三浦と窮理とブログ

主に自然科学について自分が勉強してきたことについて書いていきます.誰かの役に立ってくれれば嬉しいです.

3次元自由空間でのガウス波束の時間発展と期待値と分散

3次元自由空間でのガウス波束

\begin{equation} \psi ( \vec{x} , t = 0 ) = \langle \vec{ x } |\psi,0\rangle= \exp \left( - \frac { \alpha } { 2 } | \vec { x } | ^ { 2 } + \frac { i \vec { p } _ { 0 } \cdot \vec { x } } { \hbar } \right) \end{equation}

を考える.

この波束の時間発展は

\begin{equation} \label{eq:psi} \psi (\vec{ x } , t ) = \frac{1}{\sqrt{(1 + i \frac{\alpha\hbar }{m}t)^3}} \exp (\frac{1}{1 + i \frac{\alpha\hbar }{m}t} \left (- \frac{\alpha}{2} |\vec{ x }|^2 + i \frac{\vec{ p }_{0} \cdot \vec{ x } }{\hbar } - i\frac{|\vec{ p }_{0}|^2}{2m\hbar} t\right )) \end{equation}

となる.

$ \hat{x} _ i $ と $ \hat{p} _ i $ の期待値 $ \bar{x} _ i $ と $ \bar{p} _ i $ は

\begin{align} \bar{x} _ i &= \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \frac{p_{0i}}{ m}t \label{eq:mx}\\ \bar{p} _ i &= \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} p_{0i} \label{eq:px} \end{align}

と求まり,ガウス波束は等速で $ \vec{ x } $ 方向に進むことがわかる.

また,$x _ i$ 方向の分散$ \overline { (x _ i - \bar{x} _ i) ^ { 2 } } = \overline{{x _ i} ^ 2} - {\overline{x} _ i} ^ 2 $は

\begin{equation} \label{eq:var} \overline { (x_i - \bar{x}_i) ^ { 2 } } = \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \left [\left ( \frac{\alpha\hbar^2}{2m^2} + \left (1-\left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \right ) \frac{{p_{0i}}^2}{m^2} \right )t^2 + \frac{1}{2\alpha }\right ] \end{equation}

である.

式\eqref{eq:psi}の導出

自由粒子のハミルトニアンは $ \hat{H} = \frac{\vec{\hat{p}} ^ 2}{2m} $ である.

