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Signle particle

单粒子的情况, 参考 Sakurai & Napolitano 中的讨论, 以及 Wolfgang Ketterle 的 AMO 公开课第五讲.

Many body system Density operator $\hat{\rho}_N$, Density matrix $\rho_N$

对于一个 $N$ 粒子的多体纯态 $|\Psi_{\mathrm{N-particles}}\rangle$ , 其 density operator 定 义为

$$\begin{align} \hat{\rho}_N \equiv |\Psi_{\mathrm{N-particles}}\rangle\langle\Psi_{\mathrm{N-particles}}| \end{align}$$

选取一组基底, 写出 density opertor 在此基底下的表示矩阵, 即为 density operator. 例如有一组正交归一的单粒子本征态 $\{| \alpha\rangle\}$ , 系统的基底选取为 $N$ 个单 粒子态的直积 (Negele & Orland Eq.(1.27))

$$\begin{align} |\alpha_1 , \alpha_2, \cdots, \alpha_N) \equiv |\alpha_1\rangle \otimes |\alpha_2\rangle \otimes \cdots\otimes |\alpha_N\rangle \end{align}$$

需要注意:

  • 1. 这里等式左边不是 $|\rangle$ , 而是 $|)$ . 区别在于后者没有对费米子或是玻色子做反 对
  • 2. $N$ 是多体系统总的粒子数, 而不是单粒子本征态 $\{| \alpha\rangle\}$ 的个数. 称或者是对称化处理. 它的完备性可以写为
$$\begin{align} \sum_{\alpha_1, \cdots,\alpha_N}|\alpha_1 , \alpha_2, \cdots, \alpha_N) (\alpha_1 , \alpha_2, \cdots, \alpha_N| = 1 \end{align}$$

因此在这样一组基底下, density opertor 的矩阵表示, 也就是 density matrix 可以写为

$$\begin{align} &\rho_N(\alpha_1 , \alpha_2, \cdots, \alpha_N ;\alpha_1' , \alpha_2', \cdots, \alpha_N')\\ =& (\alpha_1 , \alpha_2, \cdots, \alpha_N| \hat{\rho}_N |\alpha_1' , \alpha_2', \cdots, \alpha_N') \\ =& (\alpha_1 , \alpha_2, \cdots, \alpha_N| \Psi_{\mathrm{N-particles}}\rangle\langle\Psi_{\mathrm{N-particles}} |\alpha_1' , \alpha_2', \cdots, \alpha_N') \end{align}$$

一种特殊的情况是将基底选为空间坐标, 就可以写成多体波函数的形式

$$\begin{align} \rho_N(r_1 , r_2, \cdots, r_N ;r_1' , r_2', \cdots, r_N') = \Psi_{\mathrm{N-particles}}(r_1 , r_2, \cdots, r_N) \Psi^{*}_{\mathrm{N-particles}}(r_1' , r_2', \cdots, r_N') \end{align}$$

其中波函数为 (Negele & Orland Eq.(1.49))

$$\begin{align} \Psi_{\mathrm{N-particles}}(r_1 , r_2, \cdots, r_N) = (r_1 , r_2, \cdots, r_N| \Psi_{\mathrm{N-particles}}\rangle \end{align}$$

如果我们对 $\rho_N$ 求 trace, 也就是对取 $r_1' r_1, r_2' r_2, \cdots, r_N' = r_N$ , 再 对所有这些指标求和, 会发现结果是 $1$ , 从多体波函数 $\Psi_{\mathrm{N-particles}}(r_1 , r_2, \cdots, r_N)$ 的归一化也可以看出这一点.

