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--- 물리:결맞는_상태_coherent_state [2024/12/24 18:44] – minwoo
+++ 물리:결맞는_상태_coherent_state [2024/12/25 08:50] (current) – minwoo
@@ Line 247: / Line 247: @@
 \\
-& \sigma_x \sigma_p =\hbar^2 \left(n+\frac{1}{2}\right)^2 \ge \left( \frac{\hbar}{2} \right)^2 \quad \left(\because \ \sigma_A = \langle A^2 \rangle - \langle A \rangle ^2 \right)
+& \sigma_x^2 \sigma_p^2 =\hbar^2 \left(n+\frac{1}{2}\right)^2 \ge \left( \frac{\hbar}{2} \right)^2 \quad \left(\because \ \sigma_A^2 = \langle A^2 \rangle - \langle A \rangle ^2 \right)
 \end{align}
@@ Line 266: / Line 266: @@
 {{:물리:number_eigenstate1.png?300|}}
-위의 폭은  $\sigma_x \sigma_p =\hbar^2 \left(n+\frac{1}{2}\right)^2 \ge \left( \frac{\hbar}{2} \right)^2$ 의 불확정성에 의한 것이다.
+위의 폭은 $ \sigma_x^2 \sigma_p^2 =\hbar^2 \left(n+\frac{1}{2}\right)^2 \ge \left( \frac{\hbar}{2} \right)^2 $의 불확정성에 의한 것이다.
 ==== '결맞는 상태' (coherent state) ====
@@ Line 277: / Line 277: @@
 \hat{p}=-i\sqrt{\frac{\hbar m\omega}{2}}(a-a^{\dagger}) ,\\
-\langle\alpha| x |\alpha\rangle=\sqrt{\frac{\hbar}{2m \omega}}\left(\ (\langle\alpha| (a^{\dagger}+a)|\alpha\rangle)\ \right) \\
+\langle\alpha| x |\alpha\rangle=\sqrt{\frac{\hbar}{2m \omega}}\left(\ \langle\alpha| (a+a^{\dagger})|\alpha\rangle\ \right) \\
 = \sqrt{\frac{\hbar}{2m \omega}}(\alpha^* +\alpha) \\
 \\
 \langle\alpha| \hat{p} |\alpha\rangle=-i\sqrt{\frac{\hbar m\omega}{2}}
-\left(\ (-\langle\alpha| (a^{\dagger}+a)|\alpha\rangle)\ \right) \\
+\left(\ \langle\alpha| (a-a^{\dagger})|\alpha\rangle\ \right) \\
 =-i\sqrt{\frac{\hbar m\omega}{2}}(-\alpha^*+\alpha)$$
@@ Line 305: / Line 305: @@
 \\
-& \sigma_x \sigma_p =(\frac{\hbar}{2m\omega})(\frac{\hbar m\omega}{2})=\left(\frac{\hbar}{2}\right)^2
+& \sigma_x^2 \sigma_p^2 =(\frac{\hbar}{2m\omega})(\frac{\hbar m\omega}{2})=\left(\frac{\hbar}{2}\right)^2
 \end{align}