# Why isn't the wave equation $\nabla^2 \psi - 1/c^2 \partial_{tt} \psi = (\frac{mc}{\hbar})^2\psi$

Special relativity was well established by the time the schrodinger equation came out. Using the correspondence of classical energy with frequency and momentum with wave number, this is the equation that comes out, and looks sensible because this is of the form of a wave equation, like the one for sound etc, except with an inhomogeneous term

$$\nabla^2 \psi - 1/c^2 \partial_{tt} \psi = (\frac{mc}{\hbar})^2\psi$$

Instead, we have schrodinger's equation which reminds of the heat equation as it is first order in time.

EDIT Some thoughts after seeing the first posted answer

What is wrong with negative energy solutions? When we do freshman physics and solve some equation quadratic in, say, time... we reject the negative time solution as unphysical for our case. We do have a reason why we get them, like, for the same initial conditions and acceleration, this equation tells usthat given these iniial velocity and acceleration, the particle would have been at the place before we started our clock, since we're interested only in what happens after the negative times don't concern us. Another example is if we have areflected wave on a rope, we get these solutions of plane progressive waves travelling at opposite velocities unbounded by our wall and rope extent. We say there is a wall, an infinite energy barrier and whatever progressive wave is travelling beyond that is unphysical, not to be considered, etc.

Now the same thing could be said about $E=\pm\sqrt{p^2c^2+(mc^2)^2}$ that negative solution can't be true because a particle can't have an energy lower than its rest mass energy $(mc^2)$ and reject that.

If we have a divergent series, and because answers must be finite, we say.. ahh! These are not real numbers in the ordinary sense, they are p-adics! And in this interpretation we get rid of divergence. IIRC Casimirt effect was the relevant phenomenon here.

My question boils down to this. I guess the general perception is that mathematics is only as convenient as long as it gives us the answers for physical phenomenon. I feel this is sensible because nature can possibly be more absolute than any formal framework we can construct to analyze it. How and when is it OK to sidestep maths and not miss out a crucial mystery in physics.

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Maths can never be sidestepped in physics, unless one only cares about approximate guesses, sloppy arguments, psychological methods to remember some patterns, or presentation. Whenever we need accurate analyses or descriptions of anything in physics, maths is totally essential. ... I find your particular physics questions that are hiding behind your question confusing, and it could be exactly because you don't really take maths seriously. All those things have very sharp and clear answers but accurate maths is needed for them. –  Luboš Motl Mar 17 '11 at 7:51
And by the way, a wave function $\psi$ obeying a second-order, rather than first-order, equation could never be interpreted as a probability amplitude because $\int |\psi|^2$ could never be conserved. But it must be conserved for the total probability of all possible answers to remain at 100 percent. It follows that the objects entering second-order equations are not wave functions but fields - either classical fields or quantum fields. Another way to see it is to notice that the 2nd order equations universally possess unphysical negative-energy solutions. Schr's equation has to be 1st order. –  Luboš Motl Mar 17 '11 at 7:52
In your EDIT, you refer to the negative frequency solutions of the KG equation as "negative energy". In classical Physics, negative frequency components contribute positively to the energy; quantum field theory uses algebraic methods to ensure that what could be called negative frequency components also in the QFT context contribute positively to the energy. The language that is used fails to do proper justice to the mathematics, IMO. –  Peter Morgan Mar 17 '11 at 12:50
@Peter Morgan, I've never really understood that argument that negative energy solutions happen to be positive energy solutions with opposite whatever; it feels like side-stepping the fact that the math gives you negative energy solutions by introducing a magical factor of -1 for them. The argument that negative energy solutions are not observed because we don't see electrons decaying to those assumes that negative solutions would be more stable, but how are we sure negative solutions aren't actually less stable or equivalent? its a thermodynamical argument? –  lurscher Mar 17 '11 at 15:44
@lurscher Sorry to say that I think it's a long, unconventional story. My Answer below points to a small part of what I think about this. The stability of a system could be ensured by other conserved quantities, in which case a lower bound for eigenvalues of the infinitesimal generator of translations would not be necessary. Stability against what is also a question. Anyway, it's not obvious enough (for it to be a universally maintained axiom) that energy bounded below is either necessary or sufficient to ensure stability, whatever that might be in obvious axiomatic terms. –  Peter Morgan Mar 17 '11 at 16:38

That's the Klein-Gordon equation, which applies to scalar fields. For fermionic fields, the appropriate relativistic equation is the Dirac equation, but that was only discovered by Dirac years after Schrödinger discovered his nonrelativistic equation. The nonrelativistic Schrödinger equation is a lot easier to solve too.

The relativistic equations admit negative energy solutions. For fermions, that was only resolved by Dirac much later with his theory of the Dirac sea. For bosons, the issue was resolved by "second quantization".

The problem with negative energy solutions is the lack of stability. A positive energy electron can radiate photons, and decay into a negative energy state, if negative energy states do exist.

