Path Integral of a Spontaneously Broken Theory

[JRudolfo]

A scalar field theory with potential $$V(\phi)=-\mu^2\phi^2+\lambda \phi^4$$ is spontaneously broken and as a consequence, for the ground state, $$\langle \phi(x) \rangle \neq 0$$.

However, the path integral, which should give ground state expectation values, looks to be zero by oddness of the integrand:

$$\langle \phi(x) \rangle=\int d\phi \, e^{i \int d^4x\, \partial^\mu \phi \partial_\mu \phi+\mu^2\phi^2-\lambda \phi^4}\phi(x)=0$$

How can this be reconciled with the fact that $$\langle \phi(x) \rangle \neq 0$$?

 

[PeterDonis]

oddness of the integrand

Where are you getting this integrand from? In particular, why are you including $\phi(x)$ in the integrand?

 

[PeterDonis]

why are you including ϕ(x)ϕ(x)\phi(x) in the integrand?

Or perhaps a better way of asking this is, why are you assuming that $\phi(x)$ is an odd function?

 

[JRudolfo]

Where are you getting this integrand from? In particular, why are you including $\phi(x)$ in the integrand?

Hi, I think it's simpler to think of a single random variable X with probabality distribution P(x), so if you want the expectation value of X, it's given by:

$$\langle X \rangle =\int \,dx P(x) x$$

To make the analogy with quantum mechanics, $X$ would be $\phi(x)$, and $P(x)$ would be the exponential. So that's why there is a $\phi(x)$in the integrand: if you want the expected value of quantity, you multiply that quantity by the probability distribution and integrate. So $\phi(x)$ is being used in two different contexts: on the LHS it's a random variable, and on the RHS it's a number. Sometimes people put a hat over $\hat{\phi}(x)$ to denote the random variable.

I suspect the path integral is not giving the ground state expectation value (the true ground state), but rather the the state that is annihilated by $\phi(x)$. But in formal derivations of the path integral, I think it always gives the true ground state expected value.

 

[PeterDonis]

To make the analogy with quantum mechanics, $X$ would be $\phi(x)$, and $P(x)$ would be the exponential.

No, that's not correct. A quantum field $\phi(x)$ is not a random variable; it's an operator. The analogy between the probability distribution $P(x)$ and the exponential is OK if you consider the probability distribution to be a kind of weighting factor (see below); but even then the analogy is limited, since the exponential in the quantum case is a complex phase factor, not a real number.

that's why there is a $\phi(x)$ in the integrand: if you want the expected value of quantity, you multiply that quantity by the probability distribution and integrate.

Ok here.

So $\phi(x)$ is being used in two different contexts: on the LHS it's a random variable, and on the RHS it's a number.

No. As above, $\phi(x)$ is an operator. But for this particular purpose, you can treat it as a simple scalar function of $x$; the expectation value is then just a weighted average value of the function.

The question is, what function of $x$ is $\phi(x)$? The answer is, the function that describes the configuration of the field in spacetime when it's in its vacuum state. Heuristically, this will be the state of lowest possible energy, which will be a state with zero kinetic energy and a minimum of the potential energy. In other words, $\phi(x)$ will be a function that is sharply peaked about a point $x$ which is a minimum of $V(x)$. If you plug such a function into the integral you wrote down, you will get a nonzero value.

(Btw, to do inline LaTeX, use ## as delimiters, not $.)

 

[PeterDonis]

I suspect the path integral is not giving the ground state expectation value (the true ground state), but rather the the state that is annihilated by $\phi(x)$.

As I said in my previous post, $\phi$ is an operator, but it's not an annihilation operator; it doesn't annihilate the vacuum state.

 

[DrDu]

A scalar field theory with potential $$V(\phi)=-\mu^2\phi^2+\lambda \phi^4$$ is spontaneously broken and as a consequence, for the ground state, $$\langle \phi(x) \rangle \neq 0$$.

The condition that $$\langle \phi(x) \rangle \neq 0$$ is sufficient for a symmetry to be broken, but not necessary.
What is sufficient and necessary is that
$\lim_{x-y \to \infty} \langle \phi(x)\phi(y) \rangle =\Phi(x)\Phi(y)$, where $\Phi(x)$ is an ordinary function, not an operator. For more complicated fields you eventually have to throw in some indices, adjoint symbols etc, so this is rather symbolical. This is called off diagonal long range order (ODLRO).

 

[vanhees71]

The OP was correct in a naive way. Indeed you can calculate expectation values of time-ordered products by path integrals, i.e.,
$$\langle \mathcal{T}_c \phi(x_1) \phi(x_2) \cdots \phi (x_N) \rangle=\int \mathrm{D} \phi \exp (\mathrm{i} S[\phi]) \phi(x_1) \phi(x_1) \phi(x_2) \cdots \phi (x_N).$$
For $N=1$ you get the formula in the OP.

However, the path integral is not well-defined, and to evaluate it you must make some approximations. The usual way is the saddle-point expansion, and for this you have to expand around a (local) minimum of the "free" action, which in this case is around $\phi \neq 0$.

