(A) Dirac notation
\ (1. \ mathinner {| \ space \ rangle} \ text {: ket, dynamic quantity eigenvectors} \ mathinner {| n \ rangle} = \ Psi_n \ text {} quantum states; \)
\(2.\mathinner{\langle\space|}\text{:左矢,}\mathinner{\langle \space |}=(\mathinner{|\space \rangle})^†,\mathinner{\langle n |}=\Psi_n^* ;\)
\(3.\mathinner{\langle \space | \space \rangle}\text{:内积,} \mathinner{\langle m | n \rangle}=\int\Psi_m^*\Psi_nd\tau\quad or\quad \mathinner{\langle \psi | \phi \rangle}=\int \psi^*\phi d\tau\)
note:
\[\begin{aligned} *\quad &\mathinner{\langle x | \psi \rangle}=\int\delta(x-x')\psi(x')dx'=\psi(x)\\ *\quad &任意态\mathinner{| \psi \rangle}=\sum_n{c_n\mathinner{| n \rangle}} \iff c_n=\mathinner{\langle n | \psi \rangle}=\int{\phi_n^* \psi}d\tau\\ &\Rarr \sum_n{\mathinner{| n \rangle}\mathinner{\langle n | \psi \rangle}}\iff\sum_n\mathinner{|n \rangle}\mathinner{\langle n |}=\hat{I}\quad\text{单位算符} \end{aligned} \]
Question: the unit operator and the probability is 1 or without contact:
\[\mathinner{\langle \psi |}\hat{I}\mathinner{| \psi \rangle}=\sum_n{\mathinner{\langle \psi | n \rangle}\mathinner{\langle n | \psi \rangle}}=\sum_n{|c_n|^2}=1 \]
In addition, the continuous form of the unit operator? It may be:
\[\int \mathinner{| q \rangle}\mathinner{\langle q |}dq\quad or \quad\int \mathinner{| q \rangle}dq\mathinner{\langle q |}\quad or\quad \int dq\mathinner{| q \rangle}\mathinner{\langle q |}? \]
Some quantum mechanical equations ( matrix elements see next section ) is the Dirac represents :
\[\begin{aligned} &\text{算符} &&\hat{F}\psi=\phi &&\iff &&\hat{F}\mathinner{| \psi \rangle}=\mathinner{| \phi \rangle}\\ &\text{归一化} &&\int{\psi^*\psi}d\tau=1 &&\iff &&\mathinner{\langle \psi | \psi \rangle}=1\\ &\text{正交性} &&\int{\Psi_m^*\Psi_n}d\tau=\delta_{mn} &&\iff &&\mathinner{\langle m | n \rangle}=\delta_{mn}\\ &&&\int\Psi_q^*\Psi_{q'}d\tau=\delta(q-q') &&\iff &&\mathinner{\langle q | q' \rangle}=\delta(q-q')\\ &\text{本征方程} &&\hat{F}\psi=\lambda\psi &&\iff &&\hat{F}\mathinner{| \psi \rangle}=\lambda\mathinner{| \psi \rangle}(\hat{F}\mathinner{| n \rangle}=\lambda_n\mathinner{| n \rangle})\\ &\text{力学量期望} &&\overline{F}=\int\Psi^*\hat{F}\Psi d\tau &&\iff &&\overline{F}=\mathinner{\langle \Psi|}\hat{F}\mathinner{|\Psi \rangle}\\ &\text{薛定谔方程} && i\hbar\frac{\partial \Psi}{\partial t}=\hat{H}\Psi &&\iff &&i\hbar\frac{\partial}{\partial t}\mathinner{| \Psi \rangle}=\hat{H}\mathinner{| \Psi \rangle}\\ &\text{F矩阵元} && F_{mn}=\int\psi_m^*\hat{F}\psi_n d\tau &&\iff && F_{mn}=\mathinner{\langle m |}\hat{F} \mathinner{| n \rangle} \end{aligned} \]
问,一般内积写为:
\[\mathinner{\langle m | n \rangle}=\int\psi_m^*(x)\psi_n(x) dx \]
是否默认处于坐标表象?而省略了其单位1,即:
\[\hat{I}=\int\mathinner{| x \rangle}\mathinner{\langle x |}dx \]
(二)矩阵力学
1.表象(态和力学量的具体表示方式)
[存疑]以力学量Q对应的算符的本征函数un(q)为基矢,构成的无限维函数空间(希尔伯特空间)称为Q表象。
2.态在表象中的表示
以下以分立的本征函数为例,设
\[\begin{aligned} \Psi(q,t)&=\sum_n{a_n(t)u_n(q)}\\ &= \begin{pmatrix} u_1(q)&u_2(q)&...&u_n(q) \end{pmatrix} \begin{pmatrix} a_1(t)\\ a_2(t)\\ ...\\ a_n(t) \end{pmatrix} \end{aligned} \]
那么有在态ψ中测力学量Q的本征值Qn的概率及概率和:
\[P_n=|a_n(t)|^2 \\\space\\ \begin{aligned} \Rarr\mathinner{\langle\Psi |\Psi \rangle}&=\sum_{mn}{a_m^*a_n}\mathinner{\langle m | n \rangle}\\ &=\sum_{mn}a_m^*a_n\delta{mm}\\ &=\sum_{n}{|a_n|^2}=1 \end{aligned} \]
也就相当于态ψ在基矢un上的投影为an(t)。那么它在1表象中可以写为一个列矢量:
\[\Psi=\begin{pmatrix} a_1(t)\\ a_2(t)\\ ...\\ a_n(t) \end{pmatrix} \Rarr \Psi^†=\begin{pmatrix} a_1^*(t)&a_2^*(t)&...&a_n^*(t) \end{pmatrix} \\\space\\ \Rarr\Psi^†\Psi=\sum_n|a_n(t)|^2=1 \]
其各分量为ψ在Q表象基矢上的投影(坐标分量):
\[a_n(t)=\mathinner{\langle n | \Psi \rangle}=\int u_n^*(q)\Psi(q,t)dq \]
特别的,Q的本征态在Q表象中的表示为:
\[\psi_n=\begin{pmatrix}0\\...\\0\\1\\0\\...\\0\end{pmatrix}\rarr第n行 \]
设Q表象的基矢为un,且:
\[\hat{F}\mathinner{| \psi \rangle}=\mathinner{| \phi \rangle} \quad \mathinner{| \psi \rangle}=\sum_n a_n \mathinner{| n \rangle}\quad \mathinner{| \phi \rangle}=\sum_n b_n \mathinner{| n \rangle} \\\space\\ \Rarr \sum_n b_n \mathinner{| n \rangle}=\sum_n a_n \hat{F}\mathinner{| n \rangle} \\\space\\ \Rarr \mathinner{\langle m|}\sum_n b_n \mathinner{| n \rangle}=b_m=\mathinner{\langle m|}\sum_n a_n \hat{F}\mathinner{| n \rangle}=\sum_n{a_n}F_{mn} \\\space\\ \mathinner{| \phi \rangle}= \begin{pmatrix} b_1\\ b_2\\ ...\\ b_n \end{pmatrix}= \begin{pmatrix} F_{11}&F_{12}&...&F_{1n}\\ F_{21}&F_{22}&...&F_{2n}\\ ... &... &...&... \\ F_{n1}&F_{n2}&...&F_{nn} \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix} =\hat{F}\mathinner{| \psi \rangle} \]
