Every wavefunction of a form $\Psi(x)$ can be described as a superposition of multiple free particle solutions. We can see the following Fourier transform: $$ \psi(x) = \int e^{ik\cdot x} \psi(k) dk $$

We can make it time dependent by replacing $e^{ik\cdot x}$ with $e^{ik\cdot x - i E_k t}$ where $E_k = \sqrt{k^2 + m^2}$. The result gives the positive energy time evolution of any initial wavefunction data.

The question is,

A. This may violate locality, as $E_k$ can be set as we want to, as $k$ can be set as we want to. Is this correct?

B. Why is this related to requiring a multi-particle theory to fix formalism in Klein-Gordon equation? (Context: Complete set and Klein-Gordon equation)

C. How is adding negative energy preserve locality?

  • $\begingroup$ Maybe I misunderstand what you are asking, but while it is true that the free particle 'states' form a basis, they are not necessarily an eigenbasis of the hamiltonian. So you can expand every wavefunction in that basis, but the time evolution will not be given by the combined time evolution of the free particle 'states' $\endgroup$ – A.O.Tell Oct 8 '12 at 11:54

Part A

A. This "simple would-be relativistic quantum theory" indeed violates locality, but not for the reason you write. You wrote that the energy i.e. the Hamiltonian is defined as $$ H =\sqrt{p^2+m^2} $$ so it cannot be "set as we want to", independently of $k$. In fact, the problem is that $H$ is too constrained, not that it is too free. By your rules, it (and even its sign that is demanded to be positive) is a function of $\vec k$. Instead, the right reason why this dynamics is non-local is that the displayed equation above – which is in the momentum representation – may be translated to the position representation. The momentum is always $$ \vec p = -i\hbar \nabla $$ which is uniquely determined by the canonical commutators and/or the fact that the momentum generates translations in space (the space-translational symmetry is equivalent to the momentum conservation by Noether's theorem). So the Hamiltonian in the position representation is $$ H = \sqrt{m^2 - \hbar^2 \nabla^2} $$ I shifted the squared mass to the first place in the square root because it makes it more obvious that we may try to Taylor-expand the formula as $$ H = \dots = m\sqrt{1 - \hbar^2/m^2 \cdot \nabla^2} = m(1-\dots) $$ The point here is that the Taylor expansion contains arbitrary powers of the $\nabla$ operators, including the 100th power. If something depends on derivatives of arbitrarily high orders, it is nonlocal. (For a well-known example, consider Taylor expansions of $f(x)$. The value $f(1)$ may be calculated from $f(0)$ and all other derivatives at $x=0$.) There are also other ways to show that $\psi(x,t+dt)$ would depend on $\psi(y,t)$ for $y$ being arbitrarily far from $x$.

Part B

B. The underlying "unnatural aspect" of the construction above is that it takes one of the values of the square root and neglects the other one. More precisely, it "violently" tries to put the coefficients of $\exp(-iEt)$ for all negative values of $E$ to zero. That's brutally constraining and it's the ultimate cause of the nonlocality because ordinary enough functions of space and time have both positive-frequency and negative-frequency components. In particular, for example, real functions $\psi(x,t)$ have "equally strong" positive- and negative-energy contributions, $\tilde\psi(-k,E)=\tilde\psi^*(k,E)$.

So all the right solutions how to write down quantum mechanical equations for particles compatible with relativity must work with objects such as wave functions that contain both positive- and negative-energy components. This doubles the number of degrees of freedom and such a doubling is only possible if we double the number of initial conditions (one has to add $\dot\Psi(t)$ next to $\Psi(t)$ as initial conditions). Your equation was "unnaturally" first-order in time, but it was infinite-order in spatial derivatives (this asymmetry between space and time is another way to see it wasn't really compatible with relativity despite superficial attempts to pretend it was). Instead, the right equation is 2nd order in both space and time, it's the Klein-Gordon equation $$ (\partial^2/\partial t^2 -\nabla^2 +m^2) \Phi = 0. $$ It's almost the same thing but it's important that we don't try to take the square root manually. (The Dirac equation and even Maxwell equations in the vacuum imply that the components obey the Klein-Gordon equations above, too: they just constrain the polarizations in various ways.)

Such Klein-Gordon equation and all of its generalizations therefore has to have both positive- and negative-energy components. However, in the real world, the energy must be bounded from below – there can't be states with energy lower than the energy of the vacuum – so it must be impossible to create particles with these negative energies. One has to treat them differently.

The only possible "different treatment" means that the negative-frequency solutions are actually linked to creation operators for the particle while the positive-frequency ones are linked to some annihilation operators. It means that the Klein-Gordon field $\Phi(x,t)$ is no longer a combination of "just creation operators", describing "one added particle" or its wave function. It's a combination of operators adding $N=+1$ and those with $N=-1$. It is able to reduce the number of particles at the same moment.

Consequently, everything that depends on such objects is inevitably able to raise and lower the number of particles and there's no way to study processes that have a fixed number of particles in isolation. This has many precise manifestations in quantum field theory. For example, if one studies the scattering of two particles, it's always possible that we create a particle-antiparticle pair out of the excess energy. The probability of such processes changing the number of particles is nonzero and in fact, it may be related to the probability of other processes that preserve the number of particles.

Part C

C. If the answer got lost, allowing the negative energy modes meant that we wrote the second-order differential equation in time with $\partial^2/\partial t^2$ in it (we had to add the first derivatives $\partial/\partial t(\Phi)$ to the list of initial conditions whose number was therefore doubled, by the way) and this allowed us to avoid the awkward square root above the $k^2+m^2$, and this square root was what produced the nonlocality (via spatial derivatives of arbitrarily high orders). Instead, the equation got symmetrically dependent on 2nd derivatives with respect to both space and time, which is really nicely relativistic, and because only finite-order spatial derivatives are used, the value of $\Phi(x,t+dt)$ only depends on $\Phi(y,t)$ for $y$ belonging to the infinitesimal neighborhood of $x$: the evolution is local.


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