5

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.

glS
  • 14,271
Quantization
  • 1,035
  • 3
    no. When it does not happen we talk of spontaneous symmetry breaking. Have also a look at this answer of Qmechanic. – glS Feb 08 '15 at 19:19
  • so is the solution that Srednicki constructed not a Lorentz invariant one if we allow arbitrary a(k)? – Quantization Feb 08 '15 at 19:28
  • 1
    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. – ACuriousMind Feb 08 '15 at 19:39
  • 1
    Actually, I find transforming integration region back still problematic.. Say I have $\phi(x) \to \phi(\Lambda^{-1} x) = \int d\tilde{k}[ a(\Lambda^{-1}\mathbf{ k})e^{ikx} + a^(\Lambda^{-1} \mathbf{k})e^{-ikx}] = \int d\tilde{k}[ a(\mathbf{k})e^{i(\Lambda k)x} + a^(\mathbf{ k})e^{-i(\Lambda k)x}] =? \phi(x)$. – Quantization Feb 09 '15 at 00:57
  • @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. – Ryan Unger Feb 09 '15 at 11:09
  • @ACuriousMind: I'm stumped. How does one show that $\phi'(x)=\phi(\Lambda^{-1}x)$ is satisfied by the standard mode expansion in the OP? – Ryan Unger Feb 09 '15 at 11:16
  • @KyleLee: I see the following in Weinberg: $$U(\Lambda)\phi(x)U^{-1}(\Lambda)=\int d\tilde k[a(\Lambda k)e^{ik(\Lambda x)}+\text{h.c.}]=\int d\tilde k[a(\Lambda (\Lambda^{-1}k))e^{i(\Lambda^{-1}k)(\Lambda x)}+\text{h.c.}]$$ $$=\int d\tilde k[a(k)e^{ikx}+\text{h.c.}]=\phi(x)$$ – Ryan Unger Feb 09 '15 at 11:28
  • @0celo7: Not sure how first equality follows. If we have $\phi(\Lambda^{-1} x)$, does $\Lambda^{-1}$ only transform $x$, or $k$ at the same time? – Quantization Feb 09 '15 at 14:18
  • I think that's a poor notation. It doesn't work when we consider fields as operators, because the modes (a) don't depend on x but are transformed. – Ryan Unger Feb 09 '15 at 14:25

1 Answers1

3

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...

Heterotic
  • 2,340
  • What about $k^0$ in the argument of exponents ? Why are they still around ? – Quantization Feb 09 '15 at 11:42
  • @KyleLee: They get put on the mass shell: $k^0=\sqrt{\mathbf{k}^2+m^2}$. – Ryan Unger Feb 09 '15 at 11:56
  • That's true. Nice ! – Quantization Feb 09 '15 at 11:58
  • 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. – Quantization Feb 09 '15 at 12:00
  • @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. – Ryan Unger Feb 09 '15 at 12:05
  • I agree with you – Quantization Feb 09 '15 at 12:12
  • I thought you were missing that $\delta(p^2-m^2)$ factor. – Ryan Unger Feb 09 '15 at 12:43
  • @Heterotic: Again, shouldn't integration over $p^0$ give $f(p)$ that dependens on $a(p)$? – Quantization Feb 09 '15 at 14:22
  • @KyleLee: At school, so I took a pic: http://gyazo.com/ed1844d7413ff428cb54b52b7ebac104. The delta function simply puts $p^0$ on the mass shell, no $p^0$ integral over $a$. – Ryan Unger Feb 09 '15 at 15:49
  • @KyleLee: Some authors scale $a$ so that one over the square root of the energy appears, rather than one over the energy. Slightly better resolution: http://imgur.com/Z07DtEY – Ryan Unger Feb 09 '15 at 15:53
  • @0celo7: Very very nice! – Quantization Feb 09 '15 at 16:19
  • I couldn't reach a conclusion but OP did. Here is what I could think: $kx$ term is Lorentz invariant, so is measure and field is a scalar, thus $a(k)$ must a Lorentz scalar (since $kx$ is invariant, $k$ transforms covariantly, thus we can talk about how $a$ transforms). Then $a(k)$ is not arbitrary but Lorentz scalar. – Mahammad Yusifov Jan 01 '24 at 19:47
  • On page 39 of Srednicki ,link, accepted answer's first equation is written. On page 40, at eq 3.21, author represents $a(k)$ in terms of solution $\phi$. From here how arbitrary $a(k)$ could be is implicitly mentioned. I think if $a(k)$ is truly arbitrary, every arbitrary function ($a(k)$)can be written as some function of solutions of KG equation (by eq 3.21) but this sounds too much strong. – Mahammad Yusifov Jan 01 '24 at 19:53