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jkb1603
jkb1603
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I have some confusion about the vacuum expectation value of scalar field in QFT. I know that the one-point function $$ \langle \Omega | \phi(x) | \Omega \rangle = \langle \Omega | \phi(0) | \Omega \rangle =: v $$ is a constant, which follows immediately by translation properties of $\phi$, that is $$ \phi(x) = e^{i P x} \phi(0) e^{-iPx}. $$ Q1: If one would like to determine $v$, should one just calculate the one-point function (that means the tadpole diagrams) in perturbation theory?

I recall hearing that it can somehow be calculated exactly from the path integral. I tried the following $$ \langle \Omega | \phi(x) | \Omega \rangle = \frac{\int \mathcal D\phi~e^{iS[\phi]} \phi(x)}{\int \mathcal D\phi~e^{iS[\phi]} }. $$ Substituting $\phi \rightarrow \phi + \phi_{\text{cl}}$ in the functional integral, where $\phi_{\text{cl}}$ is the classical solution gives $$ \langle \Omega | \phi(x) | \Omega \rangle = \frac{\int \mathcal D\phi~e^{iS[\phi] + iS[\phi_{\text{cl}} ]} (\phi(x) + \phi_{\text{cl}}(x))}{\int \mathcal D\phi~e^{iS[\phi]} } = e^{iS[\phi_{\text{cl}}]} (\langle \Omega | \phi(x) | \Omega \rangle + \phi_{\text{cl}}(x)) $$ and therefore $$ (e^{-iS[\phi_{\text{cl}}]} - 1) v = \phi_{\text{cl}}(x). $$ Now this equation can only be true of $\phi_{\text{cl}}$ Or is a constant and by the boundary conditions, that the solution must vanish at infinity, it follows thatthere an easier $\phi_{\text{cl}} = 0$.

Q2: So does this means that any solution of the classical equations of motion is identical zero? This seems incorrect(or exact) way to me. Is there some error in my approachget it?

Q3Q2: Also I have heard that one can set the one-point function to zero by making a field redefinition $\phi \rightarrow \phi - v$ (see e.g. Why can the $1$-point correlation function be made to vanish?) and therefore one can safely consider it zero while happily continuing with the same theory. I see that it sets $v \rightarrow 0$, but it also massively changes the theory, since is may generate additional interaction terms in the Lagrangian? I need some clarification.

I have some confusion about the vacuum expectation value of scalar field in QFT. I know that the one-point function $$ \langle \Omega | \phi(x) | \Omega \rangle = \langle \Omega | \phi(0) | \Omega \rangle =: v $$ is a constant, which follows immediately by translation properties of $\phi$, that is $$ \phi(x) = e^{i P x} \phi(0) e^{-iPx}. $$ Q1: If one would like to determine $v$, should one just calculate the one-point function (that means the tadpole diagrams) in perturbation theory?

I recall hearing that it can somehow be calculated exactly from the path integral. I tried the following $$ \langle \Omega | \phi(x) | \Omega \rangle = \frac{\int \mathcal D\phi~e^{iS[\phi]} \phi(x)}{\int \mathcal D\phi~e^{iS[\phi]} }. $$ Substituting $\phi \rightarrow \phi + \phi_{\text{cl}}$ in the functional integral, where $\phi_{\text{cl}}$ is the classical solution gives $$ \langle \Omega | \phi(x) | \Omega \rangle = \frac{\int \mathcal D\phi~e^{iS[\phi] + iS[\phi_{\text{cl}} ]} (\phi(x) + \phi_{\text{cl}}(x))}{\int \mathcal D\phi~e^{iS[\phi]} } = e^{iS[\phi_{\text{cl}}]} (\langle \Omega | \phi(x) | \Omega \rangle + \phi_{\text{cl}}(x)) $$ and therefore $$ (e^{-iS[\phi_{\text{cl}}]} - 1) v = \phi_{\text{cl}}(x). $$ Now this equation can only be true of $\phi_{\text{cl}}$ is a constant and by the boundary conditions, that the solution must vanish at infinity, it follows that $\phi_{\text{cl}} = 0$.

