In this case, one needs to enforce that E = − − p → → 2 + m 2 for this solution to be valid in a frame where the particle has p → → = 0 → → . The proof follows analogously to the previous case. This is the so-called negative energy solution. It can sometimes become confusing to carry around an explicitly negative energy, and so it is conventional to flip the sign on both the energy and the momentum, and to write this as ψ ψ ( − − ) = v ( χ χ ) ( p → → ) e i p ⋅ ⋅ x = E + m 2 m [ σ σ → → ⋅ ⋅ p → → E + m χ χ χ χ ] e i p ⋅ ⋅ x