Pembuktian Teorema Ehrenfest

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Kula Nuwun.

\section{Pembuktian Teorema Ehrenfest}

Laju nilai harap sebuah besaran fisis $\Omega \in \mathbb{R}$ pada swa-keadaan $\left|\psi\right>$ anggota ruang Hilbert kompleks $\mathcal{H}$ secara kuantum pada waktu $t \in \mathbb{R}$ adalah
\[ \frac{d\left<\Omega\right>_\psi}{dt} = \frac{i}{\hbar}\left<\psi\right|[\hat{H}, \hat{\Omega}]\left|\psi\right> + \left<\psi\right|\frac{d\hat{\Omega}}{dt}\left|\psi\right> \]
di mana $\left<\Omega\right>_\psi \equiv \left<\psi\right|\hat{\Omega}\left|\psi\right>$, $i := \sqrt{-1}$, $\hbar \in \mathbb{R}^+$ adalah tetapan Planck tereduksi, $[\hat{H}, \hat{\Omega}] := \hat{H}\hat{\Omega} - \hat{\Omega}\hat{H}$, $\hat{H}$ adalah operator Hamiltonian, dan $\hat{\Omega}$ adalah operator dari besaran fisis $\Omega$.

Apabila $\hat{H} := \hat{T} + \hat{V}$ di mana $\hat{T} := \sum_{j = 1}^3 \hat{p}_j^2/(2m)$, $m \in \mathbb{R}^+$ adalah massa partikel kuantum, $\hat{V} := \sum_{k, l, n = 0}^\infty (1/(k!l!n!))V_{kln}(0, 0, 0)\hat{x}_1^k\hat{x}_2^l\hat{x}_3^n$, $V \in \mathbb{R}$ adalah tenaga potensial yang bergantung pada posisi $(x_1, x_2, x_3) \in \mathbb{R}^3$,
\[ V_{kln}(0, 0, 0) := \lim_{x_1 \to 0}\lim_{x_2 \to 0}\lim_{x_3 \to 0}\frac{\partial^{k + l + n}V}{\partial x_1^k\partial x_2^l\partial x_3^n}, \]
dan $\hat{x}_k$ adalah operator posisi yang mewakili posisi $x_k$ untuk setiap $k \in \{1, 2, 3\}$.

Oleh karena itu,
\[ \frac{d\left<\vec{p}\right>_\psi}{dt} = \frac{i}{\hbar}\sum_{j = 1}^3\left<\psi\right|[\hat{H}, \hat{p}_j]\left|\psi\right>\vec{n}_j + \left<\psi\right|\frac{d\hat{\vec{p}}}{dt}\left|\psi\right> \]
di mana $\vec{n}_1 := (1, 0, 0)$, $\vec{n}_2 := (0, 1, 0)$, dan $\vec{n}_3 := (0, 0, 1)$.

Selanjutnya,
\[ \frac{d\left<\vec{p}\right>_\psi}{dt} = \frac{i}{\hbar}\sum_{j = 1}^3\sum_{k, l, n = 0}^\infty\frac{1}{k!l!n!}V_{kln}(0, 0, 0)\left<\psi\right|[\hat{x}_1^k\hat{x}_2^l\hat{x}_3^n, \hat{p}_j]\left|\psi\right>\vec{n}_j \]
Karena
\[ [\hat{x}_1^k\hat{x}_2^l\hat{x}_3^n, \hat{p}_j] = i\hbar(k\delta_{1j}\hat{x}_1^{k - 1}\hat{x}_2^l\hat{x}_3^n + l\delta_{2j}\hat{x}_1^k\hat{x}_2^{l - 1}\hat{x}_3^n + n\delta_{3j}\hat{x}_1^k\hat{x}_2^l\hat{x}_3^{n - 1}), \]
maka
\[ \frac{d\left<\vec{p}\right>_\psi}{dt} = -\left<\psi\right|\left(\frac{\partial V}{\partial x_1}\vec{n}_1 + \frac{\partial V}{\partial x_2}\vec{n}_2 + \frac{\partial V}{\partial x_3}\vec{n}_3\right)\left|\psi\right> \]
sehingga
\[ \frac{d\left<\vec{p}\right>_\psi}{dt} = -\left<\psi\right|\nabla V\left|\psi\right> = \left<\vec{F}\right>_\psi. \]

Dengan cara yang sama, kita peroleh
\[ \frac{d\left<\vec{r}\right>_\psi}{dt} = \frac{i}{\hbar}\frac{1}{2m}\sum_{j, k = 1}^3\left<\psi\right|[\hat{p}_j^2, \hat{x}_k]\left|\psi\right>\vec{n}_k. \]
Karena $[\hat{p}_j^2, \hat{x}_k] = -2i\hbar\delta_{jk}\hat{p}_j$, maka
\[ \frac{d\left<\vec{r}\right>_\psi}{dt} = \frac{1}{m}\sum_{j, k = 1}^3\left<\psi\right|\delta_{jk}\hat{p}_j\left|\psi\right>\vec{n}_k \]
alias
\[ \frac{d\left<\vec{r}\right>_\psi}{dt} = \frac{1}{m}\sum_{j = 1}^3\left<\psi\right|\hat{p}_j\left|\psi\right>\vec{n}_j \]
alias
\[ \frac{d\left<\vec{r}\right>_\psi}{dt} = \frac{\left<\psi\right|\hat{\vec{p}}\left|\psi\right>}{2m} = \frac{\left<\vec{p}\right>_\psi}{m}. \]

Alhamdulillah hirobbil alamin.