I tried to write formulas (part 8)
<Time-independent Hamiltonian>
Schrödinger's Wave Equation is described as follows.
\begin{eqnarray}i\hbar\frac{\partial}{\partial t}\Psi &=&\hat{H}\Psi\end{eqnarray}
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Schrödinger's Wave Equation is described as follows when we assume that is and that is
\begin{eqnarray}i\hbar\frac{\partial}{\partial t}\Psi &=&\hat{H}\Psi\notag\\i\hbar\frac{\partial}{\partial t}\Psi({\boldsymbol r}, t) &=&\hat{H}({\boldsymbol r}, t)\Psi({\boldsymbol r}, t) \notag\\i\hbar\frac{\partial}{\partial t}\psi({\boldsymbol r})\phi(t) &=&\hat{H}(\boldsymbol r)\psi({\boldsymbol r})\phi(t) \notag\\i\hbar\psi({\boldsymbol r})\frac{\partial}{\partial t}\phi(t) &=&\phi(t)\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) \notag\\i\hbar\frac{1}{\phi(t)}\frac{\partial}{\partial t}\phi(t) &=&\frac{1}{\psi(\boldsymbol r)}\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) \end{eqnarray}
Here, left hand side is a function of , and the right hand side is a function of . Therefore, the both sides must be equal to constant function , whose dimension is energy.
\begin{eqnarray}i\hbar\frac{1}{\phi(t)}\frac{\partial}{\partial t}\phi(t) &=&E\notag\\\frac{\partial}{\partial t}\phi(t) &=&-{i}\frac{E}{\hbar}\phi(t)\notag\\\left(\frac{\partial}{\partial t}+i\frac{E}{\hbar}\right)\phi(t) &=&0\notag\\\phi(t) &=&{\rm Const.}e^{-{i}\frac{E}{\hbar}t}\notag\\\phi(t) &=&\phi(t_0)e^{-{i}\frac{E}{\hbar}(t-t_0)}\\\notag\\\frac{1}{\psi(\boldsymbol r)}\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) &=&E\notag\\\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) &=&E\psi(\boldsymbol r)\end{eqnarray}
Therefore, can be obtained if is known by solving the equation on . is expressed as below.
\begin{eqnarray}\hat{H}&=&\hat{K} +\hat{V} \\&=&\frac{\hat{\boldsymbol{p}}^2}{2m} +\hat{V}\\&=&-\frac{\hbar^2}{2m}\nabla ^2 +\hat{V}\left(\because \hat{\boldsymbol p}=-i\hbar\frac{\partial}{\partial \boldsymbol r}\right) \end{eqnarray}
Dimensional analysis makes it to easier to memorize the equation. Also, it is important that to know conjugate variables. (Position and momentum are conjugate each other. Energy and time are conjugate each other.) Moreover, the dimension of the product of position and momentum is equal to that of energy and time.
\begin{eqnarray} {\rm \left[m\cdot \left(kg\cdot m/s\right)\right]=\space\left[m\cdot \left(kg\cdot m/s^2\right)\cdot s\right] =\space \left[m\cdot N\cdot s\right]=\space \left[J\cdot s\right]}\end{eqnarray}
<Conjugate variables>
Poisson bracket is defined as follows.
\begin{eqnarray}[\hat{A}, \hat{B}]\space{\equiv}\space\hat{A}\hat{B}-\hat{B}\hat{A}\end{eqnarray}
Poisson bracket ] is calculated as below.
\begin{eqnarray}\left[\hat{x}, \hat{p_x}\right]f&=&\left(\hat{x}\hat{p_x}-\hat{p_x}\hat{x}\right)f\notag\\ &=&\left(x\left(-i\hbar\frac{\partial}{\partial x}\right)-\left(-i\hbar\frac{\partial}{\partial x}\right)x\right)f \notag\\&=&-i\hbar\left(x\frac{\partial}{\partial x}-\frac{\partial}{\partial x}x\right)f \notag\\ &=&-i\hbar\left(x\frac{\partial}{\partial x}-\left(1+x\frac{\partial}{\partial x}\right)\right)f \notag\\&=&i\hbar f\notag\\\left[\hat{x}, \hat{p_x}\right]&=&i\hbar\hat{1}\end{eqnarray}
Here, is the identity operator.
\begin{eqnarray}\left[\hat{x}, \hat{p_x}\right]&=&i\hbar\hat{1}\notag\\\left(-i\hbar\right)\left[\hat{x}, \frac{\partial}{\partial x}\right]&=&i\hbar\hat{1}\notag\\\left[\hat{x}, \frac{\partial}{\partial x}\right]&=&-\hat{1}\end{eqnarray}
Mathematically, is the mere constant.
