坐标, 动量, 空间, 自由粒子
2024-9-28
杂项
可以用平面波描写自由粒子, 平面波的频率与波矢与自由粒子的能量和动量相联系
$$f(\boldsymbol{\vec{r}},\boldsymbol{\vec{k}})=e^{i\boldsymbol{\vec{k}}\cdot \boldsymbol{\vec{r}}}e^{-i\omega t}$$平面波的频率与波矢不随时间改变, 对应于自由粒子的能量与动量不随时间与位置改变. 一般情况下用一个函数描写粒子的波, 称该函数为波函数.
由波函数可以得到体系的各种性质, 因此波函数(也称概率幅)描写体系的量子状态(简称状态或态).
波函数在归一化后并不完全确定, 可以用一个常数 $e^{i\delta}$ 去乘以波函数. 该常数称为相因子, 归一化波函数可以含有任意相因子.
表象变换
Assume there're two complete orthonormal basis $\sum_{n}\boldsymbol{e_n}\boldsymbol{e_n}^T=1$ and $\sum_{n}\boldsymbol{u_n}\boldsymbol{u_n}^T=1$, or in Dirac notation $\sum_{n}\vert{\boldsymbol{e_n}}\rangle\langle{\boldsymbol{e_n}}\vert=1$ and $\sum_{n}\vert{\boldsymbol{u_n}}\rangle\langle{\boldsymbol{u_n}}\vert=1$, where $1$ is identity matrix. A vector $\boldsymbol{f}$ can then be represented as
$$\boldsymbol{f}=\sum_{\substack{n}}a_n \boldsymbol{e_n}$$ $$\boldsymbol{f}=\sum_{\substack{n}}b_n \boldsymbol{u_n}$$向量在某基下的坐标, 指的是在该基下的系数.
Vector $\boldsymbol{f}$ is invariant no matter what basis is used, but the coordinate can change. Then
$$a_n=\boldsymbol{e_n}\cdot\boldsymbol{f}=\boldsymbol{e_n}\cdot\sum_{m}b_m\boldsymbol{u_m}=\boldsymbol{e_n}\cdot\boldsymbol{u_1}b_1+\boldsymbol{e_n}\cdot\boldsymbol{u_2}b_2+\cdots+\boldsymbol{e_n}\cdot\boldsymbol{u_n}b_n$$ $$\begin{equation} \begin{bmatrix} \boldsymbol{e_1}\cdot\boldsymbol{u_1} & \boldsymbol{e_1}\cdot\boldsymbol{u_2} & \cdots & \boldsymbol{e_1}\cdot\boldsymbol{u_n} \\ \boldsymbol{e_2}\cdot\boldsymbol{u_1} & \boldsymbol{e_2}\cdot\boldsymbol{u_2} & \cdots & \boldsymbol{e_2}\cdot\boldsymbol{u_n} \\ \vdots & \vdots & \ddots & \vdots \\ \boldsymbol{e_n}\cdot\boldsymbol{u_1} & \boldsymbol{e_n}\cdot\boldsymbol{u_2} & \cdots & \boldsymbol{e_n}\cdot\boldsymbol{u_n} \\ \end{bmatrix} \cdot \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \\ \end{bmatrix} = \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \\ \end{bmatrix} \end{equation}$$ and $$b_n=\boldsymbol{u_n}\cdot\boldsymbol{f}=\boldsymbol{u_n}\cdot\sum_{m}a_m\boldsymbol{e_m}=\boldsymbol{u_n}\cdot\boldsymbol{e_1}a_1+\boldsymbol{u_n}\cdot\boldsymbol{e_2}a_2+\cdots+\boldsymbol{u_n}\cdot\boldsymbol{e_n}a_n$$ $$\begin{equation} \begin{bmatrix} \boldsymbol{u_1}\cdot\boldsymbol{e_1} & \boldsymbol{u_1}\cdot\boldsymbol{e_2} & \cdots & \boldsymbol{u_1}\cdot\boldsymbol{e_n} \\ \boldsymbol{u_2}\cdot\boldsymbol{e_1} & \boldsymbol{u_2}\cdot\boldsymbol{e_2} & \cdots & \boldsymbol{u_2}\cdot\boldsymbol{e_n} \\ \vdots & \vdots & \ddots & \vdots \\ \boldsymbol{u_n}\cdot\boldsymbol{e_1} & \boldsymbol{u_n}\cdot\boldsymbol{e_2} & \cdots & \boldsymbol{u_n}\cdot\boldsymbol{e_n} \\ \end{bmatrix} \cdot \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \\ \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \\ \end{bmatrix} \end{equation}$$Vector $\boldsymbol{f}$ under basis $\{\boldsymbol{e_n}\}$ has coordinate $\{a_n\}$, under basis $\{\boldsymbol{u_n}\}$ has coordinate $\{b_n\}$. They are connected by transformation matrix
$$ S=\begin{bmatrix} \boldsymbol{e_1}^T \\ \boldsymbol{e_2}^T \\ \vdots \\ \boldsymbol{e_n}^T \\ \end{bmatrix} \cdot \begin{bmatrix} \boldsymbol{u_1} & \boldsymbol{u_2} & \cdots & \boldsymbol{u_n} \\ \end{bmatrix} $$ $$ S^T=\begin{bmatrix} \boldsymbol{u_1}^T \\ \boldsymbol{u_2}^T \\ \vdots \\ \boldsymbol{u_n}^T \\ \end{bmatrix}\cdot\begin{bmatrix} \boldsymbol{e_1} & \boldsymbol{e_2} & \cdots & \boldsymbol{e_n} \\ \end{bmatrix}=S^{-1} $$Therefore transformation matrix $S$ is orthogonal (or unitary if it's complex). In Dirac notation, making use of completeness condition gives
