But why a = (q+ip)/sqrt(2)? I thought a' is the particle creation opeartor

But why a = (q+ip)/sqrt(2)? I thought a' is the particle creation opeartor

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送交者: permutation 于 2011-03-26, 20:34:34:

回答: a 是由坐标和动量算符定义的 a=(q+ip)/sqrt(2) 由 whoami 于 2011-03-26, 20:25:34:

Why it can be so simple? Considering the fundamental geometry could be complicate, there might even no proper definition of the position operator.

wiki 上有介绍 - whoami (82 bytes) 2011-03-26, 20:52:17 (568220)
- 还有谐振子是最简单的情形。为什么假设更复杂系统里面的交换关系是一样的？ (无内容) - permutation (0 bytes) 2011-03-26, 21:02:25 (568226)
- 里面是不是有个假设量子场论里的粒子行为和谐振子波函数的交换关系一样？ (无内容) - permutation (0 bytes) 2011-03-26, 20:59:04 (568225)
  - 上面的关系只是以谐振子为例， - whoami (222 bytes) 2011-03-26, 21:42:47 (568253)