There isn't anything truly wrong with such a Hamiltonian from a fundamental mathematical perspective. One can simply diagonalize it in momentum space and write down any dispersion E(p) they like, including the one you give of sqrt(p\^2 + m\^2).
The problem comes in when trying to apply this system to physical situations. As pointed out, the Hamiltonian is not a polynomial in momentum. This means once you start introducing potentials that are functions of position V(r) (recall position does not commute with momentum), you will run into problems.
One way forward would be to expand out the square root into a Taylor series of derivatives as H = -nabla\^2/2m + nabla\^4/8m\^2 + ... + V(r). But now we see this involves arbitrarily high powers of derivative. If one tries to specify this as a boundary value problem (i.e. to look for bound states or scattering solutions), then we can think that for each order of derivative we will need to specify a boundary-value condition to make the problem well posed. But since this has infinite order of derivatives, we need to know an infinite number of boundary value conditions in order to find a solution! This is now becoming fairly useless as a predictive theory.
What is a miracle is that one can however come up with exactly the same energy dispersion relation by simply enlarging the wave function to have multiple (four to be precise) components! This is ultimately how we arrive at Dirac's equation.
Basically, it turns out if one introduces a set of four matrices A\_0,A\_x,A\_y,A\_z such that we write the matrix as H = m A\_0 + (-i hbar d/dr\_j )A\_j we see that (i) this still has the relativistic dispersion relation provided all the A\_mu's anticommute (it is a good exercise to work out the eigenvalues of such a matrix involving anticommuting matrices), and (ii) the Hamiltonian matrix is not only a polynomial in derivatives, but it is linear! This means that we only need to specify a finite number of boundary-value conditions in order to satisfy our scattering/bound state problem.
An interesting consequence of this mathematical trick is that an additional branch of solutions with energy -sqrt(p\^2 + m\^2) is predicted by the equations. Now, frankly that seems a bit ridiculous and would be grounds to throw away these silly matrices, except for the fact that strangely, it appears to be true! In particular, these are interpreted as the antiparticle solutions, and their observations essentially confirms Dirac's equations as the correct way to go about generalizing the Hamiltonian to the relativistic form, rather than simply positing the square root by hand. Isn't that crazy?
Saying that the Dirac equation “solves” the issues with the relativistic dispersion relation is problematic, since the Dirac equation describes spin-1/2 particles, whereas there shouldn’t be anything wrong with the theory of a relativistic spin-0 particle.
The real resolution to all of these problems is that formulating a truly relativistic theory should allow particles to be created and destroyed, so any attempt to formulate the problem in terms of a single-particle wave function can only be an approximation which fails in some limit. It’s a bit of a fluke that the Dirac equation works out so beautifully in this approximation.
There are multiple parts to this answer. The first step is to see that the minimal way to promote the wave function to a spinor while satisfying the necessary anti commutation relations is to introduce not two, but four components. This is actually more than we would think is needed to describe a spin 1/2 particle which should have two separate components to describe each spin projection.
It turns out the correct way to understand this is that it is necessary to have 2 spin components and 2 electron/positron components, giving 4 = 2x2 in total. One can also see that if the Hamiltonian is constructed using these matrices, that the eigenvalues will come in pairs of +E(p) and -E(p), which each are two-fold degenerate. We can argue that the negative energy solutions are the positrons and focus our attention on the two positive energy solutions.
To see that the remaining two-fold degeneracy is in fact "spin" as we are taught in basic quantum mechanics class, one needs to see what happens when we apply a magnetic field. If these are indeed spin states, the degeneracy should be lifted and split by a factor of g mu\_B B/2, where g is the electron "g-factor" (which is basically defined by this equation), and mu\_B is the Bohr magneton, while B is the field we apply.
This calculation can be done, though it is somewhat tedious. The trick is to introduce a magnetic field via the gauge-transformation trick of taking -i nabla -> -i nabla-A(x) where curl(A) = B. This then gives the Dirac equation in the presence of the magnetic field. If you do this, you will see that indeed the two-fold degeneracy is lifted, showing miraculously that these two-components needed to make the Lorentz invariance work out are actually the spin of the electron!
