UA OPTI570 量子力学33 Time-dependent Perturbation基础
UA OPTI570 量子力學33 Time-dependent Perturbation基礎
背景
上一講介紹interaction picture時提到了time-dependent perturbation，假設Hamiltonian滿足
H(t)=H0+W(t)H(t)=H_0+W(t)H(t)=H0?+W(t)
也就是和時間無關的項H0H_0H0?與和時間有關的項W(t)W(t)W(t)可以分開，稱W(t)W(t)W(t)是time-dependent perturbation，基于H0H_0H0?的時間演化算符為
U(t,t0)=e?iH0(t?t0)/?U(t,t_0)=e^{-iH_0(t-t_0)/\hbar}U(t,t0?)=e?iH0?(t?t0?)/?
Interaction Picture的參考系滿足
F=U?(t,t0)=eiH0(t?t0)/?\mathbb{F}=U^{\dag}(t,t_0)=e^{iH_0(t-t_0)/\hbar}F=U?(t,t0?)=eiH0?(t?t0?)/?
簡單起見記t0=0t_0=0t0?=0，假設某個量子態可以寫成能量特征態的疊加：
∣ψ(t)?=∑ncn(t)∣?n?|\psi(t) \rangle = \sum_n c_n(t) |\phi_n \rangle∣ψ(t)?=n∑?cn?(t)∣?n??
則在interaction picture中，
∣ψI(t)?=∑ncn(t)eiH0t/?∣?n?|\psi_I(t) \rangle = \sum_n c_n(t)e^{iH_0t/\hbar} |\phi_n \rangle∣ψI?(t)?=n∑?cn?(t)eiH0?t/?∣?n??
可以發現在interaction picture中，相同能量特征態的概率不會改變，
Pn=∣cn(t)∣2=∣cn(t)eiH0t/?∣2\mathbb{P}_n=|c_n(t)|^2=|c_n(t)e^{iH_0t/\hbar} |^2Pn?=∣cn?(t)∣2=∣cn?(t)eiH0?t/?∣2
這一講我們介紹一點time-dependent perturbation theory的基礎，用來近似量子態之間的轉移概率。
問題描述
哈密頓量HHH滿足H(t)=H0+W(t)H(t)=H_0+W(t)H(t)=H0?+W(t)，假設H0H_0H0?的特征方程為
H0∣?n?=En∣?n?H_0|\phi_n \rangle = E_n |\phi_n \rangleH0?∣?n??=En?∣?n??
初始量子態為
∣ψ(t0)?=∑ncn(t0)∣?n?|\psi(t_0) \rangle = \sum_n c_n(t_0)|\phi_n \rangle∣ψ(t0?)?=n∑?cn?(t0?)∣?n??
目標是得到ttt時間后系統可能的量子態（用∣ψf?|\psi_f \rangle∣ψf??表示）及相應概率（用Pf(t)\mathbb{P}_f(t)Pf?(t)表示）。
理想情況
理想情況是∣ψ(t)?|\psi(t) \rangle∣ψ(t)?可以根據哈密頓量H(t)H(t)H(t)與薛定諤方程解析得到，那么Pf(t)\mathbb{P}_f(t)Pf?(t)就等于到∣ψf?|\psi_f \rangle∣ψf??的投影算符的均值，即
Pf(t)=?ψ(t)∣ψf??ψf∣ψ(t)?\mathbb{P}_f(t) = \langle \psi(t) | \psi_f \rangle \langle \psi_f | \psi(t) \ranglePf?(t)=?ψ(t)∣ψf???ψf?∣ψ(t)?
但是time-dependent Hamiltonian定義的量子系統通常很難有解析解，∣ψ(t)?|\psi(t) \rangle∣ψ(t)?的解析式也就找不到。
Time-dependent Perturbation Theory
Time-dependent Perturbation Theory提供了在找不到∣ψ(t)?|\psi(t) \rangle∣ψ(t)?的解析式的情況下，近似計算Pf(t)\mathbb{P}_f(t)Pf?(t)的方法。記W(t)=λW^(t)W(t)=\lambda \hat W(t)W(t)=λW^(t)，對bn(t)=cn(t)eiH0t/?b_n(t)=c_n(t)e^{iH_0t/\hbar}bn?(t)=cn?(t)eiH0?t/?做展開，
bn(t)=bn(0)(t)+λbn(1)(t)+λ2bn(2)(t)+?b_n(t)=b_n^{(0)}(t)+\lambda b_n^{(1)}(t) +\lambda^2 b_n^{(2)}(t) +\cdotsbn?(t)=bn(0)?(t)+λbn(1)?(t)+λ2bn(2)?(t)+?
