UA OPTI570 量子力学25 2-level System
UA OPTI570 量子力學25 2-level System
- 2-level System與Rabi oscillation
2-level System與Rabi oscillation
Spin-1/2的方法可以用于任意二維態空間（稱之為2-level system），考慮E2\mathcal{E}_{2}E2?，假設它的基為{∣u1?,∣u2?}\{|u_1 \rangle,|u_2 \rangle\}{∣u1??,∣u2??}，在這組基下，任意量子態可以表示為一個有兩個元素的列向量，任意算符可以表示為2×22 \times 22×2的矩陣，于是在這種與Spin-1/2類似的數學模型下，Spin-1/2的相關結果可以直接移植到2-level system中。
?∣ψ?∈E2\forall |\psi \rangle \in \mathcal{E}_2?∣ψ?∈E2?，∣ψ?=a∣u1?+b∣u2?,?a,b∈C|\psi \rangle=a|u_1 \rangle+b|u_2 \rangle,\exists a,b \in \mathbb{C}∣ψ?=a∣u1??+b∣u2??,?a,b∈C，并且?θ,?\exists \theta ,\phi?θ,?,
a=cos?θ2e?i?2,b=sin?θ2ei?2\begin{aligned} a = \cos \frac{\theta}{2} e^{-\frac{i \phi}{2}},b = \sin \frac{\theta}{2} e^{\frac{i \phi}{2}} \end{aligned}a=cos2θ?e?2i??,b=sin2θ?e2i???
注意θ,?\theta,\phiθ,?是任意參數坐標，并不一定代表真實物理空間中的角度。Pauli Spin Matrix的期望為
?σx?=sin?θcos???σy?=sin?θsin???σz?=cos?θ\langle \sigma_x \rangle = \sin \theta \cos \phi \\ \langle \sigma_y \rangle = \sin \theta \sin \phi \\ \langle \sigma_z \rangle = \cos \theta?σx??=sinθcos??σy??=sinθsin??σz??=cosθ
于是Bloch vector為
?σ?=[sin?θcos??sin?θsin??cos?θ]\langle \sigma \rangle = \left[ \begin{matrix} \sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta \end{matrix} \right]?σ?=???sinθcos?sinθsin?cosθ????
例1 假設H=E1∣u1??u1∣+E2∣u2??u2∣H=E_1|u_1 \rangle \langle u_1 |+E_2|u_2 \rangle \langle u_2 |H=E1?∣u1???u1?∣+E2?∣u2???u2?∣，它的矩陣表示為diag(E1,E2)diag(E_1,E_2)diag(E1?,E2?)，假設∣ψ(0)?=∣u1?|\psi(0) \rangle=|u_1 \rangle∣ψ(0)?=∣u1??，
∣ψ(t)?=e?iHt/?∣ψ(0)?=e?iE1t/?∣u1?|\psi(t) \rangle = e^{-iHt/\hbar}|\psi(0) \rangle=e^{-iE_1t/\hbar}|u_1 \rangle∣ψ(t)?=e?iHt/?∣ψ(0)?=e?iE1?t/?∣u1??
這是一個很有趣的結果，在有像這道題定義的哈密頓量的系統中，如果初始量子態是某個本征態，那么量子態并不會隨時間演化到另一個本征態，而是會一直停留在這個本征子空間中，也就是說P(E2)=0P(E_2)=0P(E2?)=0。
例2 假設H=E1∣u1??u1∣+E2∣u2??u2∣+?1^H=E_1|u_1 \rangle \langle u_1 |+E_2|u_2 \rangle \langle u_2 |+\epsilon \hat 1H=E1?∣u1???u1?∣+E2?∣u2???u2?∣+?1^，則延用例1的設定，
∣ψ(t)?=e?i(E1+?)t/?∣u1?|\psi(t) \rangle = e^{-i(E_1+\epsilon)t/\hbar}|u_1 \rangle∣ψ(t)?=e?i(E1?+?)t/?∣u1??
也就是同時同量改變兩個本征態的本征值不會影響例1的結論。
例3 假設H=E1∣u1??u1∣+E2∣u2??u2∣+WH=E_1|u_1 \rangle \langle u_1 |+E_2|u_2 \rangle \langle u_2 |+WH=E1?∣u1???u1?∣+E2?∣u2???u2?∣+W，其中算符W=W12∣u1??u2∣+W21∣u2??u1∣W=W_{12}|u_1 \rangle \langle u_2 |+W_{21}|u_2 \rangle \langle u_1 |W=W12?∣u1???u2?∣+W21?∣u2???u1?∣，則哈密頓量的矩陣表示為
[E1W12W21E2]=[E1W21?W21E2]=[E1+E22+E1?E22W12W21E1+E22?E1?E22]\left[ \begin{matrix} E_1 & W_{12} \\ W_{21} & E_2 \end{matrix} \right]=\left[ \begin{matrix} E_1 & W_{21}^* \\ W_{21} & E_2 \end{matrix} \right] = \left[ \begin{matrix} \frac{E_1+E_2}{2}+\frac{E_1-E_2}{2} & W_{12} \\ W_{21} & \frac{E_1+E_2}{2}-\frac{E_1-E_2}{2}\end{matrix} \right][E1?W21??W12?E2??]=[E1?W21??W21??E2??]=[2E1?+E2??+2E1??E2??W21??W12?2E1?+E2???2E1??E2???]
