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$$D(x-y)=\langle 0|\hat{\phi}(x) \hat{\phi}(y)|0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3 2 E_{\vec{p}}} \exp[-\mathrm{i} p\cdot(x-y)].$$

This is a Lorentz-scalar field, because ##\mathrm{d}^3 \vec{p}/(2 E_{\vec{p}})## is a Lorentz scalar and all other ingredients in the integral are manifestly Lorentz invariant (with Lorentz invariant here I mean invariance under the proper orthochronous Lorentz group). Since further there is no other four vector then ##z=x-y## this implies that

$$D'(z')=D(z')=D(z), \quad z'=\Lambda z, \quad \Lambda \in \mathrm{SO}(1,3)^{\uparrow}.$$

This implies that if the claim about space-like vectors is true you have

$$D(z)=D(-z) \quad \text{for} \quad z^2<0,$$

and this implies that the vev of the commutator which is ##D(x-y)-D(y-x)## by definition, indeed vanishes for space-like ##x-y##.

Now let's prove the claim about space-like vectors. So let ##z^2<0##. Then we can always use a coordinate system, where ##z^0=0##. To see this just take ##e_1=z/\sqrt{-z \cdot z}## as one of the space-like basis vectors. The basis can be obviously constructed such that the change to this coordinate system is given by a proper orthochronous Lorentz transformation. So we have

$$z=\begin{pmatrix}0\\ z^1 \\ 0 \\ 0 \end{pmatrix}.$$

Now a rotation around the ##z^3##-axis reads

$$z^{\prime 0}=z^0=0, \quad \vec{z}'=\begin{pmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}z^1 \\0 \\ 0 \end{pmatrix} = z^1 \begin{pmatrix} \cos \phi \\ -\sin \phi \\0 \end{pmatrix}.$$

For ##\phi=\pi## you get ##\vec{z}'=-\vec{z}##, as claimed.

It's a good exercise to prove that this argument doesn't work for timelike vectors, because the sign of ##z^0## cannot be changed by a proper orthochronous Lorentz transformation (that's why it's called orthochronous).

- #3

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Thank you for you answer. I am just now sure why ##D'(z')=D(z')=D(z)##. The first and the 3rd term are equal as D is a scalar field. But why does the equality in the middle holds?

$$D(x-y)=\langle 0|\hat{\phi}(x) \hat{\phi}(y)|0 \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{(2 \pi)^3 2 E_{\vec{p}}} \exp[-\mathrm{i} p\cdot(x-y)].$$

This is a Lorentz-scalar field, because ##\mathrm{d}^3 \vec{p}/(2 E_{\vec{p}})## is a Lorentz scalar and all other ingredients in the integral are manifestly Lorentz invariant (with Lorentz invariant here I mean invariance under the proper orthochronous Lorentz group). Since further there is no other four vector then ##z=x-y## this implies that

$$D'(z')=D(z')=D(z), \quad z'=\Lambda z, \quad \Lambda \in \mathrm{SO}(1,3)^{\uparrow}.$$

This implies that if the claim about space-like vectors is true you have

$$D(z)=D(-z) \quad \text{for} \quad z^2<0,$$

and this implies that the vev of the commutator which is ##D(x-y)-D(y-x)## by definition, indeed vanishes for space-like ##x-y##.

Now let's prove the claim about space-like vectors. So let ##z^2<0##. Then we can always use a coordinate system, where ##z^0=0##. To see this just take ##e_1=z/\sqrt{-z \cdot z}## as one of the space-like basis vectors. The basis can be obviously constructed such that the change to this coordinate system is given by a proper orthochronous Lorentz transformation. So we have

$$z=\begin{pmatrix}0\\ z^1 \\ 0 \\ 0 \end{pmatrix}.$$

Now a rotation around the ##z^3##-axis reads

$$z^{\prime 0}=z^0=0, \quad \vec{z}'=\begin{pmatrix} \cos \phi & \sin \phi & 0\\ -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}z^1 \\0 \\ 0 \end{pmatrix} = z^1 \begin{pmatrix} \cos \phi \\ -\sin \phi \\0 \end{pmatrix}.$$

For ##\phi=\pi## you get ##\vec{z}'=-\vec{z}##, as claimed.

It's a good exercise to prove that this argument doesn't work for timelike vectors, because the sign of ##z^0## cannot be changed by a proper orthochronous Lorentz transformation (that's why it's called orthochronous).

- #4

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$$D'(z')=D(z).$$

Now since there's no other vector involved, of which ##D## could depend, and thus you must have ##D'=D##, i.e., also ##D'(z')=D(z')##.

Another argument is that if you have no other vectors then ##z## a scalar function must be a function of ##z^2##.

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