\begin{align} \psi ( \vec{x} , t ) &= \langle \vec{x} | \psi,t \rangle = \langle \vec{x} | \exp (-i \frac{\hat{H}}{\hbar}t) |\psi,0 \rangle = \langle \vec{x} | \exp (-i \frac{\vec{\hat{p}} ^ 2}{2m\hbar}t) |\psi,0 \rangle \\ &=\int_{-\infty}^{\infty} d^3p \underbrace{\langle \vec{x} | \exp (-i \frac{\vec{\hat{p}} ^ 2}{2m\hbar}t) |\vec{ p } \rangle}_{= \exp (-i \frac{|\vec{p}| ^ 2}{2m\hbar}t)\langle \vec{x} |\vec{ p } \rangle} \underbrace{\langle \vec{ p }|\psi,0 \rangle}_{=:(a)} \\ &=\frac{1}{\sqrt{(\alpha\hbar)^3}} \frac{1}{\sqrt{(2\hbar \pi)^3}}\int_{-\infty}^{\infty} d^3p \exp \left (-i \frac{|\vec{p}| ^ 2}{2m\hbar}t \right ) \exp \left (i\frac{\vec{ p }\cdot \vec{ x }}{\hbar} \right ) \exp \left ( - \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } \right ) \\ &= \frac{1}{\sqrt{(2\alpha\hbar^2 \pi)^3}} \int_{-\infty}^{\infty} d^3p \exp (-i \frac{|\vec{p}| ^ 2}{2m\hbar}t +i\frac{\vec{ p }\cdot \vec{ x }}{\hbar} -\frac{|\vec{ p } - \vec{ p }_0|^2}{2\hbar^2 \alpha}) \\ &= \frac{1}{\sqrt{(2\alpha\hbar^2 \pi)^3}} \Bigl [\int_{-\infty}^{\infty} dp_x \exp (\underbrace{-i \frac{p_x ^ 2}{2m\hbar}t +i\frac{p_xx}{\hbar} -\frac{(p_x-p_{0x})^2}{2\hbar^2 \alpha}}_{=:(c)}) \Bigr ]^3 \\ &= \frac{1}{\sqrt{(2\alpha\hbar^2 \pi)^3}} \Bigl [\exp (\frac{1}{4A(t)} \left ( \frac{p_{0x}}{\hbar^2 \alpha} +i \frac{x}{\hbar} \right ) ^2 - \frac{{p_{0x}}^2}{2\hbar^2 \alpha} ) \underbrace{\int_{-\infty}^{\infty} dp_x \exp (-A(t) \left ({p_x}^2- \frac{1}{2A(t)} \left ( \frac{p_{0x}}{\hbar^2 \alpha} +i \frac{x}{\hbar} \right ) \right )^2 )}_{=\sqrt{\frac{\pi}{A(t)}}} \Bigr ]^3 \\ &= \frac{1}{\sqrt{(2\alpha\hbar^2 A(t))^3}} \exp (\frac{1}{4A(t)} \left | \frac{\vec{ p }_{0}}{\hbar^2 \alpha} +i \frac{\vec{ x }}{\hbar} \right | ^2 - \frac{|\vec{ p }_{0}|^2}{2\hbar^2 \alpha} ) \\ &= \frac{1}{\sqrt{(2\alpha\hbar^2 A(t))^3}} \exp (\frac{1}{4A(t)} \left ( \frac{|\vec{ p }_{0}|^2}{\hbar^4 \alpha^2} +2i \frac{\vec{ p }_{0} \cdot \vec{ x } }{\hbar^3 \alpha} - \frac{|\vec{ x }|^2}{\hbar^2} - \frac{|\vec{ p }_{0}|^2}{\hbar^2 \alpha}2A(t)\right )) \\ &= \frac{1}{\sqrt{(2\alpha\hbar^2 A(t))^3}} \exp (\frac{1}{4A(t)} \left ( \cancel{\frac{|\vec{ p }_{0}|^2}{\hbar^4 \alpha^2}} +2i \frac{\vec{ p }_{0} \cdot \vec{ x } }{\hbar^3 \alpha} - \frac{|\vec{ x }|^2}{\hbar^2} - \cancel{\frac{|\vec{ p }_{0}|^2}{\hbar^4 \alpha^2}}- i\frac{|\vec{ p }_{0}|^2}{m\hbar^3 \alpha} t\right )) \\ &= \frac{1}{\sqrt{(1 + i \frac{\alpha\hbar }{m}t)^3}} \exp (\frac{1}{4A(t)} \frac{2}{\alpha \hbar^2}\left ( i \frac{\vec{ p }_{0} \cdot \vec{ x } }{\hbar } - \frac{\alpha}{2} |\vec{ x }|^2 - i\frac{|\vec{ p }_{0}|^2}{2m\hbar} t\right )) \\ &= \frac{1}{\sqrt{(1 + i \frac{\alpha\hbar }{m}t)^3}} \exp (\frac{1}{1 + i \frac{\alpha\hbar }{m}t} \left (- \frac{\alpha}{2} |\vec{ x }|^2 + i \frac{\vec{ p }_{0} \cdot \vec{ x } }{\hbar } - i\frac{|\vec{ p }_{0}|^2}{2m\hbar} t\right )) \\ (a) &=\langle \vec{ p }|\psi,0 \rangle = \int_{-\infty}^{\infty} d^3x \langle \vec{ p }| \vec{ x } \rangle \langle \vec{ x } |\psi,0 \rangle = \frac{1}{\sqrt{(2\hbar \pi)^3}}\int_{-\infty}^{\infty} d^3x \exp (-i \frac{\vec{ p }\cdot\vec{ x }}{\hbar}) \exp \left( - \frac { \alpha } { 2 } | \vec { x } | ^ { 2 } + \frac { i \vec { p } _ { 0 } \cdot \vec { x } } { \hbar } \right) \\ &= \frac{1}{\sqrt{(2\hbar \pi)^3}}\int_{-\infty}^{\infty} d^3x \exp \left( - \frac { \alpha } { 2 } | \vec { x } | ^ { 2 } -i \frac { (\vec { p } - \vec { p } _ { 0 }) \cdot \vec { x } } { \hbar } \right) = \frac{1}{\sqrt{(2\hbar \pi)^3}} \Bigl (\int_{-\infty}^{\infty} dx \exp \underbrace{( - \frac { \alpha } { 2 } x ^ { 2 } -i \frac { (p_x - p_{0x}) x } { \hbar } )}_{=:(b)} \Bigr )^3 \\ &= \frac{1}{\sqrt{(2\hbar \pi)^3}} \Bigl (\exp \left ( - \frac{(p_x-p_{0x})^2}{2 \hbar ^2 \alpha } \right )\underbrace{\int_{-\infty}^{\infty} dx \exp ( - \frac { \alpha } { 2 } (x +i \frac{p_x-p_{0x}}{\hbar \alpha})^2 )}_{=\sqrt{\frac{2\pi}{\alpha}}} \Bigr )^3 \\ &= \frac{1}{\sqrt{(\alpha\hbar)^3}} \exp \left ( - \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } \right ) \label{eq:q0}\\ (b) &= - \frac { \alpha } { 2 } x ^ { 2 } -i \frac { (p_x - p_{0x}) x } { \hbar } = - \frac { \alpha } { 2 } ( x^2 + i \frac{2}{\hbar \alpha} (p_x-p_{0x})x) = - \frac { \alpha } { 2 } \{(x +i \frac{p_x-p_{0x}}{\hbar \alpha})^2 +(\frac{p_x-p_{0x}}{\hbar \alpha})^2 \} \\ &= - \frac { \alpha } { 2 } (x +i \frac{p_x-p_{0x}}{\hbar \alpha})^2 - \frac{(p_x-p_{0x})^2}{2 \hbar ^2 \alpha } \\ (c) &= -i \frac{p_x ^ 2}{2m\hbar}t +i\frac{p_xx}{\hbar} -\frac{(p_x-p_{0x})^2}{2\hbar^2 \alpha} = -i \frac{p_x ^ 2}{2m\hbar}t +i\frac{p_xx}{\hbar} -\frac{{p_x}^2 -2p_xp_{0x}+{p_{0x}}^2}{2\hbar^2 \alpha}\\ &= -\underbrace{\left (\frac{1}{2\hbar^2 \alpha} + i \frac{t}{2m\hbar} \right )}_{=:A(t)} {p_x}^2 +\left ( \frac{p_{0x}}{\hbar^2 \alpha} +i \frac{x}{\hbar} \right ) p_x - \frac{{p_{0x}}^2}{2\hbar^2 \alpha} \\ &= -A(t) \left ({p_x}^2- \frac{1}{2A(t)} \left ( \frac{p_{0x}}{\hbar^2 \alpha} +i \frac{x}{\hbar} \right ) \right )^2 +\frac{1}{4A(t)} \left ( \frac{p_{0x}}{\hbar^2 \alpha} +i \frac{x}{\hbar} \right ) ^2 - \frac{{p_{0x}}^2}{2\hbar^2 \alpha} \end{align}