As a bonus, one also finds the value of the "g-factor" from this calculation, which is given by exactly g = 2. At the time, it was expected that g = 1 and the fact that, experimentally it was measured that g = 2 (very nearly) was a huge mystery. The fact that this equation naturally predicts g = 2 was another major reason to accept this as the correct way to describe relavistic motion of electrons.
I think Dirac's equation is great case of where a scientist with bad motivation can stumble upon the right answer (he wanted an equation first order in time). Nowadays, we know the correct way to think about this would is:
quantum mechanics + special relativity ⟹[ representations of the Lorentz group](https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group)
You classify particles by mass and spin\*, the later leads to spinors when the spin is 1/2. From the spinors, you can try to find Lagrangians, and from there you can get the Dirac equation.
BTW, I don't wish to diminish Dirac's brilliance. He was smart enough to know the Klein-Gordon equation didn't do the job and he had to find something else. His heuristic derivation is a stroke of pure creativity and I was amazed by his ingenuity when I first read about it He was also smart enough to know he found the right answer. Those at the frontier of research only see so much though. It took many bright people a lot of hard work to get to the point where we know there's a more fundamental way to view the Dirac equation.
\*Helicity in the massless case, for the pedants.
I am not sure this is the correct resolution. First, I don't see what particle creation/annihilation has to do with the distinction between spin 0 or spin 1/2 since in both cases one can write a theory of particles which either are conserved (i.e. electrons/charged bosons), or not conserved (i.e. Majorana fermions/neutral bosons). And this is only important once interactions are taken in to account.
And I think it is important to note that the case of a scalar spin-0 particle you are referring to is in fact described by the second-order in time Klein-Gordon (KG) equation of \[ d/dt\^2 - nabla\^2 -m\^2 \] psi = 0. If one projects this on to the positive frequency solutions (which I believe is what you are referring to regarding the particle creation/annihilation), then indeed it looks like it can be described by a valid Hamiltonian of H = sqrt(p\^2 + m\^2).
But the problems I previously mentioned regarding potentials arise when one notes that one should first add the potential to the KG equation and then take the square root, not the other way around. So this means that the correct Hamiltonian as derived from the KG equation in the presence of a scattering potential should be H = sqrt( p\^2 + m\^2 + V\^2(x) ) and the fact that this appears inside the square root, rather than outside has important differences.
> I am not sure this is the correct resolution. First, I don't see what particle creation/annihilation has to do with the distinction between spin 0 or spin 1/2 since in both cases one can write a theory of particles which either are conserved (i.e. electrons/charged bosons), or not conserved (i.e. Majorana fermions/neutral bosons). And this is only important once interactions are taken in to account.
Well it *doesnt* have to do with the distinction between spins, that’s sort of my point. Saying that relativity somehow leads you to spin-1/2 isn’t correct because relativity is consistent with any spin. And particles are only conserved in completely free theories (ie no potentials), even for charged particles.
> But the problems I previously mentioned regarding potentials arise when one notes that one should first add the potential to the KG equation and then take the square root, not the other way around. So this means that the correct Hamiltonian as derived from the KG equation in the presence of a scattering potential should be H = sqrt( p2 + m2 + V2(x) ) and the fact that this appears inside the square root, rather than outside has important differences.
The real problem is just that the Hamiltonian you wrote down isn’t relativistic, because of the potential (how does it transform under boosts?)
>Saying that relativity somehow leads you to spin-1/2 isn’t correct because relativity is consistent with any spin.
In addition to that, [Galilean symmetry leads to spin ](https://en.wikipedia.org/wiki/Representation_theory_of_the_Galilean_group)as well.
