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  In the Chapter/Section 36 of Srednicki's Quantum Field Theory, he argued that the terms of $\partial^\mu\psi\partial_\mu\psi + h.c.$ cannot be included in the Lagrangians for spinor fields because of that these terms would yield a Hamiltonian with unbounded below. I cannot understand this argument, so I try to write down the Lagrangian of $$ \mathscr L = -\frac12 (\partial^\mu \psi \partial_\mu \psi + h.c.), $$ in which I ignore the quadratic terms like $\psi\psi=\psi^a\psi_a=\varepsilon^{ab}\psi_a\psi_b$ and its Hermitian conjugate, which can be interpreted as the mass terms.

  Then, I calculate the conjugate momentum of $\psi_a$ and its Hermitian conjugate $\psi_{\dot a}^\dagger$ like that $$ \begin{aligned} &\Pi^a(x) = \frac{\partial\mathscr L}{\partial(\partial_0\psi_a)} = \varepsilon^{ab}\partial_0\psi_b =: \dot\psi^a(x), \\ &\Pi^{\dagger\dot a}(x) = \frac{\partial\mathscr L}{\partial(\partial_0\psi^\dagger_\dot{a})} = \varepsilon^{\dot a \dot b}\partial_0\psi^\dagger_\dot{b} =: \dot{\psi}^{\dagger\dot a}(x). \end{aligned} $$ Therefore, we can get the Hamiltonian from Legendre transformation $$ \begin{aligned} \mathscr H & = \Pi\dot{\psi} + \Pi^\dagger\dot{\psi}^\dagger - \mathscr L \\ & = \dot\psi\dot\psi + \dot\psi^\dagger\dot\psi^\dagger - \frac12 \left[ \dot\psi\dot\psi - (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger - (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right] \\ & = \frac12 \left[ \dot\psi\dot\psi + (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger + (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right]. \end{aligned} $$ Then, how can I argue that Hamiltonian is unbounded below? I'd like to explain from $$ \psi \psi = \psi^a \psi_a = \varepsilon^{ba} \psi_a \psi_b = \psi_2 \psi_1 - \psi_1 \psi_2, \\ \psi^\dagger \psi^\dagger = \psi^\dagger_\dot{a} \psi^{\dagger\dot a} = \varepsilon^{\dot a\dot b} \psi^\dagger_\dot{a} \psi^\dagger_\dot{b} = \psi^\dagger_{\dot1} \psi^\dagger_{\dot2} - \psi^\dagger_{\dot2} \psi^\dagger_{\dot1}, $$ in which the minus signs will make the $\mathscr H \ge 0$ not correct. Is that right?

  Finally, Srednicki chose the term of $i \psi^\dagger \bar\sigma^\mu \partial_\mu\psi$ included in the Lagrangian, whose hermicity is shown in $(36.1)$. But I have the same question about it that does this term of $i \psi^\dagger \bar\sigma^\mu \partial_\mu\psi$ yield a Hamiltonian with bounded below?


New Argument: The conjugate momentum of the spinor fields may be wrong, and they should be $$ \begin{aligned} \Pi^a & = \frac{\partial\mathscr L}{\partial (\partial_0\psi_a)} \\ & = \frac12 \left[ \delta_{ac}\varepsilon^{bc}\partial_0\psi_b - \delta_{ab}\varepsilon^{bc}\partial_0\psi_c \right] \\ & = \epsilon^{ba}\partial_0\psi_b = -\dot\psi^a, \end{aligned} $$ and $$ \Pi^{\dagger\dot a}=-\dot\psi^{\dagger\dot a}. $$ Therefore the Hamiltonian of this theory is $$ \begin{aligned} \mathscr H & = \Pi\dot{\psi} + \Pi^\dagger\dot{\psi}^\dagger - \mathscr L \\ & = -\dot\psi\dot\psi - \dot\psi^\dagger\dot\psi^\dagger - \frac12 \left[ \dot\psi\dot\psi - (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger - (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right] \\ & = \frac12 \left[ -3\dot\psi\dot\psi + (\nabla\psi)\cdot(\nabla\psi) - 3\dot\psi^\dagger\dot\psi^\dagger + (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right], \end{aligned} $$ in which the Hamiltonian, I think, is no positive definite. Some useful arguments may be that put this Hamiltonian as the production of annihilation operators and creation operators.

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    $\begingroup$ @ohneVal : Which term of $\dot\psi\dot\psi + \dot\psi^\dagger\dot\psi^\dagger - \frac12 \left[ \dot\psi\dot\psi - (\nabla\psi)\cdot(\nabla\psi) + \dot\psi^\dagger\dot\psi^\dagger - (\nabla\psi^\dagger)\cdot(\nabla\psi^\dagger) \right]$? Do you mean that the terms from $\mathscr L$ should have a extra minus? The convention of the metric is $g_{\mu\nu}=\text{diag}(-1,1,1,1)$, so I don't think there is any problem. $\endgroup$ Oct 8, 2020 at 14:16
  • $\begingroup$ O so you are using the other signature, more appropriate for GR. In particle physics you usually prefer the time direction to be positive so that the dispersion relation is more natural, here perhaps you have to be careful because squaring a four momentum gives you - mass^2 and so on, just saying. $\endgroup$
    – ohneVal
    Oct 8, 2020 at 14:16
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    $\begingroup$ @ohneVal : I know what you mean. Srednicki chose this signature, and this question is from his QFT, so I chose it too. $\endgroup$ Oct 8, 2020 at 14:19
  • $\begingroup$ A stronger argument is Lorentz covariance, you can probably check that those interaction terms are not Lorentz covariant without the bar on psi, check Peskin $\endgroup$
    – ohneVal
    Oct 8, 2020 at 14:22
  • $\begingroup$ By his construction of the irreducible representation of the Lorentz group, the Lorentz covariance or invariance is not broken by the terms like $\partial_\mu\psi \partial^\mu\psi$. And $\psi$ is this term is the 2-component spinor of the left-handed representation of the Lorentz group. $\endgroup$ Oct 8, 2020 at 14:35

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I believe Srednicki's comment was made in the direction of the inner-product structure not being held. So he is just trying to motivate the construction of the Lagrangian for a spin $1/2$ fermion. Namely he mentions that before he suggests to take $\psi^\dagger$, a more naive option is the one you wrote. Now, without the dagger the product of two complex numbers is not even real (I am assuming, that we don't know yet know about a spinorial structure) so they cannot possibly define norms, so in the expression you have in your $\cal{H}$ you have squares of complex numbers. Suppose $\dot{\psi} = i K $ for some constant $K$ (while holding all spatial derivatives equal to zero w.l.g) then $$\dot{\psi}\dot{\psi} = -K^2$$ and I can make that as negative as I want. Moreover one can build such a field while satisfying Klein-Gordon's equation.

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  • $\begingroup$ Yeah, I found some other mistakes in the calculation of the conjugate momentums because of the anticommunities between two spinor fields. Therefore it will be explicit that the Hamiltonian is not positive definite. I have renewed these arguments after the origin question. $\endgroup$ Oct 9, 2020 at 15:20

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