用QuTip学量子光学(四):三原子系统(1):相干布局囚禁(CPT),绝热拉曼路径(STIRAP)

用QuTip学量子光学(四):三原子系统(1):相干布局囚禁(CPT),绝热拉曼路径(STIRAP)

三原子系统分为若干部分:

第一部分为本部分:主要是三能级原子约化为二能级原子的一个novel的方法,相干布局囚禁(CPT)与绝热拉曼过程的简要介绍.

第二部分为下一次,介绍电磁感应透明,并对三能级原子以上的所有模拟做一个讨论。

第三部分待计划。

在线阅读器容易爆炸,建议直接在我的GitHub里下载相关的notebook观看,以下是知乎中文版本。

GitHub: caidish/Quantum-Optics-with-Python

基础理论1:三能级原子与一束经典光

我们考虑下面的哈密顿量

\begin{gathered} H = {\omega _a}\left| e \right\rangle \left\langle e \right| + {\varepsilon _{g1}}\left| {g1} \right\rangle \left\langle {g1} \right| + {\varepsilon _{g2}}\left| {g2} \right\rangle \left\langle {g2} \right| \hfill \\ + \frac{\Omega }{2}\left[ {{e^{i\omega t}}\left( {\left| {g1} \right\rangle + \left| {g2} \right\rangle } \right)\left\langle e \right| + h.c} \right] \hfill \\ \end{gathered}

在相互作用表象下(或者旋转表象),采用旋转波近似,哈密顿成为

H = \Delta \left| e \right\rangle \left\langle e \right| + \left( {\Delta - \Delta '} \right)\left| {{g_2}} \right\rangle \left\langle {{g_2}} \right| + \frac{\Omega }{2}\left[ {\left( {\left| {{g_1}} \right\rangle + \left| {{g_2}} \right\rangle } \right)\left\langle e \right| + h.c.} \right]

正如上图暗示的,有着场论背景的同学们会情不自禁地把顶部的能级消掉得到一个所谓的有效理论,他们会利用费米表象勇敢的写出:

\begin{gathered} Z = \int {D\left[ {{{\bar \phi }_i},{\phi _i}} \right]} {e^{ - S\left[ {{{\bar \phi }_i},{\phi _i}} \right]}},i = e,g1,g2 \hfill \\ S\left[ {{{\bar \phi }_i},{\phi _i}} \right] = \int {d\tau \left[ \begin{gathered} \sum\limits_i {{{\bar \phi }_i}{\partial _\tau }{\phi _i}} + \Delta {{\bar \phi }_e}{\phi _e} + \left( {\Delta - \Delta '} \right){{\bar \phi }_{g2}}{\phi _{g2}} \hfill \\ + \frac{\Omega }{2}\left( {{{\bar \phi }_{g1}} + {{\bar \phi }_{g2}}} \right){\phi _e} + \frac{\Omega }{2}{{\bar \phi }_e}\left( {{\phi _{g1}} + {\phi _{g2}}} \right) \hfill \\ \end{gathered} \right]} \hfill \\ \end{gathered}

并利用下面的高斯积分:

S\left[ {\bar v,v} \right] = {e^{ - \int {d\tau \left( {\bar vMv + \bar Jv + \bar vJ} \right)} }} = {e^{Tr\ln M - \bar J{M^{ - 1}}J}}

得到:

\begin{gathered} {S_{eff}}\left[ {{{\bar \phi }_{g1}},{{\bar \phi }_{g2}},{\phi _{g1}},{\phi _{g2}}} \right] \hfill \\ = \int {d\tau \left[ \begin{gathered} \sum\limits_{i = g1,g2} {{{\bar \phi }_i}{\partial _\tau }{\phi _i}} + \Delta {{\bar \phi }_e}{\phi _e} + \left( {\Delta - \Delta '} \right){{\bar \phi }_{g2}}{\phi _{g2}} \hfill \\ - \frac{{{\Omega ^2}}}{4}\left( {{{\bar \phi }_{g1}} + {{\bar \phi }_{g2}}} \right){\Gamma ^{ - 1}}\left( {{\phi _{g1}} + {\phi _{g2}}} \right) + Tr\ln \Gamma \hfill \\ \end{gathered} \right]} \hfill \\ \Gamma = {\partial _\tau } + \Delta ,{\Gamma ^{ - 1}} = \frac{1}{{{\partial _\tau } + \Delta }} = \frac{1}{{i{\partial _t} + \Delta }} \cong {\Delta ^{ - 1}} \hfill \\ \end{gathered}

其中我们使用了大失谐条件,意思就是,顶能级相比我们的耦合光大得不可思议那么就会几乎没有布局数了,于是他如同一个介质一般——就如同班花和班草中间递纸条的你——没有啥用. 但是我们发现,有效的耦合变了。由于腔体积有限,TrLn项作为一个偏置忽略去,那么得到的有效理论就是:

