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I'v just read the solutions of Dirac equation and haven't yet known anything about the quantization of the equation, so my confusion is about this level.The problem is trivial. The solutions of Dirac equation are as follows: $$\left|\psi ^1\right\rangle =\sqrt{\frac{m+e}{2 m}} e^{-\text{ipx}} \left( \begin{array}{c} 1 \\ 0 \\ \frac{p^3}{m+e} \\ \frac{p^1+\text{ip}^2}{m+e} \\ \end{array} \right)=u_1 e^{-\text{ipx}};$$ $$\left|\psi ^2\right\rangle =\sqrt{\frac{m+e}{2 m}} e^{-\text{ipx}} \left( \begin{array}{c} 0 \\ 1 \\ \frac{p^1-\text{ip}^2}{m+e} \\ -\frac{p^3}{m+e} \\ \end{array} \right)=u_2 e^{-\text{ipx}};$$ $$\left|\psi ^3\right\rangle =\sqrt{\frac{m+e}{2 m}} e^{\text{ipx}} \left( \begin{array}{c} \frac{p^3}{m+e} \\ \frac{p^1+\text{ip}^2}{m+e} \\ 1 \\ 0 \\ \end{array} \right)=v_2 e^{\text{ipx}};$$ $$\left|\psi ^4\right\rangle =\sqrt{\frac{m+e}{2 m}} e^{\text{ipx}} \left( \begin{array}{c} \frac{p^1-\text{ip}^2}{m+e} \\ -\frac{p^3}{m+e} \\ 0 \\ 1 \\ \end{array} \right)=v_1 e^{\text{ipx}}$$

The corresponding inner product relations are as follows: $$u_{s \overset{\rightharpoonup }{p}} u_{r \overset{\rightharpoonup }{p}}{}^{\dagger }=v_{s \overset{\rightharpoonup }{p}} v_{r \overset{\rightharpoonup }{p}}{}^{\dagger }=\frac{e \delta ^{\text{rs}}}{m};$$ $$v_{s \left(-\overset{\rightharpoonup }{p}\right)} u_{r \overset{\rightharpoonup }{p}}{}^{\dagger }=0;$$

My question is:

For two particle states $\left|\psi ^1\right\rangle$and $\left|\psi ^4\right\rangle$, if they are going to have a same momentum and opposite energy, then the condition $v_{s \left(-\overset{\rightharpoonup }{p}\right)} u_{r \overset{\rightharpoonup }{p}}{}^{\dagger }=0$ is satisfied. This illustrates that the two particles cannot have their states related, therefore they do not interact. But, what if I choose to make these two particles have opposite momentum direction (i.e.$v_{s \overset{\rightharpoonup }{p}} u_{r \overset{\rightharpoonup }{p}}{}^{\dagger }\neq0$), then the inner product of the two states cannot vanish. Does this mean that under such circumstances the two particles can interact? If I am guessing right, what are the interactions? Annihilation? Sorry for being so lengthy and expatiatory.

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You are confusing the negative energy solutions of Dirac's equation (your third and forth wavefunction) with legitimate particle (or antiparticle wavefunctions). You can't learn anything about physical interactions by studying the inner products of these solutions with the positive energy solutions (which do represent real particle wavefunctions).

Dirac realized that these negative energy solutions represented a serious pathology of his theory and overcame this pathology by postulating that these states must be fully occupied and inaccessible due to the Pauli principle. This enabled him to formulate his hole theory that antiparticle states would result if one of these states in the infinite Dirac sea became unoccupied by the absorption of an energy quanta greater than twice the rest mass of an electron. Particle-antiparticle annihilation would then occur when a positive energy electron undergoes a transition that occupies the hole state. For a summary of this you might Google Dirac hole theory.

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  • $\begingroup$ oh! I was thinking of the two particles being entangled. This is obviously wrong. $\endgroup$ – ZHANG Juenjie Oct 19 '16 at 9:17

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