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拉比震蕩

2023-06-29 21:53 作者:陽既望  | 我要投稿



對于非共振場,時間平均值的推導可以利用拉格朗日中值定理和各態(tài)歷經(jīng)假設。具體步驟如下:

假設非共振場的強度為E_0%5Ccos(%5Comega%20t),其中%5Comega是遠大于原子能級間躍遷頻率的高頻場。

假設原子的二能級系統(tǒng)為%7Cg%5Crangle%7Ce%5Crangle,其中%7Cg%5Crangle是基態(tài),%7Ce%5Crangle是激發(fā)態(tài),能級差為%5Chbar%5Comega_0。

假設原子在非共振場中的哈密頓量為H%3DH_0%2BH_1(t),其中H_0%3D%5Cfrac%7B1%7D%7B2%7D%5Chbar%5Comega_0%5Csigma_z是原子的自由哈密頓量,%5Csigma_z%3D%7Ce%5Crangle%5Clangle%20e%7C-%7Cg%5Crangle%5Clangle%20g%7C是泡利矩陣,H_1(t)%3D-%5Cfrac%7B1%7D%7B2%7D%5Chbar%5COmega%5Ccos(%5Comega%20t)%5Csigma_x是與場相互作用的哈密頓量,%5COmega%3DE_0d%2F%5Chbar是拉比頻率,d%3D%5Clangle%20e%7C%5Chat%7Bd%7D%7Cg%5Crangle是偶極矩矩陣元,%5Csigma_x%3D%7Ce%5Crangle%5Clangle%20g%7C%2B%7Cg%5Crangle%5Clangle%20e%7C是泡利矩陣。

假設原子在t%3D0時處于基態(tài)%7Cg%5Crangle,求解含時薛定諤方程i%5Chbar%5Cfrac%7B%5Cmathrm%7Bd%7D%7D%7B%5Cmathrm%7Bd%7Dt%7D%7C%5Cpsi(t)%5Crangle%3DH%7C%5Cpsi(t)%5Crangle,得到原子在任意時刻t的態(tài)為%7C%5Cpsi(t)%5Crangle%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0-%5Comega)t%2F2%7D-%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0%2B%5Comega)t%2F2%7D%5Cright%5D%7Cg%5Crangle%2B%5Cfrac%7Bi%7D%7B2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0-%5Comega)t%2F2%7D%2B%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0%2B%5Comega)t%2F2%7D%5Cright%5D%7Ce%5Crangle%20

計算原子在激發(fā)態(tài)的幾率為%20P_e(t)%3D%7C%5Clangle%20e%7C%5Cpsi(t)%5Crangle%7C2%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(1%2B%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%2B%5Cfrac%7B1%7D%7B4%7D%5Cleft(1-%5Cfrac%7B%5COmega2%7D%7B%5Comega%5E2%7D%5Cright)%5Ccos(%5Comega%20t)%20

對原子在激發(fā)態(tài)的幾率進行時間平均,即在一個周期內積分并除以周期,得到 %20%5Clangle%20P_e(t)%5Crangle%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_0%5ET%20P_e(t)%5Cmathrm%7Bd%7Dt%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(1%2B%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%20

利用拉格朗日中值定理,存在一個介于$t=0$和$t=T$之間的$\tau$,使得%20%20%5Clangle%20P_e(t)%5Crangle%3DP_e(%5Ctau)%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(1%2B%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%2B%5Cfrac%7B1%7D%7B4%7D%5Cleft(1-%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%5Ccos(%5Comega%20%5Ctau)%20


%5Clangle%20P_e(t)%5Crangle%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_0%5ET%20P_e(t)%5Cmathrm%7Bd%7Dt%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_0%5ET%20%5Cfrac%7B1%7D%7B4%7D%5Cleft(1%2B%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%2B%5Cfrac%7B1%7D%7B4%7D%5Cleft(1-%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%5Ccos(%5Comega%20t)%5Cmathrm%7Bd%7Dt%5C%20%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(1%2B%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%2B%5Cfrac%7B1%7D%7B4%7D%5Cleft(1-%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%5Cfrac%7B1%7D%7BT%7D%5Cint_0%5ET%20%5Ccos(%5Comega%20t)%5Cmathrm%7Bd%7Dt%5C%20%3D%5Cfrac%7B1%7D%7B4%7D%5Cleft(1%2B%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%2B%5Cfrac%7B1%7D%7B4%7D%5Cleft(1-%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D%5Cright)%5Cfrac%7B%5Csin(%5Comega%20T)%7D%7B%5Comega%20T%7D%20