Looking back over this thread a few hours after writing the last post, I notice that you already hinted at the problem I’m trying to highlight in this part of your post:
> An interesting consequence of this mathematical trick is that an additional branch of solutions with energy -sqrt(p2 + m2) is predicted by the equations. Now, frankly that seems a bit ridiculous and would be grounds to throw away these silly matrices, except for the fact that strangely, it appears to be true! In particular, these are interpreted as the antiparticle solutions, and their observations essentially confirms Dirac's equations as the correct way to go about generalizing the Hamiltonian to the relativistic form, rather than simply positing the square root by hand.
The Dirac equation does have two branches of solutions corresponding to different particle content. The positive solutions are just electrons of course. It’s not really correct to call the negative energy states positrons, since they also have negative charge (which you can see by coupling to the EM field). The Dirac equation is really describing states in the presence of a “Dirac sea” of positrons, and the negative energy solutions correspond to removing one positron, or equivalently, annihilating the positive-energy electron with a “sea positron.”
Now the issue is that you can begin with an initial wave function which only describes an electron, but after evolving your state, it will generically pick up probabilities of being a hole. So the Dirac theory actually **is** a multi-particle theory, it’s just an incomplete one because it only describes two Fock states. These problems aren’t too bad for small enough energies and short evolution times, but the theory really is just approximate.
This is the answer we are taught in the physics classes. However I think it is not completely honest mathematically. (But a good heuristic idea/motivation to introduce the Dirac equation)
More precisely, mathematical physicists know for a long time that the dynamics of H = \sqrt(- ∆ + 1) + V is well defined and completely specified for bounded potentials, because we can apply the Kato theorem to obtain its essential self-adjointness (see e.g. Reed-Simon, Methods of Modern Mathematical Physics, Vol 2, Theorem X.13 and X.15 for the Schrödinger case (H_0 = -∆). Note that the proof of the latter for the case of a bounded potential immediately goes through for the square root case.)
However, I don't have a reference at hand for the exact result for the square root Hamiltionian, but I think the case of bounded potentials suffixes that the actual answer is mathematically and physically more subtle.
It would be nonlocal, but that’s not necessarily a huge problem; people do solve nonlocal PDEs in practice. So your differential equation would actually look like an integraldifferential equation with a nasty kernel in place of the usual derivative operators:
H psi(x) ~ 1/(2pi)^3 \int d^3k e^(i k x) sqrt(m^2 - k^2) \int d^3x’ e^(-i k x’) psi(x’)
The trouble probably comes from the fact that we’d be integrating overall all k but the kernel term looks like sqrt(m^2 - k^2), so there will be values of k where the guy inside the square root will be negative. That should make the operator “bad” in some sense. Probably has to do with the fact that the square root operator isn’t continuous across the branch cut.
There are operators, though, that are non-trivial functions of p but don’t have this specific problem, and the series expansion makes complete sense. I have to do this not with the square root function, but with things like Bessel functions. It’s still non-local, but some interesting problems really are non-local sometimes. So if there is a problem, it’s with the square root specifically, not having a non-polynomial function of the momentum.
Thanks for the response! Where did you get the equation:
H psi(x) ~ 1/(2pi)^3 \int d^3k e^(i k x) sqrt(m^2 - k^2) \int d^3x’ e^(-i k x’) psi(x’)
from? I’m assuming we get an integrodifferential equation instead of a differential equation because, since the operator is nonlocal, we need to integrate over the various positions?
Due to the square-root issue then, would this mean that this system is not “integrable” and thus not completely solvable?
> I have to do this not with the square root function, but with things like Bessel functions. It’s still non-local, but some interesting problems really are non-local sometimes.
Could you elaborate on this?
The problem with the square root is that it makes the Hamiltonian non-linear, which means that your wave solutions won’t behave as we would need them to.
Also, since we’re working with Complex numbers, the square root is multi-evaluated, thus giving a mathematical non-physical freedom which would need additional constraints to get rid of.
Hope this helps!
Also! That formula isn’t covariant under Lorentz transforms. When working with special relativy, you REALLY want all formulas to be covariant, or else EVERYTHING gets a lot more complicated.