则矩阵元Fmn构成的矩阵即为F算符在Q表象的表示。易证的如下两个性质:
\[①F_{mn}^†=F_{nm}^*=\mathinner{\langle m|}\hat{F}\mathinner{| n \rangle}=F_{mn} \\\space\\ ②if \quad \hat{F}\mathinner{| n \rangle}=\lambda_n\mathinner{| n \rangle} \Rarr F_{mn}=\mathinner{\langle m|}\hat{F}\mathinner{| n \rangle}=\lambda_n\delta_{mn} \]
即力学量对应的矩阵是厄米矩阵,在自身表象中为以本征值为对角元的对角矩阵。
4.量子力学公式的矩阵表达
(1)本征值方程
\[\hat{F}\mathinner{| \psi \rangle}=\lambda\mathinner{| \psi \rangle} \\\space\\ \Rarr \begin{pmatrix} F_{11}&F_{12}&...&F_{1n}\\ F_{21}&F_{22}&...&F_{2n}\\ ... &... &...&... \\ F_{n1}&F_{n2}&...&F_{nn} \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}=\lambda \begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix} \\\space\\ \begin{pmatrix} F_{11}-\lambda&F_{12}&...&F_{1n}\\ F_{21}&F_{22}-\lambda&...&F_{2n}\\ ... &... &...&... \\ F_{n1}&F_{n2}&...&F_{nn}-\lambda \end{pmatrix} \begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}=0 \\\space\\ as\quad\begin{pmatrix} a_1\\ a_2\\ ...\\ a_n \end{pmatrix}=\not0 \\\space\\ \begin{vmatrix} F_{11}-\lambda&F_{12}&...&F_{1n}\\ F_{21}&F_{22}-\lambda&...&F_{2n}\\ ... &... &...&... \\ F_{n1}&F_{n2}&...&F_{nn}-\lambda \end{vmatrix}=|F-\lambda I|=0 \]
其中I为单位矩阵,这个方程称为久期方程。其解λn为矩阵F的特征值,亦即力学量算符F的本征值;将λn代回得到的ψn是对应的本征函数(特征向量,或称本征矢)。
(2)薛定谔方程
\[i\hbar\frac{\partial}{\partial t}\mathinner{| \Psi \rangle}=\hat{H}\mathinner{| \Psi \rangle} \\\space\\ i\hbar\begin{pmatrix} \frac{\partial a_1(t)}{\partial t}\\ \frac{\partial a_2(t)}{\partial t}\\ ...\\ \frac{\partial a_n(t)}{\partial t} \end{pmatrix}=\begin{pmatrix} H_{11}&H_{12}&...&H_{1n}\\ H_{21}&H_{22}&...&H_{2n}\\ ... &... &...&... \\ H_{n1}&H_{n2}&...&H_{nn} \end{pmatrix} \begin{pmatrix} a_1(t)\\ a_2(t)\\ ...\\ a_n(t) \end{pmatrix} \]
(3)力学量期望
\[\overline{F}=\mathinner{\langle \Psi|}\hat{F}\mathinner{| \Psi \rangle} \\\space\\ \begin{aligned} \Rarr\overline{F}&=\begin{pmatrix} a_1^*(t)&a_2^*(t)&...&a_n^*(t) \end{pmatrix} \begin{pmatrix} F_{11}&F_{12}&...&F_{1n}\\ F_{21}&F_{22}&...&F_{2n}\\ ... &... &...&... \\ F_{n1}&F_{n2}&...&F_{nn} \end{pmatrix} \begin{pmatrix} a_1(t)\\ a_2(t)\\ ...\\ a_n(t) \end{pmatrix}\\ &=\sum_{mn}a_m^*a_nF_{mn} \end{aligned} \]
