Valores esperados de estado coherente del oscilador armónico

Dejar$\alpha \in {\Bbb C}$, y deja$\vert{n}\rangle$sea ​​el estado del oscilador armónico con energía$(n+\textstyle\frac{1}{2})\hbar\omega$. En$t=0$, el estado coherente$\vert {\alpha(0)}\rangle$es definido por

$$\vert{\alpha(0)}\rangle= e^{-\vert \alpha \vert^2/2}\,\left( \sum_{n=0}^{\infty} \displaystyle{\alpha^n\over \sqrt{n!}}\,\vert{n}\rangle\right) \tag{1}$$

Qué es$\vert{\alpha(t)}\rangle$, el estado coherente en el tiempo$t$? Comience con (1). Desde$\left\vert n\right\rangle$es un estado propio del oscilador armónico hamiltoniano$\hat{H}=\left( \hat a^{\dagger }\hat a+\frac{1}{2}\right) \hbar \omega$con valor propio$\left( n+\frac{1}{2}\right) \hbar \omega ,$la evolución temporal de$\left\vert n\right\rangle$es simple$\left\vert n(t)\right\rangle =e^{-i(n+\frac{1}{2})\omega t}\left\vert n\right\rangle$y por lo tanto

$$\begin{equation} \left\vert \alpha (t)\right\rangle =e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{% 1}{2})\omega t}\left\vert n\right\rangle \right) . \end{equation}$$ Es fácil demostrar que $\left\vert \alpha (t)\right\rangle$está normalizado.

Ahora primero tenemos que demostrar que$a\vert{\alpha(t)}\rangle=\alpha e^{i\hbar \omega t}\vert{\alpha(t)}\rangle$. Recordar que$\hat{a}\left\vert n\right\rangle =\sqrt{n}\left\vert n-1\right\rangle .$\ Entonces, desde$\hat{a}$es lineal,

$$\begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}% e^{-i(n+\frac{1}{2})\omega t}\hat{a}\left\vert n\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{1}{2})\omega t}\sqrt{n}\left\vert n-1\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{\left( n-1\right) !}}e^{-i(n+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) , \\ &=&\alpha e^{-i\omega t}e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n-1}}{\sqrt{\left( n-1\right) !}}% e^{-i(n-1+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) .\quad \end{eqnarray}$$ La suma correctamente comienza en $n=1$desde el $n=0$término no existe. Así, estableciendo $m=n-1,$podemos reescribir esta suma en términos de $m,$con $m$a partir de $m=0.$Por eso $$\begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&\alpha e^{-i\omega t}\left[ e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{m=0}^{\infty }\frac{% \alpha ^{m}}{\sqrt{m!}}e^{-i(m+\frac{1}{2})\omega t}\left\vert m\right\rangle \right) \right] \\ &=&\alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle . \end{eqnarray}$$ Un resultado secundario útil, que se sigue inmediatamente de arriba, es $$\begin{eqnarray} \left[ \hat{a}\left\vert \alpha (t)\right\rangle \right] ^{\dagger } &=&\left\langle \alpha (t)\right\vert \hat{a}^{\dagger } \\ &=&\left[ \alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle \right] ^{\dagger }=\alpha ^{\ast }e^{i\omega t}\left\langle \alpha (t)\right\vert \end{eqnarray}$$

Ahora$\langle \hat p(t) \rangle$y$\langle \hat x(t)\rangle$para$\vert{\alpha(t)}\rangle$. A partir de las definiciones

$$\hat{a} =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}+\frac{i}{% m\omega }\hat{p}\right) , \qquad \hat{a}^{\dagger } =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}-% \frac{i}{m\omega }\hat{p}\right) ,$$ tenemos $$\hat{x} =\sqrt{\frac{\hbar }{2m\omega }}\left( \hat{a}^{\dagger }+% \hat{a}\right) , \qquad \hat{p} =i\sqrt{\frac{m\omega \hbar }{2}}\left( \hat{a}^{\dagger }-% \hat{a}\right) ,$$ y por lo tanto $$\begin{eqnarray} \left\langle x(t)\right\rangle &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle +\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right]\, , \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] , \end{eqnarray}$$ que es real, como era de esperar. Podemos limpiar esto escribiendo $\alpha =\left\vert \alpha \right\vert e^{i\theta }$para obtener% $$\begin{equation} \left\langle x(t)\right\rangle =\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left( \omega t-\theta \right) \tag{2} \end{equation}$$