 

[Demystifier]

A scalar field theory with potential $$V(\phi)=-\mu^2\phi^2+\lambda \phi^4$$ is spontaneously broken and as a consequence, for the ground state, $$\langle \phi(x) \rangle \neq 0$$.

However, the path integral, which should give ground state expectation values, looks to be zero by oddness of the integrand:

$$\langle \phi(x) \rangle=\int d\phi \, e^{i \int d^4x\, \partial^\mu \phi \partial_\mu \phi+\mu^2\phi^2-\lambda \phi^4}\phi(x)=0$$

How can this be reconciled with the fact that $$\langle \phi(x) \rangle \neq 0$$?

In this path integral one does not integrate over all field configurations, but only over those that satisfy some "boundary condition". So to define the path integral, one must define the boundary condition, namely the value of $\phi$ at $t\rightarrow\pm\infty$. The choice of the boundary condition corresponds to the choice of the state one calls "vacuum". If you choose $\phi=0$ at the boundary, this corresponds to the "vacuum" in which $\langle\phi\rangle=0$. When symmetry is spontaneously broken, this is not the "vacuum" you are actually interested in. You are interested in another vacuum in which $\langle\phi\rangle=v$ is the minimum of the potential, so you must choose the appropriate boundary condition. A convenient way to do this is to write
$$\phi=v+\varphi$$
where $\varphi$ is a new integration variable which satisfies the "standard" boundary condition $\varphi=0$ at $t\rightarrow\pm\infty$.

 

[Demystifier]

The problem is also related to the following elementary problem in real analysis. Suppose one wants to calculate the integral
$$I=\frac{1}{Z}\int_{-\infty}^{\infty}dx\, x$$
where
$$Z=\int_{-\infty}^{\infty}dx$$
By symmetry, one can argue that
$$I=0$$
However, one can introduce a new integration variable $x'$ defined by
$$x=v+x'$$
where $v$ is an arbitrary constant. Thus one can argue that
$$I=v+\frac{1}{Z}\int_{-\infty}^{\infty}dx'\, x'=v$$
This demonstrates that $I$ can take any value $v$, i.e. that $I$ is not really well-defined mathematically. Hence, to determine what $I$ "really" is, one must exploit some understanding of the context of what $I$ is supposed to describe.

In general, mathematicians are better than physicists in recognizing that $I$ is meaningless, while physicists are better than mathematicians in understanding the context which gives meaning to such meaningless expressions.

 

[vanhees71]

Well, and good physicists are better in using the right regularization of the ill-defined integrals/series to get what they really want ;-).

 

[Demystifier]

Well, and good physicists are better in using the right regularization of the ill-defined integrals/series to get what they really want ;-).

Yes, and the best physicists know what they want before doing any calculation at all.

 

[JRudolfo]

The OP was correct in a naive way. Indeed you can calculate expectation values of time-ordered products by path integrals, i.e.,
$$\langle \mathcal{T}_c \phi(x_1) \phi(x_2) \cdots \phi (x_N) \rangle=\int \mathrm{D} \phi \exp (\mathrm{i} S[\phi]) \phi(x_1) \phi(x_1) \phi(x_2) \cdots \phi (x_N).$$
For $N=1$ you get the formula in the OP.

However, the path integral is not well-defined, and to evaluate it you must make some approximations. The usual way is the saddle-point expansion, and for this you have to expand around a (local) minimum of the "free" action, which in this case is around $\phi \neq 0$.

So would you say this is correct to 2nd order:

$$\int d\phi \, e^{iS[\phi]} \phi=\int d\phi \, e^{iS[\phi_c]+i/2 S^{(2)}[\phi_c](\phi-\phi_c)^2} \phi=\int d\phi\, e^{iS[\phi_c]+i/2 S^{(2)}[\phi_c](\phi)^2}(\phi_c+\phi)=\phi_c$$

where we used oddness of integrand and translational shift invariance in the last step, which to Demystifier's point, the latter is not always allowed? Isn't it odd though that the approximate answer (saddle point) gives the correct answer, but the more exact approach (straight up do the integral) indicates zero? I realize that the original integral can be regarded as saddle point expansion about $\phi=0$ taken up to fourth order, but presumably, if you were to Taylor expand the action about $\phi=\sqrt{\frac{\mu^2}{2 \lambda}}$ and keep all orders in the Taylor expansion, you would get zero for $\langle \phi(x) \rangle$? It seems natural to trust the fourth order expansion of the action about $\phi=0$ more than the 2nd order expansion about $\phi=\sqrt{\frac{\mu^2}{2 \lambda}}$, because the fourth order expansion about $\phi=0$ is exact.

In this path integral one does not integrate over all field configurations, but only over those that satisfy some "boundary condition". So to define the path integral, one must define the boundary condition, namely the value of $\phi$ at $t\rightarrow\pm\infty$. The choice of the boundary condition corresponds to the choice of the state one calls "vacuum". If you choose $\phi=0$ at the boundary, this corresponds to the "vacuum" in which $\langle\phi\rangle=0$. When symmetry is spontaneously broken, this is not the "vacuum" you are actually interested in. You are interested in another vacuum in which $\langle\phi\rangle=v$ is the minimum of the potential, so you must choose the appropriate boundary condition. A convenient way to do this is to write
$$\phi=v+\varphi$$
where $\varphi$ is a new integration variable which satisfies the "standard" boundary condition $\varphi=0$ at $t\rightarrow\pm\infty$.

