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In quantum field theory most observables $A$ do not have a definite value in the ground state (vacuum). For an observable $A$, a reasonable measure of the spread in the ground state is its variance $\operatorname{Var} A$ defined by, $$ \operatorname{Var} A=\left\langle 0\left|(A-\langle 0|A| 0\rangle)^2\right| 0\right\rangle. $$

In Klein Gordon field theory for a free real scalar consider the observable $$ A(a)=\pi^{-3 / 2} a^{-3} \int d^3 x \phi(\mathbf{x}, 0) e^{-\mathbf{x}^2 / a^2} $$

The Gaussian weighting has been normalized to unit so $A(a)$ is a smoothed out version of $\phi(\mathbf{0}, 0)$.

(a) Express the vacuum variance of $A(a)$ as an integral over a single variable.

(b) Show that in the limiting case of very large or very small $a$ that $$ \operatorname{Var} A(a)=\alpha a^\beta+\ldots . $$ $\alpha$ and $\beta$ are constants and the ellipses are less important terms. Find $\alpha$ and $\beta$. Note that $\operatorname{Var} A(a)$ goes to zero for large $a$ and blows up for small $a$, so quantum effects are most important on small scales.

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    $\begingroup$ what about using spherical coordinates... $\endgroup$ Nov 3, 2023 at 23:58
  • $\begingroup$ @GabrielPalau I just updated my attempt. I tried to substitute in r=x-y and s=x+y but that didn't seem to help $\endgroup$
    – Orion Pax
    Nov 4, 2023 at 0:07

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Starting from the field operator of a free Hermitean scalar field at time $t=0$, $$\phi(\mathbf{x},0)=\int \! \frac{d^3 k}{(2\pi)^3 2 \omega(\mathbf{k})}\left[e^{i \mathbf{k}\cdot \mathbf{x}}a(\mathbf{k})+e^{-i \mathbf{k} \cdot \mathbf{x}}a^\dagger(\mathbf{k})\right], \qquad \omega(\mathbf{k})=\sqrt{\mathbf{k}^2+m^2},$$ the commutation relations of the creation and annihilation operators are given by $$\left[ a(\mathbf{k}), a^\dagger(\mathbf{k}^\prime) \right]= (2 \pi)^3 2 \omega(\mathbf{k}) \delta^{(3)}(\mathbf{k}-\mathbf{k}^\prime), \qquad \left[a(\mathbf{k}),a(\mathbf{k}^\prime) \right]=0.$$ As $a(\mathbf{k}) |0\rangle=0$ implies $\langle 0 |A(a) |0\rangle=0$, the determination of the variance of the smeared operator $A(a)$ in the vacuum state boils down to the computation of $$\langle 0 | A(a)^2 |0\rangle= \pi^{-3}a^{-6}\int \! d^3x \, e^{-\mathbf{x}^2/a^2}\int \! d^3y \, e^{-\mathbf{y}^2/a^2}\langle 0 |\phi(\mathbf{x},0) \phi(\mathbf{y},0) | 0\rangle, $$where $$ \langle 0 | \phi(\mathbf{x}, 0) \phi(\mathbf{y}, 0) | 0\rangle = \int \!\frac{d^3k}{(2\pi)^3 2 \omega(\mathbf{k})} e^{i \mathbf{k}\cdot (\mathbf{x}-\mathbf{y})}. $$ Using the Gauss integral formula $$\int \! d^3x \, e^{-\mathbf{x}^2/a^2}e^{i \mathbf{k} \cdot \mathbf{x}} =\pi^{3/2} a^3 e^{-a^2 \mathbf{k}^2/4},$$ the integration over the variables $\mathbf{x}$ and $\mathbf{y}$ can be performed, yielding $$\langle 0 |A(a)^2 |0\rangle=\frac{1}{4 \pi^2}\int\limits_0^\infty \frac{dk \, k^2}{\sqrt{k^2+m^2}}e^{-a^2k^2/2}=\frac{1}{4\pi^2 a^2}\int\limits_0^\infty\frac{ds s^2}{\sqrt{s^2+(ma)^2}} e^{-s^2/2}.$$ The result is dimensionally correct as $[\phi]=[A]=[m]=1$ and $[a]=-1$. It is now an easy task to work out the requested asymptotic expansions in the dimensionless variable $ma$ of the remaining integral.

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  • $\begingroup$ Thank you so so much! A little detail follow up: when I calculate the Gaussian, I got $\pi^{3/2}a^3e^{-a^2\mathbf{k}^2/4}$. So in other words, I have the denominator of 4 instead of 2 in the exponent of e, do you mind confirming if this is correct or wrong, please? $\endgroup$
    – Orion Pax
    Nov 5, 2023 at 21:53
  • $\begingroup$ @Orion Pax Thank you, this was a typo! Already corrected. $\endgroup$
    – Hyperon
    Nov 6, 2023 at 2:38

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