Example: 2 Fermions $2$ spin $1/2$ Fermions, 一个在 spin up, 一个在 spin down, 并且空间波函数是交换 对称的. 考虑其自旋部分的 density opertor. 因为 Fermion 总的波函数是交换反对称的, 而这里已经假设空间波函数是交换对称的, 所 以自旋部分只能是交换反对称的. 反对称化(只有单个粒子处于的态正交时才能够这样反对 称化, 详见 Negele, J. W. & Orland)后的 $2$ 粒子态的自旋部分为(也就是自旋单态) \begin{align} |\Psi_{\mathrm{2-particles}} \rangle = \frac{1}{\sqrt{2}}(|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle\otimes|\uparrow\rangle) \end{align} 选取单粒子态基底为 $\{|\uparrow\rangle, |\downarrow\rangle\}$ . 在此基底下, spin up 和 spin down 表示为 \begin{align} |\uparrow\rangle \sim& \begin{pmatrix} \langle\uparrow|\uparrow\rangle \\\langle\downarrow|\uparrow\rangle \end{pmatrix} = \begin{pmatrix} 1 \\0 \end{pmatrix} \\ |\downarrow\rangle \sim& \begin{pmatrix} \langle\downarrow|\uparrow\rangle \\\langle\downarrow|\uparrow\rangle \end{pmatrix} = \begin{pmatrix} 0 \\1 \end{pmatrix} \end{align} 其中 $\sim$ 表示 'represented by' (参见 Sakurai & Napolitano). 直积 \begin{align} |\uparrow \downarrow ) = |\uparrow\rangle \otimes |\downarrow\rangle &\sim \begin{pmatrix} 1 \\0 \end{pmatrix}\otimes \begin{pmatrix} 0 \\1 \end{pmatrix} = \begin{pmatrix} 1\times 0 \\1\times 1\\ 0\times 0\\0\times 1 \end{pmatrix} = \begin{pmatrix} 0 \\1\\ 0\\0 \end{pmatrix} \\ |\downarrow \uparrow) = |\downarrow\rangle\otimes|\uparrow\rangle& \sim\begin{pmatrix} 0 \\0\\ 1\\0 \end{pmatrix} \end{align} 也就是说 \begin{align} |\Psi_{\mathrm{2-particles}} \rangle \sim\begin{pmatrix} (\uparrow \uparrow |\Psi_{\mathrm{2-particles}} \rangle\\ (\uparrow \downarrow |\Psi_{\mathrm{2-particles}} \rangle\\ (\downarrow \uparrow |\Psi_{\mathrm{2-particles}} \rangle\\ (\downarrow \downarrow |\Psi_{\mathrm{2-particles}} \rangle \end{pmatrix} =\frac{1}{\sqrt{2}}\left(\begin{pmatrix} 0 \\1\\ 0\\0 \end{pmatrix} - \begin{pmatrix} 0 \\ 0 \\ 1\\0 \end{pmatrix}\right) = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\1\\ -1\\0 \end{pmatrix} \end{align} 也可以这样算, 比如 \begin{align} (\uparrow \uparrow |\Psi_{\mathrm{2-particles}} \rangle = \langle \uparrow |\otimes \langle\uparrow|\Psi_{\mathrm{2-particles}} \rangle = \frac{1}{\sqrt{2}} (\langle \uparrow |\otimes \langle\uparrow|)\cdot(|\uparrow\rangle \otimes |\downarrow\rangle - |\downarrow\rangle\otimes|\uparrow\rangle) \end{align} 其中 \begin{align} \langle \uparrow |\otimes \langle\uparrow|\cdot|\uparrow\rangle \otimes |\downarrow\rangle = \langle\uparrow|\uparrow \rangle\langle\uparrow|\downarrow\rangle = 0 \end{align} 如此这般, 所有的项都可以算出来. Density operator and density matrix \begin{align} \hat{\rho}_N =& |\Psi_{\mathrm{2-particles}} \rangle\langle\Psi_{\mathrm{2-particles}} |\sim \rho \\ =& \frac{1}{2} \begin{pmatrix} 0 \\1\\ -1\\0 \end{pmatrix} \begin{pmatrix} 0 &1 & -1 &0 \end{pmatrix} = \frac{1}{2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0& 1 & -1 & 0\\ 0& -1 & 1 & 0\\ 0& 0 & 0 & 0 \end{pmatrix} \end{align} 很明显, 它的迹是 $1$ .

Reduced density matrix

按照 C. N. Yang 1962 中单体密度算符的定义

$$\begin{align} \langle j | \rho_1 | i\rangle \equiv \mathrm{Sp} \, a_j \rho a_i^{\dagger} \end{align}$$

其中 $i, j$ 表示单粒子态.

用这篇博客里的符号来表达就是

$$\begin{align} \rho_1(\alpha; \alpha') \equiv \sum_{\alpha_2, \alpha_3,\cdots \alpha_N} (\alpha_2, \alpha_3, \cdots \alpha_N |\hat{a}_{\alpha} \hat{\rho}_N \hat{a}_{\alpha'}^{\dagger}| \alpha_2, \alpha_3, \cdots, \alpha_N) \end{align}$$

也就是

$$\begin{align} \rho_1(\alpha, \alpha')=& \sum_{\alpha_2, \alpha_3,\cdots \alpha_N} (\alpha_2, \alpha_3, \cdots \alpha_N |\hat{a}_{\alpha} |\Psi_{\mathrm{N-particles}}\rangle \langle\Psi_{\mathrm{N-particles}}|\hat{a}_{\alpha'}^{\dagger}| \alpha_2, \alpha_3, \cdots, \alpha_N) \\ =& \langle \Psi_{\mathrm{N-particles}} |\hat{a}_{\alpha'}^{\dagger} \hat{a}_{\alpha} | \Psi_{\mathrm{N-particles}}\rangle \end{align}$$

