UA OPTI570 量子力学21 Atom Trapping

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

讨论atom trapping一般用下面两种设定：

Stark Shift：E-field shifting atomic energy levels
Zeeman Shift: H-field shifting atomic energy levels

在实验室中可以用Off-resonance laser light做atom trapping的实验，Off-resonance laser light的好处是几乎可以忽略H-field，只保留oscillating E-field，于是可以观测到electric dipole interaction energy：−d⋅E-\textbf d \cdot \textbf E−d⋅E，其中d\textbf dd是induced dipole moment，E\textbf EE是laser E-field。Off-resonant的重要特征是laser detuning Δ=wlaser−watom\Delta=w_{laser}-w_{atom}Δ=wlaser−watom，这里watomw_{atom}watom表示resonant frequency，如果Δ<0\Delta<0Δ<0，称这种情况为red detuning（因为atom frequency向红光方向偏移），这时可以构造potential well for atom（如下图）
如果Δ>0\Delta>0Δ>0，称这种情况为blue detuning，这时可以构造potential barrier for atom（potential well的形状上下翻转一下就是barrier）。用UUU表示势能，在上述实验设定中，
U∝I(x,y,z)ΔU \propto \frac{I(x,y,z)}{\Delta}U∝ΔI(x,y,z)

其中III是spatially dependent laser beam indensity。

1-D Gaussian Laser Beam

考虑在z^\hat zz^上传播的Gaussian Laser Beam，假设
I(x,y,z)=I(x)=I0e−2x2d2I(x,y,z)=I(x)=I_0e^{-\frac{2x^2}{d^2}}I(x,y,z)=I(x)=I0e−d22x2

则势能满足
U(x)=−U0e−2x2d2,U0∝I0ΔU(x) =-U_0e^{-\frac{2x^2}{d^2}},U_0\propto \frac{I_0}{\Delta}U(x)=−U0e−d22x2,U0∝ΔI0

这是一个Gaussian Potential Well，取x=0x=0x=0为零势点，也就是
U(x)=U0−U0e−2x2d2U(x) = U_0-U_0e^{-\frac{2x^2}{d^2}}U(x)=U0−U0e−d22x2

考虑一个有趣的问题：怎么构造一个量子谐振子，它的势能可以近似U(x)U(x)U(x)？简单设想一下，Q.H.O的势能关于xxx是二次的，在这个势能阱的底部可以很好地近似，但无法近似势能阱的顶部，也就是说Q.H.O近似Gaussian势能阱对那些低能量粒子才会有比较好的效果。

U(x)=U0(1−e−2x2d2)=U0[1−(∑n=0+∞(−1)n2nx2nn!d2n)]=U0∑n=1+∞(−1)n−12nx2nn!d2n\begin{aligned}U(x) & = U_0(1-e^{-\frac{2x^2}{d^2}}) \\ & = U_0 \left[ 1 - \left(\sum_{n=0}^{+\infty}(-1)^n \frac{2^nx^{2n}}{n! d^{2n}} \right)\right] \\ & = U_0 \sum_{n=1}^{+\infty}(-1)^{n-1}\frac{2^nx^{2n}}{n! d^{2n}}\end{aligned}U(x)=U0(1−e−d22x2)=U0[1−(n=0∑+∞(−1)nn!d2n2nx2n)]=U0n=1∑+∞(−1)n−1n!d2n2nx2n

当xxx足够小时，可以用一阶展开近似，
U(x)≈2U0x2d2U(x) \approx \frac{2U_0x^2}{d^2}U(x)≈d22U0x2

用能量守恒
12mwx2=2U0x2d2w=4U0md2\frac{1}{2}mwx^2=\frac{2U_0x^2}{d^2} \\ w = \sqrt{\frac{4U_0}{md^2}}21mwx2=d22U0x2w=md24U0

