User:Ripe

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de-2 Dieser Benutzer hat fortgeschrittene Deutschkenntnisse.
zh-2 該用戶能以一般的中文進行交流。
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la-1 Hic usor simplici latinitate contribuere potest.
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[edit] mathml scratchpad

dx^2 = \frac {\int_{-\infty}^{\infty} {x^2f(x)f^*(x)dx} } { \int_{-\infty}^{\infty} {f(x)f^*(x)dx} }

ds^2 = \frac {\int_{-\infty}^{\infty} {s^2f(s)f^*(s)dx} } { \int_{-\infty}^{\infty} {f(s)f^*(s)ds} }


dx^2 \cdot ds^2 = \frac {\int_{-\infty}^{\infty} {x^2f(x)f^*(x)dx} } { \int_{-\infty}^{\infty} {f(x)f^*(x)dx} } \cdot \frac {\int_{-\infty}^{\infty} {s^2f(s)f^*(s)dx} } { \int_{-\infty}^{\infty} {f(s)f^*(s)ds} }

dx \cdot ds \ge \frac {1} {4\pi}


dx^2 ds^2 \ge \frac {|\int_{-\infty}^{\infty} dx x f^* f' + x f f'^* |^2} {16 \pi (\int_{-infty}^{\infty} f f^* dx)^2 }

integrate by parts with u=x and dv = d/dx:

dx^2 ds^2\ge \frac {|\int_{-\infty}^{\infty} dx x f^* f' + x f f'^* |^2} {16 \pi (\int_{-\infty}^{\infty} f f^* dx)^2 }

\frac {|\int_{-\infty}^{\infty} dx f f^* | ^2} {16 \pi (\int_{-\infty}^{\infty} dx f f^*) ^2 } \sim \frac {1}{16\pi} \le dx^2ds^2

\frac {1} {4\pi} \le dx ds

\int_0^\infty \Delta p \ge \frac{\hbar}{2}

Gabor function: G(t) \propto e^{[\frac{-t^2}{2a^2} + i(kt + \theta)]}