5.表象变换(幺正变换)
设力学量F在两个不同表象P和Q中的表述为:
\[F_{mn}=\mathinner{\langle m |}\hat{F}\mathinner{| n \rangle}\quad F_{\alpha\beta}=\mathinner{\langle \alpha |}\hat{F}\mathinner{| \beta \rangle} \]
又有:
\[\mathinner{| \beta \rangle}=\sum_n\mathinner{| n \rangle}S_{n\beta}\quad\mathinner{\langle \alpha |}=\sum_m S_{m\alpha}^* \mathinner{\langle m |} \\\space\\ as\quad S_{n\beta}=\mathinner{\langle n | \beta \rangle}\quad S_{m\alpha}^*=\mathinner{\langle \alpha | m \rangle} \]
(1)易证得,由Snβ构成的变换矩阵S,把P表象基矢转换为Q表象基矢(表象变换):
\[\vec{Q}= \begin{pmatrix} \mathinner{| 1' \rangle}\\ \mathinner{| 2' \rangle}\\ ...\\ \mathinner{| n' \rangle} \end{pmatrix}^T = \begin{pmatrix} \mathinner{| 1 \rangle}\\ \mathinner{| 2 \rangle}\\ ...\\ \mathinner{| n \rangle} \end{pmatrix}^T \begin{pmatrix} S_{11}&S_{21}&...&S_{n1}\\ S_{12}&S_{22}&...&S_{n2}\\ ... &... &...&... \\ S_{1n}&S_{n2}&...&S_{nn} \end{pmatrix} =\vec{P}S \\\space\\ \iff\vec{Q}^T=\widetilde{S}\vec{P}^T \iff\vec{Q}^†=S^†\vec{P}^† (=S^{-1}\vec{P}^†) \]
又有:
\[\begin{aligned} \delta_{\alpha\beta} &=\mathinner{\langle \alpha}\mathinner{| \beta \rangle}\\ &=\sum_{m,n} S_{m\alpha}^*\mathinner{\langle m | n \rangle}S_{n\beta}\\ &=\sum_{m,n} S_{m\alpha}^*S_{n\beta}\delta_{mn}\\ &=\sum_n S_{n\alpha}^*S_{n\beta}\\ &=\sum_n S_{\alpha n}^†S_{n\beta}\\ &=(S^†S)_{\alpha\beta}\\ \end{aligned} \\\space\\ \Rarr S^†S=I \]
同理,有:
\[\delta_{mn}=(SS^†)_{mn}\Rarr SS^†=I \]
满足以上条件的矩阵称为幺正矩阵,故而表象变换是幺正变换。
那么态在不同表象中的变换为:
\[\vec{P}\Psi=\vec{P}SS^{-1}\Psi=\vec{Q}\Psi' \\\space\\ \Rarr \Psi'=S^{-1}\Psi\iff S\Psi'=\Psi \]
(2)易知:
\[\begin{aligned} F_{\alpha\beta}&=\mathinner{\langle \alpha | \hat{F} | \beta \rangle}\\ &=\sum_{m,n} S_{m\alpha}^* \mathinner{\langle m |\hat{F}| n \rangle}S_{n\beta}\\ &=\sum_{m,n}{S_{\alpha m}^†F_{mn}S_{n\beta}}\\ &=(S^†FS)_{\alpha\beta} \end{aligned} \]
则有力学量在不同表象中的变换:
\[F'=S^†FS=S^{-1}FS \]
易证得表象变换不改变力学量的本征值,本征值的和(也即矩阵的迹)不变:
\[F'\Psi'=S^{-1}FSS^{-1}\Psi=S^{-1}F\Psi=S^{-1}\lambda\Psi=\lambda\Psi' \\\space\\ \begin{aligned} Sp(AB)&=\sum_{m}(AB)_{mm}\\ &=\sum_m{\sum_n A_{mn} B_{nm}}\\ &=\sum_n\sum_m B_{nm} A_{mn}\\ &=\sum_n(BA)_{nn}\\ &=Sp(BA) \end{aligned} \\\space\\ \Rarr Sp(S^{-1}FS)=Sp(S^{-1}SF)=Sp(F) \]
(三)表象实例——占有数表象