$$a_n=\langle{e_n}\vert{f}\rangle=\langle{e_n}\vert{\sum_{m}\vert{u_m}\rangle\langle{u_m}}\vert{f}\rangle=\sum_{m}\langle{e_n}\vert{u_m}\rangle b_m$$ $$ \begin{bmatrix} \langle{e_1}\vert{u_1}\rangle & \langle{e_1}\vert{u_2}\rangle & \cdots & \langle{e_1}\vert{u_n}\rangle \\ \langle{e_2}\vert{u_1}\rangle & \langle{e_2}\vert{u_2}\rangle & \cdots & \langle{e_2}\vert{u_n}\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle{e_n}\vert{u_1}\rangle & \langle{e_n}\vert{u_2}\rangle & \cdots & \langle{e_n}\vert{u_n}\rangle \\ \end{bmatrix}\cdot\begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \\ \end{bmatrix}=\begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \\ \end{bmatrix} $$ $$ S=\begin{bmatrix} \langle{e_1}\vert \\ \langle{e_2}\vert \\ \vdots \\ \langle{e_n}\vert \end{bmatrix}\cdot\begin{bmatrix} \vert{u_1}\rangle & \vert{u_2}\rangle & \cdots \vert{u_n}\rangle \\ \end{bmatrix} $$Which furnishes the result that transformation matrix $S$ is orthogonal (or unitary) also.
算符的表象变换
Assume two basis $\{\boldsymbol{e_n}\}$ and $\{\boldsymbol{u_n}\}$, operator $H$ under basis $\{\boldsymbol{e_n}\}$ is $H_{ij}=\langle{e_i}\vert{H}\vert{e_j}\rangle=\sum_{\substack{mn}}\langle{e_i}\vert{u_m}\rangle\langle{u_m}\vert{H}\vert{u_n}\rangle\langle{u_n}\vert{e_j}\rangle$, then
$$ \begin{bmatrix} \langle{e_1}\vert{u_1}\rangle & \langle{e_1}\vert{u_2}\rangle & \cdots & \langle{e_1}\vert{u_n}\rangle \\ \langle{e_2}\vert{u_1}\rangle & \langle{e_2}\vert{u_2}\rangle & \cdots & \langle{e_2}\vert{u_n}\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle{e_n}\vert{u_1}\rangle & \langle{e_n}\vert{u_2}\rangle & \cdots & \langle{e_n}\vert{u_n}\rangle \\ \end{bmatrix} \cdot \begin{bmatrix} \langle{u_1}\vert{H}\vert{u_1}\rangle & \langle{u_1}\vert{H}\vert{u_2}\rangle & \cdots & \langle{u_1}\vert{H}\vert{u_n}\rangle \\ \langle{u_2}\vert{H}\vert{u_1}\rangle & \langle{u_2}\vert{H}\vert{u_2}\rangle & \cdots & \langle{u_2}\vert{H}\vert{u_n}\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle{u_n}\vert{H}\vert{u_1}\rangle & \langle{u_n}\vert{H}\vert{u_2}\rangle & \cdots & \langle{u_n}\vert{H}\vert{u_n}\rangle \\ \end{bmatrix} \cdot \begin{bmatrix} \langle{u_1}\vert{e_1}\rangle & \langle{u_1}\vert{e_2}\rangle & \cdots & \langle{u_1}\vert{e_n}\rangle \\ \langle{u_2}\vert{e_1}\rangle & \langle{u_2}\vert{e_2}\rangle & \cdots & \langle{u_2}\vert{e_n}\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle{u_n}\vert{e_1}\rangle & \langle{u_n}\vert{e_2}\rangle & \cdots & \langle{u_n}\vert{e_n}\rangle \\ \end{bmatrix} = \begin{bmatrix} \langle{e_1}\vert{H}\vert{e_1}\rangle & \langle{e_1}\vert{H}\vert{e_2}\rangle & \cdots & \langle{e_1}\vert{H}\vert{e_n}\rangle \\ \langle{e_2}\vert{H}\vert{e_1}\rangle & \langle{e_2}\vert{H}\vert{e_2}\rangle & \cdots & \langle{e_2}\vert{H}\vert{e_n}\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle{e_n}\vert{H}\vert{e_1}\rangle & \langle{e_n}\vert{H}\vert{e_2}\rangle & \cdots & \langle{e_n}\vert{H}\vert{e_n}\rangle \\ \end{bmatrix} $$ Let $$S=\begin{bmatrix} \langle{e_1}\vert{u_1}\rangle & \langle{e_1}\vert{u_2}\rangle & \cdots & \langle{e_1}\vert{u_n}\rangle \\ \langle{e_2}\vert{u_1}\rangle & \langle{e_2}\vert{u_2}\rangle & \cdots & \langle{e_2}\vert{u_n}\rangle \\ \vdots & \vdots & \ddots & \vdots \\ \langle{e_n}\vert{u_1}\rangle & \langle{e_n}\vert{u_2}\rangle & \cdots & \langle{e_n}\vert{u_n}\rangle \\ \end{bmatrix} $$ then $SH_{u}S^T=H_e$. Here $S$ is also orthogonal or unitary matrix.实空间与 $k$空间的三维 $\delta$ 函数
This content can also be found in "Mathematical Physics Integral Transformation".
