看板 Physics 關於我們 聯絡資訊
先幫你做幾題好了 剩下自己做喔^^ + If f(a,a )is a function which can be expanded in a power series + of a and a ,then show that + + (a)[a,f(a,a )]=∂f/∂a + (b)[f(a,a ),a]=∂f/∂a + + (c)exp[-αa a]f(a,a )exp[αa a]=f(a*exp[α],a*exp[-α]) ----- + n + n-1 [a,(a ) ]=n(a )------(1) n + n-1 [a , a ]=n(a )------(2) 2 exp[iA] B exp[-iA]=B+i[A,B]+(i)/(2!)[A,[A,B]] n +...+(i)/(n!)[A,[A,...[A,B]...]B]------(3) m + n let f=Σg_mn a (a ) (a) + m + n [a,f(a,a )]==Σg_mn [a,a (a ) ] m + n =Σg_mn a [a,(a ) ] m + n-1 =Σg_mn a (n)(a ) + ==∂f/∂a (b) + + m + n + [f(a,a ),a ]=Σg_mn [a (a ) ,a ] m + + n =Σg_mn[a ,(a )](a ) =∂f/∂a (c) + + exp[-αa a]f(a,a )exp[αa a] + m + n + =Σg_mn exp[-αa a]a (a ) exp[αa a] m + n =Σg_mn exp[-αN]a (a ) exp[αN] m + =Σg_mn exp[-αN]a exp[αN]exp[-αN](a )exp[αN] --- exp[iA] B exp[-iA]=B+i[A,B]+(i)/(2!)[A,[A,B]] n +...+(i)/(n!)[A,[A,...[A,B]...]B]------(3) n +n let A=iN B=a or B=(a ) and then using [N,a]=-a + + [N,a ]=a then you get the answer 第二題 你用第一題用道的等式就oK^^ ---- show that + [a,exp[-αa a]]=(exp[-]) ---- 第三題 你用exp的定義去展開 第二條式子 有計算上問題可以再提出來問版友喔! --
nightkid:電子檔可接受嗎?08/16 12:18
Griffiths1:......樓上也太高調了點08/16 12:34
xgcj:你完蛋了nightkid~原作者來了 08/16 12:44
nightkid:作者抱歉(_ _)08/16 13:03
Entropy1988:原來物理版其實也是很歡樂的 XD08/16 13:16
-- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.66.194 ※ 編輯: xgcj 來自: 140.113.66.194 (03/04 21:09)
murai111:請問一下第二題一開始怎麼下手比較好 03/05 00:07
murai111:第三題還是不太懂...唉 真的是量力新手 03/05 00:35