sci.physics.research

On 2008-02-29, Jay R. Yablon wrote:
> I have a question: Consider the path integral:
>
> Z = DA exp[-iS(A)] = C exp [iW(J)]
^ missing an integration sign here?
>
> where S(A) = $d^4x L.
>
> Above, the field variable is A, the source of the field is J, the
> amplitude is W(J), the action is S(A), the Lagrangian density is L, and
> C is an overall product factor independent of J and often containing a
> product of inverse determinants.
>
> Let's posit that L=0, everywhere. We take that as a supposition.
> Perhaps 0 = some other expression involving the sources and fields, but
> nonetheless, this expression L is always = 0.
>
> Would the following deductions be true / permissible?

As true as saying that 0 = 1 - 1 and 0 = 2 - 2 are both true.

> 1) S(A) = 0, because the integral over a volume of anything which is
> zero, is itself zero. No constants of integration come into play. For
> example, thinking about Maxwell's equations in integral form, if there
> is a three volume within which the charge is zero everywhere, then the
> total enclosed charge is zero.
>
> if 1) is true, then:
>
> 2) Z = DA, because exp[-iS(A)] = 1
>
> if 2) is true, then:
>
> 3) given Z=DA, we can select C such that DA = C. Then, exp [iW(J)] =1.

Note that you are allowed to make a normalization choice for the path
integral measure DA precisely because this choice does not affect any
physical observables, such as correlation functions and scattering
amplitudes computed from the path integral.

> if 3) is true, then
>
> 4) W(J) = 2pi n and so is quantized.
>
> Please evaluate and advise if there is any flaw in this logic.

Let me try this too. In step 3, I'm equally allowed to make the choice
int DA = C*exp(i*b), where b is a real number. In step 4, I would have
W(J) = b. But, b can be any real number. Therefore W(J) is continuous!

Can we both be right? If not, what is the flaw in my logic?

Igor