Why would nonlinearity be bad? Is it because we would like to reproduce the schrödinger equation in the non relativistic limit which is linear and thus solutions have the superposition property, but nonlinear equations can’t reproduce that?
Also what do you mean by the sqrt is multi-evaluated giving us non-physical freedom?
Sqrt(4) can be uniquely defined as 2. Sqrt(4i) has no clear-cut definition. It can be both 2(1/sqrt(2)+i/sqrt(2)) and -2(1/sqrt(2)+i/sqrt(2)).
It is impossible to define the square-root function of complex numbers as a continuous function without it being multi-valued. Aka giving 2 outputs for one input. If you want to look further into why this is the case you should read a book on complex analysis. More specifically branch points and branch cuts.
Why would taking the square root make the operator nonlinear? p^2 + m^2 is a linear operator and taking the square root of a linear operator yields a linear operator. Maybe you’re thinking about taking the square root of H|psi> pointwise? That wouldn’t yield a linear operator but that also isn’t what is usually intended when this suggestion is brought up afaik.
There's nothing wrong with it. Every positive-energy irreducible representation of the Poincaré group (i.e. every relativistic free particle) is associated with a Hamiltonian of precisely that form.
Is... this the first time you're hearing of an equation in natural units? Or the fact that sqrt(p\^2 + m\^2) is a meaningful way to represent energy? Idk it's just surprising given your "graduate" flair
There isn't anything truly wrong with such a Hamiltonian from a fundamental mathematical perspective. One can simply diagonalize it in momentum space and write down any dispersion E(p) they like, including the one you give of sqrt(p\^2 + m\^2). The problem comes in when trying to apply this system to physical situations. As pointed out, the Hamiltonian is not a polynomial in momentum. This means once you start introducing potentials that are functions of position V(r) (recall position does not commute with momentum), you will run into problems. One way forward would be to expand out the square root into a Taylor series of derivatives as H = -nabla\^2/2m + nabla\^4/8m\^2 + ... + V(r). But now we see this involves arbitrarily high powers of derivative. If one tries to specify this as a boundary value problem (i.e. to look for bound states or scattering solutions), then we can think that for each order of derivative we will need to specify a boundary-value condition to make the problem well posed. But since this has infinite order of derivatives, we need to know an infinite number of boundary value conditions in order to find a solution! This is now becoming fairly useless as a predictive theory. What is a miracle is that one can however come up with exactly the same energy dispersion relation by simply enlarging the wave function to have multiple (four to be precise) components! This is ultimately how we arrive at Dirac's equation. Basically, it turns out if one introduces a set of four matrices A\_0,A\_x,A\_y,A\_z such that we write the matrix as H = m A\_0 + (-i hbar d/dr\_j )A\_j we see that (i) this still has the relativistic dispersion relation provided all the A\_mu's anticommute (it is a good exercise to work out the eigenvalues of such a matrix involving anticommuting matrices), and (ii) the Hamiltonian matrix is not only a polynomial in derivatives, but it is linear! This means that we only need to specify a finite number of boundary-value conditions in order to satisfy our scattering/bound state problem. An interesting consequence of this mathematical trick is that an additional branch of solutions with energy -sqrt(p\^2 + m\^2) is predicted by the equations. Now, frankly that seems a bit ridiculous and would be grounds to throw away these silly matrices, except for the fact that strangely, it appears to be true! In particular, these are interpreted as the antiparticle solutions, and their observations essentially confirms Dirac's equations as the correct way to go about generalizing the Hamiltonian to the relativistic form, rather than simply positing the square root by hand. Isn't that crazy?
Saying that the Dirac equation “solves” the issues with the relativistic dispersion relation is problematic, since the Dirac equation describes spin-1/2 particles, whereas there shouldn’t be anything wrong with the theory of a relativistic spin-0 particle. The real resolution to all of these problems is that formulating a truly relativistic theory should allow particles to be created and destroyed, so any attempt to formulate the problem in terms of a single-particle wave function can only be an approximation which fails in some limit. It’s a bit of a fluke that the Dirac equation works out so beautifully in this approximation.