{H_{eff}} = \left( {\Delta - \Delta '} \right)\left| {g2} \right\rangle \left\langle {g2} \right| - \frac{{{\Omega ^2}}}{{4\Delta }}\left( {\left| {g1} \right\rangle \left\langle {g1} \right| + \left| {g2} \right\rangle \left\langle {g2} \right|} \right) - \frac{{{\Omega ^2}}}{{4\Delta }}\left( {\left| {g1} \right\rangle \left\langle {g2} \right| + \left| {g2} \right\rangle \left\langle {g1} \right|} \right)

这个有效哈密顿成立的条件是 e能级的布局数远远小于其它的,而且 \left| {\Delta {c_e}} \right| \gg \left| {{\partial _t}{c_e}} \right|

这个有效哈密顿也可以用微分方程显然的得到(作为练习)

现在我们写一个数值程序来验证一下:

from qutip import *
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
g1 = basis(3,0)
g2 = basis(3,1)
e = basis(3,2)
Ee=10*2*np.pi
Eg1 = 0 * 2*np.pi
Eg2 = 0.4 *2*np.pi
omega = 1*2*np.pi
wa = Ee-Eg1
wa_n = Ee-Eg2
Delta = wa-omega
Delta_n = wa_n-omega
Omega = 0.4 * 2*np.pi #Omega should match the gap between Eg1 and Eg2
H = Delta * e * e.dag() + (Delta-Delta_n)*g2*g2.dag()+Omega/2.0*(g1+g2)*e.dag()+Omega/2.0*e*(g1.dag()+g2.dag())
H_eff = (Delta-Delta_n)*g2*g2.dag() - Omega**2.0/(4*Delta)*(g1*g2.dag()+g2*g1.dag())
psi0 = (g1 + g2).unit()
t = np.linspace(0,10,10000)
result = mesolve(H,psi0,t,[],[g1*g1.dag(),g2*g2.dag()])
result_eff = mesolve(H_eff,psi0,t,[],[g1*g1.dag(),g2*g2.dag()])
fig,ax = plt.subplots(1,2,figsize=(16,9))
ax[0].plot(t,result.expect[0],label='Real Result:g1')
ax[0].plot(t,result_eff.expect[0],label = 'Effective Result:g1')
ax[0].legend()
ax[1].plot(t,result.expect[0],label='Real Result:g2')
ax[1].plot(t,result_eff.expect[0],label = 'Effective Result:g2')
ax[1].legend()

结果如图:

自己看图,不必讨论了.

基础理论2:三能级与两个经典场耦合

哈密顿写作

H = \sum\limits_{i = a,b,c} {{\omega _i}\left| i \right\rangle \left\langle i \right|} + {\Omega _1}\cos \left( {{\nu _1}t + {\phi _1}} \right)\left( {\left| a \right\rangle \left\langle b \right| + \left| b \right\rangle \left\langle a \right|} \right) + {\Omega _2}\cos \left( {{\nu _2}t + {\phi _2}} \right)\left( {\left| a \right\rangle \left\langle c \right| + \left| c \right\rangle \left\langle a \right|} \right)

在相互作用表象中,旋波近似后:

{H_I} = - \Delta \left( {\left| b \right\rangle \left\langle b \right| + \left| c \right\rangle \left\langle c \right|} \right) + \left( {\frac{{{\Omega _1}}}{2}{e^{ - i{\phi _1}}}\left| a \right\rangle \left\langle b \right| + \frac{{{\Omega _2}}}{2}{e^{ - i{\phi _2}}}\left| a \right\rangle \left\langle c \right| + h.c.} \right)

相干布局囚禁

注意到这个哈密顿有一个特殊的本征态:

\left| D \right\rangle = {e^{i{\phi _1}}}\cos \theta \left| b \right\rangle - {e^{i{\phi _2}}}\sin \theta \left| c \right\rangle

这里 \theta = \arctan \Omega_1/\Omega_2

注意到顶能级a被偶然的消除了,但是这和我们上面通过设置初始条件故意消掉顶能级是不一样的.

