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This question was previously asked here http://physics.stackexchange.com/questions/251927/two-point-function-of-a-free-massless-scalar-field-in-euclidean-space-time though I did not get there an answer.

Let $\phi(x)$ be a quantum free massless scalar field (i.e. operator valued function) on a Euclidean space-time of dimension $D>2$. Thus it satisfies the equation $$\Delta\phi(x)=0,$$ where $\Delta =\sum_{i=1}^D(\partial_i)^2$ is the Laplacian.

I would like to compute the function $$D(x):=\langle 0|\phi(x)\phi(0)|0\rangle$$ (there is no time ordering!).

The argument below shows that $D(x)$ is a constant function; that does not sound to make sense. Indeed one has $$\Delta D(x)=\langle 0|\left(\Delta \phi(x)\right)\phi(0)|0\rangle=0.$$ Since any harmonic generalized function is smooth (elliptic regularity), $D(x)$ is a smooth function. Moreover $D(x)$ is $SO(D)$-invariant. Hence on $\mathbb{R}^D\backslash\{0\}$ as a function of $r=|x|$, $D(r)$ satisfies some linear second order differential equation (which can easily be written down). The space of solutions is 2-dimensional. Hence $D(x)$ is a linear combination of the constant function and of $\frac{1}{|x|^{D-2}}$. Since $D(x)$ is smooth, it must be constant.

Remark. To compare with, if instead of Euclidean metric one considers Minkowski metric on $\mathbb{R}^{D}=\mathbb{R}^{1+d}$, then the above argument shows $\Box D(x)=0$ where $\Box =\partial_0^2-\sum_{i=1}^d\partial_i^2$, and $D(x)$ is Lorentz-invariant. In fact the correct answer is $D(x)=\int \frac{d^dp}{(2\pi)^d}\frac{1}{2|p|}e^{-i(|p|x^0-\sum_{i=1}^dp_ix^i)}=\int \frac{d^{d+1}p}{(2\pi)^d}\delta(p^2)\theta(p_0)e^{-i\sum_{\mu=0}^dp_\mu x^\mu}$; it is obtained by an explicit construction of the quantum field $\phi(x)$. Here $\delta(p^2)\theta(p_0)$ can be considered as the only (up to proportionality) Lorentz invariant measure on the light cone; it is finite in a neighborhood of 0 for $D>2$ hence its Fourier transform is well defined in sense of distributions.

This question was previously asked here https://physics.stackexchange.com/questions/251927/two-point-function-of-a-free-massless-scalar-field-in-euclidean-space-time though I did not get there an answer.

Let $\phi(x)$ be a quantum free massless scalar field (i.e. operator valued function) on a Euclidean space-time of dimension $D>2$. Thus it satisfies the equation $$\Delta\phi(x)=0,$$ where $\Delta =\sum_{i=1}^D(\partial_i)^2$ is the Laplacian.

I would like to compute the function $$D(x):=\langle 0|\phi(x)\phi(0)|0\rangle$$ (there is no time ordering!).

The argument below shows that $D(x)$ is a constant function; that does not sound to make sense. Indeed one has $$\Delta D(x)=\langle 0|\left(\Delta \phi(x)\right)\phi(0)|0\rangle=0.$$ Since any harmonic generalized function is smooth (elliptic regularity), $D(x)$ is a smooth function. Moreover $D(x)$ is $SO(D)$-invariant. Hence on $\mathbb{R}^D\backslash\{0\}$ as a function of $r=|x|$, $D(r)$ satisfies some linear second order differential equation (which can easily be written down). The space of solutions is 2-dimensional. Hence $D(x)$ is a linear combination of the constant function and of $\frac{1}{|x|^{D-2}}$. Since $D(x)$ is smooth, it must be constant.

Remark. To compare with, if instead of Euclidean metric one considers Minkowski metric on $\mathbb{R}^{D}=\mathbb{R}^{1+d}$, then the above argument shows $\Box D(x)=0$ where $\Box =\partial_0^2-\sum_{i=1}^d\partial_i^2$, and $D(x)$ is Lorentz-invariant. In fact the correct answer is $D(x)=\int \frac{d^dp}{(2\pi)^d}\frac{1}{2|p|}e^{-i(|p|x^0-\sum_{i=1}^dp_ix^i)}=\int \frac{d^{d+1}p}{(2\pi)^d}\delta(p^2)\theta(p_0)e^{-i\sum_{\mu=0}^dp_\mu x^\mu}$; it is obtained by an explicit construction of the quantum field $\phi(x)$. Here $\delta(p^2)\theta(p_0)$ can be considered as the only (up to proportionality) Lorentz invariant measure on the light cone; it is finite in a neighborhood of 0 for $D>2$ hence its Fourier transform is well defined in sense of distributions.