I thought the the selection of the true ground state automatically followed from either the $i \epsilon$ prescription (e.g. adding $+i\epsilon \phi^2(x)$ to the potential) or a Wick rotation. That is, if I have a good enough computer, then for any scalar field theory in Minkowski space, if I add $i\epsilon \phi^2(x)$ to the potential and do the path integral with no restrictions on the integration range of $\phi(x)$, then only the ground state survives the path integration when $\epsilon\rightarrow 0$. So in addition to the $i \epsilon$ prescription, the fields must go to their global classical minimum at infinity?

What about the true global minimum (I believe, but I'm not sure, that in the true ground state, the expectation value of the field operator is the minimum of the effective action) - shouldn't we be expanding about the true minimum instead of the classical minimum?

Also, you mentioned that the vacuum in which $\langle\phi\rangle=0$ is not the one I'm interested in. What's interesting is that in Bose-Einstein condensation, the *vacuum* $|0\rangle$ is the state for which $\langle\phi\rangle=0$, which is not the true *ground* state $|\Omega\rangle$ which has particles in it (the condensate, i.e., $a^\dagger a |\Omega\rangle=n|\Omega\rangle$ whereas $a^\dagger a|0\rangle=0$), which is a coherent state $|\Omega \rangle=e^{i \phi_c a^\dagger}|0\rangle$. I'm not as familiar with particle physics, but is there something similar with the Higgs field in that the ground state is filled with particles, or is the ground state empty and only excitations above the ground state are Higgs particles?

Thanks everyone.

 

[Demystifier]

I thought the the selection of the true ground state automatically followed from either the $i \epsilon$ prescription (e.g. adding $+i\epsilon \phi^2(x)$ to the potential) or a Wick rotation. That is, if I have a good enough computer, then for any scalar field theory in Minkowski space, if I add $i\epsilon \phi^2(x)$ to the potential and do the path integral with no restrictions on the integration range of $\phi(x)$, then only the ground state survives the path integration when $\epsilon\rightarrow 0$. So in addition to the $i \epsilon$ prescription, the fields must go to their global classical minimum at infinity?

Is there any rational why one would add $+i\epsilon \phi^2$ and not $+i\epsilon \varphi^2$ (with $\varphi=\phi-v$)?

What about the true global minimum (I believe, but I'm not sure, that in the true ground state, the expectation value of the field operator is the minimum of the effective action) - shouldn't we be expanding about the true minimum instead of the classical minimum?

In principle, one should be allowed to expand around any point one wishes. However I would expect that, for low excited states, expansion around the true minimum has faster convergence than around the classical minimum.

Also, you mentioned that the vacuum in which $\langle\phi\rangle=0$ is not the one I'm interested in. What's interesting is that in Bose-Einstein condensation, the *vacuum* $|0\rangle$ is the state for which $\langle\phi\rangle=0$, which is not the true *ground* state $|\Omega\rangle$ which has particles in it (the condensate, i.e., $a^\dagger a |\Omega\rangle=n|\Omega\rangle$ whereas $a^\dagger a|0\rangle=0$), which is a coherent state $|\Omega \rangle=e^{i \phi_c a^\dagger}|0\rangle$. I'm not as familiar with particle physics, but is there something similar with the Higgs field in that the ground state is filled with particles, or is the ground state empty and only excitations above the ground state are Higgs particles?

In principle, particle physics is not much different from condensed matter physics in this regard. In a sense, the usual Higgs vacuum in which $\langle\phi\rangle=v$ is a Bose-Einstein condensate. The massive Higgs particle discovered at LHC is only an excitation around this low-temperature condensate. At high temperature one has a restoration of symmetry, i.e. the ground state of Higgs field at high temperature has $\langle\phi\rangle=0$.

The only real difference between condensed matter physics and particle physics is that typical energy scales are different. This means that in condensed matter the default state is a high-temperature state and we must work hard to prepare a low-temperature state, while in particle physics it's the exact opposite.

 

[vanhees71]

Expansion around an arbitrary point is not a good idea since it makes the calculation much more complicated. Also note that the idea behind perturbation theory (in the formal power series in $\hbar$) is a saddle-point evaluation of the path integral, which as the name suggests, should be made around a saddle point. For more details (also concerning the counting of $\hbar$ powers), see Sect. 4.6.6, page 118 in

https://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf

 

[JRudolfo]

Thanks everyone. Just a quick note - it seems in particle physics you use the counter-term formalism, so that the Lagrangian splits into a renormalized part plus a counter-term part. The renormalized parameters are chosen so that the classical solution to the renormalized Lagrangian is the quantum solution, i.e., $\phi_c=\sqrt{\frac{\mu_R^2}{2\lambda_R}} =\langle \phi \rangle$ to all orders, since the parameters are renormalized so that the classical minimum is the quantum minimum (i.e. counter-terms are set to impose the condition that the two are equal).