記Em=E1+E22,δ=E1?E22E_m=\frac{E_1+E_2}{2},\delta = \frac{E_1-E_2}{2}Em?=2E1?+E2??,δ=2E1??E2??, 則
H=Em1^+[δW21?W21?δ]=Em1^+δ2+∣W21∣2σuσu=1δ2+∣W21∣2[W12+W212iW12?W212δ]H = E_m \hat 1+\left[ \begin{matrix} \delta & W_{21}^* \\ W_{21} & -\delta\end{matrix} \right]=E_m \hat 1+\sqrt{\delta^2+|W_{21}|^2}\sigma_u \\ \sigma_u = \frac{1}{\sqrt{\delta^2+|W_{21}|^2}}\left[ \begin{matrix} \frac{W_{12}+W_{21}}{2} \\ i\frac{W_{12}-W_{21}}{2} \\ \delta \end{matrix} \right]H=Em?1^+[δW21??W21???δ?]=Em?1^+δ2+∣W21?∣2?σu?σu?=δ2+∣W21?∣2?1????2W12?+W21??i2W12??W21??δ????
HHH的本征值為
E+=Em+δ2+∣W21∣2E1=Em?δ2+∣W21∣2E_+=E_m+\sqrt{\delta^2+|W_{21}|^2} \\ E_1 = E_m - \sqrt{\delta^2+|W_{21}|^2}E+?=Em?+δ2+∣W21?∣2?E1?=Em??δ2+∣W21?∣2?
本征態為
∣ψ+?=cos?θ2e?i?2∣u1?+sin?θ2ei?2∣u2?∣ψ??=?sin?θ2e?i?2∣u1?+cos?θ2ei?2∣u2?θ=arctan?∣W21∣δ,?=Arg(W12)|\psi_+ \rangle= \cos \frac{\theta}{2} e^{-\frac{i \phi}{2}}|u_1 \rangle+ \sin \frac{\theta}{2} e^{\frac{i \phi}{2}} |u_2 \rangle \\ |\psi_- \rangle=- \sin \frac{\theta}{2} e^{-\frac{i \phi}{2}}|u_1 \rangle+ \cos \frac{\theta}{2} e^{\frac{i \phi}{2}} |u_2 \rangle \\ \theta = \arctan \frac{|W_{21}|}{\delta},\phi = Arg(W_{12})∣ψ+??=cos2θ?e?2i??∣u1??+sin2θ?e2i??∣u2??∣ψ???=?sin2θ?e?2i??∣u1??+cos2θ?e2i??∣u2??θ=arctanδ∣W21?∣?,?=Arg(W12?)
下面討論任意量子態的演化規律：假設初始態為
∣ψ(0)?=a1(0)∣u1?+a2(0)∣u2?|\psi(0) \rangle = a_1(0)|u_1 \rangle + a_2(0)|u_2 \rangle∣ψ(0)?=a1?(0)∣u1??+a2?(0)∣u2??
目標是得到
∣ψ(t)?=a1(t)∣u1?+a2(t)∣u2?|\psi(t) \rangle = a_1(t)|u_1 \rangle + a_2(t)|u_2 \rangle∣ψ(t)?=a1?(t)∣u1??+a2?(t)∣u2??
這里的系數含義是狀態轉移概率幅，從0時刻到ttt時刻，由量子態∣u1?|u_1 \rangle∣u1??轉移到∣u2?|u_2 \rangle∣u2??與量子態∣u2?|u_2 \rangle∣u2??轉移到∣u1?|u_1 \rangle∣u1??的概率為
P1→2(t)=∣a2(t)∣2,P2→1(t)=∣a1(t)∣2P_{1 \to 2} (t)=|a_2(t)|^2,P_{2 \to 1}(t)=|a_1(t)|^2P1→2?(t)=∣a2?(t)∣2,P2→1?(t)=∣a1?(t)∣2
要做這個計算有下面兩種方法：
結果為
{a1(t)=a1(0)(cos?2θ2e?iΩt/2+sin?2θ2eiΩt/2)?ia2(0)sin?θe?i?sin?Ωt2a2(t)=a2(0)(sin?2θ2e?iΩt/2+cos?2θ2eiΩt/2)?ia1(0)sin?θei?sin?Ωt2\begin{cases} a_1(t)=a_1(0) \left( \cos^2 \frac{\theta}{2} e^{-i \Omega t/2}+ \sin^2 \frac{\theta}{2} e^{i \Omega t/2}\right)-ia_2(0)\sin \theta e^{-i \phi}\sin \frac{\Omega t}{2} \\ a_2(t)=a_2(0) \left( \sin^2 \frac{\theta}{2} e^{-i \Omega t/2}+ \cos^2 \frac{\theta}{2} e^{i \Omega t/2}\right)-ia_1(0)\sin \theta e^{i \phi}\sin \frac{\Omega t}{2}\end{cases}{a1?(t)=a1?(0)(cos22θ?e?iΩt/2+sin22θ?eiΩt/2)?ia2?(0)sinθe?i?sin2Ωt?a2?(t)=a2?(0)(sin22θ?e?iΩt/2+cos22θ?eiΩt/2)?ia1?(0)sinθei?sin2Ωt??