因此低能量的粒子可以用频率为www的Q.H.O近似。

例 Consider atom in ground state ∣ϕ0⟩|\phi_0 \rangle∣ϕ0⟩ of red-detuned Gaussian laser beam trap, assuming Q.H.O is good approximation. Let x=0x=0x=0 be the center of the Q.H.O and the Gaussian laser beam trap. Let σ=ℏmw\sigma=\sqrt{\frac{\hbar}{mw}}σ=mwℏ be the Q.H.O length scale. Given w=4U0md2w = \sqrt{\frac{4U_0}{md^2}}w=md24U0,
σ2=ℏd2mU0\sigma^2 = \frac{\hbar d}{2 \sqrt{mU_0}}σ2=2mU0ℏd Assume the wave function corresponding to the ground state is
ϕ0(x)=(πσ2)−1/4e−x2σ2\phi_0(x) = (\pi \sigma^2)^{-1/4}e^{-\frac{x^2}{\sigma^2}}ϕ0(x)=(πσ2)−1/4e−σ2x2 In the following figure, the y-axis represents the potential energy devided by U0U_0U0 and the x-axis represents x/σx/\sigmax/σ. The red curve is U(x)U(x)U(x) and the blue curve shows the Q.H.O approximation.

If the mirror bumped at t=0t=0t=0, U(x)U(x)U(x) becomes U(x−b)U(x-b)U(x−b) and then the ground state moves to a coherent state
D(α0′)∣ψ0⟩=∣α0′⟩,α0′=−b2σD(\alpha_0')|\psi_0 \rangle = |\alpha_0' \rangle,\alpha_0'=-\frac{b}{\sqrt{2} \sigma}D(α0′)∣ψ0⟩=∣α0′⟩,α0′=−2σb

such that
A^∣α0′⟩=α0′∣α0′⟩\hat A |\alpha_0' \rangle = \alpha_0' |\alpha_0 ' \rangleA^∣α0′⟩=α0′∣α0′⟩

Let A^∣α′(t)⟩=α′(t)∣α′(t)⟩\hat A |\alpha'(t) \rangle = \alpha'(t) |\alpha'(t) \rangleA^∣α′(t)⟩=α′(t)∣α′(t)⟩, using the Schroedinger time-evolving operator
α′(t)=α0′e−iwt=−b2σe−iwt\alpha'(t)=\alpha_0'e^{-iwt} = -\frac{b}{\sqrt{2} \sigma}e^{-iwt} α′(t)=α0′e−iwt=−2σbe−iwt

The maximum magnitude of momentum is given by
∣α′(t)∣=σPmax2ℏ,Pmax=ℏbσ2|\alpha'(t)|=\frac{\sigma P_{max}}{\sqrt{2} \hbar}, P_{max}=\frac{\hbar b}{\sigma^2}∣α′(t)∣=2ℏσPmax,Pmax=σ2ℏb

But how to transform back to the ground state?

D(−α0′)U^(t=2πw)D(α0′)D(-\alpha_0')\hat U(t=\frac{2 \pi }{w})D(\alpha'_0)D(−α0′)U^(t=w2π)D(α0′)
D(α0′)U^(t=πw)D(α0′)D(\alpha_0')\hat U(t=\frac{ \pi }{w})D(\alpha'_0)D(α0′)U^(t=wπ)D(α0′)
D(iα0′)U^(t=π2w)D(α0′)D(i\alpha_0')\hat U(t=\frac{ \pi }{2w})D(\alpha'_0)D(iα0′)U^(t=2wπ)D(α0′)

Red-detuned Laser

考虑
I(x,y,z)=I0e−2x2d2e−2y2d2cos⁡2(kz)I(x,y,z)=I_0e^{-\frac{2x^2}{d^2}}e^{-\frac{2y^2}{d^2}}\cos^2(kz)I(x,y,z)=I0e−d22x2e−d22y2cos2(kz)

其中k=2π/λk=2 \pi /\lambdak=2π/λ，沿z^\hat zz^方向，假设x=y=0x=y=0x=y=0，则
I(z)=I0cos⁡2(kz),U(z)∝I0Δcos⁡2(kz)I(z) = I_0 \cos^2(kz), U(z) \propto \frac{I_0}{\Delta}\cos^2(kz)I(z)=I0cos2(kz),U(z)∝ΔI0cos2(kz)

