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When deriving LSZ formulae, we assume asymptotic particles’ creation/annihilation operators as: $$a_\text{g,in/out}\ \ (\mathbf{p})\equiv \int d^3k \ g(\mathbf{k}) a_\text{in/out}(\mathbf{k}), \ \text{where}\ \ g(\mathbf{k}) =\exp\left(-\frac{(\mathbf{p} - \mathbf{k})^2}{2\sigma^{2}}\right).$$

This is because to get such normalized initial/final states as can define weak convergence of asymptotic creation/annihilation operators, and to ignore the interaction between different particles in initial/final states. However, after computing the LSZ, $g(\mathbf k)$ term is ignored by taking the limit of $\sigma \rightarrow 0$ & integrating about $\mathbf{k}$.

Here are some questions.

1. Why can we ignore $g(\mathbf{k})$ in LSZ? I think such the limit abandons spacial localization of particles at initial and final state and it makes particles interact even in the initial state.
2. Even if such a limit is physically correct, what is the difference between assuming the localized operator $a_\text{g,in/out}(\mathbf{k})$ or not. In other word, why should we introduce the $a_\text{g,in/out}(\mathbf{k})$ even though we make wave packets collapse anyway by taking the limit $\sigma \rightarrow 0$ at the end of derivation?
3. What is the difference between simply taking plane waves as an asymptotic state and to get plane waves by collapsing wave packets? When we take the limit $\sigma \rightarrow 0$, I think wave packets correspond to delta function, so it seems meaningless to define $a_\text{g,in/out}(\mathbf{k})$ and its spatial locality.

I have already read this post and this post, yet never understood clearly.

References

1. M. Srednicki, QFT ; chapter 5.

2. Peskin & Schroeder, QFT; sections 7.1-7.2

3. Wikipedia, LSZ reduction formula

When deriving LSZ formulae, we assume asymptotic particles’ creation/annihilation operators as: $$a_\text{g,in/out}\ \ (\mathbf{p})\equiv \int d^3k \ g(\mathbf{k}) a_\text{in/out}(\mathbf{k}), \ \text{where}\ \ g(\mathbf{k}) =\exp\left(-\frac{(\mathbf{p} - \mathbf{k})^2}{2\sigma^{2}}\right).$$

This is because to get such normalized initial/final states as can define weak convergence of asymptotic creation/annihilation operators, and to ignore the interaction between different particles in initial/final states. However, after computing the LSZ, $g(\mathbf k)$ term is ignored by taking the limit of $\sigma \rightarrow 0$ & integrating about $\mathbf{k}$.

Here are some questions.

1. Why can we ignore $g(\mathbf{k})$ in LSZ? I think such the limit abandons spacial localization of particles at initial and final state and it makes particles interact even in the initial state.
2. Even if such the limit is physically correct, what is the difference between assuming the localized operator $a_\text{g,in/out}(\mathbf{k})$ or not. In other word, why should we introduce the $a_\text{g,in/out}(\mathbf{k})$ even though we make wave packets collapse anyway by taking the limit $\sigma \rightarrow 0$ at the end of derivation?
3. What is the difference between simply taking plane waves as an asymptotic state and to get plane waves by collapsing wave packets? When we take the limit $\sigma \rightarrow 0$, I think wave packets correspond to delta function, so it seems meaningless to define $a_\text{g,in/out}(\mathbf{k})$ and its spatial locality.

I have already read this post and this post, yet never understood clearly.

References

1. M. Srednicki, QFT ; chapter 5.

2. Peskin & Schroeder, QFT; sections 7.1-7.2

3. Wikipedia, LSZ reduction formula

When deriving LSZ formulae, we assume asymptotic particles’ creation/annihilation operators as: $$a_\text{g,in/out}\ \ (\mathbf{p})\equiv \int d^3k \ g(\mathbf{k}) a_\text{in/out}(\mathbf{k}), \ \text{where}\ \ g(\mathbf{k}) =\exp\left(-\frac{(\mathbf{p} - \mathbf{k})^2}{2\sigma^{2}}\right).$$

This is because to get such normalized initial/final states as can define weak convergence of asymptotic creation/annihilation operators, and to ignore the interaction between different particles in initial/final states. However, after computing the LSZ, $g(\mathbf k)$ term is ignored by taking the limit of $\sigma \rightarrow 0$ & integrating about $\mathbf{k}$.

Here are some questions.

1. Why can we ignore $g(\mathbf{k})$ in LSZ? I think such the limit abandons spacial localization of particles at initial and final state and it makes particles interact even in the initial state.
2. Even if such a limit is physically correct, what is the difference between assuming the localized operator $a_\text{g,in/out}(\mathbf{k})$ or not. In other word, why should we introduce the $a_\text{g,in/out}(\mathbf{k})$ even though we make wave packets collapse anyway by taking the limit $\sigma \rightarrow 0$ at the end of derivation?
3. What is the difference between plane waves and the limit of wave packets? When we take the limit $\sigma \rightarrow 0$, I think wave packets correspond to plane waves, so it seems meaningless to define $a_\text{g,in/out}(\mathbf{k})$.

I have already read this post and this post, yet never understood clearly.

