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If I have a Lorentz invariant equation of motion, like Klein-Gordon equation, is the solution automatically guaranteed to be Lorentz invariant?

I ask this question because of the discussion from Mark Srednicki's Quantum Field Theory section 3 from equations (3.11) to (3.14). If I have a K-G equation, $$ \tag 1 \partial^\mu\partial_\mu\phi -m^2\phi=0,$$ we have a solution of the form $$ \tag 2\exp (i \mathbf{k} \cdot \mathbf{x} \pm i\omega t),$$ which I do not think is Lorentz invariant for solution with $i \mathbf{k} \cdot \mathbf{x} + i\omega t$ as an argument, unless we allow $k^\mu = (-\omega, \mathbf{k})$.

However, he starts constructing a Lorentz invariant solution, and comes up with $$ \tag 3 \phi(\mathbf{x},t) = \int d\tilde{k}[ a(\mathbf{k})e^{ikx} + a^*(\mathbf{k})e^{-ikx}],$$ where $kx = \mathbf{k}\cdot\mathbf{x} - \omega t$. $d\tilde{k}$ is a Lorentz invariant measure and argument of each exponents are Lorentz invariant as well.

However, he says in the beginning that $a(\mathbf{k})$ is an arbitrary function of the wave vector $\mathbf{k}$, which does not sound Lorentz invariant to me. So I am not sure how $\phi(\mathbf{x},t)$ is Lorentz invariant.

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    $\begingroup$ no. When it does not happen we talk of spontaneous symmetry breaking. Have also a look at this answer of Qmechanic. $\endgroup$
    – glS
    Feb 8, 2015 at 19:19
  • $\begingroup$ so is the solution that Srednicki constructed not a Lorentz invariant one if we allow arbitrary a(k)? $\endgroup$ Feb 8, 2015 at 19:28
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    $\begingroup$ It would help if you explicitly showed where you think this violates Lorentz covariance/invariance, because I don't see it - $a(k)$ goes to $a(\Lambda^{-1} k)$ under a Lorentz transformation, but Lorentz invariance of the measure means that you can just transform the integration region back to still only have $a(k)$ in there. $\endgroup$
    – ACuriousMind
    Feb 8, 2015 at 19:39
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    $\begingroup$ Actually, I find transforming integration region back still problematic.. Say I have $\phi(x) \to \phi(\Lambda^{-1} x) = \int d\tilde{k}[ a(\Lambda^{-1}\mathbf{ k})e^{ikx} + a^*(\Lambda^{-1} \mathbf{k})e^{-ikx}] = \int d\tilde{k}[ a(\mathbf{k})e^{i(\Lambda k)x} + a^*(\mathbf{ k})e^{-i(\Lambda k)x}] =? \phi(x)$. $\endgroup$ Feb 9, 2015 at 0:57
  • $\begingroup$ @KyleLee: The transformation of a scalar field is $\phi'(x)=\phi(\Lambda^{-1}x)$. That said, I don't know what you're doing wrong. $\endgroup$
    – Ryan Unger
    Feb 9, 2015 at 11:09

1 Answer 1

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In the spirit of the original post, let $k,x$ be 4-vectors and $\mathbf{k}$, $\mathbf{x}$ the spatial components. Then a quantity of the form $$\phi(x) \propto \int dk[ a(k)e^{ikx} + a^*(k)e^{-ikx}]$$ is manifestly Lorentz invariant because it does not explicit contain any free Lorentz indices. What Srednicki does is that he performs the $k^0$ integration, resulting in $$\phi(\mathbf{x},t) = \int \frac{d\mathbf{k}}{f(\mathbf{k})}[ a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}} + a^*(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}}],$$ which only includes spatial components. This expression is Lorentz invariant because it is just a different form of the previous one, but it does not manifestly look Lorentz invariant which I assume what causes the confusion. For an explicit form of the function $f$ which will of course be related to the energy as it is the integral over $k^0$, see for example Peskin and Schroeder eqn (2.47).


EDIT: Some more justification:

The Klein-Gordon eqn is $$\partial^\mu\partial_\mu\phi -m^2\phi=0.$$ To solve it, we Fourier transform to momentum space and we get: $$(p^\mu p_\mu -m^2)\tilde\phi=0.$$ The general solution of this eqn is $$\tilde\phi(p)=a(p)\delta(p^\mu p_\mu -m^2),$$ which means that the general solution for the Klein-Gordon is: $$\phi(x)=\frac{1}{(2\pi)^4}\int d^4pe^{ipx}\tilde\phi(p)=\frac{1}{(2\pi)^4}\int d^4pe^{ipx}a(p)\delta(p^\mu p_\mu -m^2)$$ which is manifestly Lorentz invariant. You can then perform the $p^0$ integration as claimed above. I have ignored the complex conjugate term everywhere but should be trivial to restore...

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  • $\begingroup$ What about $k^0$ in the argument of exponents ? Why are they still around ? $\endgroup$ Feb 9, 2015 at 11:42
  • $\begingroup$ @KyleLee: They get put on the mass shell: $k^0=\sqrt{\mathbf{k}^2+m^2}$. $\endgroup$
    – Ryan Unger
    Feb 9, 2015 at 11:56
  • $\begingroup$ That's true. Nice ! $\endgroup$ Feb 9, 2015 at 11:58
  • $\begingroup$ Well, one thing that im still confused about is I feel like f(k) should then be appropriately set to make the whole thing Lorentz invariant, not just the measure. I don't see how the coefficients don't play any role in determining f(k) if I integrate the first integral. $\endgroup$ Feb 9, 2015 at 12:00
  • $\begingroup$ @KyleLee: What worries me more is the first equation. What does $a(k)$ mean? Does that imply it has $k^0$ dependence? I can't recall seeing this expansion before. It's normally the $d\mathbf{k}$ one. $\endgroup$
    – Ryan Unger
    Feb 9, 2015 at 12:05

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