其中λrbn(r)(t)\lambda^r b_n^{(r)}(t)λrbn(r)?(t)表示bn(t)b_n(t)bn?(t)的展開式中的第rrr階項，將其代入到interaction picture的Effective薛定諤方程中，
i???t∣ψI(t)?=HI(t)∣ψI(t)?=λeiH0(t?t0)/?W^(t)e?iH0(t?t0)/?∣ψI(t)?\begin{aligned} i\hbar \frac{\partial}{\partial t}|\psi_I(t) \rangle = H_I(t)|\psi_I(t) \rangle = \lambda e^{iH_0(t-t_0)/\hbar} \hat W(t)e^{-iH_0(t-t_0)/\hbar} |\psi_I(t) \rangle\end{aligned}i??t??∣ψI?(t)?=HI?(t)∣ψI?(t)?=λeiH0?(t?t0?)/?W^(t)e?iH0?(t?t0?)/?∣ψI?(t)??
其中
∣ψI(t)?=∑nbn(t)∣?n?=∑n∑rλrbn(r)(t)∣?n?|\psi_I(t) \rangle=\sum_n b_n(t)|\phi_n \rangle = \sum_n \sum _r \lambda^r b_n^{(r)}(t) |\phi_n \rangle∣ψI?(t)?=n∑?bn?(t)∣?n??=n∑?r∑?λrbn(r)?(t)∣?n??
所以
i???t∣ψI(t)?=i?∑n∣?n?∑rλr??tbn(r)(t)λeiH0(t?t0)/?W^(t)e?iH0(t?t0)/?∣ψI(t)?=λeiH0(t?t0)/?W^(t)e?iH0(t?t0)/?∑n∑rλrbn(r)(t)∣?n?i\hbar \frac{\partial}{\partial t}|\psi_I(t) \rangle = i\hbar \sum_n |\phi_n \rangle \sum _r \lambda^r \frac{\partial}{\partial t} b_n^{(r)}(t) \\ \lambda e^{iH_0(t-t_0)/\hbar} \hat W(t)e^{-iH_0(t-t_0)/\hbar} |\psi_I(t) \rangle \\ = \lambda e^{iH_0(t-t_0)/\hbar} \hat W(t)e^{-iH_0(t-t_0)/\hbar} \sum_n \sum _r \lambda^r b_n^{(r)}(t) |\phi_n \ranglei??t??∣ψI?(t)?=i?n∑?∣?n??r∑?λr?t??bn(r)?(t)λeiH0?(t?t0?)/?W^(t)e?iH0?(t?t0?)/?∣ψI?(t)?=λeiH0?(t?t0?)/?W^(t)e?iH0?(t?t0?)/?n∑?r∑?λrbn(r)?(t)∣?n??
化簡可得
??tbn(r)(t)∣?n?=bn(r?1)(t)eiH0(t?t0)/?W^(t)e?iH0(t?t0)/?∣?n?bn(r)(t)∣?n?=∫t0tbn(r?1)(t′)eiH0(t′?t0)/?W^(t′)e?iH0(t′?t0)/?∣?n?dt′??k∣bn(r)(t)∣?n?=∫t0tbn(r?1)(t′)??k∣eiH0(t′?t0)/?W^(t′)e?iH0(t′?t0)/?∣?n?dt′=1i?∫t0tbk(r?1)(t′)??k∣W^(t′)dt′∣?n?eiwnkt′dt′∑k??k∣bn(r)(t)∣?n?=bn(r)(t)=1i?∫t0t∑keiwnkt′W^nk(t′)bk(r?1)(t′)dt′\frac{\partial}{\partial t} b_n^{(r)}(t) |\phi_n\rangle =b_n^{(r-1)}(t) e^{iH_0(t-t_0)/\hbar} \hat W(t)e^{-iH_0(t-t_0)/\hbar}|\phi_n \rangle \\ b_n^{(r)}(t) |\phi_n\rangle =\int_{t_0}^t b_n^{(r-1)}(t') e^{iH_0(t'-t_0)/\hbar} \hat W(t')e^{-iH_0(t'-t_0)/\hbar}|\phi_n \rangle dt' \\ \begin{aligned} \langle \phi_k |b_n^{(r)}(t) |\phi_n\rangle & =\int_{t_0}^t b_n^{(r-1)}(t') \langle \phi_k |e^{iH_0(t'-t_0)/\hbar} \hat W(t')e^{-iH_0(t'-t_0)/\hbar}|\phi_n \rangle dt' \\ & = \frac{1}{i\hbar} \int_{t_0}^t b_k^{(r-1)}(t') \langle \phi_k | \hat W(t')dt' | \phi_n \rangle e^{iw_{nk}t'} dt'\end{aligned} \\ \sum_k \langle \phi_k |b_n^{(r)}(t) |\phi_n\rangle = b_n^{(r)}(t)=\frac{1}{i\hbar} \int_{t_0}^t \sum_k e^{iw_{nk}t'}\hat W_{nk}(t')b_k^{(r-1)}(t')dt'?t??bn(r)?