<時間に依存しないハミルトニアン>
\begin{eqnarray}i\hbar\frac{\partial}{\partial t}\Psi &=&\hat{H}\Psi\end{eqnarray}
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がであり、がであると仮定すると、シュレーディンガーの波動方程式は次のように記述される。
\begin{eqnarray}i\hbar\frac{\partial}{\partial t}\Psi &=&\hat{H}\Psi\notag\\i\hbar\frac{\partial}{\partial t}\Psi({\boldsymbol r}, t) &=&\hat{H}({\boldsymbol r}, t)\Psi({\boldsymbol r}, t) \notag\\i\hbar\frac{\partial}{\partial t}\psi({\boldsymbol r})\phi(t) &=&\hat{H}(\boldsymbol r)\psi({\boldsymbol r})\phi(t) \notag\\i\hbar\psi({\boldsymbol r})\frac{\partial}{\partial t}\phi(t) &=&\phi(t)\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) \notag\\i\hbar\frac{1}{\phi(t)}\frac{\partial}{\partial t}\phi(t) &=&\frac{1}{\psi(\boldsymbol r)}\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) \end{eqnarray}
ここで、左辺はの関数であり、右辺はの関数であるから、両辺は定数関数に等しくなければならない。はエネルギーの次元を持つ。
\begin{eqnarray}i\hbar\frac{1}{\phi(t)}\frac{\partial}{\partial t}\phi(t) &=&E\notag\\\frac{\partial}{\partial t}\phi(t) &=&-{i}\frac{E}{\hbar}\phi(t)\notag\\\left(\frac{\partial}{\partial t}+i\frac{E}{\hbar}\right)\phi(t) &=&0\notag\\\phi(t) &=&{\rm Const.}e^{-{i}\frac{E}{\hbar}t}\notag\\\phi(t) &=&\phi(t_0)e^{-{i}\frac{E}{\hbar}(t-t_0)}\\\notag\\\frac{1}{\psi(\boldsymbol r)}\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) &=&E\notag\\\hat{H}(\boldsymbol r)\psi({\boldsymbol r}) &=&E\psi(\boldsymbol r)\end{eqnarray}
従って、に関する方程式を解くことでが分かれば、は得られる。は次のように表現できる。
\begin{eqnarray}\hat{H}&=&\hat{K} +\hat{V} \\&=&\frac{\hat{\boldsymbol{p}}^2}{2m} +\hat{V}\\&=&-\frac{\hbar^2}{2m}\nabla ^2 +\hat{V}\left(\because \hat{\boldsymbol p}=-i\hbar\frac{\partial}{\partial \boldsymbol r}\right) \end{eqnarray}
次元解析は方程式を覚えるのに役立つ。また、共役変数について知ることが重要である(位置と運動量は互いに共役であり、エネルギーと時間は互いに共役である)。さらに位置と運動量の積の次元は、エネルギーと時間の積の次元に等しい。
\begin{eqnarray} {\rm \left[m\cdot \left(kg\cdot m/s\right)\right]=\space\left[m\cdot \left(kg\cdot m/s^2\right)\cdot s\right] =\space \left[m\cdot N\cdot s\right]=\space \left[J\cdot s\right]}\end{eqnarray}
<共役変数>
ポアソン括弧は次のように定義される。
\begin{eqnarray}[\hat{A}, \hat{B}]\space{\equiv}\space\hat{A}\hat{B}-\hat{B}\hat{A}\end{eqnarray}
ポアソン括弧]は次のように計算される。
\begin{eqnarray}\left[\hat{x}, \hat{p_x}\right]f&=&\left(\hat{x}\hat{p_x}-\hat{p_x}\hat{x}\right)f\notag\\ &=&\left(x\left(-i\hbar\frac{\partial}{\partial x}\right)-\left(-i\hbar\frac{\partial}{\partial x}\right)x\right)f \notag\\&=&-i\hbar\left(x\frac{\partial}{\partial x}-\frac{\partial}{\partial x}x\right)f \notag\\ &=&-i\hbar\left(x\frac{\partial}{\partial x}-\left(1+x\frac{\partial}{\partial x}\right)\right)f \notag\\&=&i\hbar f\notag\\\left[\hat{x}, \hat{p_x}\right]&=&i\hbar\hat{1}\end{eqnarray}
ここで、は恒等演算子である。
\begin{eqnarray}\left[\hat{x}, \hat{p_x}\right]&=&i\hbar\hat{1}\notag\\\left(-i\hbar\right)\left[\hat{x}, \frac{\partial}{\partial x}\right]&=&i\hbar\hat{1}\notag\\\left[\hat{x}, \frac{\partial}{\partial x}\right]&=&-\hat{1}\end{eqnarray}
数学的には, はただの定数である。