现在, 有一个问题, 我们想知道 $\rho_1(\alpha; \alpha')$ 与 $\rho_N(\alpha_1, \alpha_2,\cdots, \alpha_N; \alpha_1', \alpha_{2}',\cdots \alpha_N')$ 的关系. 注意

$$\begin{align} \langle\Psi_{\mathrm{N-particles}}|\hat{a}_{\alpha'}^{\dagger}| \alpha_2, \alpha_3, \cdots, \alpha_N) \neq \langle\Psi_{\mathrm{N-particles}}|\sqrt{n_{\alpha'} +1}| \alpha', \alpha_2, \alpha_3, \cdots, \alpha_N) \end{align}$$

其中 $n_{\alpha'}$ 表示 $\alpha_2, \alpha_3, \cdots, \alpha_N$ 中有多少个与 $\alpha'$ 相同的态. 产生算符应 该作用在对称化过的态上, 即

$$\begin{align} \hat{a}_{\alpha'}^{\dagger}| \alpha_2, \alpha_3, \cdots, \alpha_N\rangle = \sqrt{n_{\alpha'} +1}| \alpha', \alpha_2, \alpha_3, \cdots, \alpha_N\rangle \end{align}$$

二者之间的关系为 (Negele & Orland Eq.(1.46))

$$\begin{align} |\alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_N\rangle = \frac{1}{\sqrt{N! \prod_\alpha n_{\alpha}!}} \sum_P\zeta^P |\alpha_{P1} , \alpha_{P2}, \alpha_{P3}, \cdots , \alpha_{PN} ) \end{align}$$

其中 $P$ 表示置换, $\zeta = \pm 1$ (Boson $+1$ , Fermion $-1$ ) . 因此有

$$\begin{align} &\langle\Psi_{\mathrm{N-particles}}| \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_N\rangle \\ =& \frac{1}{\sqrt{N! \prod_\alpha n_{\alpha}!}} \sum_P\zeta^P\langle\Psi_{\mathrm{N-particles}}|\alpha_{P1} , \alpha_{P2}, \alpha_{P3}, \cdots , \alpha_{PN}) \\ =& \frac{1}{\sqrt{N! \prod_\alpha n_{\alpha}!}} \sum_P\zeta^P\cdot \zeta^P\langle\Psi_{\mathrm{N-particles}}|\alpha_{1} , \alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) \\ =& \frac{1}{\sqrt{N! \prod_\alpha n_{\alpha}!}} N! \langle\Psi_{\mathrm{N-particles}}|\alpha_{1} , \alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) \\ =& \sqrt{\frac{N!}{ \prod_\alpha n_{\alpha}!}} \langle\Psi_{\mathrm{N-particles}}|\alpha_{1} , \alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) \end{align}$$

同理

$$\begin{align} &\langle\Psi_{\mathrm{N-particles}}| \alpha_1, \alpha_2, \alpha_3, \cdots, \alpha_N\rangle \\ =&\langle\Psi_{\mathrm{N-particles}}|\frac{1}{\sqrt{n_{\alpha_1}}}\hat{a}^{\dagger}_{\alpha_1} |\alpha_2, \alpha_3, \cdots, \alpha_N\rangle \\ =& \frac{1}{\sqrt{n_{\alpha_1}}}\sqrt{\frac{(N - 1)!}{\frac{1}{n_{\alpha_1}} \prod_\alpha n_{\alpha}!}} \langle\Psi_{\mathrm{N-particles}}|\hat{a}^{\dagger}_{\alpha_1}|\alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) \\ =& \sqrt{\frac{(N - 1)!}{ \prod_\alpha n_{\alpha}!}} \langle\Psi_{\mathrm{N-particles}}|\hat{a}^{\dagger}_{\alpha_1}|\alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) \end{align}$$

这样就得到

$$\begin{align} \sqrt{N}\langle\Psi_{\mathrm{N-particles}}|\alpha_{1} , \alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) = \langle\Psi_{\mathrm{N-particles}}|\hat{a}^{\dagger}_{\alpha_1}|\alpha_{2}, \alpha_{3}, \cdots , \alpha_{N}) \end{align}$$