This question was previously asked here http://physics.stackexchange.com/questions/251927/two-point-function-of-a-free-massless-scalar-field-in-euclidean-space-time though I did not get there an answer.

Let $\phi(x)$ be a quantum free massless scalar field (i.e. operator valued function) on a Euclidean space-time of dimension $D>2$. Thus it satisfies the equation $$\Delta\phi(x)=0,$$ where $\Delta =\sum_{i=1}^D(\partial_i)^2$ is the Laplacian.

I would like to compute the function $$D(x):=\langle 0|\phi(x)\phi(0)|0\rangle$$ (there is no time ordering!).

The argument below shows that $D(x)$ is a constant function; that does not sound to make sense. Indeed one has $$\Delta D(x)=\langle 0|\left(\Delta \phi(x)\right)\phi(0)|0\rangle=0.$$ Since any harmonic generalized function is smooth (elliptic regularity), $D(x)$ is a smooth function. Moreover $D(x)$ is $SO(D)$-invariant. Hence on $\mathbb{R}^D\backslash\{0\}$ as a function of $r=|x|$, $D(r)$ satisfies some linear second order differential equation (which can easily be written down). The space of solutions is 2-dimensional. Hence $D(x)$ is a linear combination of the constant function and of $\frac{1}{|x|^{D-2}}$. Since $D(x)$ is smooth, it must be constant.

REMARK. To compare with, if instead of Euclidean metric one considers Minkowski metric on $\mathbb{R}^{D}=\mathbb{R}^{1+d}$, then the above argument shows $\Box D(x)=0$ where $\Box =\partial_0^2-\sum_{i=1}^d\partial_i^2$, and $D(x)$ is Lorentz-invariant. In fact the correct answer is $D(x)=\int \frac{d^dp}{(2\pi)^d}\frac{1}{2|p|}e^{-i(|p|x^0-\sum_{i=1}^dp_ix^i)}=\int \frac{d^{d+1}p}{(2\pi)^d}\delta(p^2)\theta(p_0)e^{-i\sum_{\mu=0}^dp_\mu x^\mu}$; it is obtained by an explicit construction of the quantum field $\phi(x)$. Here $\delta(p^2)\theta(p_0)$ can be considered as the only (up to proportionality) Lorentz invariant measure on the light cone; it is finite in a neighborhood of 0 for $D>2$ hence its Fourier transform is well defined in sense of distributions.

This question was previously asked here http://physics.stackexchange.com/questions/251927/two-point-function-of-a-free-massless-scalar-field-in-euclidean-space-time though I did not get there an answer.

Let $\phi(x)$ be a quantum free massless scalar field (i.e. operator valued function) on a Euclidean space-time of dimension $D>2$. Thus it satisfies the equation $$\Delta\phi(x)=0,$$ where $\Delta =\sum_{i=1}^D(\partial_i)^2$ is the Laplacian.

I would like to compute the function $$D(x):=\langle 0|\phi(x)\phi(0)|0\rangle$$ (there is no time ordering!).

The argument below shows that $D(x)$ is a constant function; that does not sound to make sense. Indeed one has $$\Delta D(x)=\langle 0|\left(\Delta \phi(x)\right)\phi(0)|0\rangle=0.$$ Since any harmonic generalized function is smooth (elliptic regularity), $D(x)$ is a smooth function. Moreover $D(x)$ is $SO(D)$-invariant. Hence on $\mathbb{R}^D\backslash\{0\}$ as a function of $r=|x|$, $D(r)$ satisfies some linear second order differential equation (which can easily be written down). The space of solutions is 2-dimensional. Hence $D(x)$ is a linear combination of the constant function and of $\frac{1}{|x|^{D-2}}$. Since $D(x)$ is smooth, it must be constant.

Remark. To compare with, if instead of Euclidean metric one considers Minkowski metric on $\mathbb{R}^{D}=\mathbb{R}^{1+d}$, then the above argument shows $\Box D(x)=0$ where $\Box =\partial_0^2-\sum_{i=1}^d\partial_i^2$, and $D(x)$ is Lorentz-invariant. In fact the correct answer is $D(x)=\int \frac{d^dp}{(2\pi)^d}\frac{1}{2|p|}e^{-i(|p|x^0-\sum_{i=1}^dp_ix^i)}=\int \frac{d^{d+1}p}{(2\pi)^d}\delta(p^2)\theta(p_0)e^{-i\sum_{\mu=0}^dp_\mu x^\mu}$; it is obtained by an explicit construction of the quantum field $\phi(x)$. Here $\delta(p^2)\theta(p_0)$ can be considered as the only (up to proportionality) Lorentz invariant measure on the light cone; it is finite in a neighborhood of 0 for $D>2$ hence its Fourier transform is well defined in sense of distributions.

