물리:결맞는_상태_coherent_state

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물리:결맞는_상태_coherent_state [2024/12/19 15:03] – minwoo물리:결맞는_상태_coherent_state [2024/12/25 08:50] (current) – minwoo
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 $$ $$
-\langle N \rangle = \langle \alpha | a^\dagger a | \alpha \rangle =  |\alpha|^2 \langle \alpha || \alpha \rangle = |\alpha|^2.+\langle N \rangle = \langle \alpha | a^\dagger a | \alpha \rangle =  |\alpha|^2 \langle \alpha | \alpha \rangle = |\alpha|^2.
 $$ $$
  
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 \\ \\
  
-& \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} \end{align}
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 {{:물리:number_eigenstate1.png?300|}} {{:물리:number_eigenstate1.png?300|}}
  
-위의 폭은 σxσp=ℏ2(n+12)2≥(ℏ2)2의 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) ==== ==== '결맞는 상태' (coherent state) ====
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 \hat{p}=-i\sqrt{\frac{\hbar m\omega}{2}}(a-a^{\dagger}) ,\\ \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) \\ = \sqrt{\frac{\hbar}{2m \omega}}(\alpha^* +\alpha) \\
 \\ \\
 \langle\alpha| \hat{p} |\alpha\rangle=-i\sqrt{\frac{\hbar m\omega}{2}} \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)$$ =-i\sqrt{\frac{\hbar m\omega}{2}}(-\alpha^*+\alpha)$$
  
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 \\ \\
  
-& \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} \end{align}
  
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 $$|\alpha\rangle = \sum_{n=0}^\infty c_n(\alpha) |n\rangle = \sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-|\alpha|^2/2}|n\rangle\\ $$|\alpha\rangle = \sum_{n=0}^\infty c_n(\alpha) |n\rangle = \sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-|\alpha|^2/2}|n\rangle\\
-= \sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-\alpha^* \alpha/2}|n\rangle $$+= \sum_{n=0}^\infty \frac{\alpha^{n}}{\sqrt{n!}}e^{-\alpha^* \alpha/2}|n\rangle $$
  
 '시간 변화 연산자(time evolution operator)'인 e−iHt/ℏ 를 걸어보자. '시간 변화 연산자(time evolution operator)'인 e−iHt/ℏ 를 걸어보자.
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 = &\sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-i\hbar \omega (n+\frac{1}{2})t/\hbar} |n\rangle e^{-\alpha^* \alpha/2} \\ = &\sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-i\hbar \omega (n+\frac{1}{2})t/\hbar} |n\rangle e^{-\alpha^* \alpha/2} \\
 = &\sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-in\omega t} |n\rangle e^{-\alpha^* \alpha/2}e^{-i\omega t/2 } \\ = &\sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-in\omega t} |n\rangle e^{-\alpha^* \alpha/2}e^{-i\omega t/2 } \\
-= &e^{-\alpha^* \alpha/2}e^{-i\omega t/2 }\sum_{n=0}^\infty\frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}}|n\rangle  \\+= &e^{-i\omega t/2 }\sum_{n=0}^\infty e^{-\alpha^* \alpha/2} \frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}}|n\rangle  \\
 = &e^{-i\omega t/2 }|\alpha e^{-i\omega t }\rangle  = &e^{-i\omega t/2 }|\alpha e^{-i\omega t }\rangle 
 \end{align} \end{align}
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