其中
Ω=E+?E??=2?δ2+∣W21∣2\Omega = \frac{E_+-E_-}{\hbar}=\frac{2}{\hbar}\sqrt{\delta^2+|W_{21}|^2}Ω=?E+??E???=?2?δ2+∣W21?∣2?
考慮一個特例，比如a1(0)=1,a2(0)=0a_1(0)=1,a_2(0)=0a1?(0)=1,a2?(0)=0，則
P1→2(t)=sin?2θsin?2Ωt2,P1→1=1?P1→2(t)P_{1 \to 2}(t) = \sin^2 \theta \sin^2 \frac{\Omega t}{2},P_{1 \to 1}=1-P_{1 \to 2}(t)P1→2?(t)=sin2θsin22Ωt?,P1→1?=1?P1→2?(t)
定義
Δ=E1?E2?=2δ?Ω0=2?W21=∣Ω0∣ei?\Delta = \frac{E_1-E_2}{\hbar} = \frac{2 \delta }{\hbar} \\ \Omega_0 = \frac{2}{\hbar}W_{21} = |\Omega_0|e^{i \phi}Δ=?E1??E2??=?2δ?Ω0?=?2?W21?=∣Ω0?∣ei?
則哈密頓量為
H{u}=[EmEm]+?2[ΔΩ0?Ω0?Δ]E±=Em±?2Ω,Ω=Δ2+∣Ω0∣2H_{\{u\}} = \left[ \begin{matrix} E_m \\ & E_m \end{matrix} \right] +\frac{\hbar}{2}\left[ \begin{matrix} \Delta & \Omega_0^* \\ \Omega_0 & - \Delta \end{matrix} \right] \\ E_{\pm} = E_m \pm \frac{\hbar}{2}\Omega,\ \Omega = \sqrt{\Delta^2+|\Omega_0|^2}H{u}?=[Em??Em??]+2??[ΔΩ0??Ω0???Δ?]E±?=Em?±2??Ω,?Ω=Δ2+∣Ω0?∣2?
狀態轉移概率為
P1→2(t)=∣Ω0∣2Ω2sin?2Ωt2P_{1 \to 2}(t)=\frac{|\Omega_0|^2}{\Omega^2}\sin^2 \frac{\Omega t}{2}P1→2?(t)=Ω2∣Ω0?∣2?sin22Ωt?
這個公式被稱為Rabi公式，在2-level system中處理狀態轉移時這個公式具有通用性，其中Ω0\Omega_0Ω0?被稱為Resonant Rabi frequency；Ω\OmegaΩ被稱為Rabi frequency或者generalized Rabi frequency；Δ\DeltaΔ被稱為detuning；這個公式是Rabi Oscillation模型的一部分；∣Ω0∣2Ω2\frac{|\Omega_0|^2}{\Omega^2}Ω2∣Ω0?∣2?被稱為Rabi oscillations的振幅；在Δ=0,Ω=Ω0\Delta=0,\Omega=\Omega_0Δ=0,Ω=Ω0?時，稱P1→2(t)P_{1 \to 2}(t)P1→2?(t)的半個周期，t=π∣Ω0∣t=\frac{\pi}{|\Omega_0|}t=∣Ω0?∣π?為π\piπ-pulse，整個周期為2π2\pi2π-pulse。
Bloch vector為
?σ?=(?σx?,?σy?,?σz?)\langle \sigma \rangle = (\langle \sigma_x \rangle,\langle \sigma_y \rangle,\langle \sigma_z \rangle)?σ?=(?σx??,?σy??,?σz??)
在量子態∣ψ?=a1∣u1?+a2∣u2?|\psi \rangle=a_1|u_1 \rangle+a_2 |u_2 \rangle∣ψ?=a1?∣u1??+a2?∣u2??中，
?σz?=[a1?a2?][100?1][a1a2]=∣a1∣2?∣a2∣2\langle \sigma_z \rangle = \left[\begin{matrix} a_1^* & a_2^* \end{matrix} \right]\left[\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right]\left[\begin{matrix} a_1 \\ a_2 \end{matrix} \right]=|a_1|^2-|a_2|^2?σz??=[a1???a2???][10?0?1?][a1?a2??]=∣a1?∣2?∣a2?∣2
類似地，
?σx?=a1?a2+a1a2?,?σy?=?ia1?a2+ia1a2?\langle \sigma_x \rangle=a_1^*a_2+a_1a_2^*,\langle \sigma_y \rangle=-ia_1^*a_2+ia_1a_2^*?σx??=a1??a2?+a1?a2??,?σy??=?ia1??a2?+ia1?a2??
總結
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