$$\delta(x)=\frac{1}{2\pi}\int e^{ikx}dk$$ $$\left(\frac{1}{2\pi}\int e^{ik_x x}dx\right) \left(\frac{1}{2\pi}\int e^{ik_y y}dy\right) \left(\frac{1}{2\pi}\int e^{ik_z z}dz\right)=\left(\frac{1}{2\pi}\right)^3\int e^{i\boldsymbol{k}\cdot\boldsymbol{r}}d^3\boldsymbol{r}=\delta(\boldsymbol{k})$$ $$\delta(\boldsymbol{r})=\delta(x)\delta(y)\delta(z)$$ $$\delta(\boldsymbol{k})=\delta(k_x)\delta(k_y)\delta(k_z)$$ $$\begin{equation} \begin{split} \delta(\boldsymbol{k}) & =\delta(k_x)\delta(k_y)\delta(k_z)=\left(\frac{1}{2\pi}\right)^3\int e^{i\boldsymbol{k}\cdot\boldsymbol{r}}d^3\boldsymbol{r} \\ \delta(\boldsymbol{k}) & =\left(\frac{1}{2\pi}\right)^3\int e^{ik_x x}dx\int e^{ik_y y}dy\int e^{ik_z z}dz \\ \delta(\hbar\boldsymbol{k}) & =\left(\frac{1}{2\pi}\right)^3\int e^{i\hbar k_x x}dx\int e^{i\hbar k_y y}dy\int e^{i\hbar k_z z}dz \\ & = \left(\frac{1}{\hbar}\right)^3\left(\frac{1}{2\pi}\right)^3\int e^{ik_x \hbar x}d\hbar x\int e^{ik_y \hbar y}d\hbar y\int e^{ik_z \hbar z}d\hbar z \\ & =\left(\frac{1}{\hbar}\right)^3\left(\frac{1}{2\pi}\right)^3 \int e^{ik_x X}dX\int e^{ik_y Y}dY\int e^{ik_z Z}dZ \\ & = \left(\frac{1}{\hbar}\right)^3\delta(\boldsymbol{k}) \\ \Rightarrow & \delta(\hbar\boldsymbol{k})=\left(\frac{1}{\hbar}\right)^3\delta(\boldsymbol{k}) \\ \end{split} \end{equation}$$This means the $\hbar^3\delta(\boldsymbol{p})=\delta(\boldsymbol{k})$.
$\delta$ 函数的导数
$$\begin{equation} \begin{split} \int f(x)\delta^\prime(-x)dx & = -\int f(x)d\delta(-x) \\ & = -f(x)\delta(-x)\vert_{-\infty}^\infty +\int \delta(-x)f^\prime(x)dx \\ & = f^\prime(0) \\ \int f(x)\delta^\prime(x)dx & = -f^\prime(0) \\ \Rightarrow & -\delta^\prime(x)=\delta^\prime(-x) \\ \end{split} \end{equation}$$ $$\begin{equation} \begin{split} \int f(x)\delta^{(2)}(-x)dx & = -\int f(x)\frac{d}{dx}\delta^\prime(-x)dx \\ & = \int f(x)\frac{d}{dx}\delta^\prime(x)dx \\ & = \int f(x)\delta^{(2)}(x)dx \\ & = f^{(2)}(0) \\ \int f(x)\delta^{(2)}dx & = f^{(2)}(0) \\ \Rightarrow\:\: & \delta^{(2)}(-x) =\delta^{(2)}(x) \\ \\ \int f(x)\delta^{(3)}(-x)dx & = -\int f(x)\frac{d}{dx}\delta^{(2)}(-x)dx \\ & = -\int f(x)\frac{d}{dx}\delta^{(2)}(x)dx \\ & = -\int f(x)\delta^{(3)}(x)dx \\ & = f^{(3)}(0) \\ \int f(x)\delta^{(3)}(x)dx & = -f^{(3)}(0) \\ \Rightarrow\:\: & \delta^{(3)}(-x) =-\delta^{(3)}(x) \\ \vdots \\ \delta^{(n)}(-x) & = (-1)^n \delta^{(n)}(x) \\ \end{split} \end{equation}$$A example that makes use of the derivative of $\delta$ function
$$\begin{equation} \begin{split} \frac{1}{2\pi}\int k^2 e^{ikx}dk & = \frac{1}{2\pi}\left(-\frac{\partial^2}{\partial x^2}\int e^{ikx}dk\right) \\ & = -\frac{\partial^2}{\partial x^2}\delta(x) \\ & = -\delta^{(2)}(x) \\ \end{split} \end{equation}$$坐标表象与动量表象
归一化为 $\delta$ 函数
The position operator is just position itself $\hat{r}=r$, the momentum operator is $\hat{\boldsymbol{p}}=-i\hbar\boldsymbol{\nabla}$, for a single component is $\hat{p_x}=-i\hbar\frac{d}{dx}$. The eigenfunction for momentum operator is $f_k(x)=e^{ik_x x}$, and $f_{\boldsymbol{k}}(\boldsymbol{r})=Ae^{i\boldsymbol{k}\cdot\boldsymbol{r}}$. $f_k(\boldsymbol{r})$ 是动量算符的特征函数, 但是其本身是位矢的函数. $\hat{\boldsymbol{p}}f_k(\boldsymbol{r})=\hbar\boldsymbol{k}f_{k}(\boldsymbol{r})$, $\hat{\boldsymbol{p}}f_{k^\prime}(\boldsymbol{r})=\hbar\boldsymbol{k^\prime}f_{k^\prime}(\boldsymbol{r})$. $k$ and $k^\prime$ indicates they are eigenfunctions of different eigenvalues ($k$ and $k^\prime$) of momentum operator. 其本身当然满足正交性 (考察的是动量特征函数(对应不同的特征值)的正交性, 积掉的是空间坐标!)