有谐振子的哈密顿量:
\[\begin{aligned} \hat{H}&=-\frac{\hbar^2}{2m}\nabla^2+\frac{1}{2}m\omega^2x^2\\ &=(\sqrt{\frac{m}{2}}\omega x-\frac{\hbar}{\sqrt{2m}}\nabla)(\sqrt{\frac{m}{2}}\omega x+\frac{\hbar}{\sqrt{2m}}\nabla)+\frac{1}{2}\hbar\omega\\ &=\frac{\hbar\omega}{2}[(\sqrt{\frac{m\omega}{\hbar}}x-\sqrt{\frac{\hbar}{m\omega}}\frac{\partial}{\partial x})(\sqrt{\frac{m\omega}{\hbar}}x+\sqrt{\frac{\hbar}{m\omega}}\frac{\partial}{\partial x})+1]\\ &=\hbar\omega\{[\sqrt{\frac{1}{2}}(\xi-\frac{\partial}{\partial \xi})][\sqrt{\frac{1}{2}}(\xi+\frac{\partial}{\partial \xi})]+\frac{1}{2}\}\quad(\xi=\sqrt{\frac{m\omega}{\hbar}}x)\\ &=\hbar\omega(\hat{a}^†\hat{a}+\frac{1}{2})=\hbar\omega(\hat{n}+\frac{1}{2}) \end{aligned} \]
其中:
\[\hat{a}=\sqrt{\frac{1}{2}}(\xi+\frac{\partial}{\partial \xi})\quad\hat{a}^†=\sqrt{\frac{1}{2}}(\xi-\frac{\partial}{\partial \xi}) \\\space\\ \hat{x}=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}+\hat{a}^†)\quad\hat{p}=-i\hbar\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}-\hat{a}^†) \]
由谐振子薛定谔方程:
\[\hat{H}\psi_n=\hbar\omega(\hat{n}+\frac{1}{2})\psi_n=\hbar\omega(n+\frac{1}{2})\psi_n=E_n\psi_n \\\space\\ \Rarr \hat{n}\psi_n=n\psi_n \]
即算符n有本征值n,称为占有数(声子数)。其本征函数为谐振子本征函数,构成占有数表象。
由谐振子本征函数的递推关系:
\[\xi\psi_n(\xi)=\sqrt{\frac{n}{2}}\psi_{n-1}(\xi)+\sqrt{\frac{n+1}{2}}\psi_{n+1}(\xi) \\\space\\ \frac{d}{d\xi}\psi_n(\xi)=\sqrt{\frac{n}{2}}\psi_{n-1}(\xi)-\sqrt{\frac{n+1}{2}}\psi_{n+1}(\xi) \\\space\\ \begin{aligned} \Rarr \hat{a}\mathinner{| n \rangle}&=\sqrt{n}\mathinner{| n-1 \rangle} &\text{湮灭算符}\\ \hat{a}^†\mathinner{| n \rangle} &=\sqrt{n+1}\mathinner{| n+1 \rangle} &\text{产生算符}\\ \Rarr\hat{n}\mathinner{| n \rangle} &=\hat{a}^†\hat{a}\mathinner{| n \rangle}\\ &=\sqrt{n}\hat{a}^†\mathinner{| n-1 \rangle}\\ &=n\mathinner{| n \rangle}\\ \hat{a}\hat{a}^†\mathinner{| n \rangle} &=\sqrt{n+1}\hat{a}\mathinner{| n+1 \rangle}\\ &=(n+1)\mathinner{| n \rangle}\\ \Rarr \hat{a}\hat{a}^† &= \hat{a}^†\hat{a}+1\iff[\hat{a},\hat{a}^†]=1 \end{aligned} \]
容易证得,在n表象中,产生算符和湮灭算符的矩阵元为:
\[(\hat{a})_{mn}=\mathinner{\langle m| \hat{a} | n \rangle}=\sqrt{n}\delta_{m,n-1} \\\space\\ (\hat{a}^†)_{mn}=\mathinner{\langle m| \hat{a}^† | n \rangle}=\sqrt{n+1}\delta_{m,n+1} \]