What is negative energy and how does it predict antimatter?

According to multiple online sources, antimatter was discovered through the Dirac equation because there were multiple solutions; a positive energy solution, to be expected and a negative energy solution. What does this mean. Also I read something about this giving rise to a symmetry called CPT symmetry, however that is not the main focus of this question.

I would just like an explanation of what negative energy is and why it results in antimatter.

In brief :

From relativity, the Dirac equation gives the following relation : $$\tag{1} E^2 = p^2 c^2 + m_0^2 \, c^4,$$ which is a second order algebraic equation, with two roots : $$\tag{2} E = \pm \, \sqrt{p^2 c^2 + m_0^2 \, c^4}.$$ Now, quantum mechanics requires that all solutions be considered, since any superposition of solutions is another solution to the linear Dirac equation. Thus, you have negative solutions to consider. You can't just throw them away just because they don't have "physical" sense to you. Now, the Dirac equation admits an operation (complex conjugate and a matrix multiplication) which can convert a negative solution to a positive solution traveling in the opposite direction, and reversed spin and electric charge. This operation gives another solution of the Dirac equation : $$\tag{3} \psi_{\text{c}}(t, x) = \gamma^2 \, \psi^{\ast}(t, x).$$ This suggest that the negative solution may be interpreted as an "anti-particle", i.e. one with a reversed spin and electric charge.

• One comment - the equation (1) is pure special relativity, isnt it? Any special relativity compatible system ends up with that, desnt it? – jaromrax Apr 6 '17 at 11:05
• Yes, equ (1) is imposed by special relativity, for any real particle. This is the relation that gives the Klein-Gordon equation after quantisation. The "square root" of equ. (1) gives the Dirac equation : $$E \, \psi = \alpha_i \, p_i \, c \, \psi + \beta \, m c^2 \, \psi,$$ where $\alpha_i$ and $\beta$ are the four Dirac matrices. – Cham Apr 6 '17 at 13:07

Negative Energy States

Following on from Cham's answer, in Quantum Mechanics the negative matter will still have energy $E = \hbar \nu$ since $E \rightarrow - E$ and $\hbar \rightarrow - \hbar$, meaning the de Broglie wavelength is positive. The Uncertainty principle is $$(\Delta x)^2 (\Delta P_x)^2 \geq \hbar^2/4$$ Whereas another relation is $$(\Delta x) (-\Delta P_x) \geq -\hbar/2$$

The Heisenberg equation is also invariant, mass becomes negative in the Schrodinger equation and since energy-momentum operators are invariant. Hence $$-E=\frac{p^2}{-2m}+U(r)$$ Corresponding to $$i \hbar \frac{\partial \psi}{\partial t}=\frac{\hbar^2}{2m}\nabla^2 \psi -U(r) \psi$$ Here only $U \rightarrow -U$. The Klein-Gordon equations and Dirac equations are also invariant.

Negative matter in Inflationary Cosmology

The Standard model of Inflationary Cosmology has problems. They relate to expansion, antimatter, flatness etc. The Schwarzschild metric of negative matter should be $$ds^2 = \left(1+\frac{2m}{r} \right)dt^2 - \frac{dr^2}{1-(2m/r)}-r^2 (d \theta^2 + \sin^2 d \phi^2)$$

Einstein and Rosen investigated the particle problem in GR, proposing a new variable, namely $u^2=r+2m$ or, $u= \pm \sqrt{r+2m}$ gives two corresponding sheets to $u >0$ and $u < 0$ are then joined by the Einstein-Rosen bridge to $r=-2m$ (or $u=0$) for which $g_{\mu \nu}=0$.

• i didn downvote, but i humbly think, that you cannot have a negative mass $m$. And i have never seen Schrodinger with $-$ signs. I have never seen $-\hbar$ The other answer seems to me more reasonable. – jaromrax Apr 6 '17 at 11:09
• @jaromrax the negative signs in front of $\hbar$ and $m$ arise in consideration of so-called "phantom matter" which can serve as a model for dark energy (which, mathematically is modelled through a negative energy density). There there are a number of problems with this notion however, such as the violation of the energy conditions and that of a related negative energy density, the lack of a plausible explanation of the origin of such phantom matter etc. Because of those, such possibility of existence becomes rather non-realistic. I offer the above answer, with those caveats in mind. – Hyperkähler Apr 7 '17 at 9:36