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UA OPTI570 量子力学22 2-D Isotropic Q.H.O.简介

發布時間:2025/4/14 编程问答 28 豆豆
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UA OPTI570 量子力學22 2-D Isotropic Q.H.O.簡介

    • 基本概念
    • 2-D state的矩陣表示

基本概念

這一講介紹一個2-D量子諧振子的例子作為量子諧振子這部分的結尾。考慮2-D Isotropic Q.H.O.,
V=Vx+Vy=12mw2(X2+Y2)H=Hx+Hy=Px2+Py22m+12mw2(X2+Y2)V=V_x+V_y=\frac{1}{2}mw^2(X^2+Y^2) \\ H = H_x+H_y=\frac{P_x^2+P_y^2}{2m}+\frac{1}{2}mw^2(X^2+Y^2)V=Vx?+Vy?=21?mw2(X2+Y2)H=Hx?+Hy?=2mPx2?+Py2??+21?mw2(X2+Y2)

定義湮滅算符
ax=12(Xσ+iσPx?)ay=12(Yσ+iσPy?)a_x=\frac{1}{\sqrt{2}}\left( \frac{X}{\sigma}+\frac{i\sigma P_x}{\hbar}\right) \\ a_y=\frac{1}{\sqrt{2}}\left( \frac{Y}{\sigma}+\frac{i\sigma P_y}{\hbar}\right)ax?=2?1?(σX?+?iσPx??)ay?=2?1?(σY?+?iσPy??)

Coherent state滿足
ax∣αx?=αx∣αx?,αx=12(?X?σ+iσ?Px??),∣αx?=e?∣αx∣22∑nx=0+∞αxnxnx!∣nx?ay∣αy?=αy∣αy?,αy=12(?Y?σ+iσ?Py??),∣αy?=e?∣αy∣22∑ny=0+∞αynyny!∣ny?a_x |\alpha_x \rangle = \alpha_x | \alpha_x \rangle,\alpha_x=\frac{1}{\sqrt{2}}\left( \frac{\langle X \rangle }{\sigma}+\frac{i\sigma \langle P_x \rangle }{\hbar}\right),|\alpha_x \rangle = e^{-\frac{|\alpha_x|^2}{2}}\sum_{n_x=0}^{+\infty} \frac{\alpha_x^{n_x}}{\sqrt{n_x!}}|n_x \rangle \\ a_y |\alpha_y \rangle = \alpha_y | \alpha_y\rangle,\alpha_y=\frac{1}{\sqrt{2}}\left( \frac{\langle Y \rangle }{\sigma}+\frac{i\sigma \langle P_y \rangle }{\hbar}\right),|\alpha_y \rangle = e^{-\frac{|\alpha_y|^2}{2}}\sum_{n_y=0}^{+\infty} \frac{\alpha_y^{n_y}}{\sqrt{n_y!}}|n_y \rangleax?αx??=αx?αx??,αx?=2?1?(σ?X??+?iσ?Px???),αx??=e?2αx?2?nx?=0+?nx?!?αxnx???nx??ay?αy??=αy?αy??,αy?=2?1?(σ?Y??+?iσ?Py???),αy??=e?2αy?2?ny?=0+?ny?!?αyny???ny??

記x與y方向的energy eigenstate分別為Ex={∣nx?}\mathcal{E}_x=\{|n_x \rangle\}Ex?={nx??}Ey={∣ny?}\mathcal{E}_{y}=\{|n_y \rangle\}Ey?={ny??},則系統的energy eigenstate為
E={∣nx,ny?}=Ex?Ey\mathcal{E} =\{|n_x,n_y \rangle\}= \mathcal{E}_x \otimes \mathcal{E}_yE={nx?,ny??}=Ex??Ey?

其中?\otimes?表示張量積。所以
∣αx,αy?=e?∣αx∣22e?∣αy∣22∑nx,ny=0+∞αxnxαynynx!ny!∣nx,ny?|\alpha_x,\alpha_y \rangle = e^{-\frac{|\alpha_x|^2}{2}}e^{-\frac{|\alpha_y|^2}{2}}\sum_{n_x,n_y=0}^{+\infty} \frac{\alpha_x^{n_x}\alpha_y^{n_y}}{\sqrt{n_x!}\sqrt{n_y!}}|n_x,n_y \rangleαx?,αy??=e?2αx?2?e?2αy?2?nx?,ny?=0+?nx?!?ny?!?αxnx??αyny???nx?,ny??

某一方向上的哈密頓量的作用為
Hx∣nx,ny?=Hx1y∣nx,ny?=(Hx∣nx?)?(1∣ny?)=?w(n+1/2)∣nx,ny?H_x|n_x,n_y \rangle = H_x1_y |n_x,n_y \rangle=(H_x|n_x \rangle)\otimes(1|n_y \rangle) = \hbar w(n+1/2)|n_x,n_y \rangleHx?nx?,ny??=Hx?1y?nx?,ny??=(Hx?nx??)?(1ny??)=?w(n+1/2)nx?,ny??

湮滅算符的作用為
ax∣αx,αy?=αx∣αx,αy?axay∣αx,αy?=αx∣αx,αy?axay?∣nx,ny?=nxny+1∣nx?1,ny+1?,nx≥1a_x|\alpha_x,\alpha_y \rangle=\alpha_x | \alpha_x,\alpha_y \rangle \\ a_xa_y|\alpha_x,\alpha_y \rangle=\alpha_x | \alpha_x,\alpha_y \rangle \\ a_xa_y^{\dag}|n_x,n_y \rangle=\sqrt{n_x}\sqrt{n_y+1} | n_x-1,n_y+1 \rangle,n_x \ge 1ax?αx?,αy??=αx?αx?,αy??ax?ay?αx?,αy??=αx?αx?,αy??ax?ay??nx?,ny??=nx??ny?+1?nx??1,ny?+1?,nx?1

Displacement Operator的作用為
Dx(αx)Dy(αy)∣nx=0,ny=0?=∣αx,αy?D_x(\alpha_x)D_y(\alpha_y)|n_x=0,n_y=0 \rangle = |\alpha_x,\alpha_y \rangleDx?(αx?)Dy?(αy?)nx?=0,ny?=0?=αx?,αy??

2-D state的矩陣表示

按先y后x的順序,比如nx,ny=0,1,2n_x,n_y=0,1,2nx?,ny?=0,1,2
E={∣00?,∣01?,∣10?,∣02?,∣11?,∣20?,∣12?,∣21?,∣22?}\mathcal{E}=\{|00 \rangle,|01 \rangle,|10 \rangle,|02 \rangle,|11 \rangle,|20 \rangle,|12 \rangle,|21 \rangle,|22\rangle\}E={00?,01?,10?,02?,11?,20?,12?,21?,22?}

這幾個狀態的向量表示為e1,e2,?,e8e_1,e_2,\cdots,e_8e1?,e2?,?,e8?,哈密頓量的矩陣表示為
?w?diag(1,2,2,3,3,3,4,4,5)\hbar w \cdot diag(1,2,2,3,3,3,4,4,5)?w?diag(1,2,2,3,3,3,4,4,5)

如果某個量子態滿足
∣ψ?=∑nx,nycnx,ny∣nx,ny?|\psi \rangle = \sum_{n_x,n_y}c_{n_x,n_y}|n_x,n_y \rangleψ?=nx?,ny??cnx?,ny??nx?,ny??

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