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I was going through Chapter 9 of Schwartz's QFT book and one of the results bothers me.

Suppose we have a complex scalar field theory, and we want to find the propagator associated to the complex scalar field $\phi$. In principle, there are two ways of going about finding propagators, we could take the time-ordered product and manually calculate it, like the book did previously for a real scalar field. In which case I understand why the propagator must be $\langle 0|T\{\phi^*(x)\phi(y)\}|0\rangle$, and why this coincides with the real scalar field propagator, which, in momentum space, is given by

$$ \frac{i}{p^2-m^2+i\epsilon}. $$ This propagates both $\phi$ and $\phi^*$, you cannot disentangle them.

Alternatively, we could do what the book does for massless and massive spin 1 particles, which is taking the momentum space equations of motion, inverting them, and manually adding $+i\epsilon$ to make it time ordered. This is where the issue lies, I understand that the equations of motion for both $\phi$ and $\phi^*$ are Klein Gordon equations, justifying the fact that the propagator must be $\sim1/(p^2-m^2+i\epsilon)$. My issue is that, through this method, I cannot see why they must be entangled, since we get independent equations for both $\phi$ and $\phi^*$.

EDIT: I think I understand now, the fact that the propagator obtained from each equation of motion is the same, means that at the level of free theories, you can't distinguish one from the other (clearly once interactions are turned on they couple with $A_\mu$ with opposite charges). Which in turn means you can't separate propagating one from propagating the other. If this is correct, then a follow-up would be what if I have multiple complex scalar fields? Then they'd be distinguished by their mass in the propagator, correct?

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    $\begingroup$ What do you mean exactly by "they must be entangled"? $\endgroup$
    – Gold
    Commented Feb 16, 2023 at 22:16
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    $\begingroup$ Not in the actual formal sense of entanglement, but in the sense that in the first method, because we're calculating $ \langle 0| T\{\phi^*(x)\phi(x)\}|0\rangle $, we clearly see we cannot separate one from the other, whereas in the EoM method, this isnt clear. $\endgroup$
    – FranDahab
    Commented Feb 16, 2023 at 22:37

2 Answers 2

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Let's see what is meant by "inverting the EOMs". Time-ordered products can be written as expectation values weighted by $e^{iS}$ in path integrals:

$$ \begin{aligned} \left<0|T\{\phi^*(x)\phi(y)\}|0\right> &= \int D\phi^*D\phi\; \phi^*(x)\phi(y) e^{iS}\\ &= \frac{\delta}{i\delta J(x)}\frac{\delta}{i\delta J^*(y)} \int D\phi^* D\phi\; e^{i\int d^dx (\mathcal{L} + J\phi^* + J^*\phi)} \Bigg\vert_{J=0, J^*=0}\\ &\equiv \frac{\delta}{i\delta J(x)}\frac{\delta}{i\delta J^*(y)} Z[J^*, J]\Bigg\vert_{J=0, J^*=0} \end{aligned} $$

$Z$ is just a Gaussian integral and is easily evaluated: $$ \begin{aligned} Z[J,J^*] &= \int D\phi^*D\phi\;\exp\left[i\int d^dx [-\phi^*(x)(\partial^2+m^2)\phi(x) + J(x)\phi^*(x) + J^*(x)\phi(x)]\right]\\ &= \exp \left[i\int d^dx d^dy J^*(x)\Delta(x-y)J(y)\right] \end{aligned} $$ So you can immediately see that to get something that does not vanish, you have to take $\frac{\delta}{i\delta J}\frac{\delta}{i\delta J^*}$, not $\frac{\delta}{i\delta J}\frac{\delta}{i\delta J}$ (which would correspond to $\left<0|T\{\phi^*\phi^*\}|0\right>$) or its hermitian conjugate.

The Feynman propagator $\Delta(x-y)$, which is the inverse of $\partial^2+m^2$ shows up when you complete the square in the exponent of the integrand, just as $$ \int dz^*dz e^{-az^*z + b^*z + bz^*} = \left(\int dz^*dz e^{-a(z-\frac{b}{a})^*(z-\frac{b}{a})}\right)e^{b^*a^{-1}b}\propto e^{b^*a^{-1}b}. $$

The fact that $J^*$ and $J$ appear together in the exponent can be traced back to the Lagrangian density $\mathcal{L} = \partial^\mu\phi^*\partial_\mu\phi - m^2\phi^*\phi$, each term of which has both $\phi$ and $\phi^*$. You could ask why we don't write $\mathcal{L} = \frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2 + h.c.$, which is hermitian and also gives us the Klein-Gordon equation, and would result in a $Z$ in which $J$ and $J^*$ appear separately. However, this would yield a Hamiltonian that is not bounded from below.

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  • $\begingroup$ Unfortunately Schwartz's book takes a while (chapter 14 I believe) to introduce the QFT path integral formalism, so I'm not able to fully understand every step of this answer. However, I think I get the gist of the argument, thank you. $\endgroup$
    – FranDahab
    Commented Feb 17, 2023 at 5:20
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    $\begingroup$ No problem. If you can accept the first equation (using path integrals to calculate time-ordered products), then the rest is really just mathematics. $\endgroup$
    – Mike
    Commented Feb 17, 2023 at 5:51
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  1. A complex scalar field $\phi$ has 2 real DOF. Equivalently, one can to some extend treat $\phi$ and $\phi^{\ast}$ as independent fields, cf .e.g. this Phys.SE post. Therefore the 2-point function/propagator $$ \begin{pmatrix}\langle \phi\phi\rangle &\langle\phi\phi^{\ast}\rangle \cr \langle\phi^{\ast}\phi\rangle &\langle\phi^{\ast}\phi^{\ast}\rangle\end{pmatrix}\tag{*}$$ is really a $2\times 2$ matrix. However due to a $U(1)$ phase symmetry the matrix elements $\langle\phi\phi\rangle=0$ and $\langle\phi^{\ast}\phi^{\ast}\rangle=0$ vanish, cf. e.g. this related Phys.SE post. Hence only the mixed/off-diagonal matrix elements in eq. (*) survive, cf. OP's observation that $\phi$ and $\phi^{\ast}$ are "entangled".

  2. In fact, due to the conservation of $U(1)$ charge, the propagator lines become directed, i.e. we can consistently introduce an arrow, say from $\phi^{\ast}$ to $\phi$, cf. Ref. [1]. As a result, a complex scalar line can never start or end inside a Feynman diagram.

  3. The free propagator can be calculated by standard methods, see e.g. my Phys.SE answer here.

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References:

  1. M. Srednicki, QFT, 2007; problem 9.3. A prepublication draft PDF file is available here.
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