$$\begin{equation} \begin{split} f_k(\boldsymbol{r})=Ae^{i\boldsymbol{k}\cdot\boldsymbol{r}} \\ f_{k^\prime}(\boldsymbol{r})=Ae^{i\boldsymbol{k^\prime}\cdot\boldsymbol{r}} \\ \end{split} \end{equation}$$ $$\begin{equation} \begin{split} \langle{f_k(\boldsymbol{r})}\vert{f_{k^\prime}(\boldsymbol{r})}\rangle & = A^2\int_{\mathbb{R}^3} e^{i(\boldsymbol{k^\prime-k})\cdot\boldsymbol{r}} d^3\boldsymbol{r} \\ & = (2\pi)^3A^2\delta(\boldsymbol{k^\prime}-\boldsymbol{k}) \\ \end{split} \end{equation}$$这里的积分范围为全空间.
已知 $\delta(\hbar\boldsymbol{k})=\frac{\delta(\boldsymbol{k})}{\hbar^3}$, and $\hbar^3\delta(\boldsymbol{p})=\delta(\boldsymbol{k})=\delta(\frac{\boldsymbol{p}}{\hbar})$.
$$\begin{equation} \begin{split} f_k(\boldsymbol{r}) & = Ae^{i\boldsymbol{k}\cdot\boldsymbol{r}} \\ f_{k^\prime}(\boldsymbol{r}) & = Ae^{i\boldsymbol{k^\prime}\cdot\boldsymbol{r}} \\ \int f_{k^\prime}^\ast(\boldsymbol{r})f_{k}(\boldsymbol{r})d^3\boldsymbol{r} & = \int A^2 e^{i\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)\cdot{\boldsymbol{r}}}d^3\boldsymbol{r} \\ & = \int A^2 e^{i\frac{\left(\boldsymbol{p}-\boldsymbol{p^\prime}\right)}{\hbar}\cdot{\boldsymbol{r}}}d^3\boldsymbol{r} \\ & = A^2\left(2\pi\right)^3\delta(\boldsymbol{k}-\boldsymbol{k^\prime}) \\ & = A^2\left(2\pi\right)^3\delta(\frac{\boldsymbol{p}-\boldsymbol{p^\prime}}{\hbar}) \\ & = A^2\left(2\pi\hbar\right)^3\delta(\boldsymbol{p}-\boldsymbol{p^\prime}) \\ \end{split} \end{equation}$$特征函数的正交性表现为 Dirac $\delta$ function, 原因在于这里的积分范围是全空间, from $-\infty$ to $\infty$. 对这里的特征函数要求其归一化为 $\delta$ 函数, 为非归一化为1
$$A^2\left(2\pi\hbar\right)^3=1\Rightarrow A=\left(\frac{1}{2\pi\hbar}\right)^{\frac{3}{2}}$$$f_k(\boldsymbol{r})$ 的 $\boldsymbol{r}$ 定义在无穷区域, 动量的特征值取连续谱 (或者说没有周期边界条件的限制, 或者说周期为无限大).
箱归一化
如果考虑周期边界条件. 注意这里自由粒子并不是因为无限深势被限制在边长为 $L$ 的盒子中, 这里只是要求周期边界条件 $$\begin{equation} \begin{cases} f_k(x+L,y,z)=f_k(x,y,z) \\ f_k(x,y+L,z)=f_k(x,y,z) \\ f_k(x,y,z+L)=f_k(x,y,z) \\ \end{cases} \end{equation}$$ Consider the $x$-component $e^{ik_x x}=e^{ik_x\left(x+L\right)}$, then $k_x L=2\pi n_x\Rightarrow k_x=\frac{2\pi}{L}n_x$. So $\boldsymbol{k}=\frac{2\pi}{L}\boldsymbol{n}$, where $\boldsymbol{n}=\begin{bmatrix}n_x & n_y & n_z\end{bmatrix}^{T}$, $n_i\in\mathbb{Z}$. The eigenvalue $\boldsymbol{k}$ is no longer continous, but discrete. \par 在周期边界条件下动量算符的特征函数要求归一到1 $$\begin{equation} \begin{split} f_k(\boldsymbol{r}) & = Ae^{i\boldsymbol{k}\cdot\boldsymbol{r}} \\ f_{k^\prime}(\boldsymbol{r}) & = Ae^{i\boldsymbol{k^\prime}\cdot\boldsymbol{r}} \\ k_x & = \frac{2\pi}{L}n_x \\ k_x-k_{x^\prime} & = \frac{2\pi}{L}n \\ \langle{f_{k^\prime}(\boldsymbol{r})}\vert{f_k(\boldsymbol{r})}\rangle & = A^2\int e^{i\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)\cdot\boldsymbol{r}}d^3\boldsymbol{r} \\ & = A^2\left(\int_{0}^{L} e^{i\left(k_x-k_{x^\prime}\right)x}dx\right) \left(\int_{0}^{L} e^{i\left(k_y-k_{y^\prime}\right)y}dy\right) \left(\int_{0}^{L} e^{i\left(k_z-k_{z^\prime}\right)z}dz\right) \end{split} \end{equation}$$ 注意到指数上 $k_i-k_{i^\prime}=\frac{2\pi}{L}n$, $\frac{2\pi}{\omega}=\frac{L}{n}$. 盒子的周期 $L$ 是平面波周期 $\frac{L}{n}$ 的整数倍. 又 $\frac{1}{2\pi}\int_0^{2\pi}e^{i\left(n-m\right)\theta}d\theta=\delta_{mn}$, 所以 $\int_0^L e^{i\left(k_x-k_{x^\prime}\right)x}dx=L\delta_{k_xk_{x^\prime}}$. 带入上式得到 $A^2L^3\delta_{k_xk_{x^\prime}} \delta_{k_yk_{y^\prime}} \delta_{k_zk_{z^\prime}}$, 归一化有 $A=\left(\frac{1}{L}\right)^{\frac{3}{2}}$. 最后得到归一化的满足周期边界条件的动量算符特征函数 $f_k(\boldsymbol{r})=\left(\frac{1}{L}\right)^{3/2}e^{i\boldsymbol{k}\cdot\boldsymbol{r}}$. \par几点评论
几点注意事项, 积分上下限只需要差值为 $L$ 即可. 由于周期边界条件的限制, 动量特征值不再构成连续谱, 而只能取分立的值, 因此这里出现的是 Kronecker $\delta$, 而动量特征值取连续谱时出现的是 Dirac $\delta$. Kronecker $\delta$ is discrete version of Dirac $\delta$.
在 $\hat{O}$ 的特征值构成分立谱的情况下, 特征函数归一化为 Kronecker $\delta$
$$\langle{f_k}\vert{f_{k^\prime}}\rangle=\delta_{kk^\prime}$$在 $\hat{O}$ 的特征值构成连续谱的情况下, 特征函数归一化为 Dirac $\delta$
$$\langle{f_k}\vert{f_{k^\prime}}\rangle=\delta(k-k^\prime)$$坐标算符与动量算符在非周期边界条件下的特征值构成连续谱, 他们的特征函数正交归一化为 Dirac $\delta$ 函数. 在坐标表象下(意指自变量为坐标)的动量算符的特征函数(为简单起见, 接下来只考虑一维情况, 三维情形只是一维的结果乘起来而已)为
$$f_k(x)=\sqrt{\frac{1}{2\pi}}e^{ikx}$$ 或者 $$f_p(x)=\sqrt{\frac{1}{2\pi\hbar}}e^{i\frac{p}{\hbar}x}$$该记号指的是在有确定波矢 $k$ (或确定动量 $p$) 时的动量算符的特征函数在坐标表象下的表现形式.
动量特征函数归一化为 $\delta$ 函数 (积掉坐标)
$$\langle{f_k(x)}\vert{f_{k^\prime}(x)}\rangle=\delta(k-k^\prime)$$ or $$\langle{f_p(x)}\vert{f_{p^\prime}(x)}\rangle=\delta(p-p^\prime)$$在动量表象下, 动量算符就是动量本身 $\hat{p}=p$. 动量特征函数的自变量在动量表象下是动量 $p$
$$\hat{p}f_{p_i}(p)=p_i f_{p_i}(p)$$$f_{p_i}(p)$ 指的是在有确定动量为 $p_i$ 时的动量的特征函数, $p$ 是未定自变量. 注意到有 $x\delta(x-x_i)=x_i$, 得到在动量表象下动量算符的特征函数为
$$f_{p_i}(p)=\delta(p-p_i)$$其归一化到 Dirac $\delta$ 函数 (在动量表象下自变量为动量, 积掉动量)
$$\langle{f_{p_i}(p)}\vert{f_{p_j}(p)}\rangle=\int \delta(p-p_i)\delta(p-p_j)dp=\delta(p_i-p_j)$$同理, 在坐标表象下, 坐标算符就是坐标本身, 坐标算符的特征函数在坐标表象下同样也是 Dirac $\delta$ 函数
$$x\delta(x-x^\prime)=x^\prime\delta(x-x^\prime)$$在自身表象下, 动量与坐标的特征函数都是对角的.
