備忘録

\phi\left(\mathbf{x}\right)=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}+b_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}}\right)

\phi^{*}\left(\mathbf{x}\right)=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\frac{1}{\sqrt{2E_{\mathbf{p}}}}\left(b_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}+a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}}\right)

\pi\left(\mathbf{x}\right)=\frac{\partial}{\partial t}\phi^{*}\left(\mathbf{x},t\right)=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\left(-i\right)\sqrt{\frac{E_{\mathbf{p}}}{2}}\left(b_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}-a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}}\right)

\pi^{*}\left(\mathbf{x}\right)= \frac{\partial}{\partial t}\phi\left(\mathbf{x},t\right)=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}\left(-i\right)\sqrt{\frac{E_{\mathbf{p}}}{2}}\left(a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}}-b_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}}\right)

Hamitonianは

H=\int d^{3}x\left(\pi^{*}\pi+\nabla\phi^{*}\cdot\nabla\phi+m^{2}\phi^{*}\phi\right)

正準量子化で対角化すると、

H=\int\frac{d^{3}p}{\left(2\pi\right)^{3}}E_{\mathbf{p}}\left(a_{\mathbf{p}}^{\dagger}a_{\mathbf{p}}+b_{\mathbf{p}}^{\dagger}b_{\mathbf{p}}+\frac{1}{2}\left[a_{\mathbf{p}},a_{\mathbf{p}}^{\dagger}\right]+\frac{1}{2}\left[b_{\mathbf{p}},b_{\mathbf{p}}^{\dagger}\right]\right)