如果初态是这个态,那么这个态会只旋转,不会跳到顶能级a去.这个技术叫做相干布局囚禁.

from qutip import *
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
a = basis(3,0)
b = basis(3,1)
c = basis(3,2)
n_a = a*a.dag()
n_b = b*b.dag()
n_c = c*c.dag()
w_a = 0 * 2*np.pi
w_b = 0.4 * 2*np.pi
w_c = 1* 2*np.pi
Delta = 0.1* 2*np.pi
nu1 = w_a-w_b-Delta
nu2 = w_a-w_c-Delta
O_1 = 1* 2*np.pi
O_2 = 2* 2*np.pi
theta = np.arctan(O_1/O_2)
H0 = w_a*n_a+w_b*n_b+w_c*n_c
def Omega_1(t,args):
    return O_1/2.0*np.exp(-1j*nu1*t)
def Omega_1_dag(t,args):
    return O_1/2.0*np.exp(1j*nu1*t)
def Omega_2(t,args):
    return O_2/2.0*np.exp(-1j*nu2*t)
def Omega_2_dag(t,args):
    return O_2/2.0*np.exp(1j*nu2*t)
H=[H0,[a*b.dag(),Omega_1],[b*a.dag(),Omega_1_dag],[a*c.dag(),Omega_2],[c*a.dag(),Omega_2_dag]]
psi0 = np.cos(theta)*b-np.sin(theta)*c
t=np.linspace(0,10,1000)
result = mesolve(H,psi0,t,[],[n_a,n_b,n_c])
fig,ax = plt.subplots(figsize=(16,9))
plt.plot(t,result.expect[0],label='N_a')
plt.plot(t,result.expect[1],label='N_b')
plt.plot(t,result.expect[2],label='N_c')
result_state = mesolve(H,psi0,t)
result_b = [b.overlap(t.dag()) for t in result_state.states]
plt.plot(t,np.real(result_b),label='Real part of Cb')
plt.legend(fontsize='large')

看到b能级的布局数不变,只是系数的实部在转.

受激拉曼绝热路径

通过应用量子绝热定理,一个简单的操作b,c态的方式就是利用绝热变化的的脉冲.例如从b演化到c:

\begin{gathered} \frac{{{\Omega _1}\left( { - \infty } \right)}}{{{\Omega _2}\left( { - \infty } \right)}} = 0,\frac{{{\Omega _1}\left( { + \infty } \right)}}{{{\Omega _2}\left( { + \infty } \right)}} = \infty \hfill \\ \theta \left( { - \infty } \right) = 0,\theta \left( { + \infty } \right) = \frac{\pi }{2} \hfill \\ \end{gathered}

上面例子的具体构造是:

{\Omega _1} = {\Omega _0}\exp \left[ { - \frac{{{{\left( {t - \Delta t} \right)}^2}}}{{{T^2}}}} \right],{\Omega _2} = {\Omega _0}\exp \left[ { - \frac{{{{\left( {t + \Delta t} \right)}^2}}}{{{T^2}}}} \right]

from qutip import *
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
a = basis(3,0)
b = basis(3,1)
c = basis(3,2)
n_a = a*a.dag()
n_b = b*b.dag()
n_c = c*c.dag()
w_a = 0 * 2*np.pi
w_b = 0.4 * 2*np.pi
w_c = 1* 2*np.pi
Delta = 0.1* 2*np.pi
nu1 = w_a-w_b-Delta
nu2 = w_a-w_c-Delta
O_1 = 10* 2*np.pi
O_2 = 10* 2*np.pi
theta = np.arctan(O_1/O_2)
Delta_t = 1.1
T=4
H0 = w_a*n_a+w_b*n_b+w_c*n_c
def Gas_1(t):
    return np.exp(-(t-Delta_t)**2/T**2)
def Gas_2(t):    
    return np.exp(-(t+Delta_t)**2/T**2)
def Omega_1(t,args):
    return Gas_1(t)*O_1/2.0*np.exp(-1j*nu1*t)
def Omega_1_dag(t,args):
    return Gas_1(t)*O_1/2.0*np.exp(1j*nu1*t)
def Omega_2(t,args):
    return Gas_2(t)* O_2/2.0*np.exp(-1j*nu2*t)
def Omega_2_dag(t,args):
    return Gas_2(t)*O_2/2.0*np.exp(1j*nu2*t)
H=[H0,[a*b.dag(),Omega_1],[b*a.dag(),Omega_1_dag],[a*c.dag(),Omega_2],[c*a.dag(),Omega_2_dag]]
psi0 = b
t=np.linspace(-20,20,1000)
result = mesolve(H,psi0,t,[],[n_a,n_b,n_c])
fig,ax = plt.subplots(figsize=(16,9))
plt.plot(t,result.expect[0],label='N_a',linewidth=4.0)
plt.plot(t,result.expect[1],label='N_b',linewidth=4.0)
plt.plot(t,result.expect[2],label='N_c',linewidth=4.0)
plt.legend(fontsize='large')
T=4
T=2

通用三能级量子理论:三能级系统能被约化成2维吗?

参考:PRA,54,1586.

REF:

上面有几个内容的详细推导可以参考 @梁昊:

https://chaoli.club/index.php/4037​chaoli.club

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