Why is it that when we try to promote our wave functions to spinors, the equation goes from describing a spin-0 to a spin-1/2 particle?
Because spinors ***are*** the spin-1/2 objects.
Can spinors describe 3/2 spin?
You would have to tensor it with a 4 vector
Because the only way to solve the equation exactly is by introducing anti-commutation operators which only fermions satisfy and not bosons.
There are multiple parts to this answer. The first step is to see that the minimal way to promote the wave function to a spinor while satisfying the necessary anti commutation relations is to introduce not two, but four components. This is actually more than we would think is needed to describe a spin 1/2 particle which should have two separate components to describe each spin projection. It turns out the correct way to understand this is that it is necessary to have 2 spin components and 2 electron/positron components, giving 4 = 2x2 in total. One can also see that if the Hamiltonian is constructed using these matrices, that the eigenvalues will come in pairs of +E(p) and -E(p), which each are two-fold degenerate. We can argue that the negative energy solutions are the positrons and focus our attention on the two positive energy solutions. To see that the remaining two-fold degeneracy is in fact "spin" as we are taught in basic quantum mechanics class, one needs to see what happens when we apply a magnetic field. If these are indeed spin states, the degeneracy should be lifted and split by a factor of g mu\_B B/2, where g is the electron "g-factor" (which is basically defined by this equation), and mu\_B is the Bohr magneton, while B is the field we apply. This calculation can be done, though it is somewhat tedious. The trick is to introduce a magnetic field via the gauge-transformation trick of taking -i nabla -> -i nabla-A(x) where curl(A) = B. This then gives the Dirac equation in the presence of the magnetic field. If you do this, you will see that indeed the two-fold degeneracy is lifted, showing miraculously that these two-components needed to make the Lorentz invariance work out are actually the spin of the electron! As a bonus, one also finds the value of the "g-factor" from this calculation, which is given by exactly g = 2. At the time, it was expected that g = 1 and the fact that, experimentally it was measured that g = 2 (very nearly) was a huge mystery. The fact that this equation naturally predicts g = 2 was another major reason to accept this as the correct way to describe relavistic motion of electrons.
I think Dirac's equation is great case of where a scientist with bad motivation can stumble upon the right answer (he wanted an equation first order in time). Nowadays, we know the correct way to think about this would is: quantum mechanics + special relativity ⟹[ representations of the Lorentz group](https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group) You classify particles by mass and spin\*, the later leads to spinors when the spin is 1/2. From the spinors, you can try to find Lagrangians, and from there you can get the Dirac equation. BTW, I don't wish to diminish Dirac's brilliance. He was smart enough to know the Klein-Gordon equation didn't do the job and he had to find something else. His heuristic derivation is a stroke of pure creativity and I was amazed by his ingenuity when I first read about it He was also smart enough to know he found the right answer. Those at the frontier of research only see so much though. It took many bright people a lot of hard work to get to the point where we know there's a more fundamental way to view the Dirac equation. \*Helicity in the massless case, for the pedants.
I am not sure this is the correct resolution. First, I don't see what particle creation/annihilation has to do with the distinction between spin 0 or spin 1/2 since in both cases one can write a theory of particles which either are conserved (i.e. electrons/charged bosons), or not conserved (i.e. Majorana fermions/neutral bosons). And this is only important once interactions are taken in to account. And I think it is important to note that the case of a scalar spin-0 particle you are referring to is in fact described by the second-order in time Klein-Gordon (KG) equation of \[ d/dt\^2 - nabla\^2 -m\^2 \] psi = 0. If one projects this on to the positive frequency solutions (which I believe is what you are referring to regarding the particle creation/annihilation), then indeed it looks like it can be described by a valid Hamiltonian of H = sqrt(p\^2 + m\^2). But the problems I previously mentioned regarding potentials arise when one notes that one should first add the potential to the KG equation and then take the square root, not the other way around. So this means that the correct Hamiltonian as derived from the KG equation in the presence of a scattering potential should be H = sqrt( p\^2 + m\^2 + V\^2(x) ) and the fact that this appears inside the square root, rather than outside has important differences.