假设
U(z)/U0=−cos⁡2(kz),U0∝I0/ΔU(z)/U_0 = -\cos^2(kz),U_0 \propto I_0 / \DeltaU(z)/U0=−cos2(kz),U0∝I0/Δ

U(z)/U0U(z)/U_0U(z)/U0关于kzkzkz的图像如下，

取z=0z=0z=0为零势点，当zzz足够小时
U(z)=−U0cos⁡2(kz)+U0≈U0k2z2U(z)=-U_0\cos^2(kz)+U_0 \approx U_0 k^2 z^2U(z)=−U0cos2(kz)+U0≈U0k2z2

根据能量守恒，
U0k2z2=12mwzz2wz=2π2U0mλ2U_0 k^2 z^2 = \frac{1}{2}mw_z z^2 \\ w_z = 2 \pi \sqrt{\frac{2U_0}{m\lambda^2}}U0k2z2=21mwzz2wz=2πmλ22U0

用Q.H.O近似U(z)U(z)U(z)，能级为
En=ℏwz(n+1/2)≈2π2U0mλ2ℏ(n+1/2)E_n=\hbar w_z(n+1/2) \approx 2 \pi \sqrt{\frac{2U_0}{m\lambda^2}} \hbar (n+1/2)En=ℏwz(n+1/2)≈2πmλ22U0ℏ(n+1/2)

这个近似成立的条件是
U0ℏwz>>1\frac{U_0}{\hbar w_z} >> 1ℏwzU0>>1

也就是不能只有几个低能级的状态可以被近似。

3-D Q.H.O. Approximation

考虑用3-D Q.H.O.近似3-D Gaussian Laser Beam Trap，假设Laser Beam关于zzz轴对称，则Q.H.O.的eigen-energy满足
E=Enx,ny,nz=ℏwxy(nx+ny+1)+ℏwz(nz+1/2)E=E_{n_x,n_y,n_z}=\hbar w_{xy}(n_x+n_y+1)+\hbar w_z(n_z+1/2)E=Enx,ny,nz=ℏwxy(nx+ny+1)+ℏwz(nz+1/2)

记对应的energy eigenstate为∣nx,ny,nz⟩|n_x,n_y,n_z \rangle∣nx,ny,nz⟩，Ground state为∣nx=0,ny=0,nz=0⟩|n_x=0,n_y=0,n_z=0\rangle∣nx=0,ny=0,nz=0⟩。When t=0t=0t=0, atom in ground state, shortly the mirror bumped and the y^\hat yy^ direction got shifted of +1μ+1 \mu+1μm. For the y^\hat yy^ direction,
αy=12(⟨y⟩σy+iσy⟨Py⟩ℏ)=12−1μmσy\alpha_y = \frac{1}{\sqrt{2}} \left( \frac{\langle y \rangle}{\sigma_y}+\frac{i \sigma_y \langle P_y \rangle}{\hbar} \right)=\frac{1}{\sqrt{2}} \frac{-1 \mu m}{\sigma_y}αy=21(σy⟨y⟩+ℏiσy⟨Py⟩)=21σy−1μm

Assume σy=0.25μm\sigma_y =0.25 \mu mσy=0.25μm, α=−2.8\alpha = -2.8α=−2.8. And ⟨Hy⟩ℏwy=∣−2.8∣2+1/2=8.5\frac{\langle H_y \rangle}{\hbar w_y}=|-2.8|^2+1/2=8.5ℏwy⟨Hy⟩=∣−2.8∣2+1/2=8.5. Now we know the initial state of the atom is
∣nx=0,ny=−2.8,nz=0⟩=∣ψ(0)⟩|n_x = 0,n_y = -2.8,n_z=0 \rangle = | \psi(0) \rangle∣nx=0,ny=−2.8,nz=0⟩=∣ψ(0)⟩

Applying the Schroedinger time-evolution operator,
∣ψ(t)⟩=∣nx=0,ny=−2.8e−iwt,nz=0⟩|\psi(t) \rangle = |n_x=0 ,n_y = -2.8 e^{-iwt},n_z=0 \rangle∣ψ(t)⟩=∣nx=0,ny=−2.8e−iwt,nz=0⟩