This question was previously asked here http://physics.stackexchange.com/questions/251927/two-point-function-of-a-free-massless-scalar-field-in-euclidean-space-time though I did not get there an answer.

Let $\phi(x)$ be a quantum free massless scalar field (i.e. operator valued function) on a Euclidean space-time of dimension $D>2$. Thus it satisfies the equation $$\Delta\phi(x)=0,$$ where $\Delta =\sum_{i=1}^D(\partial_i)^2$ is the Laplacian.

I would like to compute the function $$D(x):=\langle 0|\phi(x)\phi(0)|0\rangle$$ (there is no time ordering!).

The argument below shows that $D(x)$ is a constant function; that does not sound to make sense. Indeed one has $$\Delta D(x)=\langle 0|\left(\Delta \phi(x)\right)\phi(0)|0\rangle=0.$$ Since any harmonic generalized function is smooth (elliptic regularity), $D(x)$ is a smooth function. Moreover $D(x)$ is $SO(D)$-invariant. Hence on $\mathbb{R}^D\backslash\{0\}$ as a function of $r=|x|$, $D(r)$ satisfies some linear second order differential equation (which can easily be written down). The space of solutions is 2-dimensional. Hence $D(x)$ is a linear combination of the constant function and of $\frac{1}{|x|^{D-2}}$. Since $D(x)$ is smooth, it must be constant.

REMARK. To compare with, if instead of Euclidean metric one considers Minkowski metric on $\mathbb{R}^{D}=\mathbb{R}^{1+d}$, then the above argument shows $\Box D(x)=0$ where $\Box =\partial_0^2-\sum_{i=1}^d\partial_i^2$, and $D(x)$ is Lorentz-invariant. In fact the correct answer is $D(x)=\int \frac{d^dp}{(2\pi)^d}\frac{1}{2|p|}e^{-i(|p|x^0-\sum_{i=1}^dp_ix^i)}=\int \frac{d^{d+1}p}{(2\pi)^d}\delta(p^2)\theta(p_0)e^{-ip\cdot x}$; it is obtained by an explicit construction of the quantum field $\phi(x)$. Here $\delta(p^2)\theta(p_0)$ can be considered as the only (up to proportionality) Lorentz invariant measure on the light cone; it is finite in a neighborhood of 0 for $D>2$ hence its Fourier transform is well defined in sense of distributions.

This question was previously asked here http://physics.stackexchange.com/questions/251927/two-point-function-of-a-free-massless-scalar-field-in-euclidean-space-time though I did not get there an answer.

Let $\phi(x)$ be a quantum free massless scalar field (i.e. operator valued function) on a Euclidean space-time of dimension $D>2$. Thus it satisfies the equation $$\Delta\phi(x)=0,$$ where $\Delta =\sum_{i=1}^D(\partial_i)^2$ is the Laplacian.

I would like to compute the function $$D(x):=\langle 0|\phi(x)\phi(0)|0\rangle$$ (there is no time ordering!).

The argument below shows that $D(x)$ is a constant function; that does not sound to make sense. Indeed one has $$\Delta D(x)=\langle 0|\left(\Delta \phi(x)\right)\phi(0)|0\rangle=0.$$ Since any harmonic generalized function is smooth (elliptic regularity), $D(x)$ is a smooth function. Moreover $D(x)$ is $SO(D)$-invariant. Hence on $\mathbb{R}^D\backslash\{0\}$ as a function of $r=|x|$, $D(r)$ satisfies some linear second order differential equation (which can easily be written down). The space of solutions is 2-dimensional. Hence $D(x)$ is a linear combination of the constant function and of $\frac{1}{|x|^{D-2}}$. Since $D(x)$ is smooth, it must be constant.

REMARK. To compare with, if instead of Euclidean metric one considers Minkowski metric on $\mathbb{R}^{D}=\mathbb{R}^{1+d}$, then the above argument shows $\Box D(x)=0$ where $\Box =\partial_0^2-\sum_{i=1}^d\partial_i^2$, and $D(x)$ is Lorentz-invariant. In fact the correct answer is $D(x)=\int \frac{d^dp}{(2\pi)^d}\frac{1}{2|p|}e^{-i(|p|x^0-\sum_{i=1}^dp_ix^i)}=\int \frac{d^{d+1}p}{(2\pi)^d}\delta(p^2)\theta(p_0)e^{-i\sum_{\mu=0}^dp_\mu x^\mu}$; it is obtained by an explicit construction of the quantum field $\phi(x)$. Here $\delta(p^2)\theta(p_0)$ can be considered as the only (up to proportionality) Lorentz invariant measure on the light cone; it is finite in a neighborhood of 0 for $D>2$ hence its Fourier transform is well defined in sense of distributions.