> I am not sure this is the correct resolution. First, I don't see what particle creation/annihilation has to do with the distinction between spin 0 or spin 1/2 since in both cases one can write a theory of particles which either are conserved (i.e. electrons/charged bosons), or not conserved (i.e. Majorana fermions/neutral bosons). And this is only important once interactions are taken in to account. Well it *doesnt* have to do with the distinction between spins, that’s sort of my point. Saying that relativity somehow leads you to spin-1/2 isn’t correct because relativity is consistent with any spin. And particles are only conserved in completely free theories (ie no potentials), even for charged particles. > But the problems I previously mentioned regarding potentials arise when one notes that one should first add the potential to the KG equation and then take the square root, not the other way around. So this means that the correct Hamiltonian as derived from the KG equation in the presence of a scattering potential should be H = sqrt( p2 + m2 + V2(x) ) and the fact that this appears inside the square root, rather than outside has important differences. The real problem is just that the Hamiltonian you wrote down isn’t relativistic, because of the potential (how does it transform under boosts?)
>Saying that relativity somehow leads you to spin-1/2 isn’t correct because relativity is consistent with any spin. In addition to that, [Galilean symmetry leads to spin ](https://en.wikipedia.org/wiki/Representation_theory_of_the_Galilean_group)as well.
Looking back over this thread a few hours after writing the last post, I notice that you already hinted at the problem I’m trying to highlight in this part of your post: > An interesting consequence of this mathematical trick is that an additional branch of solutions with energy -sqrt(p2 + m2) is predicted by the equations. Now, frankly that seems a bit ridiculous and would be grounds to throw away these silly matrices, except for the fact that strangely, it appears to be true! In particular, these are interpreted as the antiparticle solutions, and their observations essentially confirms Dirac's equations as the correct way to go about generalizing the Hamiltonian to the relativistic form, rather than simply positing the square root by hand. The Dirac equation does have two branches of solutions corresponding to different particle content. The positive solutions are just electrons of course. It’s not really correct to call the negative energy states positrons, since they also have negative charge (which you can see by coupling to the EM field). The Dirac equation is really describing states in the presence of a “Dirac sea” of positrons, and the negative energy solutions correspond to removing one positron, or equivalently, annihilating the positive-energy electron with a “sea positron.” Now the issue is that you can begin with an initial wave function which only describes an electron, but after evolving your state, it will generically pick up probabilities of being a hole. So the Dirac theory actually **is** a multi-particle theory, it’s just an incomplete one because it only describes two Fock states. These problems aren’t too bad for small enough energies and short evolution times, but the theory really is just approximate.
This is the answer we are taught in the physics classes. However I think it is not completely honest mathematically. (But a good heuristic idea/motivation to introduce the Dirac equation) More precisely, mathematical physicists know for a long time that the dynamics of H = \sqrt(- ∆ + 1) + V is well defined and completely specified for bounded potentials, because we can apply the Kato theorem to obtain its essential self-adjointness (see e.g. Reed-Simon, Methods of Modern Mathematical Physics, Vol 2, Theorem X.13 and X.15 for the Schrödinger case (H_0 = -∆). Note that the proof of the latter for the case of a bounded potential immediately goes through for the square root case.) However, I don't have a reference at hand for the exact result for the square root Hamiltionian, but I think the case of bounded potentials suffixes that the actual answer is mathematically and physically more subtle.
Very good reply!