动量表象下的位置算符
在纯粹数学领域, 傅里叶变换的系数的选择是人为的, 只要正向与反向的系数的乘积得到 $\frac{1}{2\pi}$ 即可 (1D)
$$\begin{equation} \begin{split} F(k) & = A\int f(x)e^{-ikx}dx \\ f(x) & = B\int F(k)e^{ikx}dk \\ A\cdot B & = \frac{1}{2\pi} \end{split} \end{equation}$$在量子力学领域, 由于波函数归一化的限制, 傅里叶变换的系数的选择不是随意的. 位置算符 $\hat{x}$ 的特征函数在位置表象 (实空间表象), 与在动量表象 ($k$ 空间表象) 构成傅里叶变换对
$$\begin{equation} \begin{split} \begin{cases} f(k) & = \sqrt{\frac{1}{2\pi}}\int f(x)e^{-ikx}dx \\ f(x) & = \sqrt{\frac{1}{2\pi}}\int f(k)e^{ikx}dk \\ \end{cases} \\ \begin{cases} f(p) & = \sqrt{\frac{1}{2\pi\hbar}}\int f(x)e^{-i\frac{p}{\hbar}x}dx \\ f(x) & = \sqrt{\frac{1}{2\pi\hbar}}\int f(k)e^{i\frac{p}{\hbar}x}dk \\ \end{cases} \\ \end{split} \end{equation}$$相应的归一化要求
$$\begin{equation} \begin{split} \langle{f(k)}\vert{f(k)}\rangle & = \int\frac{1}{2\pi}\int f^\ast(x)e^{ikx}dx\cdot \int f(x^\prime)e^{-ikx^\prime}dx^\prime dk \\ & = \int\left(\frac{1}{2\pi}\int e^{ik(x-x^\prime)}dk\right)f^\ast(x)f(x^\prime)dx^\prime dx \\ & = \int \delta(x-x^\prime) f^\ast(x)f(x^\prime) dx^\prime dx \\ & = \int \vert{f(x)}\vert^2 dx\:\:=1 \\ \langle{f(x)}\vert{f(x)}\rangle & = 1 \\ \end{split} \end{equation}$$出于归一化的要求, 傅里叶变换的系数对称地分布在傅里叶变换对上.
位置算符在位置表象下就是坐标本身, 位置算符地期望值在位置表象下为
$$\langle{x}\rangle=\int f^\ast(x)\hat{x}f(x)dx=\int f^\ast(x)xf(x)dx$$同理, 动量算符在动量表象下就是动量本身. 但是位置算符在动量表象, 以及动量算符在位置表象下具有不同地形式. 在位置表象下动量算符的表达式为 $\hat{p}=i\hbar\frac{d}{dx}$.
由于期望值不因表象的改变而改变, 故而位置算符的期望值在动量表象下应与在位置表象下等同, 而位置算符在动量表象下并不等于位置本身 (动量表象下的自变量为动量)
$$\begin{equation} \begin{split} \langle{x}\rangle & = \int f^\ast(k)\hat{x}f(k)dp \\ & = \int \left(f^\ast(x)e^{ikx}\right)\hat{x}\left(f(x^\prime)e^{-ikx^\prime}\right)dxdx^\prime dk \\ & = \int f^\ast(x)\cdot \left(\int x^\prime f(x^\prime)\delta(x-x^\prime) dx^\prime\right) dx \\ & = \int f^\ast(x) xf(x) dx \\ \Rightarrow\:\: & \hat{x} e^{-ikx^\prime}=x^\prime e^{-ikx^\prime} \\ & \text{Note that $\hat{x}$ does not act on $f(x^\prime)$, for $f(x^\prime)$ if not a function of $k$} \\ \Rightarrow\:\: & \hat{x}=i\frac{d}{dk} \\ \text{or } & \hat{x}=i\hbar\frac{d}{dp} \end{split} \end{equation}$$ $$\begin{equation} \begin{split} \langle{x}\rangle & = \frac{1}{2\pi\hbar}\int f^\ast(p)e^{i\frac{p}{\hbar}x} \hat{x}f(x^\prime)e^{-i\frac{p}{\hbar}x^\prime} dx^\prime dx \\ & = \int f^\ast(x)x^\prime f(x^\prime)\delta(x-x^\prime)dx^\prime dx \\ \Rightarrow\:\: & \hat{x}=i\hbar\frac{d}{dp} \\ \end{split} \end{equation}$$For completeness, the momentum operator under position representation is $\hat{p}=-i\hbar\frac{d}{dx}$.
关于 $\delta$ 函数的一点评论
Full description of Dirac $\delta$ function can be referred to another notes.
$$\begin{equation} \begin{split} \frac{1}{2\pi}\int e^{i\frac{p-p^\prime}{\hbar}x}dx & = \hbar\delta(p-p^\prime)\:\:\:\text{(scaling property)} \\ \frac{1}{2\pi\hbar}\int e^{i\frac{p-p^\prime}{\hbar}x}dx & = \delta(p-p^\prime) \\ \frac{1}{2\pi\hbar}\int e^{i\frac{p}{\hbar}x}dx & = \delta(p) \\ \frac{1}{2\pi\hbar}\int e^{i\frac{p}{\hbar}x}dp & = \delta(x) \\ \end{split} \end{equation}$$自由粒子正则系综的密度矩阵
For brevity, here only 1D situation is considered. For a single particle (therefore no particle indistingusable issue) in cubic PBC box with length $L$, the Hamiltonian and corresponding eigenfunctions are
