Re: [問題] 量子力學中的角動量
※ 引述《slime036 (slime036)》之銘言:
: 今天吃飯和同學討論到角動量能不能為零
: 我同學的觀點是古典力學中的角動量不能為零 因為這樣就是靜止不動(...他因該是這
: 樣說啦) 但量子力學中角動量可以為0 他是用L^2=L(L+1) L受主量子數影響可以為0
: 但是我是用測不準原理去想 角動量為0的話 角度的不準度就會無限大 這樣物理就不合
: 就這樣討論了滿久 也不知道誰對誰錯
: 另外一個疑問是零點能量 量力中的零點能量是指主量子數為零那項 還是基態能量
: 某孫有說過 以波粒二像性來看 所有粒子都有零點能量
: 快被搞混了....
古典力學中,角動量可以為零。 角動量為零僅表示物體"不轉動",並非靜止不動。
量子力學中,角動量也可以為零,只是很怪,因為你"很難想像"一個繞著原子核在轉的
電子,其軌道角動量"可以為零"!(L=0)。
此外,我不清楚你如何利用Heisenberg's uncertainty principle推得出,當角動量為零
時,其角度的不準確度就會無限大。
換個說法,當角動量不為零時,角度的不準確度就不會無限大囉!?
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就我對Heisenberg's uncertainty principle的認知
(Delta_Theta)*(Delta_angular momentum) >= some constant
--(感謝chungweitw 大大 & mantour大大)
我看不出來為什麼角動量等於零時,角度的不準確度才會無限大。
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另一方面,即使當角動量為零時,其角度的不準確度為無限大,那又怎樣?
為什麼你會覺得不符合物理呢? (是因為角度的不準確度為無限大嗎??)
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當一個實驗"很精確地"測得一個粒子(這裡請容許我用"粒子"...)所擁有的動量時,其位置
的測量誤差就會趨向無限大,不是嗎?
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Appendix:
(我認真的找了一下那篇paper... 以下是那篇paper的Abstract(他寫summary啦...)
,大家可以參考看看。看起來蠻有趣的~)
The uncertainty relation between the orbital angular
momentum component Lz and the corresponding angle "phi" is discussed.
The uncertainty for an angle variable is defined. The formula
(delta_Lz)*(delta_phi) >= h/4pi,
which is sometimes quoted, is shown to be incorrect, and an alternative
relation, in full accord with the Heisenberg Uncertainty Principle, is
derived.
PS: 因為有些數學符號無法正常顯示,所以我對原文作了些修正。 有興趣的人可以參考
http://www.springerlink.com/content/x4x83103v5nk2g24/fulltext.pdf?page=1
--
" Research is to see what everybody else has seen,
and think nobody has thought." - Albert Szent-Gyorgyi
--
※ 發信站: 批踢踢實業坊(ptt.cc)
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