How can the Klein-Gordon equation have negative-energy solution if its Hamiltonian is positive-definite?

Question

In a lesson about the introduction of classical field theory it was mentioned the Klein-Gordon equation $$(\Box + m^2) \phi(x) = 0. \tag{1}$$ But before we got this equation, we studied the Hamiltonian density that we obtained from the Lagrangian density and defined as $$\mathcal{H}=\frac{1}{2} \left(\dot\phi^2+|\nabla\phi|^2+m^{2}\phi^2\right), \tag{2}$$ and it was said that $\mathcal{H}$ is the total energy of the field. What I can't understand is when it was commented that the solutions of (1) can have negative energy, while the Hamiltonian is defined positive, how? Am I missing something?

Answer 1

If you interpret the Klein-Gordon equation as describing a field theory (which is the modern way to look at it), then yes, the energy of the field is positive definite.

The problem is if you try to interpret the Klein-Gordon equation as a relativistic generalization of the Schrodinger equation, describing the motion of a single quantum particle. Then the energy of the wavefunction is related to its frequency by $E=\hbar \omega$. However, there are positive and negative frequency solutions to the Klein-Gordon equation, $\sim \exp\left[\pm i \left(\sqrt{\vec{k}^2 + m^2}\right) t\right]$. The problem is resolved by understanding that the Klein-Gordon equation should not be interpreted as a "relativistic Schrodinger equation for one particle", but rather as the classical equations of motion for a field we are going to quantize (or, as the Heisenberg equations of motion for a quantum field).

Answer 2

Briefly speaking, there are at least two notions of energy $E$ and momentum ${\bf p}$:

1. One notion stems from the Fourier-transform $({\bf x},t)\to ({\bf p},E)$ of spacetime, or equivalently, a plane-wave expansion $$\phi({\bf x},t)~\sim~\exp\left(\frac{i}{\hbar} ({\bf p}\cdot {\bf x}-E t)\right).\tag{1}$$ Here the energy $E$ can in principle take both positive and negative values.

2. Another notion arises from the energy-momentum tensor.

For the Klein-Gordon field, the latter energy happens to be the square of the former, and is therefore positive.

Answer 3

You have a valid point. 'Negative energy solution' is a misnomer. Negative as well as positive frequency solutions have positive energy, as the Klein-Gordon hamiltonian correctly implies. As an example, electrons and positrons both have positive rest energy mc$^2$.

Answer 4

I sometimes think that the whole negative energy thing is a bit overrated. For example one can solve the classical Klein-Gordon equation and find via the appropriate ansatz that the solutions will be plane waves: $$\phi(x)=e^{i(k^0t-k_ik^ix)}.$$ Further we know that the equation implies that $k^0=\pm\sqrt{k_ik^i+m^2}$, which is natural to interpret as the relativistic energy. However, as you note in your post, the Hamiltonian density, which physically corresponds to the total energy of the field since $\mathcal H=T^{00}$, is positive. Thus, it seems reasonable to choose the positive branch for $E=k^0=\sqrt{k_ik^i+m^2}$, if the solutions are to be physically significant. However, I must have two linearly independent solutions of the Klein Gordon operator in order to span the solution space, so the general solution must be something like: $$\phi(x)=\int_{-\infty}^{+\infty} d^3k\bigg[\alpha^*(\vec k)e^{i(k^0t-k_ix^i )}+\alpha(\vec k)e^{-i(k^0t-k_ix^i )}\bigg].$$ Here, I have a general solution without any assumption that the energy of any mode is negative. One can argue that the $\alpha$ term has a negative energy or frequency because of the minus sign in the exponential, however, this is just a result of taking the complex conjugate for the sake of constructing a linearly independent solution from $e^{i(k^0t-k_ix^i)}$, as opposed to a negative energy mode, or a mode traveling backwards in time.

I think folks would have been fine with choosing the positive branch for $k^0$, however, even if one does this, a "first quantized" Klein-Gordon theory does not make a suitable relativistic generalization of the Schrodinger equation because the probability density lacks positive definiteness as a result of the Klein Gordon equation being second order in time as well as space. As a classical field, however, there does not seem to be any problem.