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  1. Is there a well-known Lagrangian that, writing the corresponding equation of motion, gives the Klein-Gordon equation in QFT? If so, what is it?

  2. What is the canonical conjugate momentum? I derive the same result as in two sources separately, but with opposite sign, and I am starting to suspect that the error could be in the Lagrangian I am departing from.

  3. Is there any difference in the answers to these two questions if you choose $(+---)$ or $(-+++)$? If so, which one?

Urb
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Eduardo Guerras Valera
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1 Answers1

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  1. Yes. The standard scalar field which all QFT books (e.g. Peskin & Schroeder, Zee) start with yields the KG equation. For that reason it is also called the Klein-Gordon field. The Lagrangian (density) is \begin{align} \mathcal{L} = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{2} m^2 \phi^2. \end{align} Here the metric is $(+---)$.

  2. By definition it is $\pi = \frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}$. This gives $\pi = \partial^0 \phi$.

  3. It is purely convention, there is no right choice. The only difference in using a different metric will be in how we write things down - any quantities that involve contraction with the metric $\eta_{\mu \nu}$ will change by a minus sign. For example in the Lagrangian, using the metric (- + + +), the first term is changed to $-\frac{1}{2} \partial_\mu \phi \partial^\mu \phi$. But this is still equal to $\frac{1}{2}(\partial_t^2 \phi - \nabla^2 \phi)$ regardless of which metric we use.

J.G.
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nervxxx
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  • Thanks a lot for your answer, it has put me on the right track. Eqs (1.14) and (1.15) in the preprint of Srednicki have the key to the changed sign of question 2. I found it thank to your indications. – Eduardo Guerras Valera Feb 03 '13 at 13:24
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    The answer above is unclear to me. The Lagrangian above is (1) 1/2 (ħ ∂μφ) (ħ ∂νφ) g^μν – 1/2 m^2 φ^2 If you change the sign of the metric, the first term changes sign and the second term does not change. To get the same Lagrangian, the sign of the first or second term must change. No? – rjpetti Sep 10 '21 at 15:21
  • @rjpetti : It seems you are right. But I think that we must not expect any respond from the answerer "nervxxx" since I feel he/she is inactive (no recent answers,questions, comments etc). – Frobenius Sep 10 '21 at 17:46
  • ...the same is valid for the OP "Eduardo Guerras Valera"... – Frobenius Sep 10 '21 at 18:37
  • The assertion of the answerer that the term $,\partial_\mu \phi \partial^\mu \phi,$ is the same for any sign convention is wrong. More exactly $$ \partial_\mu \phi \partial^\mu \phi\boldsymbol= \left. \begin{cases} \boldsymbol+\partial^2\phi/c^2\partial t^2\boldsymbol-\nabla^2\phi \texttt{ for } (+,-,-,-)\ \ \boldsymbol-\partial^2\phi/c^2\partial t^2\boldsymbol+\nabla^2\phi \texttt{ for } (-,+,+,+) \end{cases} \right} $$ – Frobenius Sep 10 '21 at 19:10
  • So we have two Lagrangian densities $$ \left. \begin{cases} \mathcal L_1\left(\phi,\partial_\mu \phi\right)\boldsymbol=\frac12\left(\partial^2\phi/c\partial t^2\boldsymbol-\nabla^2\phi\right)-\frac12 m^2\phi^2 \text{ for } (+,-,-,-)\ \mathcal L_2\left(\phi,\partial_\mu \phi\right)\boldsymbol=\frac12\left(\nabla^2\phi\boldsymbol-\partial^2\phi/c\partial t^2\right)-\frac12 m^2\phi^2 \text{ for } (-,+,+,+) \end{cases} \right} $$ – Frobenius Sep 10 '21 at 20:46
  • @Frobenius much ado about nothing. The original answer from 8 years ago already says: upon switching metric conventions, "any quantities that involve contraction with the metric $\eta_{\mu\nu}$ will change by a minus sign." In particular, if you change your convention of the metric to (-+++), then the first term in $\mathcal{L}$ written in my answer is changed to $-\frac{1}{2}\partial_\mu\phi \partial^\mu \phi$, which if you expand out is equal to $\frac{1}{2} (\partial_t^2 \phi - \nabla^2 \phi)$, identical to the case with metric (+---). Reading comprehension fail? – nervxxx Sep 13 '21 at 21:39
  • @nervxxx : I disagree with you. The two Lagrangians $\mathcal L_1$ and $\mathcal L_2$ yield the same equation of motion, that is the Klein-Gordon equation. This is due to the fact that they differ by a 4-divergence. – Frobenius Sep 13 '21 at 21:51
  • @Frobenius Please see accepted answer of https://physics.stackexchange.com/questions/589964/sign-convention-in-field-theory I guess you would disagree with it too? – nervxxx Sep 14 '21 at 03:46
  • @Frobenius actually if you do a bit of digging, stackexchange has tons of posts on this sign issue. Here's one (https://physics.stackexchange.com/questions/100557/sign-in-front-of-qft-kinetic-terms) where if you use your Lagrangian $\mathcal{L}_2$, inconsistencies happen where the spectrum becomes bounded from above, the relativistics energy-momentum relation is wrong etc. – nervxxx Sep 14 '21 at 03:59
  • @nervxxx : Be patient. I'll post an answer in a few days. If during the elaboration of this answer I would find that I am wrong I will agree with you. In summary I don't like the plus/minus signs in the Lagrangian and in the equation of motion in the linked Qmechanic's answer. Things are working automatically with tensor calculus notation. The opposite sign in the Lagrangian on changing signature is compensated by a changing sign in the first term of the Euler-Lagrange equation which after that yields the Klein-Gordon equation. – Frobenius Sep 14 '21 at 04:12
  • @nervxxx : You are right. Elaborating an answer I realized that it was a fault of mine to think that the opposite sign in the first item of the Lagrangian on changing signature is compensated by a changing sign in the first term of the Euler-Lagrange equation. In order to retract my downvote on your answer I made a pseudo-edit to your answer. – Frobenius Sep 14 '21 at 08:09
  • I prefer my comments to be as they are in order to remember my faults (from which I learn much too). – Frobenius Sep 14 '21 at 08:14
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    The Lagrangian is invariant up to an overall scalar, so the point about the spectrum being bounded or unbounded when you allow negative kinetic energy terms can always be avoided by a trivial re-definition, the point about not satisfying the relativistic energy-momentum relation is the non-trivial fact which fixes the overall sign of the kinetic terms with respect to the potential terms for each choice of metric. – bolbteppa Sep 14 '21 at 08:27