8
$\begingroup$

I'd like to know if the sign in front of a kinetic term in QFT important. For the scalar field we conventionally write (in the $ + --- $ metric), \begin{equation} {\cal L} _{ kin} = \frac{1}{2} \partial _\mu \phi \partial ^\mu \phi \end{equation} Based on the answer given here, this makes perfect sense since we want to have positive kinetic energy $\propto \dot{\phi}^2$. So would the Hamiltonian with a negative in front of the kinetic term be unbounded?

Does this logic extend to the Dirac Lagrangian typically given by, \begin{equation} \bar{\psi} i \partial _\mu \gamma ^\mu \psi \quad ? \end{equation} i.e., would having a negative in front of the Dirac Lagrangian make the Hamiltonian unbounded?

$\endgroup$
4
$\begingroup$

Yes. Though the energy will not be unbounded, but bounded from above, if my calculation is correct.

For real scalar field under $(+---)$ metric, besides the negative classical kinetic energy for the Lagrangian $$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi - \frac{1}{2} m^2 \phi^2 \tag{1} $$, the classical equation of motion will be $$ (\square - m^2 )\phi=0 . \tag{2}$$ For plane wave $\phi ~\sim e^{ipx} $, it gives $p^2+m^2 = (p^0)^2 - \mathbf{p}^2+m^2=0$ which is inconsistent with relativistic energy momentum relation. I am not sure if it is necessary to quantize it.

Though the energy-momentum-relation argument will not work for the Dirac field, we can quantize it to see the energy will be negative definite. $$\mathcal{L} = \bar{\psi}( -i \gamma^{\mu} \partial_{\mu} - m ) \psi \tag{3}$$

The classical equation of motion is $$ (i \gamma^{\mu} \partial_{\mu} +m) \psi=0 \tag{4}$$

To preserve all properties of $u(\mathbf{p})$ and $v(\mathbf{p})$, we define $$ \psi =: u(\mathbf{p}) e^{ipx} $$ $$ \psi =: v(\mathbf{p}) e^{-ipx} $$
Thus we can replace the $u(\mathbf{p})$ as $v(\mathbf{p})$ and $v(\mathbf{p})$ as $u(\mathbf{p})$ in the expansion of $\psi$ and $\bar{\psi}$. By $$\pi = -i \bar{\psi} \gamma^0 $$ then $$H= \int d^3 x \bar{\psi} ( i \gamma^i \partial_i \psi + m ) \psi $$

Plug in expansions of spinors in the Schrodinger picture $$ \psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x} } + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{i \mathbf{p} \cdot \mathbf{x} } \right) $$ $$ \bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x}} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{i\mathbf{p} \cdot \mathbf{x}} \right) $$ we have

$$ H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p}) $$ $$ = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p}) $$ $$ = \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) $$ $$ = \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty $$

Changing anticommutator into commutator will make the spectrum unbounded.

$\endgroup$
0
$\begingroup$

The sign is indeed important for ensuring that the energy spectrum is bounded from below. Also, the potential well for a scalar needs to "open up" for stability. So there are really two things you could mean when you talk about changing the sign of the kinetic term.

Changing the relative sign between the kinetic term and the highest power of $\phi$ in the potential (mass term in the free field case) is not allowed as can be seen from the classical equation of motion in the answer above. Changing the sign of everything is allowed provided that you also change the conventions of how you quantize the theory. $[\phi(t, x), \dot{\phi}(t, y)] = i\delta(x-y)$ is only the right commutation relation to impose when the conjugate momentum is $\dot{\phi}$. It could of course be $-\dot{\phi}$ if we wanted.

For spinors, the situation is similar when we change the overall sign. That is, it only makes sense if we go back and reverse all the anti-commutation relations that we usually take for granted. However, the relative sign between the kinetic and mass terms is no longer physical as explained in this answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.