5 added 5 characters in body edited Feb 26 '14 at 9:15 Rodriguez 15622 silver badges77 bronze badges 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 $$u(p)$$$$u(\mathbf{p})$$ and $$v(p)$$$$v(\mathbf{p})$$, 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 expansionexpansions 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 $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p})$$$$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p})$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p})$$$$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} a_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p})$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 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 $$u(p)$$ and $$v(p)$$, 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 expansion of spinors in 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 $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p})$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p})$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 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. 4 revise quantization procedure in Schrodinger picture edited Feb 26 '14 at 9:08 Rodriguez 15622 silver badges77 bronze badges 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 $$u(p)$$ and $$v(p)$$, we define $$\psi =: u(p) e^{ipx}$$$$\psi =: u(\mathbf{p}) e^{ipx}$$ $$\psi =: v(p) e^{-ipx}$$$$\psi =: v(\mathbf{p}) e^{-ipx}$$ Thus we can replace the $$u(p)$$$$u(\mathbf{p})$$ as $$v(p)$$$$v(\mathbf{p})$$ and $$v(p)$$$$v(\mathbf{p})$$ as $$u(p)$$$$u(\mathbf{p})$$ 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 $$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-ipx} + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{ipx} \right)$$expansion of spinors in Schrodinger picture $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-ipx} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{ipx} \right)$$$$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x} } + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{i \mathbf{p} \cdot \mathbf{x} } \right)$$ $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-i\mathbf{p} \cdot \mathbf{x}} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{i\mathbf{p} \cdot \mathbf{x}} \right)$$ we have $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(p)$$$$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p})$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(p)$$$$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p})$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 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 $$u(p)$$ and $$v(p)$$, we define $$\psi =: u(p) e^{ipx}$$ $$\psi =: v(p) e^{-ipx}$$ Thus we can replace the $$u(p)$$ as $$v(p)$$ and $$v(p)$$ as $$u(p)$$ 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 $$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-ipx} + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{ipx} \right)$$ $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-ipx} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{ipx} \right)$$ we have $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(p)$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(p)$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 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 $$u(p)$$ and $$v(p)$$, 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 expansion of spinors in 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 $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(\mathbf{p})$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(\mathbf{p}) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(\mathbf{p})$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 3 added 115 characters in body edited Feb 26 '14 at 8:27 Rodriguez 15622 silver badges77 bronze badges 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 $$u(p)$$ and $$v(p)$$, we define This implies$$\psi =: u(p) e^{ipx}$$ $$\psi =: v(p) e^{-ipx}$$ Thus we can replace the $$u(p)$$ as $$v(p)$$ and $$v(p)$$ as $$u(p)$$ 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 $$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-ipx} + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{ipx} \right)$$ $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-ipx} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{ipx} \right)$$ we have $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(p)$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(p)$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 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 $$(i \gamma^{\mu} \partial_{\mu} +m) \psi=0 \tag{4}$$ This implies we can replace the $$u(p)$$ as $$v(p)$$ and $$v(p)$$ as $$u(p)$$ 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 $$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-ipx} + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{ipx} \right)$$ $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-ipx} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{ipx} \right)$$ we have $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(p)$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(p)$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 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 $$u(p)$$ and $$v(p)$$, we define $$\psi =: u(p) e^{ipx}$$ $$\psi =: v(p) e^{-ipx}$$ Thus we can replace the $$u(p)$$ as $$v(p)$$ and $$v(p)$$ as $$u(p)$$ 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 $$\psi = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( a_{\mathbf{p}}^s v^s (\mathbf{p}) e^{-ipx} + b_{\mathbf{p}}^{s\dagger} u^s(\mathbf{p}) e^{ipx} \right)$$ $$\bar{\psi} = \int \frac{ d^3 p }{ (2\pi)^3} \frac{1}{ \sqrt{2 E_{\mathbf{p}}}} \sum_s \left( b_{\mathbf{p}}^s \bar{u}^s (\mathbf{p}) e^{-ipx} + a_{\mathbf{p}}^{s\dagger} \bar{v}^s(\mathbf{p}) e^{ipx} \right)$$ we have $$H = \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( - \gamma^i p_i +m) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( \gamma^i p_i +m) v^s(p)$$ $$= \sum_{ss'} \int \frac{d^3p}{ (2\pi)^3 2E_{\mathbf{p}} } b_{\mathbf{p}}^{s'} b_{\mathbf{p}}^{s\dagger} \bar{u}^{s'}(\mathbf{p}) ( \gamma^0 p_0 ) u^s(p) + a_{\mathbf{p}}^{s'\dagger} s_{\mathbf{p}}^{s} \bar{v}^{s'}(\mathbf{p}) ( - \gamma^0 p_0 ) v^s(p)$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} E_{\mathbf{p}} ( b_{\mathbf{p}}^{s} b_{\mathbf{p}}^{s\dagger} - a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} )$$ $$= \sum_s \int \frac{ d^3p}{ (2\pi)^3} - E_{\mathbf{p}} (b_{\mathbf{p}}^{s\dagger} b_{\mathbf{p}}^{s} + a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^{s} ) - \infty$$ Changing anticommutator into commutator will make the spectrum unbounded. 2 added 34 characters in body edited Feb 26 '14 at 4:31 Rodriguez 15622 silver badges77 bronze badges 1 answered Feb 26 '14 at 3:16 Rodriguez 15622 silver badges77 bronze badges