利用各態(tài)歷經(jīng)假設,即認為在一個較長的時間內,原子的狀態(tài)可以遍歷所有可能的狀態(tài),那么可以用時間平均代替系綜平均,即

%5Clangle%20P_e(t)%5Crangle%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_0%5ET%20P_e(t)%5Cmathrm%7Bd%7Dt%3D%5Cfrac%7B1%7D%7BT%7D%5Cint_0%5ET%20P_e(%5Ctau)%5Cmathrm%7Bd%7D%5Ctau%3DP_e(%5Ctau)

由此得到?

%5Ccos(%5Comega%20%5Ctau)%3D%5Cfrac%7B%5COmega2%7D%7B%5Comega2%7D

由于%5Comega%20%5Cgg%20%5COmega,所以%5Ccos(%5Comega%20%5Ctau)接近于1,因此可以近似得到 %20%5Comega%20%5Ctau%3D2n%5Cpi%2B%5Cdelta%20其中$n$是整數(shù),%5Cdelta是一個很小的角度。

將上式代入原子的態(tài),得到

%20%7C%5Cpsi(%5Ctau)%5Crangle%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0-%5Comega)(2n%5Cpi%2B%5Cdelta)%2F2%7D-%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0%2B%5Comega)(2n%5Cpi%2B%5Cdelta)%2F2%7D%5Cright%5D%7Cg%5Crangle%2B%5Cfrac%7Bi%7D%7B2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0-%5Comega)(2n%5Cpi%2B%5Cdelta)%2F2%7D%2B%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0%2B%5Comega)(2n%5Cpi%2B%5Cdelta)%2F2%7D%5Cright%5D%7Ce%5Crangle

由于%5Cdelta很小,可以忽略其對指數(shù)函數(shù)的影響,同時利用歐拉公式,得到?

%7C%5Cpsi(%5Ctau)%5Crangle%3D%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0-%5Comega)2n%5Cpi%2F2%7D-%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0%2B%5Comega)2n%5Cpi%2F2%7D%5Cright%5D%7Cg%5Crangle%2B%5Cfrac%7Bi%7D%7B2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0-%5Comega)2n%5Cpi%2F2%7D%2B%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)e%7B-i(%5Comega_0%2B%5Comega)2n%5Cpi%2F2%7D%5Cright%5D%7Ce%5Crangle%5C%20%3D%5Cfrac%7B1%7D%7B2%7De%7B-in(%5Comega_0-%5Comega)%5Cpi%2F2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)-(-1)n%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)%5Cright%5D%7Cg%5Crangle%2B%5Cfrac%7Bi%7D%7B2%7De%7B-in(%5Comega_0-%5Comega)%5Cpi%2F2%7D%5Cleft%5B%5Cleft(1%2B%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)%2B(-1)n%5Cleft(1-%5Cfrac%7B%5COmega%7D%7B%5Comega%7D%5Cright)%5Cright%5D%7Ce%5Crangle%5C%20%3D%5Cbegin%7Bcases%7D%20%7Cg%5Crangle%20%26%20n%3D4k%5C%20%7Ce%5Crangle%20%26%20n%3D4k%2B1%5C%20-%7Cg%5Crangle%20%26%20n%3D4k%2B2%5C%20-%7Ce%5Crangle%20%26%20n%3D4k%2B3%5C%20%5Cend%7Bcases%7D

其中$k$是整數(shù)。

這個結果說明,在非共振場中,原子的態(tài)會在基態(tài)和激發(fā)態(tài)之間周期性地跳躍,每個周期為T%3D4%5Cpi%2F%5Comega,這種現(xiàn)象稱為拉比振蕩。



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