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Just to be sure, the Klein-Gordon equation was discovered before the non-relativistic Schrödinger equation - and it was discovered by Schrödinger himself who was efficient enough and realized that the KG equation disagreed with the Hydrogen atom. –  Luboš Motl Mar 17 '11 at 7:48
"A positive energy electron can radiate photons, and decay into a negative energy state, if negative energy states do exist" only if all other conserved discrete and continuous quantities are conserved. –  Peter Morgan Mar 17 '11 at 12:47

"My question boils down to this. I guess the general perception is that mathematics is only as convenient as long as it gives us the answers for physical phenomenon. I feel this is sensible because nature can possibly be more absolute than any formal framework we can construct to analyze it. How and when is it OK to sidestep maths and not miss out a crucial mystery in physics."

Your question just took a huge leap sideways. Probably into Soft-question. You don't "sidestep maths" in Physics. You introduce "different math" and look carefully to see whether it fits the experimental data better. People generally pick a particular mathematical structure and try to characterize its differences from the existing best theories, which for the best alternatives takes people decades. There are other issues, such as whether your different maths is more tractable, whether people think it's beautiful, whether it gives a better explanation, whether it suggests other interesting maths, whether it suggests interesting engineering, whether it's simple enough for it to be used for engineering. The wish-list for a successful theory in Physics is quite long and not very articulated. Philosophy of Physics tries to say something about the process and the requirements for theory acceptance, which I've found it helpful to read but ultimately rather unsatisfying. "miss[ing] out a crucial mystery in physics" would be bad, but it's arguably the case that if it's not Mathematics it's not Physics, which in the end of hard practical use will be because if it's not mathematics you'll be hard pressed to do serious quantitative engineering.

For your original question, I've been pursuing why negative frequency components are so bad for about 5 years. If you feel like wasting your time, by all means look at my published papers (you can find them through my SE links) and at my Questions on SE, all of which revolve around negative frequency components (although I think you won't see very clearly why they do in many cases, even if you know more about QFT than your Question reveals). I don't recommend it. I can't at this stage of my research give you a concise Answer to your original Question.

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It seems to me that the questioner is asking about Quantum Mechanics, not Quantum Field Theory. So what one can or cannot do in QFT is evading the question.

The short answer to the question as posed is « Yes, it is the wave equation.»

As prof. Motl pointed out, it was discovered by Schroedinger first, following exactly the reasoning the OP presents. It does not describe the electron, but it does describe one kind of meson, and so it has to be said that it agrees with experiment. I emphasise that this is qua relativistic one-particle QM equation, not second-quantised. Both Pauli's lectures in wave mechanics and Greiner's Relativistic Quantum Mechanics treat the K.-G. equation at length as a one-particle relativistic wave equation. Furthermore, the negative energy states can be eliminated by taking the positive square root: $$\sqrt{-\nabla ^2 + m^2} \psi = i{\partial \over\partial t}\psi.$$ Every solution of this equation is a solution of the original K.-G. equation, so if the latter is physical, so is this one.

Now we have an equation that agrees with experiment, does not need any fiddling with the maths, but does not allow the Born-rule probability interpretation of the non-relativistic Schroedinger equation. What one says about this depends on whether one thinks philosophy, the Born interpretation, should trump experimental data, namely the « mesonic hydrogen » energy levels, or the other way round....

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The square root equation you wrote down admits a Born rule interpretation-- it is just the one-particle fock space evolution equation. –  Ron Maimon Dec 25 '11 at 7:32
The $\sqrt{-\nabla^2+m^2}$ operator is non-local, which is usually taken as something of a strike against it, right? I don't think this necessarily kills this approach, because the ways in which different aspects of QM might be local or non-local is a very long running saga, but it does give a certain pause. In contrast, usually taken to be better, the Dirac operator is local. I imagine you have a definite response to this particular cavil, Joseph, and I'm curious what it is. –  Peter Morgan Dec 25 '11 at 16:21
Hi, Peter. Well, every solution of this non-local eq. is also a solution of the original K.G. eq., and it is only the solutions which are physical, the linguistic form of the equation has no physical significance. So one has to criticise the original eq. too, if there is something wrong with the solutions of the « positive energy K.G. eq.» My question is, what does « non-locality » mean as far as a given, concrete, solution is concerned? And then, after all, there is the experimental support for the eq., ... –  joseph f. johnson Dec 25 '11 at 18:10

You can throw away the negative energy solution if you're just doing relativity and not QM, just like in the examples you mentioned. But if you add QM you can no longer do that: remember the energy spectrum are the eigenvalues of the Hamiltonian (operator), so you just can't throw away some eigenvalues without also throwing away the linear algebra the theory is based upon.

So no, it is never ok to sidestep math. When you eliminate negative roots of cuadratic equations in classical physics you're not sidestepping math, after all you need to apply the rules of algebra to get the solutions, you're just applying physical considerations to the solutions to get rid of (physiccally) spurious ones that can be ignored without altering the mathematical structure of the theory. In the RQM case the negative energy solutions are unphysical but you can't just ignore them, you have to deal with them.

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If I'm not mistaken, people have shown that in a scattering problem, even if we start with a wave purely made of positive energies, after the scattering there will be negative components popping up, so it does not make sense to simply throw away the negative energy states. But I can't find the reference now.

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This is true for local coupling only, you can make up artificial coupling for the square-root equation which allows scattering. –  Ron Maimon Sep 3 '12 at 2:57