References

1. M. Srednicki, QFT ; chapter 5.

2. Peskin & Schroeder, QFT; sections 7.1-7.2

3. Wikipedia, LSZ reduction formula

When deriving LSZ formulae, we assume asymptotic particles’ creation/annihilation operators as: $$a_\text{g,in/out}\ \ (\mathbf{p})\equiv \int d^3k \ g(\mathbf{k}) a_\text{in/out}(\mathbf{k}), \ \text{where}\ \ g(\mathbf{k}) =\exp\left(-\frac{(\mathbf{p} - \mathbf{k})^2}{2\sigma^{2}}\right).$$

This is because to get such normalized initial/final states as can define weak convergence of asymptotic creation/annihilation operators, and to ignore the interaction between different particles in initial/final states. However, after computing the LSZ, $g(\mathbf k)$ term is ignored by taking the limit of $\sigma \rightarrow 0$ & integrating about $\mathbf{k}$.

Here are some questions.

1. Why can we ignore $g(\mathbf{k})$ in LSZ? I think such the limit abandons spacial localization of particles at initial and final state and it makes particles interact even in the initial state.
2. Even if such a limit is physically correct, what is the difference between assuming the localized operator $a_\text{g,in/out}(\mathbf{k})$ or not. In other word, why should we introduce the $a_\text{g,in/out}(\mathbf{k})$ even though we make wave packets collapse anyway by taking the limit $\sigma \rightarrow 0$ at the end of derivation?
3. What is the difference between simply taking plane waves as an asymptotic state and to get plane waves by collapsing wave packets? When we take the limit $\sigma \rightarrow 0$, I think wave packets correspond to delta function, so it seems meaningless to define $a_\text{g,in/out}(\mathbf{k})$ and its spatial locality.

I have already read this post and this post, yet never understood clearly.

References

1. M. Srednicki, QFT ; chapter 5.

2. Peskin & Schroeder, QFT; sections 7.1-7.2

3. Wikipedia, LSZ reduction formula

When deriving LSZ formulae, we assume asymptotic particles’ creation/annihilation operators as: $$a_\text{g,in/out}\ \ (\mathbf{p})\equiv \int d^3k \ g(\mathbf{k}) a_\text{in/out}(\mathbf{k}), \ \text{where}\ \ g(\mathbf{k}) =\exp\left(\frac{(\mathbf{p} - \mathbf{k})^2}{2\sigma^{2}}\right).$$

This is because to get such normalized initial/final states as can define weak convergence of asymptotic creation/annihilation operators, and to ignore the interaction between different particles in initial/final states. However, after computing the LSZ, the term of $g(\mathbf k)$ is ignored by taking the limit of $\sigma \rightarrow 0$ & integrating about $\mathbf{k}$.

Here are some questions.

1. Why can we ignore $g(\mathbf{k})$ in LSZ? I think such the limit abandons spacial localization of particles at initial and final state and it makes particles interact even in the initial state.
2. Even if such a limit is physically correct, what is the difference between assuming the localized operator $a_\text{g,in/out}(\mathbf{k})$ or not. In other word, why should we introduce the $a_\text{g,in/out}(\mathbf{k})$ even though we make wave packets collapse anyway by taking the limit $\sigma \rightarrow 0$ at the end of derivation?
3. What is the difference between plane waves and the limit of wave packets? When we take the limit $\sigma \rightarrow 0$, I think wave packets correspond to plane waves, so it seems meaningless to define $a_\text{g,in/out}(\mathbf{k})$.

I have already read this post and this post, yet never understood clearly.

References

1. M. Srednicki, QFT ; chapter 5.

2. Peskin & Schroeder, QFT; sections 7.1-7.2

3. Wikipedia, LSZ reduction formula

When deriving LSZ formulae, we assume asymptotic particles’ creation/annihilation operators as: $$a_\text{g,in/out}\ \ (\mathbf{p})\equiv \int d^3k \ g(\mathbf{k}) a_\text{in/out}(\mathbf{k}), \ \text{where}\ \ g(\mathbf{k}) =\exp\left(-\frac{(\mathbf{p} - \mathbf{k})^2}{2\sigma^{2}}\right).$$

This is because to get such normalized initial/final states as can define weak convergence of asymptotic creation/annihilation operators, and to ignore the interaction between different particles in initial/final states. However, after computing the LSZ, $g(\mathbf k)$ term is ignored by taking the limit of $\sigma \rightarrow 0$ & integrating about $\mathbf{k}$.

Here are some questions.

1. Why can we ignore $g(\mathbf{k})$ in LSZ? I think such the limit abandons spacial localization of particles at initial and final state and it makes particles interact even in the initial state.
2. Even if such a limit is physically correct, what is the difference between assuming the localized operator $a_\text{g,in/out}(\mathbf{k})$ or not. In other word, why should we introduce the $a_\text{g,in/out}(\mathbf{k})$ even though we make wave packets collapse anyway by taking the limit $\sigma \rightarrow 0$ at the end of derivation?
3. What is the difference between plane waves and the limit of wave packets? When we take the limit $\sigma \rightarrow 0$, I think wave packets correspond to plane waves, so it seems meaningless to define $a_\text{g,in/out}(\mathbf{k})$.

I have already read this post and this post, yet never understood clearly.

References

1. M. Srednicki, QFT ; chapter 5.

2. Peskin & Schroeder, QFT; sections 7.1-7.2

3. Wikipedia, LSZ reduction formula