因此我们就知道了 $\rho_1(\alpha, \alpha')$ 与 $\rho_N(\alpha_1, \alpha_2,\cdots, \alpha_N; \alpha_1', \alpha_{2}',\cdots \alpha_N')$ 的关系, 即

$$\begin{align} \rho_1(\alpha; \alpha') =& \sum_{\alpha_2, \alpha_3,\cdots \alpha_N} (\alpha_2, \alpha_3, \cdots \alpha_N |\hat{a}_{\alpha} |\Psi_{\mathrm{N-particles}}\rangle \langle\Psi_{\mathrm{N-particles}}|\hat{a}_{\alpha'}^{\dagger}| \alpha_2, \alpha_3, \cdots, \alpha_N) \\ = & N \sum_{\alpha_2, \alpha_3,\cdots \alpha_N} (\alpha, \alpha_2, \alpha_3, \cdots \alpha_N |\Psi_{\mathrm{N-particles}}\rangle \langle\Psi_{\mathrm{N-particles}}|\alpha', \alpha_2, \alpha_3, \cdots, \alpha_N) \\ = & N \sum_{\alpha_2, \alpha_3,\cdots \alpha_N} \rho_N(\alpha, \alpha_2,\cdots, \alpha_N; \alpha', \alpha_{2},\cdots \alpha_N) \end{align}$$

也就是说, 多体系统的单体密度矩, 就是将多体系统的密度矩阵求偏迹 (Partial trace), 只留下一对指标. 如果我们对 $\rho_1(\alpha, \alpha')$ 求 trace, 会发现

$$\begin{align} \sum_{\alpha} \rho_1(\alpha; \alpha) = N \end{align}$$

此即 C. N. Yang 1962 中的 Eq.(4) .

我们可以 inductive 地定义多体系统的两体算符, 三体算符等等.

Example 1: 3 Bosons $2$ 个本征态, $\{|\alpha\rangle , |\beta\rangle\}$ , 3 个 Boson, 全都处于 $|\alpha\rangle$ 态. 在 $\{|\alpha\rangle , |\beta\rangle\}$ 基底下 \begin{align} |\Psi_{\mathrm{3-particles}} \rangle &= \frac{1}{\sqrt{3! 3!}} \cdot 3! |\alpha, \alpha, \alpha) \\ &\sim (1, 0, 0, 0, 0, 0, 0, 0)^T \end{align} density opertor and density matrix \begin{align} \hat{\rho}_3 \sim \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{align} \begin{align} \hat{\rho}_1 \sim 3 \begin{pmatrix} 1 & 0\\ 0 & 0\\ \end{pmatrix} \end{align} 可以发现 $\rho_1$ 的对角元表示每个本征态上的粒子数
Example 2: 3 Bosons $2$ 个本征态, $\{|\alpha\rangle , |\beta\rangle\}$ , 3 个 Boson (记为 $A, B, C$ ), 两个处于 $|\alpha\rangle$ 态, 一个处于 $|\beta\rangle$ 在 $\{|\alpha\rangle , |\beta\rangle\}$ 基底下(如果我没算错的话) \begin{align} \hat{\rho}_3 \sim\rho_3= \frac{2!}{3!}\begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix} \end{align} 两体密度矩阵 \begin{align} \hat{\rho}_2 \sim\rho_2 = 3(3-1)\mathrm{Tr}_C\rho_{3} = 3(3-1)\frac{2!}{3!}\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \end{align} 单体密度矩阵 \begin{align} \hat{\rho}_1 \sim\rho_1 = 3\mathrm{Tr}_{BC}\rho_{3} = 3\frac{2!}{3!}\begin{pmatrix} 2 & 0 \\ 0 & 1 \\ \end{pmatrix} \end{align}

再次验证了 $\rho_1$ 的对角元表示每个本征态上的粒子数. 此即 Hui Zhai 书上的 Eq.(3.7)

$$\begin{align} \rho(\vec{r}, \vec{r}') = \sum_i N_i \psi_i^{*} (\vec{r}) \psi_i(\vec{r}') \end{align}$$

Mixed state

上述内容均在纯态中讨论, 可以容易地推广到混合态.

Reference

  • J. J. Sakurai & Jim Napolitano, Modern Quantum Mechanics, second edition. Chap 3.4, dneisyt operators and pure versus mixed ensembles
  • MIT Open Courses: Atomic and Optical Physics I
  • Negele, J. W. & Orland, H. Quantum many-particle systems. (Perseus Books, 1998). Chapter 1
  • $|\rangle\langle|$ 对应 [[eww:https://en.wikipedia.org/wiki/Outer_product][Outer product]] , $\otimes$ 对应 Tensor product of linear maps
  • Yang, C. N. Concept of Off-Diagonal Long-Range Order and the Quantum Phases of Liquid He and of Superconductors. Rev. Mod. Phys. 34, 694–704 (1962). 文章中 $\mathrm{Sp}$ 即为迹 $\mathrm{Tr}$ , 参见 Spur .
  • Partial trace
  • Zhai, H. Ultracold atomic physics. (Cambridge University Press, 2020).