Why are negative energy solutions problematic in the Klein-Gordon and Dirac equations?

[reilly]

Strj-- Read.

Regards,
Reilly Atkinson

 

[Sterj]

Ok, ok. I read a few pdf. files. There are files that say:
The total energy of an electromagnetic field is E=h(bar)*w/2+h(bar)*w+h(bar)*w+...
and there are some files where it is written:
The total energy of an electromagnetic fiel is E=h(bar)*w+h(bar)*w+...
The second is without the ground oscillation.

Now, what is true?

 

[Edgardo]

Ok, ok. I read a few pdf. files. There are files that say:
The total energy of an electromagnetic field is E=h(bar)*w/2+h(bar)*w+h(bar)*w+...
and there are some files where it is written:
The total energy of an electromagnetic fiel is E=h(bar)*w+h(bar)*w+...
The second is without the ground oscillation.

Now, what is true?

Hello Sterj,

I found this document from http://iftia9.univ.gda.pl/~sjk/skok/om03.pdf
Have a look at page 37, "We renormalize the energy by dropping the term 1/2"

Do you know the book "Quantum Field Theory" by Mandl and Shaw?
In chapter 1.2.3 the text says: "This constant ( $\frac{1}{2} \sum_{k} \sum_{r} \hbar \omega_{k}$) is of no physical significance. Just scale the energy by replacing equation (1.30) by $H_{rad} = \sum_{\vec{k}} \sum_{r} \hbar \omega_{k} a^{\dagger}_{r}(\vec{k}) a_{r}(\vec{k})$, where (1.30) is
$H_{rad} = \sum_{\vec{k}} \sum_{r} \hbar \omega_{k} \left( a^{\dagger}_{r}(\vec{k}) a_{r}(\vec{k}) +\frac{1}{2} \right)$

 

[Sterj]

ok, but in reality the term is there, but we don't use it.
thanks

 

[Cinderella]

Antiparticles have positive kinetic energie

In the case of electrons, you are mixing this with potential energy. Indeed the latter is negative (Coulombic potential). However, electrons obey the positive square root relation for energy.

marlon

Thanks a lot. My problem is really quite basic, I´m afraid. Are you saying something like the following?
" Traditionally one would have thought that potential energy can be negative but kinetic energy cannot. And indeed kinetic energy cannot be negative for the electron. However, the fact that there are solutions for free particles displaying negative (kinetic) energies has led to the discovery of anti-particles, such as the positron. And they can have negative kinetic energies."
Is this a correct understanding?[/QUOTE]


Antiparticles do not have negative kinetic energy, because if a particel with energy E and an antiparticel wirh energy (-)E collide the resulting energy, which may be set free as radaiation, is not zero, its 2E, this is an experimental fact. Out from the relativistic wave equations e.g. the Klein-Gordon equation it does not follow necessarily that that kinetic energy is negative. The energy occures there in terms of the product Et, so that an altermative interpretation of a negaive product -Et is that antiparticles move backwards in time with positive energy.

 

[Antiphon]

Cinderella has it right.

Energy in a tunneling problem going negative is the result of that particular
problem's boudnary conditions and reference energy levels. It does not mean
that the energy density surrounding the particle (or the majority of its wavefunction)
has become negative.

An open region of space with a true negative energy density would repel ordinary
matter and be "antigravitational". To form a very loose analogy to charge polarity,
negative energy density is the opposite gravitational "charge" as compared with
ordinary matter. (For staunch relativists who are offended by the concept of
gravitational "charge", a negative energy density will curve spacetime in the
opposite way that a positive energy density does.)