把叉乘寫(xiě)成 ε_(tái){abc} C_b C_c,_{abc}表示下指標(biāo)abc,這里用到愛(ài)因斯坦求和約定,即重復(fù)指標(biāo)求和.
然后C_b= p_b + eA_b,乘C_c后得：p_b p_c+ p_b (eA_c)+(eA_b)p_c+e^2 A_b A_c.由于p和p對(duì)易,對(duì)于電磁場(chǎng)是阿貝爾規(guī)范場(chǎng),即A與A對(duì)易,所以p和p的叉乘還有A和A的叉乘都是0.有貢獻(xiàn)的是中間的兩項(xiàng),而選擇空間表象時(shí)動(dòng)量算符可以寫(xiě)成 -i h-bar D_b,這里打不出偏微分符號(hào)就用D表示吧,同樣的下指標(biāo)b是取1,2,3表示三個(gè)方向.
現(xiàn)在求ε_(tái){abc} p_b (eA_c)作用在態(tài)ψ上后為
-i h-bar ε_(tái){abc} D_b [ (eA_c)ψ]=-i h-bar ε_(tái){abc} [D_b(eA_c)] ψ -i h-bar ε_(tái){abc} (eA_c)D_b(ψ)
而ε_(tái){abc} (eA_b)p_c作用在態(tài)ψ上后給出：
-i h-bar ε_(tái){abc} (eA_b)D_c(ψ)
對(duì)比第一個(gè)式子的等號(hào)后面第二項(xiàng)和第二個(gè)式子,把第二個(gè)式子的ε_(tái){abc}的bc指標(biāo)交換成cb,給出一個(gè)符號(hào),則求和后和第一式子等號(hào)后第二項(xiàng)消掉,所以最終結(jié)果就是：
-i h-bar ε_(tái){abc} [D_b(eA_c)] =-i h-bar ∇× A= -i h-bar B
總結(jié)一下關(guān)鍵的地方就是,對(duì)于對(duì)易的量,自身叉乘為0,但對(duì)于非對(duì)易的量（本題指的是算符C）自身叉乘不會(huì)給出0.