Skip to main content
Post Closed as "Duplicate" by AccidentalFourierTransform, CommunityBot
Tweeted twitter.com/StackPhysics/status/1286496552429719553
added 235 characters in body
Source Link
jcai
  • 654
  • 1
  • 5
  • 9

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not p = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not p$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not p = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not p - m_P) \frac{i}{\not p - m_R + \Sigma_R(\not p)} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not p)} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not p$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

Edit: It did occur to me that maybe $\not p = m_P$ is a shorthand for $m_P = \sqrt{p^2} = \sqrt{\not p^2}$. However, for there to be a pole at $\not p = m_P$ they would have to coincide fully. $m_P = \sqrt{p^2}$ seems insufficient.

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not p = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not p$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not p = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not p - m_P) \frac{i}{\not p - m_R + \Sigma_R(\not p)} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not p)} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not p$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not p = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not p$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not p = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not p - m_P) \frac{i}{\not p - m_R + \Sigma_R(\not p)} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not p)} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not p$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

Edit: It did occur to me that maybe $\not p = m_P$ is a shorthand for $m_P = \sqrt{p^2} = \sqrt{\not p^2}$. However, for there to be a pole at $\not p = m_P$ they would have to coincide fully. $m_P = \sqrt{p^2}$ seems insufficient.

deleted 8 characters in body; edited title
Source Link
G. Smith
  • 52.2k
  • 4
  • 85
  • 154

Spinor indices in $\not{p} p = m_P$ (mass renormalization)

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not{p} = m_P$$\not p = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not{p}$$\not p$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not{p} = m_P \mathbf{1}$$\not p = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not{p} - m_P) \frac{i}{\not{p} - m_R + \Sigma_R(\not{p})} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not{p})} \tag{18.41}$$$$i = \lim_{\not{p}\to m_P} (\not p - m_P) \frac{i}{\not p - m_R + \Sigma_R(\not p)} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not p)} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not{p}$$\not p$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

Spinor indices in $\not{p} = m_P$ (mass renormalization)

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not{p} = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not{p}$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not{p} = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not{p} - m_P) \frac{i}{\not{p} - m_R + \Sigma_R(\not{p})} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not{p})} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not{p}$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

Spinor indices in $\not p = m_P$ (mass renormalization)

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not p = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not p$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not p = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not p - m_P) \frac{i}{\not p - m_R + \Sigma_R(\not p)} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not p)} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not p$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?

Source Link
jcai
  • 654
  • 1
  • 5
  • 9

Spinor indices in $\not{p} = m_P$ (mass renormalization)

I'm reading Schwartz QFT, Chapter 18 (mass renormalization) and I'm confused about the equations about on-shell subtraction/pole mass. He writes:

The renormalized propagator should have a single pole at $\not{p} = m_P$ with residue $i$. The location of the pole is a definition of mass.

But $\not{p}$ has two spinor indices (i.e. a 4x4 matrix) while $m_P$ is just a number, so how does this make sense? I thought maybe it means $\not{p} = m_P \mathbf{1}$, but in the Weyl representation for instance, the $\gamma^\mu$ have zeros on the main diagonal so $p_\mu \gamma^\mu$ can't be proportional to $\mathbf{1}$ (unless it's all zeros).

He also goes on to write the equation,

$$i = \lim_{\not{p}\to m_P} (\not{p} - m_P) \frac{i}{\not{p} - m_R + \Sigma_R(\not{p})} = \lim_{\not{p} \to m_P} \frac{i}{1 + \frac{d}{d\not{p}} \Sigma_R(\not{p})} \tag{18.41}$$

I don't understand what $\lim_{\not{p}\to m_P}$ and $\frac{d}{d\not{p}}$ mean if $\not{p}$ is a 4x4 matrix. Is it a limit/derivative in the 16-dimensional space of the matrix entries?