$ \langle \vec{x} | \exp (-i \frac{\vec{\hat{p}} ^ 2}{2m\hbar}t) |\psi,0 \rangle = \exp (-\frac{\hbar}{2m}t\Delta ) \langle \vec{x} | \psi,0 \rangle$ とした場合の計算の進め方が分からなかったので運動量空間での計算(フーリエ変換)で進めた.

$ \langle \vec{x} |\vec{ p } \rangle = \frac{1}{\sqrt{(2\pi \hbar) ^ 3}} \exp (i\frac{\vec{ p }\cdot \vec{ x }}{\hbar} ) $

式\eqref{eq:mx},\eqref{eq:px}の導出

\begin{align} \bar{x} _ i &= \langle \psi ,t | \hat{x} _ i | \psi ,t \rangle = \int_{-\infty}^{\infty} d^3p \langle \psi ,t |\vec{ p } \rangle \langle \vec{ p } | \hat{x} _ i | \psi ,t \rangle = \int_{-\infty}^{\infty} d^3p \langle \psi ,t |\vec{ p } \rangle i\hbar \frac{\partial}{\partial p_i} \underbrace{\langle \vec{ p } | \psi ,t \rangle}_{=:(d)} \\ &= \frac{i\hbar}{(\alpha\hbar)^3} \int_{-\infty}^{\infty} d^3p \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } +i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \frac{\partial}{\partial p_i} \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } -i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \\ &= \frac{i\hbar}{(\alpha\hbar)^3} \int_{-\infty}^{\infty} d^3p \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } +i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \left ( - \frac{p_i -p _{0i}}{ \hbar ^2 \alpha } -i\frac{p_i}{m\hbar}t \right ) \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } -i\frac{|\vec{p}|^2}{2m\hbar}t \right )\\ &= \frac{i\hbar}{(\alpha\hbar)^3} \int_{-\infty}^{\infty} d^3p \left ( - \frac{p_i -p _{0i}}{ \hbar ^2 \alpha } -i\frac{p_i}{m\hbar}t \right ) \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{ \hbar ^2 \alpha } \right ) \\ &= \frac{i\hbar}{(\alpha\hbar)^3} \left (\int_{-\infty}^{\infty} dp_i \left ( - \frac{p_i -p _{0i}}{ \hbar ^2 \alpha } -i\frac{p_i}{m\hbar}t \right ) \exp \left (- \frac{(p_i -p _{0i})^2}{ \hbar ^2 \alpha } \right )\right ) \Bigl (\underbrace{\int_{-\infty}^{\infty} dp_j \exp \left (- \frac{(p_j -p _{0j})^2}{ \hbar ^2 \alpha } \right )}_{=\sqrt{\hbar^2 \alpha \pi}}\Bigr )^2 \\ &(s = p_i -p _{0i} で置換積分) \\ &= \frac{i \pi}{\alpha^2} \int_{-\infty}^{\infty} ds \left ( - \frac{s}{ \hbar ^2 \alpha } -i\frac{s+p_{0i} }{m\hbar}t \right ) \exp \left (- \frac{s^2}{ \hbar ^2 \alpha } \right ) = \frac{i \pi}{\alpha^2} \int_{-\infty}^{\infty} ds \left (-i\frac{p_{0i} }{m\hbar}t \right ) \exp \left (- \frac{s^2}{ \hbar ^2 \alpha } \right ) \\ &= \frac{\pi p_{0i}}{\alpha^2 m\hbar}t \sqrt{\hbar^2\alpha \pi} = \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \frac{p_{0i}}{ m}t\\ (d) &= \langle \vec{ p } | \psi ,t \rangle = \langle \vec{ p } |\exp (-i\frac{\hat{H}}{\hbar}t)| \psi ,0 \rangle = \langle \vec{ p } |\exp (-i\frac{\vec{\hat{p}}^2}{2m\hbar}t)| \psi ,0 \rangle = \exp (-i\frac{|\vec{p}|^2}{2m\hbar}t)\langle \vec{ p } | \psi ,0 \rangle \\ &\underset{\eqref{eq:q0}}{=} \frac{1}{\sqrt{(\alpha\hbar)^3}} \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } -i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \\ \bar{p} _ i &= \langle \psi ,t | \hat{p} _ i | \psi ,t \rangle = \int_{-\infty}^{\infty} d^3p \langle \psi ,t |\vec{ p } \rangle \langle \vec{ p } | \hat{p} _ i | \psi ,t \rangle = \frac{1}{(\alpha\hbar)^3} \int_{-\infty}^{\infty} d^3p p_i \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{ \hbar ^2 \alpha } \right ) \\ &= \frac{1}{(\alpha\hbar)^3} \left (\int_{-\infty}^{\infty} dp_i p_i \exp \left (- \frac{(p_i-p_{0i})^2}{ \hbar ^2 \alpha } \right )\right ) \Bigl (\underbrace{\int_{-\infty}^{\infty} d^2p_j \exp \left (- \frac{(p_j-p_{0j})^2}{ \hbar ^2 \alpha } \right )}_{=\sqrt{\hbar^2 \alpha \pi}}\Bigr )^2 \\ &(s = p_i -p _{0i} で置換積分) \\ &= \frac{\pi}{\alpha^2\hbar} \int_{-\infty}^{\infty} ds (s+p_{0i}) \exp \left (- \frac{s^2}{ \hbar ^2 \alpha } \right ) = \frac{\pi p_{0i}}{\alpha^2\hbar} \sqrt{\hbar^2 \alpha \pi} = \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} p_{0i} \end{align}