$$\begin{equation} \begin{split} \hat{H} & = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \\ f_k(x) & = \sqrt{\frac{1}{L}}e^{ikx} \\ \epsilon_k & = \frac{\hbar^2k^2}{2m} \\ k & = \frac{2\pi}{L}n\:\:\:,\:n\in\mathbb{Z} \end{split} \end{equation}$$$f_k(x)$ 指的是动量 (波矢)为 $k$的, 在位置表象下的, 动量算符 (哈密顿算符)的特征函数. 动量算符的特征函数在位置算符特征函数上的投影就是位置表象下动量算符的特征函数
$$\langle{r}\vert{k}\rangle=f_{k}(r)=\sqrt{\frac{1}{L}}e^{ikx}$$正则系综下的密度算符表达式
$$\begin{equation} \hat{\rho} = \frac{e^{-\beta\hat{H}}}{Tr\{e^{-\beta\hat{H}}\}} \end{equation}$$接下来考察密度算符在位置表象下的密度算符矩阵元
$$\begin{equation} \begin{split} \langle{r}\vert{e^{-\beta\hat{H}}}\vert{r^\prime}\rangle & = \sum_k \langle{r}\vert{e^{-\beta\hat{H}}}\vert{k}\rangle\langle{k}\vert{r^\prime}\rangle \\ & = \sum_k e^{-\frac{\beta\hbar^2k^2}{2m}}\langle{r}\vert{k}\rangle\langle{k}\vert{r^\prime}\rangle \\ & = \sum_k e^{-\frac{\beta\hbar^2 k^2}{2m}}f_k(r)f^\ast_k(r^\prime) \\ & = \sum_k e^{-\frac{\beta\hbar^2 k^2}{2m}}\frac{1}{L}e^{ik(x-x^\prime)} \\ \end{split} \end{equation}$$考虑态密度的计算 (箱归一化, 波矢只能取分立值)
$$\begin{equation} \begin{split} \int_{\xi}^{\xi+L}\frac{dxdp}{h} & = \frac{L}{h}\hbar dk = \frac{L}{2\pi}dk \\ \sum_k & \approx \frac{L}{2\pi}\int dk \end{split} \end{equation}$$对动量的求和过渡到对动量的积分要注意有系数的差异! 将上式带入
$$\begin{equation} \begin{split} \sum_k e^{-\frac{\beta\hbar^2k^2}{2m}}\frac{1}{L}e^{ik(x-x^\prime)} \approx & \frac{L}{2\pi}\int e^{-\frac{\beta\hbar^2k^2}{2m}}\frac{1}{L}e^{ik(x-x^\prime)} dk \\ & = \frac{1}{2\pi}\int e^{-\frac{\beta\hbar^2k^2}{2m}}e^{ik(x-x^\prime)} dk \\ & = \sqrt{\frac{m}{2\pi\beta\hbar^2}}e^{-\frac{m}{2\beta\hbar^2}(x-x^\prime)^2} \\ \langle{r}\vert{e^{-\beta\hat{H}}}\vert{r^\prime}\rangle & = \sqrt{\frac{m}{2\pi\beta\hbar^2}}e^{-\frac{m}{2\beta\hbar^2}(x-x^\prime)^2} \\ \end{split} \end{equation}$$三维情况下则是
$$\begin{equation} \langle{r}\vert{e^{-\beta\hat{H}}}\vert{r^\prime}\rangle=\left(\frac{m}{2\pi\beta\hbar^2}\right)^{3/2}e^{-\frac{m}{2\beta\hbar^2}\vert{\boldsymbol{r}-\boldsymbol{r^\prime}}\vert^2} \end{equation}$$The trace of it is then
$$\begin{equation} \begin{split} \text{Tr}\{e^{-\beta\hat{H}}\} & = \int_\xi^{\xi+L} \langle{r}\vert{e^{-\beta\hat{H}}}\vert{r}\rangle dx \\ & = \sqrt{\frac{m}{2\pi\beta\hbar^2}}\int_\xi^{\xi+L}dx \\ & = L\sqrt{\frac{m}{2\pi\beta\hbar^2}} \\ \end{split} \end{equation}$$Therefore the matrix element of density matrix is
$$\begin{equation} \begin{split} \langle{r}\vert{\rho}\vert{r^\prime}\rangle & = \frac{\langle{r}\vert{e^{-\beta\hat{H}}}\vert{r^\prime}\rangle}{\text{Tr}\{e^{-\beta\hat{H}}\}} \\ & = \frac{1}{L}e^{-\frac{m}{2\beta\hbar^2}(x-x^\prime)^2} \\ \end{split} \end{equation}$$Finally let's evaluate the expectation value of Hamiltonian
$$\begin{equation} \begin{split} \langle{H}\rangle & = \text{Tr}\{\hat{H}\hat{\rho}\} = \sum_{r_i}\langle{r_i}\vert{\hat{H}\hat{\rho}}\vert{r_i}\rangle \\ \langle{r_i}\vert{\hat{H}\hat{\rho}}\vert{r_i}\rangle & = \sum_{r_j} \langle{r_i}\vert{\hat{H}}\vert{r_j}\rangle \langle{r_j}\vert{\hat{\rho}}\vert{r_i}\rangle \\ \vert{r_j}\rangle & = \delta(x-x_j) \\ \langle{r_i}\vert{\hat{H}}\vert{r_j}\rangle & = \int \delta(x-x_i)\left(-\frac{\hbar^2}{2m}\right)\frac{d^2}{dx^2} \delta(x-x_j) dx \\ & = -\frac{\hbar^2}{2m}\int \delta(x-x_i)\delta^{(2)}(x-x_j) dx \\ & = -\frac{\hbar^2}{2m}\delta^{(2)}(x_j-x_i) \\ \langle{r_j}\vert{\hat{\rho}}\vert{r_i}\rangle & = \frac{1}{L}e^{-\frac{m}{2\beta\hbar^2}(x_j-x_i)^2} \\ \sum_{r_j} \langle{r_i}\vert{\hat{H}}\vert{r_j}\rangle \langle{r_j}\vert{\hat{\rho}}\vert{r_i}\rangle & = \int -\frac{\hbar^2}{2m}\delta^{(2)}(x_j-x_i)\frac{1}{L}e^{-\frac{m}{2\beta\hbar^2}(x_j-x_i)^2}dx_j \\ & = -\frac{\hbar^2}{2mL}\int \delta^{(2)}(x_j-x_i) e^{-\frac{m}{2\beta\hbar^2}(x_j-x_i)^2}dx_j \\ & = -\frac{\hbar^2}{2mL}\frac{d^2}{dx_j^2}e^{-\frac{m}{2\beta\hbar^2}(x_j-x_i)^2}\Bigg\vert_{x_j=x_i} \\ & = -\frac{\hbar^2}{2mL}\left[\frac{m^2(x_j-x_i)^2}{\beta^2\hbar^4}e^{-\frac{m}{2\beta\hbar^2}(x_j-x_i)^2}-\frac{m}{\beta\hbar^2}e^{-\frac{m}{2\beta\hbar^2}(x_j-x_i)^2}\right]\Bigg\vert_{x_j=x_i} \\ & = \frac{1}{2\beta L} \\ \langle{\hat{H}}\rangle & = \sum_{r_i}\sum_{r_j} \langle{r_i}\vert{\hat{H}}\vert{r_j}\rangle \langle{r_j}\vert{\hat{\rho}}\vert{r_i}\rangle \\ & = \int_{\xi}^{\xi+L} \frac{1}{2\beta L} dx_i=\frac{k_B T}{2} \\ \end{split} \end{equation}$$上述结果正是能量均分定理. 自由粒子在一维情况只有一个动能二次项, 对总能量的贡献为 $k_B T/2$. 通过内插动量算符在位置表象下的特征函数完备集 ($\{\vert{k}\rangle\}$), 可以从另一个角度计算哈密顿量的期望值
$$\begin{equation} \begin{split} \langle{x_i}\vert{\rho}\vert{x_j}\rangle & = \frac{1}{L}e^{-\frac{m}{2\beta\hbar^2}(x_i-x_j)^2} \\ \langle{x_i}\vert{\hat{H}\hat{\rho}}\vert{x_i}\rangle & = \sum_{x_j}\langle{x_i}\vert{\hat{H}}\vert{x_j}\rangle \langle{x_j}\vert{\hat{\rho}}\vert{x_i}\rangle \\ \langle{x_i}\vert{\hat{H}}\vert{x_j}\rangle & = \sum_k\langle{x_i}\vert{\hat{H}}\vert{k}\rangle \langle{k}\vert{x_j}\rangle \\ & = \sum_k \frac{\hbar^2k^2}{2m}\langle{x_i}\vert{k}\rangle\langle{k}\vert{x_j}\rangle \\ & = \frac{\hbar^2k^2}{2m}\sum_k \frac{1}{L}e^{ik(x_i-x_j)} \\ \approx & \frac{\hbar^2}{2m} \frac{L}{2\pi} \frac{1}{L} \int k^2e^{ik(x_i-x_j)} dk \\ & = -\frac{\hbar^2}{2m}\frac{1}{2\pi} \int\frac{d^2}{d{x_i}^2} e^{ik(x_i-x_j)} dk \\ & = -\frac{\hbar^2}{2m}\frac{d^2}{dx_i^2}\frac{1}{2\pi} \int e^{ik(x_i-x_j)}dk \\ & = -\frac{\hbar^2}{2m}\frac{d^2}{dx_i^2}\delta(x_i-x_j) \\ \end{split} \end{equation}$$剩余步骤同上.
如果在动量表象下计算则会大大简化, 因为在动量表象下密度算符和哈密顿算符都是对角的
$$\begin{equation} \begin{split} \langle{k}\vert{e^{-\beta\hat{H}}}\vert{k^\prime}\rangle & = e^{-\frac{\beta\hbar^2{k^\prime}^2}{2m}} \delta_{kk^\prime} \\ & = e^{-\frac{\beta\hbar^2k^2}{2m}} \\ \sum_k e^{-\frac{\beta\hbar^2k^2}{2m}} & \approx \frac{L}{2\pi}\int e^{-\frac{\beta\hbar^2k^2}{2m}} dk \\ \text{Tr}\{e^{-\beta\hat{H}}\} & = \frac{L}{2\pi}\sqrt{\frac{2\pi m}{\beta\hbar^2}} \\ \langle{\hat{H}}\rangle & = \sum_k \langle{k}\vert{\hat{H}\hat{\rho}}\vert{k}\rangle \\ & = \sum_k\sum_{k^\prime} \langle{k}\vert{\hat{H}}\vert{k^\prime}\rangle \langle{k^\prime}\vert{\rho}\vert{k}\rangle \\ & = \sum_k\sum_{k^\prime} \frac{\hbar^2{k^\prime}^2}{2m}\delta_{kk^\prime} e^{-\frac{\beta\hbar^2{k^\prime}^2}{2m}}\delta_{kk^\prime} \frac{2\pi}{L}\sqrt{\frac{\beta\hbar^2}{2\pi m}} \\ & = \sum_k \frac{\hbar^2k^2}{2m} e^{-\frac{\beta\hbar^2k^2}{2m}} \frac{1}{L} \sqrt{\frac{2\pi\beta\hbar^2}{m}} \\ \approx & \frac{L}{2\pi} \int \frac{\hbar^2k^2}{2m} e^{-\frac{\beta\hbar^2k^2}{2m}}\frac{1}{L}\sqrt{\frac{2\pi\beta\hbar^2}{m}} \\ & = \frac{1}{2\beta} \\ \end{split} \end{equation}$$