Did I Misinterpret the Change of Sign in the Feynman Propagator Integral?

[binbagsss]

Homework Statement

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Hi in the first attachment I am stuck on the sign change argument used to get from line 2 to 3 , see below

Homework Equations

above

The Attempt at a Solution

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Q1) please correct me if I'm wrong but :

$d^3 p \neq d\vec{p}$ since $d^3 p = dp_x dp_y dp_z$ and $\vec{p}=\sqrt{p_x^2 + p_y^2 + p_z^2}$

So and $\vec{p} \to -\vec{p} \to -\sqrt{p_x^2 + p_y^2 + p_z^2}$ ?

I am unsure how to explicitly relate these two integrals to show that the effect of changing the sign in the exponent changes the integral factor such to leave the integral invariant ?

 

[MathematicalPhysicist]

Well, you have $p\cdot (y-x)= E_{\vec{p}}(y_0-x_0)-\vec{p}\cdot (\vec{y}-\vec{x})$ from the pseudo inner product of Minkowski space, where the LHS is the 4-momentum pseudo inner product with 4-displacement vector.

 

[binbagsss]

What's this got to do with the integral variables sorry ? Do I need to compute a Jacobian or?

Well, you have $p\cdot (y-x)= E_{\vec{p}}(y_0-x_0)-\vec{p}\cdot (\vec{y}-\vec{x})$ from the pseudo inner product of Minkowski space, where the LHS is the 4-momentum pseudo inner product with 4-displacement vector.

 

[binbagsss]

Well, you have $p\cdot (y-x)= E_{\vec{p}}(y_0-x_0)-\vec{p}\cdot (\vec{y}-\vec{x})$ from the pseudo inner product of Minkowski space, where the LHS is the 4-momentum pseudo inner product with 4-displacement vector.

Sorry yes I'm aware of that, the pseudo product But not the part why it is valid 'since its an integration variable '

 

[MathematicalPhysicist]

The integration works on $\vec{p}$ and not on the 4-momentum $p$.

Edit: erased an error.

 

[MathematicalPhysicist]

I think there's an abuse of notation $d^3 p$ is as you wrote dp_x dp_y dp_z, so there are two mistakes here one is the abuse of notation I just remarked and the second is that in the pic you gave there's a missing plus sign, it should be $-E_p(y^0-x^0)+\vec{p}\cdot(\vec{y}-\vec{x})$ and not as it's written with two minuses.

 

[binbagsss]

I think there's an abuse of notation $d^3 p$ is as you wrote dp_x dp_y dp_z, so there are two mistakes here one is the abuse of notation I just remarked and the second is that in the pic you gave there's a missing plus sign, it should be $-E_p(y^0-x^0)+\vec{p}\cdot(\vec{y}-\vec{x})$ and not as it's written with two minuses.

Yhepp agree with the sign Change and notinf the integral is only over three momentum not four momentum

Not sure what you mean about abuse of notation though sorry ? And, still don't we need to relate $d^3 p$ and $d \vec p$ ?

Am I being stupid sorry, ta

 

[stevendaryl]

Yhepp agree with the sign Change and notinf the integral is only over three momentum not four momentum

Not sure what you mean about abuse of notation though sorry ? And, still don't we need to relate $d^3 p$ and $d \vec p$ ?

Am I being stupid sorry, ta

$d^3 p$ means the triple integral $dp_1\ dp_2\ dp_3$. The expression $\overrightarrow{p} \cdot (\overrightarrow{y} - \overrightarrow{x})$ means $p_1 (y_1 - x_1) + p_2 (y_2 - x_2) + p_3 (y_3 - x_3)$. They're using $p_0 = E_p = \sqrt{p_1^2 + p_2^2 + p_3^2 + m^2}$

 

[binbagsss]

$d^3 p$ means the triple integral $dp_1\ dp_2\ dp_3$. The expression $\overrightarrow{p} \cdot (\overrightarrow{y} - \overrightarrow{x})$ means $p_1 (y_1 - x_1) + p_2 (y_2 - x_2) + p_3 (y_3 - x_3)$. They're using $p_0 = E_p = \sqrt{p_1^2 + p_2^2 + p_3^2 + m^2}$

I'm aware of that
And that it's clear that E_p is invariant under this, but still can't see why we don't need to relate the integrals, my question still holds..

 

[binbagsss]

and $\vec{p} \to -\vec{p}$ is $\to -\sqrt{p_x^2 + p_y^2 + p_z^2}$ ?

Bump

 

[stevendaryl]

I'm aware of that
And that it's clear that E_p is invariant under this, but still can't see why we don't need to relate the integrals, my question still holds..