式\eqref{eq:var}の導出

\begin{align} \overline { (x_i - \bar{x}_i) ^ { 2 } } &= \overline{{x_i} ^ 2} - {\overline{x}_i} ^ 2 =\left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \left [\left ( \frac{\alpha\hbar^2}{2m^2} + \frac{{p_{0i}}^2}{m^2} \right )t^2 + \frac{1}{2\alpha }\right ] - \left (\frac{\pi}{\alpha} \right )^3 \frac{{p_{0i}}^2}{ m^2}t^2 \\ &= \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \left [\left ( \frac{\alpha\hbar^2}{2m^2} + \left (1-\left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \right ) \frac{{p_{0i}}^2}{m^2} \right )t^2 + \frac{1}{2\alpha }\right ] \\ \overline{{x_i} ^ 2} &= \langle \psi ,t | {\hat{x}_i} ^ {~2} | \psi ,t \rangle = \int_{-\infty}^{\infty} d^3 p \langle \psi ,t |\vec{ p } \rangle \langle \vec{ p }| {\hat{x}_i} ^ {~2} | \psi ,t \rangle =-\hbar^2 \int_{-\infty}^{\infty} d^3 p \langle \psi ,t |\vec{ p } \rangle \frac{\partial^2}{\partial {p_i } ^ 2} \langle \vec{ p }| \psi ,t \rangle \\ &= -\hbar^2 \frac{1}{(\alpha\hbar)^3} \int_{-\infty}^{\infty} d^3 p \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } +i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \frac{\partial^2}{\partial { p_i }^2} \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } -i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \\ &= - \frac{1}{\alpha^3\hbar} \int_{-\infty}^{\infty} d^3 p \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } +i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \left ( - \frac{1}{ \hbar ^2 \alpha } -i\frac{t}{m\hbar} +\left ( - \frac{p_i-p_{0i}}{ \hbar ^2 \alpha } -i\frac{p_i}{m\hbar}t\right )^2 \right ) \\ & \hspace{3cm} \times \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{2 \hbar ^2 \alpha } -i\frac{|\vec{p}|^2}{2m\hbar}t \right ) \\ &= - \frac{1}{\alpha^3\hbar} \int_{-\infty}^{\infty} d^3 p \exp \left (- \frac{|\vec{p}-\vec{ p }_0|^2}{ \hbar ^2 \alpha }\right ) \left ( - \frac{1}{ \hbar ^2 \alpha } -i\frac{t}{m\hbar} +\left ( \frac{p_i-p_{0i}}{ \hbar ^2 \alpha } +i\frac{p_i}{m\hbar}t\right )^2 \right ) \\ &= - \frac{ 1}{\alpha^3\hbar} \hbar^2 \alpha \pi \int_{-\infty}^{\infty} dp_i \exp \left (- \frac{(p_i-p_{0i})^2}{ \hbar ^2 \alpha }\right ) \left ( - \frac{1}{ \hbar ^2 \alpha } -i\frac{t}{m\hbar} +\frac{(p_i-p_{0i})^2}{ \hbar ^4 \alpha^2 } +2i \frac{p_i-p_{0i}}{ \hbar ^3 \alpha m } p_i t -\frac{{p_i}^2}{m^2\hbar^2}t^2 \right ) \\ &(s = p_i -p _{0i} で置換積分) \\ &= - \frac{ \hbar \pi}{\alpha^2} \int_{-\infty}^{\infty} d