Lagragians for spin 1/2 fields

[LAHLH]

Hi,

Why would having [tex] \partial\psi\partial\psi [/tex] lead to a Hamiltonian that is unbounded below? Srednicki states that in order to have a bounded Hamiltonian one must include [tex]\psi^{\dag} [/tex] in the combination too.

Also why exactly do we require or Lagrangian to be Hermitian, is this somehow to give real eigenvalues for observables like in QM?

cheers

 

[haushofer]

I'm also puzzled by this statement. The Lagrangian however has to be Hermitian to ensure CPT-invariance.

 

[LAHLH]

I'm also puzzled by this statement. The Lagrangian however has to be Hermitian to ensure CPT-invariance.

Thanks, could you say anymore about Hermiticity ensures CPT invariance or anywhere I could look this up?

Anyone know about why the term from Srednicki leads to unbounded Hamiltonians from below?

 

[Avodyne]

The hamiltonian constructed from this term would be
[tex](\dot\psi)^2 + (\nabla \psi)^2[/tex]
which is not positive-definitie because [itex]\psi[/itex] is complex. The square of a complex number is just another complex number.

 

[LAHLH]

The hamiltonian constructed from this term would be
LaTeX Code: (\\dot\\psi)^2 + (\\nabla \\psi)^2
which is not positive-definitie because LaTeX Code: \\psi is complex. The square of a complex number is just another complex number.

Isn't [tex] \psi [/tex] just an operator not a complex number?

how can you talk about positive and negativeness for the sum of two complex numbers? and how is this situation different from the term [tex] \psi\psi [/tex] being included which is legal? not sure I follow your reasong for why this term is unbounded below.

 

[naima]

I and how is this situation different from the term [tex] \psi\psi [/tex] being included which is legal?

This term is only legal with its hermitian conjugate.

 

[LAHLH]

This term is only legal with its hermitian conjugate

Yeah, I'm aware of that, sorry, was just being sloppy for the sake of brevity. The Lagragian must be Hermitian so my previous post should read [tex] \psi\psi+\psi^{\dag}\psi^{\dag} [/tex]

 

[dextercioby]

[tex] \psi\psi + \psi^{\dagger}\psi^{\dagger} = 0 + 0 = 0 [/tex]

because the Dirac/Weyl spinor fiels are fermionic variables (Grassmann parity = 1).

The mass term is proportional with [itex] \psi^{\dagger}\psi [/itex].

And an interesting question i/o the one from the original post would be:

'Why isn't the <kinetic> term in the Dirac/Weyl Lagrangian second order in the derivatives

[tex] \mathcal{L}_{kin}= k \left(\partial^{\mu}\psi^{\dagger}\right)\left(\partial_{\mu}\psi\right) [/tex]

?

 

[LAHLH]

[tex] \psi\psi + \psi^{\dagger}\psi^{\dagger} = 0 + 0 = 0 [/tex]

because the Dirac/Weyl spinor fiels are fermionic variables (Grassmann parity = 1).

The mass term is proportional with [itex] \psi^{\dagger}\psi [/itex].

This is not true, for Weyl spinors at least the mass term is proportional to [tex] \psi\psi + \psi^{\dagger}\psi^{\dagger} [/tex]

It is true that the spinors anticommute in the sense [tex] \chi_{a}\psi_{b}=-\psi_{b}\chi_{a}[/tex].

But [tex] \chi\psi=\chi^{a}\psi_{a}=-\psi_{a}\chi^{a}=+\psi^{a}\chi_{a}=\psi\chi [/tex]
The third equality follows from anticommutation, the next by properties of [tex]\epsilon[/tex] contraction.

It is in fact precisely because of the anticommutation that [tex] \psi\psi=\epsilon^{ab}\psi_{b}\psi_{a}=\psi_{1}\psi_{2}-\psi_{2}\psi_{1}\neq 0[/tex]

However if one is considering the Dirac field then the mass term is proportional to [tex] \bar{\Psi}\Psi[/tex]

I'm interested in the answer to the question you posed about 'why not second order' too, and why this leads to unbounded Hamiltonian