Q2: So does this means that any solution of the classical equations of motion is identical zero? This seems incorrect to me. Is there some error in my approach?

Q3: Also I have heard that one can set the one-point function to zero by making a field redefinition $\phi \rightarrow \phi - v$ (see e.g. Why can the $1$-point correlation function be made to vanish?) and therefore one can safely consider it zero while happily continuing with the same theory. I see that it sets $v \rightarrow 0$, but it also massively changes the theory, since is may generate additional interaction terms in the Lagrangian? I need some clarification.

I have some confusion about the vacuum expectation value of scalar field in QFT. I know that the one-point function $$ \langle \Omega | \phi(x) | \Omega \rangle = \langle \Omega | \phi(0) | \Omega \rangle =: v $$ is a constant, which follows immediately by translation properties of $\phi$, that is $$ \phi(x) = e^{i P x} \phi(0) e^{-iPx}. $$ Q1: If one would like to determine $v$, should one just calculate the one-point function (that means the tadpole diagrams) in perturbation theory? Or is there an easier (or exact) way to get it?

Q2: Also I have heard that one can set the one-point function to zero by making a field redefinition $\phi \rightarrow \phi - v$ (see e.g. Why can the $1$-point correlation function be made to vanish?) and therefore one can safely consider it zero while happily continuing with the same theory. I see that it sets $v \rightarrow 0$, but it also massively changes the theory, since is may generate additional interaction terms in the Lagrangian? I need some clarification.

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jkb1603
jkb1603
  • 1.2k
  • 6
  • 15

Vacuum expectation value of scalar field in QFT

I have some confusion about the vacuum expectation value of scalar field in QFT. I know that the one-point function $$ \langle \Omega | \phi(x) | \Omega \rangle = \langle \Omega | \phi(0) | \Omega \rangle =: v $$ is a constant, which follows immediately by translation properties of $\phi$, that is $$ \phi(x) = e^{i P x} \phi(0) e^{-iPx}. $$ Q1: If one would like to determine $v$, should one just calculate the one-point function (that means the tadpole diagrams) in perturbation theory?

I recall hearing that it can somehow be calculated exactly from the path integral. I tried the following $$ \langle \Omega | \phi(x) | \Omega \rangle = \frac{\int \mathcal D\phi~e^{iS[\phi]} \phi(x)}{\int \mathcal D\phi~e^{iS[\phi]} }. $$ Substituting $\phi \rightarrow \phi + \phi_{\text{cl}}$ in the functional integral, where $\phi_{\text{cl}}$ is the classical solution gives $$ \langle \Omega | \phi(x) | \Omega \rangle = \frac{\int \mathcal D\phi~e^{iS[\phi] + iS[\phi_{\text{cl}} ]} (\phi(x) + \phi_{\text{cl}}(x))}{\int \mathcal D\phi~e^{iS[\phi]} } = e^{iS[\phi_{\text{cl}}]} (\langle \Omega | \phi(x) | \Omega \rangle + \phi_{\text{cl}}(x)) $$ and therefore $$ (e^{-iS[\phi_{\text{cl}}]} - 1) v = \phi_{\text{cl}}(x). $$ Now this equation can only be true of $\phi_{\text{cl}}$ is a constant and by the boundary conditions, that the solution must vanish at infinity, it follows that $\phi_{\text{cl}} = 0$.

Q2: So does this means that any solution of the classical equations of motion is identical zero? This seems incorrect to me. Is there some error in my approach?

Q3: Also I have heard that one can set the one-point function to zero by making a field redefinition $\phi \rightarrow \phi - v$ (see e.g. Why can the $1$-point correlation function be made to vanish?) and therefore one can safely consider it zero while happily continuing with the same theory. I see that it sets $v \rightarrow 0$, but it also massively changes the theory, since is may generate additional interaction terms in the Lagrangian? I need some clarification.