s \exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right ) \left ( - \frac{1}{ \hbar ^2 \alpha } -i\frac{t}{m\hbar} +\frac{s^2}{ \hbar ^4 \alpha^2 } +2i \frac{s}{ \hbar ^3 \alpha m }(s+p_{0i}) t -\frac{(s+p_{0i})^2}{m^2\hbar^2}t^2 \right ) \\ &= - \frac{ \hbar \pi}{\alpha^2} \int_{-\infty}^{\infty} d s \exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right ) \left ( - \frac{1}{ \hbar ^2 \alpha } -i\frac{t}{m\hbar} +\frac{s^2}{ \hbar ^4 \alpha^2 } +2i \frac{s^2}{ \hbar ^3 \alpha m } t -\frac{s^2 +{p_{0i}}^2}{m^2\hbar^2}t^2 \right ) \\ &= - \frac{ \hbar \pi}{\alpha^2} \Bigl [ \left ( \frac{1}{ \hbar ^4 \alpha^2 } +2i \frac{1}{ \hbar ^3 \alpha m } t -\frac{1}{m^2\hbar^2}t^2 \right ) \int_{-\infty}^{\infty} d s \underbrace{s^2 \exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right )}_{\rlap{=-\frac{\hbar^2\alpha}{2} s \frac{d}{ds} \exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right ) }} \\ &\hspace{4cm} + \left ( - \frac{1}{ \hbar ^2 \alpha } -i\frac{t}{m\hbar} -\frac{{p_{0i}}^2}{m^2\hbar^2}t^2 \right ) \underbrace{\int_{-\infty}^{\infty} d s \exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right ) }_{=\sqrt{\hbar^2\alpha \pi}} \Bigr ]\\ &= \frac{ \hbar \pi}{\alpha^2} \Bigl [ \left ( \frac{1}{ \hbar ^4 \alpha^2 } +2i \frac{1}{ \hbar ^3 \alpha m } t -\frac{1}{m^2\hbar^2}t^2 \right )\frac{\hbar^2\alpha}{2} \Bigl \{\underbrace{[s\exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right ) ]^\infty_{-\infty}}_{=0} - \underbrace{\int_{-\infty}^{\infty} d s \exp \left (- \frac{s^2}{ \hbar ^2 \alpha }\right )}_{=\sqrt{\hbar^2\alpha \pi}} \Bigr \} \\ &\hspace{4cm} + \left ( \frac{1}{ \hbar ^2 \alpha } +i\frac{t}{m\hbar} +\frac{{p_{0i}}^2}{m^2\hbar^2}t^2 \right )\sqrt{\hbar^2\alpha \pi} \Bigr ]\\ &= \frac{ \hbar \pi}{\alpha^2} \sqrt{\hbar^2\alpha \pi} \left [ -\left ( \frac{1}{ \hbar ^4 \alpha^2 } +2i \frac{1}{ \hbar ^3 \alpha m } t -\frac{1}{m^2\hbar^2}t^2 \right )\frac{\hbar^2\alpha}{2} + \left ( \frac{1}{ \hbar ^2 \alpha } +i\frac{t}{m\hbar} +\frac{{p_{0i}}^2}{m^2\hbar^2}t^2 \right ) \right ] \\ &= \hbar^2 \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \left [\left ( \frac{\alpha}{2m^2} + \frac{{p_{0i}}^2}{m^2\hbar^2} \right )t^2 + \frac{1}{2 \hbar ^2 \alpha }\right ] = \left (\frac{\pi}{\alpha} \right )^\frac{3}{2} \left [\left ( \frac{\alpha\hbar^2}{2m^2} + \frac{{p_{0i}}^2}{m^2} \right )t^2 + \frac{1}{2\alpha }\right ] \end{align}