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In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

Edit 3: Another reference that uses $1=\sum_n|n\rangle\langle n|$: arXiv link (page 5) and one that uses $1=\int\frac{d^3 p}{(2\pi)^3}|p\rangle\langle p|$: arxiv link (page 19).

Edit 4: Found a reference (warning, large PDF file size) that uses the off-shell completeness relation $1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$ and they don't run into the trouble that the energies cancel.

In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

Edit 3: Another reference that uses $1=\sum_n|n\rangle\langle n|$: arXiv link (page 5) and one that uses $1=\int\frac{d^3 p}{(2\pi)^3}|p\rangle\langle p|$: arxiv link (page 19).

Edit 4: Found a reference (warning, large PDF file size) that uses the completeness relation $1=\sum_n \int \frac{\text{d}^3 p}{(2\pi)^3} |\vec p \rangle\langle \vec p |$ in eq. (3.2) as well (Thanks to @ChiralAnomaly for pointing this out!).

In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

Edit 3: Another reference that uses $1=\sum_n|n\rangle\langle n|$: arXiv link (page 5) and one that uses $1=\int\frac{d^3 p}{(2\pi)^3}|p\rangle\langle p|$: arxiv link (page 19).

In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

Edit 3: Another reference that uses $1=\sum_n|n\rangle\langle n|$: arXiv link (page 5) and one that uses $1=\int\frac{d^3 p}{(2\pi)^3}|p\rangle\langle p|$: arxiv link (page 19).

Edit 4: Found a reference (warning, large PDF file size) that uses the off-shell completeness relation $1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$ and they don't run into the trouble that the energies cancel.

In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

In chapter 11-2-2, I&Z discuss Goldstone's theorem. They start by claiming that if an operator $A$ exists, such that $$ \delta a(t) \equiv \langle 0| [Q(t),A]|0\rangle \neq 0 \tag{11-30} $$ the symmetry is spontaneously broken. In Eq. (11-31), they insert a completeness relation as $1=\sum_n |n\rangle\langle n|$: $$\begin{align} \delta a(t) &= \sum_n \int d^3\vec x [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i P\cdot x}-c.c. ] \tag{11-31} \\ &= \sum_n (2\pi)^3 \delta^3(\vec P) [ \langle 0|j_0(0)|n\rangle\langle n|A|0\rangle e^{-i E t}-c.c. ] \end{align}$$ After taking the time derivative in Eq. (11-33), which brings down a $E(\vec p)$, they conclude that $\delta^3(\vec p) E(\vec p)$ must be zero, which leads to massless states.

However, if we were to use the following completeness relation, $$1=\sum_n \int \frac{\text{d}^3 \vec p}{(2\pi)^3 2E(\vec p)} |n,\vec p \rangle\langle n,\vec p, |$$ then the energy in the denominator and the energy coming from the time derivative of the exponential function would cancel and the conclusion from above wouldn't work anymore because there's no $E(\vec p)$ anymore!

Edit 1: I found that Ryder's QFT textbook (p. 292) as well as Nair's QFT textbook (p. 246) use the same completeness relation $1=\sum_n |n\rangle\langle n|$, i.e. the proof goes along the same lines as the one in I&Z. But why do they choose this completeness relation?

Edit 2: Maybe the answer is to take the following (off-shell) completeness relation: $$1=\sum_n \int \frac{\text{d}^4 p}{(2\pi)^4} |p \rangle\langle p |$$ since this would not give an $E(\vec p)$ in the denominator...?

Edit 3: Another reference that uses $1=\sum_n|n\rangle\langle n|$: arXiv link (page 5) and one that uses $1=\int\frac{d^3 p}{(2\pi)^3}|p\rangle\langle p|$: arxiv link (page 19).