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Haifeng Yang deleted file Quantize K-G Field.tex
over 10 years ago
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\subsection{Quantize Klein-Gordon}
\subsubsection{Fourier Transformation}
$$\hat{\phi}(\mathbf{x}) = \int \frac{d^3 k}{(2\pi)^3 } \hat{\tilde{\phi}}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}$$
$$\hat{\pi}(\mathbf{y}) = \int \frac{d^3 k}{(2\pi)^3 } \hat{\tilde{\pi}}(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{y}}$$
So (excise: Prove this):
$$\hat{H} = \int \frac{d^3 k}{(2\pi)^3 } \left[
\frac{1}{2}\hat{\tilde{\pi}}(\mathbf{k})^\dagger \hat{\tilde{\pi}}(\mathbf{k})
+\frac{1}{2}\omega_\mathbf{k}^2 \hat{\tilde{\phi}}(\mathbf{k})^\dagger \hat{\tilde{\phi}}(\mathbf{k})
\right]$$
Note: These are complex harmonic oscillators. $\tilde{\phi}(-\mathbf{k})=\tilde{\phi}(\mathbf{k})^*$. All $\mathbf{k}$'s are not independent.
\subsubsection{Commutation Relations}
$$[\hat{\tilde{\pi}}(\mathbf{k'}), \hat{\tilde{\phi}}(\mathbf{k})] = - i (2\pi)^3 \delta^{(3)} (\mathbf{k} + \mathbf{k'})$$
Or:
$$[\hat{\tilde{\pi}}(\mathbf{k'})^\dagger, \hat{\tilde{\phi}}(\mathbf{k})] = - i (2\pi)^3 \delta^{(3)} (\mathbf{k} - \mathbf{k'})$$
Defining uppering and lowering operators, we get:
$$\hat{\phi}(\mathbf{x}) = \int \frac{d^3 k}{(2\pi)^3 } \left[
\hat{a}_\mathbf{k} e^{i\mathbf{k}\cdot\mathbf{x}} + \hat{a}_\mathbf{k}^\dagger e^{-i\mathbf{k}\cdot \mathbf{x}}
\right]$$
