3
$\begingroup$

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?

$\endgroup$
2
  • 3
    $\begingroup$ When I saw that square symbol at the beginning of the equation I thought my browser couldn't render some weird Unicode character. As it turns out, no, it really looks like that! According to Wikipedia it's called the d'Alembert operator, or wave operator, or box operator. $\endgroup$ Feb 16 at 10:44
  • $\begingroup$ @FabiosaysReinstateMonica LOL! Thats funny!I have never liked to write the D'Alembertian that way, a square just does not look good! I always just use either $\nabla^2$ (keeping in mind that $dim=3+1$) or $\partial_\mu\partial^\mu$. $\endgroup$ Feb 16 at 12:46

4 Answers 4

9
$\begingroup$

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).

$\endgroup$
1
  • 1
    $\begingroup$ The Klein-Gordon equation is perfectly acceptable as a relativistic quantum equation for a single particle. In 'QFT' by Itzykson&Zuber, the Klein-Gordon equation is solved for the hydrogen atom. The answer is off because spin is not included. When spin ís included, in the so-called squared Dirac equation, which really is the relativistic form of the Pauli equation, the solution is correct. $\endgroup$
    – my2cts
    Feb 16 at 12:10
6
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Of course these notions should reduce to one and the same thing. The volume integral of the energy-momentum tensor is the conserved energy-momentum vector. $\endgroup$
    – my2cts
    Feb 16 at 11:48
  • 3
    $\begingroup$ @my2cts they really need not, $E$ in $e^{-iEt/\hbar}$ is eigenvalue of some Hamiltonian. In general, Hamiltonian value is not necessarily the system energy, this is seen already in classical mechanics of systems with finite number of degrees of freedom. $\endgroup$ Feb 16 at 12:21
  • 1
    $\begingroup$ @JánLalinský If they don't an error was made. Note: this question so far is about free particles. $\endgroup$
    – my2cts
    Feb 16 at 12:27
  • $\begingroup$ @Qmechanic I see. It all depends on which definition of $E$ that one deems fundamental. At the end of the day, either view will work; the difference is in the way one reconciles $\pm E$ or sidesteps the issue altogether by utilizing the stress-energy tensor. $\endgroup$ Feb 16 at 17:24
5
$\begingroup$

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$^2$.

$\endgroup$
1
  • 2
    $\begingroup$ I agree, just because there are minus signs in the exponentials of KG (bosons) or Dirac(fermions) solutions doesn't mean that there are negative energies. "Misnomer" is a good way of putting it. $\endgroup$ Feb 16 at 12:25
3
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Energy of a field outside of classical mechanics is not "mechanical"; it is just energy of the field, or field energy. It can be defined based on some appropriate Hamiltonian function. Mechanics/mechanical usually refers to systems with finite number of degrees of freedom. $\endgroup$ Feb 16 at 10:23
  • $\begingroup$ @JánLalinský Isn't the OP talking about a classical field? At least that was what I was talking about, as I didn't get into the KG field as a QM equation until later and never mentioned its role in QFT. $\endgroup$ Feb 16 at 12:20
  • $\begingroup$ Yes, but "classical" there refers to mathematical property that we deal with c-number fields, not that the field is a mechanical system, or a model of a mechanical system of finite number of degrees of freedom. $\endgroup$ Feb 16 at 12:25
  • 1
    $\begingroup$ I see, I will edit that off the post, thanks! $\endgroup$ Feb 16 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.