Asimismo,

$$\begin{eqnarray} \left\langle p(t)\right\rangle &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle -\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right] \\ &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&-\sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left( \omega t-\theta \right) \tag{3} \end{eqnarray}$$ que vuelve a ser real.

En su caso específico, está comenzando con un estado coherente para el cual, en$t=0$, tenemos

$$\langle x(0)\rangle= b\sqrt{2}x_0\, ,\qquad \langle p(0)\rangle=0$$ entonces esto implica de (2) y (3) evaluados en $t=0$eso $$b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left(\theta \right)\, , \qquad 0= \sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left(\theta \right)$$ Comparando con sus condiciones iniciales da $\theta=0$y $b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }} \alpha$con $\alpha$real.

Finalmente,$\hat{x}^{2}$y$\hat{p}^{2}.$De$\hat{x}$y$\hat{p},$encontramos

$$\begin{eqnarray} \hat{x}^{2} &=&\frac{\hbar }{2m\omega }\left( \hat{a}^{\dagger }+ \hat{a}\right) ^{2}=\frac{\hbar }{2m\omega }\left( \left( \hat{a} ^{\dagger }\right) ^{2}+\hat{a}^{\dagger }\hat{a}+\hat{a}\hat{a} ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&\frac{\hbar }{2m\omega }\left( \left( \hat{a}^{\dagger }\right) ^{2}+2 \hat{a}^{\dagger }\hat{a}+1+\left( \hat{a}\right) ^{2}\right) , \\ \hat{p}^{2} &=&-\frac{m\omega \hbar }{2}\left( \hat{a}-\hat{a} ^{\dagger }\right) ^{2}=-\frac{m\omega \hbar }{2}\left( \left( \hat{a} ^{\dagger }\right) ^{2}-\hat{a}^{\dagger }\hat{a}-\hat{a}\hat{a}% ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&-\frac{m\omega \hbar }{2}\left( \left( \hat{a}^{\dagger }\right) ^{2}-2% \hat{a}^{\dagger }\hat{a}-1+\left( \hat{a}\right) ^{2}\right) , \end{eqnarray}$$ dónde $$\begin{equation} \hat{a}\hat{a}^{\dagger }=\hat{a}\hat{a}^{\dagger }-\hat{a}% ^{\dagger }\hat{a}+\hat{a}^{\dagger }\hat{a}=\left[ \hat{a},% \hat{a}^{\dagger }\right] +\hat{a}^{\dagger }\hat{a}=1+\hat{a}% ^{\dagger }\hat{a} \end{equation}$$ ha sido usado. De este modo, $$\begin{eqnarray} \left\langle x^{2}(t)\right\rangle &=&\frac{\hbar }{2m\omega }\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle +2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad +1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}+2\alpha ^{\ast }\alpha +1+\left( \alpha e^{-i\omega t}\right) ^{2}% \right]\, ,\\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right) ^{2}+1\right] , \\ &=&\frac{\hbar }{2m\omega }\left[ 4\left\vert \alpha \right\vert ^{2}\cos ^{2}\left( \omega t-\theta \right) +1\right] . \\ \left\langle p^{2}(t)\right\rangle &=&-\frac{m\omega \hbar }{2}\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle -2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad -1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}-2\alpha ^{\ast }\alpha -1+\left( \alpha e^{-i\omega t}\right) ^{2}\right]\, ,\\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right) ^{2}-1\right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ -4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) -1\right]\, ,\\ &=&\frac{m\omega \hbar }{2}\left[ 4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) +1% \right] . \end{eqnarray}$$