Let's look at the integral $\int_A^B f(p) dp$. Now, if we change variables to $p' = -p$, then we get:

$\int_A^B f(p) dp = - \int_{-A}^{-B} f(p') dp' = \int_{-B}^{-A} f(p') dp'$

We get a minus sign from the coordinate change, but then we get another minus sign from switching the limits of the integral. Now, in the particular case $A = -\infty, B = +\infty$, we have:

$\int_{-\infty}^{+\infty} f(p) dp = - \int_{+\infty}^{-\infty} f(p') dp' = \int_{-\infty}^{+\infty} f(p') dp'$

So when you're integrating over all space, there is no minus sign associated with changing variables from $p$ to $-p$.

 

[binbagsss]

Let's look at the integral $\int_A^B f(p) dp$. Now, if we change variables to $p' = -p$, then we get:

$\int_A^B f(p) dp = - \int_{-A}^{-B} f(p') dp' = \int_{-B}^{-A} f(p') dp'$

We get a minus sign from the coordinate change, but then we get another minus sign from switching the limits of the integral. Now, in the particular case $A = -\infty, B = +\infty$, we have:

$\int_{-\infty}^{+\infty} f(p) dp = - \int_{+\infty}^{-\infty} f(p') dp' = \int_{-\infty}^{+\infty} f(p') dp'$

So when you're integrating over all space, there is no minus sign associated with changing variables from $p$ to $-p$.

That's all fine
But, as in my op, I thought we do not have $d \vec{p} \to d \ved{-p}$

But are analyzing how a change in $\vec{p}$ affects $d^3p \neq d\vec{p}$

Since, again , as said before, $d^3p = dp_x dp_y dp_z$ , and $d\vec{p} = d ( \sqrt{p_x^2+...)$

So are you saying $d^3p = d\vec{p}$ ??

 

[stevendaryl]

That's all fine
But, as in my op, I thought we do not have $d \vec{p} \to d \ved{-p}$

I don't know what you mean by $d \vec{p}$. There is no integration over $\vec{p}$, there are three separate integrations over $dp_1 dp_2 dp_3$ (or $dp_x dp_y dp_z$, except that the original post uses $x$ and $y$ as position vectors).

Under the change $\vec{p} \rightarrow -\vec{p}$, we have:

$p_1 \rightarrow - p_1$
$p_2 \rightarrow - p_2$
$p_3 \rightarrow - p_3$

The integral $\int dp^3$ means $\int_{-\infty}^{+\infty} dp_1 \int_{-\infty}^{+\infty} dp_2 \int_{-\infty}^{+\infty} dp_3$

So under the change $\vec{p} \rightarrow -\vec{p}$, this becomes:

$\int_{+\infty}^{-\infty} (- dp_1) \int_{+\infty}^{-\infty} (- dp_2) \int_{+\infty}^{-\infty} (- dp_3)$

If you swap the upper and lower indices, you get 3 more minus signs, to give:

$\int_{-\infty}^{+\infty} dp_1 \int_{-\infty}^{+\infty} dp_2 \int_{-\infty}^{+\infty} dp_3$

So there is no overall minus sign from the coordinate change.

But are analyzing how a change in $\vec{p}$ affects $d^3p \neq d\vec{p}$

Yes, and I'm telling you, it doesn't affect it.

 

[MathematicalPhysicist]

In books we have sometimes the notations $\int d\textbf{p}$ and $\int d^3p$, both mean the same as @stevendaryl wrote.

 

[stevendaryl]

The integral $\int dp^3$ means $\int_{-\infty}^{+\infty} dp_1 \int_{-\infty}^{+\infty} dp_2 \int_{-\infty}^{+\infty} dp_3$

I should probably say that in differential geometry, there is a more sophisticated way to look at volume integrals which doesn't view them as three separate line integrals. It uses the language of differential forms, so you would say $\int f(p_1, p_2, p_3) dp_1 \wedge dp_2 \wedge dp_3$. I am not familiar with doing integrals this way, so I can't comment on it, but I assume that it's still true that the integration over all space will leave the sign unchanged.

 

[binbagsss]

I don't know what you mean by $d \vec{p}$. There is no integration over $\vec{p}$, there are three separate integrations over $dp_1 dp_2 dp_3$ (or $dp_x dp_y dp_z$, except that the original post uses $x$ and $y$ as position vectors).

Under the change $\vec{p} \rightarrow -\vec{p}$, we have:

$p_1 \rightarrow - p_1$
$p_2 \rightarrow - p_2$
$p_3 \rightarrow - p_3$

.

ok ta so why didn't somone simplly reply to me misinterpreting this in theop...