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Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

Now suppose $\phi(x)$ is a scalar multiplet, so that we have a column of $\phi_i(x)$ transforming into one another by the global symmetry. If instead of (2) have $$\langle 0|\phi(x)|0\rangle=\alpha\tag{3}$$

where $\alpha$ is some non-zero column vector then the only way (1) and (3) can hold at the same time is if $a_i(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=\alpha+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value".

Because of this we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.

Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

Now suppose $\phi(x)$ is a scalar multiplet, so that we have a column of $\phi_i(x)$ transforming into one another by the global symmetry. Suppose instead of (2) have $$\langle 0|\phi(x)|0\rangle=\alpha\tag{3}$$

where $\alpha$ is some non-zero constant column vector. Since $\alpha$ is constant, once we approach the asymptotic region and try to relate $\phi(x)$ to an in/out field like (1) as usual, the only way (1) and (3) can hold at the same time is if $a_i(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=\alpha+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value".

Because of this we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.

Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

Now suppose $\phi(x)$ is a scalar multiplet, so that we have a column of $\phi_i(x)$ transforming into one another by the global symmetry. If instead of (2) have $$\langle 0|\phi(x)|0\rangle=\alpha\tag{3}$$

where $\alpha$ is some non-zero column vector then the only way (1) and (3) can hold at the same time is if $a_i(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=\alpha+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value" and therefore we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.

Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

Now suppose $\phi(x)$ is a scalar multiplet, so that we have a column of $\phi_i(x)$ transforming into one another by the global symmetry. If instead of (2) have $$\langle 0|\phi(x)|0\rangle=\alpha\tag{3}$$

where $\alpha$ is some non-zero column vector then the only way (1) and (3) can hold at the same time is if $a_i(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=\alpha+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value".

Because of this we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.

Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

So if we instead have $$\langle 0|\phi(x)|0\rangle=a\tag{3}$$

then the only way (1) and (3) can hold at the same time is if $a(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=a+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value" and therefore we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.

Let me share the way I currently understand this (which may be flawed since I'm also studying it right now - corrections are welcome!). IMHO this is all about how we quantize a field and relate it to particles. Recall that if $\phi(x)$ is a free scalar field we quantize it by expanding it into creation and annihilation operators $$\phi(x)=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{2\omega_p}\bigg(a(\mathbf{p})e^{ipx}+a^\dagger(\mathbf{p})e^{-ipx}\bigg)\tag{1}.$$

If the field is not free this still holds asymptotically with in/out fields and this reveals the spectrum of particles in the in/out states of scattering experiments.

Now take this and evaluate the vacuum expectation value. As you will immediately recognize because $a(\mathbf{p})$ annihilates the vacuum we have: $$\langle 0|\phi(x)|0\rangle=0\tag{2}.$$

Now suppose $\phi(x)$ is a scalar multiplet, so that we have a column of $\phi_i(x)$ transforming into one another by the global symmetry. If instead of (2) have $$\langle 0|\phi(x)|0\rangle=\alpha\tag{3}$$

where $\alpha$ is some non-zero column vector then the only way (1) and (3) can hold at the same time is if $a_i(\mathbf{p})|0\rangle\neq 0$, because this being zero already implies (2).

The Physics behind this is that as you can see from (1) in a sense we are viewing the field as a collection of harmonic oscilators. We are quantizing excitations about a ground state.

Then what we do is to realize that if $\phi(x)$ is our dynamical variable writing it as $$\phi(x)=\alpha+\chi(x)\tag{4}$$

we loose no information at all. We are just shifting to a more convenient variable. In fact $\langle 0|\chi(x)|0\rangle=0$ by definition and $\chi(x)$ can be quantized as we often do, according to (1), without problem.

As Goldstone, Salam and Weinberg put in their 1962 paper ("Broken Symmetries"): "It is inconvenient to work with fields with nonzero vacuum expectation value" and therefore we perform that shift working with an equivalent, but more convenient variable, whose quantization can be understood as "quantizing excitations about a ground state". In particular observe that once this shift is done and the Lagrangian is expressed in terms of the more convenient fields with zero vacuum expectation value one readily identifies the implications of SSB upon the particle content of the theory: the Goldstone bosons.