What Happens When the Hamiltonian in Dirac's Equation Isn't Linear in Momentum?

[Tio Barnabe]

When constructing a relativistic quantum mechanical equation, namely Dirac equation, what would happen if we choose the Hamiltonian so that it's not linear in the momentum operator and the rest energy?

You could say, why don't try it yourself and see what happens? That's because my knowledge is not enough to do that, but as I love QM I want to know what the result would be.

 

[MisterX]

By linear I assume you mean it's raised to the first power. The Hamiltonian has to be linear in momentum. This is because the Schrödinger equation is linear in the time derivative $H\psi = i\hbar\partial_t \psi$. Now for relativity time an space are interchangeable, so the time and space derivatives should have the same order. Thus if we want only the first time derivative, we must have an equation linear in momentum.

 

[dextercioby]

When constructing a relativistic quantum mechanical equation, namely Dirac equation, what would happen if we choose the Hamiltonian so that it's not linear in the momentum operator and the rest energy? [...]

You end up with the Hamiltonian for the Klein-Gordon field.

 

[Tio Barnabe]

thank you guys

 

[PeterDonis]

The Hamiltonian has to be linear in momentum. This is because the Schrödinger equation is linear in the time derivative $H\psi = i\hbar\partial_t \psi$. Now for relativity time an space are interchangeable, so the time and space derivatives should have the same order. Thus if we want only the first time derivative, we must have an equation linear in momentum.

You have this backwards. What relativity forces on us is not a Hamiltonian linear in momentum; it's having to accept a second time derivative instead of a first time derivative. The Schrödinger equation is non-relativistic; when you try to make a relativistic analogue, you end up, as @dextercioby has pointed out, with the Klein-Gordon equation, which involves only second derivatives.