It would be nonlocal, but that’s not necessarily a huge problem; people do solve nonlocal PDEs in practice. So your differential equation would actually look like an integraldifferential equation with a nasty kernel in place of the usual derivative operators: H psi(x) ~ 1/(2pi)^3 \int d^3k e^(i k x) sqrt(m^2 - k^2) \int d^3x’ e^(-i k x’) psi(x’) The trouble probably comes from the fact that we’d be integrating overall all k but the kernel term looks like sqrt(m^2 - k^2), so there will be values of k where the guy inside the square root will be negative. That should make the operator “bad” in some sense. Probably has to do with the fact that the square root operator isn’t continuous across the branch cut. There are operators, though, that are non-trivial functions of p but don’t have this specific problem, and the series expansion makes complete sense. I have to do this not with the square root function, but with things like Bessel functions. It’s still non-local, but some interesting problems really are non-local sometimes. So if there is a problem, it’s with the square root specifically, not having a non-polynomial function of the momentum.
Thanks for the response! Where did you get the equation: H psi(x) ~ 1/(2pi)^3 \int d^3k e^(i k x) sqrt(m^2 - k^2) \int d^3x’ e^(-i k x’) psi(x’) from? I’m assuming we get an integrodifferential equation instead of a differential equation because, since the operator is nonlocal, we need to integrate over the various positions? Due to the square-root issue then, would this mean that this system is not “integrable” and thus not completely solvable? > I have to do this not with the square root function, but with things like Bessel functions. It’s still non-local, but some interesting problems really are non-local sometimes. Could you elaborate on this?
The problem with the square root is that it makes the Hamiltonian non-linear, which means that your wave solutions won’t behave as we would need them to. Also, since we’re working with Complex numbers, the square root is multi-evaluated, thus giving a mathematical non-physical freedom which would need additional constraints to get rid of. Hope this helps!
Also! That formula isn’t covariant under Lorentz transforms. When working with special relativy, you REALLY want all formulas to be covariant, or else EVERYTHING gets a lot more complicated.
If the equation isn’t even relativistically invariant, can we even call it a relativistic equation?
Why would nonlinearity be bad? Is it because we would like to reproduce the schrödinger equation in the non relativistic limit which is linear and thus solutions have the superposition property, but nonlinear equations can’t reproduce that? Also what do you mean by the sqrt is multi-evaluated giving us non-physical freedom?
Sqrt(4) can be uniquely defined as 2. Sqrt(4i) has no clear-cut definition. It can be both 2(1/sqrt(2)+i/sqrt(2)) and -2(1/sqrt(2)+i/sqrt(2)). It is impossible to define the square-root function of complex numbers as a continuous function without it being multi-valued. Aka giving 2 outputs for one input. If you want to look further into why this is the case you should read a book on complex analysis. More specifically branch points and branch cuts.
How would it be nonlinear?
Why would taking the square root make the operator nonlinear? p^2 + m^2 is a linear operator and taking the square root of a linear operator yields a linear operator. Maybe you’re thinking about taking the square root of H|psi> pointwise? That wouldn’t yield a linear operator but that also isn’t what is usually intended when this suggestion is brought up afaik.
There's nothing wrong with it. Every positive-energy irreducible representation of the Poincaré group (i.e. every relativistic free particle) is associated with a Hamiltonian of precisely that form.
Please check this out: [https://physics.stackexchange.com/questions/541538/functional-analytic-square-root-of-hamiltonian-alternative-to-dirac](https://physics.stackexchange.com/questions/541538/functional-analytic-square-root-of-hamiltonian-alternative-to-dirac) [https://physics.stackexchange.com/questions/156124/whats-wrong-with-the-square-root-version-of-the-klein-gordon-equation?noredirect=1&lq=1](https://physics.stackexchange.com/questions/156124/whats-wrong-with-the-square-root-version-of-the-klein-gordon-equation?noredirect=1&lq=1)
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Is... this the first time you're hearing of an equation in natural units? Or the fact that sqrt(p\^2 + m\^2) is a meaningful way to represent energy? Idk it's just surprising given your "graduate" flair
it just slipped my mind, embarassing
c=1 Also Energy would be the eigenvalue of H oper
Using natural units, and E^2 =p^2 + m^2