UA OPTI570 量子力學21 Atom Trapping

• • 1-D Gaussian Laser Beam
  • Red-detuned Laser
  • 3-D Q.H.O. Approximation

討論atom trapping一般用下面兩種設定：

Stark Shift：E-field shifting atomic energy levels

Zeeman Shift: H-field shifting atomic energy levels

在實驗室中可以用Off-resonance laser light做atom trapping的實驗，Off-resonance laser light的好處是幾乎可以忽略H-field，只保留oscillating E-field，于是可以觀測到electric dipole interaction energy：$?\text{d}?\text{E}$，其中$\text{d}$是induced dipole moment，$\text{E}$是laser E-field。Off-resonant的重要特征是laser detuning $\Delta =w_{laser}?w_{atom}$，這里$w_{atom}$表示resonant frequency，如果$\Delta <0$，稱這種情況為red detuning（因為atom frequency向紅光方向偏移），這時可以構造potential well for atom（如下圖）
如果$\Delta >0$，稱這種情況為blue detuning，這時可以構造potential barrier for atom（potential well的形狀上下翻轉一下就是barrier）。用$U$表示勢能，在上述實驗設定中，
$$U\propto \frac{I(x,y,z)}{\Delta }$$

其中$I$是spatially dependent laser beam indensity。

---

1-D Gaussian Laser Beam

考慮在$\hat{z}$上傳播的Gaussian Laser Beam，假設
$$I(x,y,z)=I(x)=I_0{e}^{?\frac{2x^2}{d^2}}$$

則勢能滿足
$$U(x)=?U_0{e}^{?\frac{2x^2}{d^2}},U_0\propto \frac{I_0}{\Delta }$$

這是一個Gaussian Potential Well，取$x=0$為零勢點，也就是
$$U(x)=U_0?U_0{e}^{?\frac{2x^2}{d^2}}$$

考慮一個有趣的問題：怎么構造一個量子諧振子，它的勢能可以近似$U(x)$？簡單設想一下，Q.H.O的勢能關于$x$是二次的，在這個勢能阱的底部可以很好地近似，但無法近似勢能阱的頂部，也就是說Q.H.O近似Gaussian勢能阱對那些低能量粒子才會有比較好的效果。

$$\begin{array}{rl}U(x)& =U_0(1?e^{?\frac{2x^2}{d^2}})\\ & =U_0\left[1?\left(\sum _{n=0}^{+\infty }(?1{)}^n\frac{2^n{x}^{2n}}{n!d^{2n}}\right)\right]\\ & =U_0\sum _{n=1}^{+\infty }(?1{)}^{n?1}\frac{2^n{x}^{2n}}{n!d^{2n}}\end{array}$$

當$x$足夠小時，可以用一階展開近似，
$$U(x)\approx \frac{2U_0{x}^2}{d^2}$$

用能量守恒
$$\begin{gathered} \frac{1}{2}mw{x}^2=\frac{2U_0{x}^2}{d^2} \\ w=\sqrt{\frac{4U_0}{m{d}^2}} \end{gathered}$$

因此低能量的粒子可以用頻率為$w$的Q.H.O近似。

例 Consider atom in ground state $|{?}_0?$ of red-detuned Gaussian laser beam trap, assuming Q.H.O is good approximation. Let $x=0$ be the center of the Q.H.O and the Gaussian laser beam trap. Let $\sigma =\sqrt{\frac{?}{mw}}$ be the Q.H.O length scale. Given $w=\sqrt{\frac{4U_0}{m{d}^2}}$,
$${\sigma }^2=\frac{?d}{2\sqrt{m{U}_0}}$$ Assume the wave function corresponding to the ground state is
$${?}_0(x)=(\pi {\sigma }^2{)}^{?1/4}{e}^{?\frac{x^2}{{\sigma }^2}}$$ In the following figure, the y-axis represents the potential energy devided by $U_0$ and the x-axis represents $x/\sigma$. The red curve is $U(x)$ and the blue curve shows the Q.H.O approximation.