(t)∣?n??=bn(r?1)?(t)eiH0?(t?t0?)/?W^(t)e?iH0?(t?t0?)/?∣?n??bn(r)?(t)∣?n??=∫t0?t?bn(r?1)?(t′)eiH0?(t′?t0?)/?W^(t′)e?iH0?(t′?t0?)/?∣?n??dt′??k?∣bn(r)?(t)∣?n???=∫t0?t?bn(r?1)?(t′)??k?∣eiH0?(t′?t0?)/?W^(t′)e?iH0?(t′?t0?)/?∣?n??dt′=i?1?∫t0?t?bk(r?1)?(t′)??k?∣W^(t′)dt′∣?n??eiwnk?t′dt′?k∑???k?∣bn(r)?(t)∣?n??=bn(r)?(t)=i?1?∫t0?t?k∑?eiwnk?t′W^nk?(t′)bk(r?1)?(t′)dt′
所以它的解為
bn(r)(t)=1i?∫t0t∑keiwnkt′W^nk(t′)bk(r?1)(t′)dt′b_n^{(r)}(t) =\frac{1}{i\hbar} \int_{t_0}^t \sum_k e^{iw_{nk}t'}\hat W_{nk}(t')b_k^{(r-1)}(t')dt'bn(r)?(t)=i?1?∫t0?t?k∑?eiwnk?t′W^nk?(t′)bk(r?1)?(t′)dt′
比如，
bn(0)=bn(t0)bn(1)(t)=1i?∫t0t∑keiwnkt′W^nk(t′)bk(t′)dt′bn(2)(t)=1i?∫t0t∑k′eiwnkt′′W^nk(t′′)bk′(1)(t′′)dt′′b_n^{(0)}=b_n(t_0) \\ b_n^{(1)}(t)=\frac{1}{i\hbar} \int_{t_0}^t \sum_k e^{iw_{nk}t'}\hat W_{nk}(t')b_k(t')dt' \\ b_n^{(2)}(t)=\frac{1}{i\hbar} \int_{t_0}^t \sum_{k'} e^{iw_{nk}t''}\hat W_{nk}(t'')b_{k'}^{(1)}(t'')dt'' bn(0)?=bn?(t0?)bn(1)?(t)=i?1?∫t0?t?k∑?eiwnk?t′W^nk?(t′)bk?(t′)dt′bn(2)?(t)=i?1?∫t0?t?k′∑?eiwnk?t′′W^nk?(t′′)bk′(1)?(t′′)dt′′
其中wnkw_{nk}wnk?是Bohr frequency，可以用Einstein-Planck關系得到
wnk=En?Ek?w_{nk}=\frac{E_n-E_k}{\hbar}wnk?=?En??Ek??
W^nk(t′)\hat W_{nk}(t')W^nk?(t′)由perturbation計算得到
W^nk(t′)=??n∣W^(t′)∣?k?\hat W_{nk}(t') = \langle \phi_n |\hat W(t')|\phi_k \rangleW^nk?(t′)=??n?∣W^(t′)∣?k??
例：假設∣ψ(0)?=∣?i?|\psi(0) \rangle=|\phi_i \rangle∣ψ(0)?=∣?i??，則bi(0)=bi(0)=1b_i^{(0)}=b_i(0)=1bi(0)?=bi?(0)=1，所以
bn(1)(t)=1i?∫t0t∑keiwnkt′W^nk(t′)bk(t′)dt′=1i?∫t0teiwnit′W^ni(t′)dt′b_n^{(1)}(t)=\frac{1}{i\hbar} \int_{t_0}^t \sum_k e^{iw_{nk}t'}\hat W_{nk}(t')b_k(t')dt'=\frac{1}{i\hbar} \int_{t_0}^t e^{iw_{ni}t'}\hat W_{ni}(t')dt'bn(1)?(t)=i?1?∫t0?t?k∑?eiwnk?t′W^nk?(t′)bk?(t′)dt′=i?1?∫t0?t?eiwni?t′W^ni?(t′)dt′
綜上，如果用一階近似，則
Pf(t)=∣bf(0)+λbf(1)(t)∣2\mathbb{P}_f(t)=|b_f(0)+\lambda b_f^{(1)}(t)|^2Pf?(t)=∣bf?(0)+λbf(1)?(t)∣2
如果初始量子態為∣?i?|\phi_i \rangle∣?i??，ttt時間后的量子態為∣?f?|\phi_f \rangle∣?f??，則
Pi→f(t)=λ2∣bf(1)∣2=λ2?2∣∫t0teiwnit′W^ni(t′)dt′∣2\mathbb{P}_{i \to f}(t)=\lambda^2|b_f^{(1)}|^2 = \frac{\lambda^2}{\hbar^2}|\int_{t_0}^t e^{iw_{ni}t'}\hat W_{ni}(t')dt'|^2Pi→f?(t)=λ2∣bf(1)?∣2=?2λ2?∣∫t0?t?eiwni?t′W^ni?(t′)dt′∣2
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