccum\: energy\: density} = \frac{i\sum_{all\: disconnected\: diagramms}amplitude}{(2\pi)^4\delta^{(4)}(0)}

The authors do not comment really this result, but could someone tell me at least if this is finite, calculable, ...


Thank you.

I'm not sure I understand the result either. To me it would make more sense if all the diagrams are connected. Also I'm a bit unsure about factors of the imaginary number 'i', but those always get me.

Basically the path integral is e^-iHt, and the amplitude is exp[i*connected diagrams without sources], so setting the argument of the exponentials equal and setting VT=(2pi)⁴ delta⁴(0) should give a result that look similar, except with connected diagrams and not disconnected ones.

 

[RedX]

He is equating the exponents on 4.55, and using what he discusses in 4.49

Does he use connected diagrams or disconnected ones? I don't have a copy of Peskin and Schroeder and my library doesn't have it so I can't look it up.

 

[byzheng]

Actually your plain old e(-iS) gives only diagrams without endpoints; i.e. if you were to associate particles with these diagrams, you would see nothing turning into something back into nothing. These are typically called vacuum bubbles, and their sum represents the vacuum(or ground state) energy E0(peskin uses 'disconnected diagrams' in place of vacuum bubbles). He is dividing by the delta function because from his delta function normalization(discussed in 4.49) the delta(0) factor represents the volume of space(up to 2 pis).

 

[RedX]

Actually your plain old e(-iS) gives only diagrams without endpoints; i.e. if you were to associate particles with these diagrams, you would see nothing turning into something back into nothing. These are typically called vacuum bubbles, and their sum represents the vacuum(or ground state) energy E0(peskin uses 'disconnected diagrams' in place of vacuum bubbles). He is dividing by the delta function because from his delta function normalization(discussed in 4.49) the delta(0) factor represents the volume of space(up to 2 pis).

Does Peskin use e(-iS), or e(iS)?

I think I see what you mean. In the path integral you usually have:

e^{-iS+\int J\phi d^4x}

So now you are setting J=0.

Still, you can have J=0 diagrams that are the product of other J=0 diagrams. For example, you can have two bubbles side by side, unconnected to each other. I don't think you should count those diagrams in the formula for the vacuum energy density, but I'm not sure.

Also, what is the meaning of the vacuum energy density? Isn't an additive constant to the energy not only arbitrary, but also unmeasurable?