물리:결맞는_상태_coherent_state

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물리:결맞는_상태_coherent_state [2024/12/19 14:42] – 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 $ \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) ==== ==== '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 $e^{-iHt/\hbar}$ 를 걸어보자. '시간 변화 연산자(time evolution operator)'인 $e^{-iHt/\hbar}$ 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 } \\
-= &\sum_{n=0}^\infty\frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}}|n\rangle e^{-\alpha^* \alpha/2}e^{-i\omega t/2 } \\ += &e^{-i\omega t/2 }\sum_{n=0}^\infty e^{-\alpha^* \alpha/2} \frac{(\alpha e^{-i\omega t})^n}{\sqrt{n!}}|n\rangle  \\ 
-= &|\alpha e^{-i\omega t }\rangle e^{-i\omega t/2 }+= &e^{-i\omega t/2 }|\alpha e^{-i\omega t }\rangle 
 \end{align} \end{align}
  
 $$ \\ $$ $$ \\ $$
-즉, $e^{-iHt/\hbar} | \alpha \rangle  = |\alpha e^{-i\omega t }\rangle e^{-i\omega t/2 }$의 관계식이 성립한다.+즉, $e^{-iHt/\hbar} | \alpha \rangle  = e^{-i\omega t/2 } |\alpha e^{-i\omega t }\rangle $의 관계식이 성립한다.
  
 $$ \\ $$ $$ \\ $$
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 $$ \\ $$ $$ \\ $$
 +더 일반적으로 기술하기 위해서, 실수 $\theta$에 대하여 $\hat{U}(\theta)=e^{-i\theta N}$을 결맞는 상태인 $|\alpha\rangle$에 걸어보자. 이때 $N=a^\dagger a$는 앞서 살펴본 'number operator'이다.
 +
 +$|\alpha \rangle$은 $|n \rangle$의 선형결합으로 표현 되므로, 우선 $\hat{U}(\theta)|n \rangle$를 풀어보면 다음과 같다.
 +
 +\begin{align}
 +\hat{U}(\theta) | n \rangle &= e^{-i\theta N}| n \rangle \\
 +&= \left(1+ (-i\theta N) + \frac{(-i\theta N)^2}{2!} + ... \right)| n \rangle \\
 +&= \left(1+ (-i\theta n) + \frac{(-i\theta)^2 n^2}{2!} + ... \right)| n \rangle\\
 +&= e^{-i\theta n }| n \rangle.
 +\end{align}
 +
 +따라서, 결맞는 상태 $|\alpha \rangle =\sum_{n=0}^\infty\frac{\alpha^{n}}{\sqrt{n!}}e^{-\frac{|\alpha|^2}{2}}|n\rangle $에 $\hat{U}(\theta)$를 걸어주면:
 +
 +$$
 +\hat{U}(\theta) | \alpha\rangle = e^{-\frac{|\alpha|^2}{2}}\sum_{n=0}^\infty \frac{(\alpha e^{-i\theta})^n}{\sqrt{n!}} |n\rangle = |\alpha e^{-i\theta} \rangle.
 +
 +$$
 +
 +따라서, 결맞는 상태 $|\alpha\rangle $에 $\theta$가 실수인 $e^{-i\theta N}$의 연산자를 걸어주면, 그 결과가 되는 상태의 고유 값은 $\alpha$에서 $e^{-i\theta}$만큼 곱해진 만큼의 값에 해당한다.
 +
 +
 +$$ \\ $$
 +
 ===== 참고 문헌 ===== ===== 참고 문헌 =====
   * Hitoshi Murayama, Jan27 151 Coherent state, QFT on 1D lattice, 2021. (lecture of Prof. Hitoshi Murayama)   * Hitoshi Murayama, Jan27 151 Coherent state, QFT on 1D lattice, 2021. (lecture of Prof. Hitoshi Murayama)
   * Lancaster and Blundell, Quantum field theory for the gifted amateur, 2014.   * Lancaster and Blundell, Quantum field theory for the gifted amateur, 2014.
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