If the mirror bumped at $t=0$, $U(x)$ becomes $U(x?b)$ and then the ground state moves to a coherent state
$$D({\alpha }_0^{\prime })|{\psi }_0?=|{\alpha }_0^{\prime }?,{\alpha }_0^{\prime }=?\frac{b}{\sqrt{2}\sigma }$$

such that
$$\hat{A}|{\alpha }_0^{\prime }?={\alpha }_0^{\prime }|{\alpha }_0^{\prime }?$$

Let $\hat{A}|{\alpha }^{\prime }(t)?={\alpha }^{\prime }(t)|{\alpha }^{\prime }(t)?$, using the Schroedinger time-evolving operator
$${\alpha }^{\prime }(t)={\alpha }_0^{\prime }{e}^{?iwt}=?\frac{b}{\sqrt{2}\sigma }{e}^{?iwt}$$

The maximum magnitude of momentum is given by
$$|{\alpha }^{\prime }(t)|=\frac{\sigma P_{max}}{\sqrt{2}?},P_{max}=\frac{?b}{{\sigma }^2}$$

But how to transform back to the ground state?

$D(?{\alpha }_0^{\prime })\hat{U}(t=\frac{2\pi }{w})D({\alpha }_0^{\prime })$

$D({\alpha }_0^{\prime })\hat{U}(t=\frac{\pi }{w})D({\alpha }_0^{\prime })$

$D(i{\alpha }_0^{\prime })\hat{U}(t=\frac{\pi }{2w})D({\alpha }_0^{\prime })$

Red-detuned Laser

考慮
$$I(x,y,z)=I_0{e}^{?\frac{2x^2}{d^2}}{e}^{?\frac{2y^2}{d^2}}{\cos }^2(kz)$$

其中$k=2\pi /\lambda$，沿$\hat{z}$方向，假設$x=y=0$，則
$$I(z)=I_0{\cos }^2(kz),U(z)\propto \frac{I_0}{\Delta }{\cos }^2(kz)$$

假設
$$U(z)/U_0=?{\cos }^2(kz),U_0\propto I_0/\Delta$$

$U(z)/U_0$關于$kz$的圖像如下，

取$z=0$為零勢點，當$z$足夠小時
$$U(z)=?U_0{\cos }^2(kz)+U_0\approx U_0{k}^2{z}^2$$

根據能量守恒，
$$\begin{gathered} U_0{k}^2{z}^2=\frac{1}{2}m{w}_z{z}^2 \\ {w}_z=2\pi \sqrt{\frac{2U_0}{m{\lambda }^2}} \end{gathered}$$

用Q.H.O近似$U(z)$，能級為
$$E_n=?w_z(n+1/2)\approx 2\pi \sqrt{\frac{2U_0}{m{\lambda }^2}}?(n+1/2)$$

這個近似成立的條件是
$$\frac{U_0}{?w_z}>>1$$

也就是不能只有幾個低能級的狀態可以被近似。

3-D Q.H.O. Approximation

考慮用3-D Q.H.O.近似3-D Gaussian Laser Beam Trap，假設Laser Beam關于$z$軸對稱，則Q.H.O.的eigen-energy滿足
$$E=E_{n_x,n_y,n_z}=?w_{xy}(n_x+n_y+1)+?w_z(n_z+1/2)$$

記對應的energy eigenstate為$|n_x,n_y,n_z?$，Ground state為$|n_x=0,n_y=0,n_z=0?$。When $t=0$, atom in ground state, shortly the mirror bumped and the $\hat{y}$ direction got shifted of $+1\mu$m. For the $\hat{y}$ direction,
$${\alpha }_y=\frac{1}{\sqrt{2}}\left(\frac{?y?}{{\sigma }_y}+\frac{i{\sigma }_y?P_y?}{?}\right)=\frac{1}{\sqrt{2}}\frac{?1\mu m}{{\sigma }_y}$$

Assume ${\sigma }_y=0.25\mu m$, $\alpha =?2.8$. And $\frac{?H_y?}{?w_y}=|?2.8{|}^2+1/2=8.5$. Now we know the initial state of the atom is
$$|n_x=0,n_y=?2.8,n_z=0?=|\psi (0)?$$

Applying the Schroedinger time-evolution operator,
$$|\psi (t)?=|n_x=0,n_y=?2.8e^{?iwt},n_z=0?$$

總結

以上是生活随笔為你收集整理的UA OPTI570 量子力学21 Atom Trapping的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。