UA OPTI570 量子力学恒等算符在算符计算中的应用

恒等算符代换是量子力学中一个很有用的技巧：考虑算符A^:E→E\hat A:\mathcal{E} \to \mathcal{E}A^:E→E的函数f(A^)f(\hat A)f(A^)，假设存在一个左矢∣ψ⟩∈E|\psi \rangle \in \mathcal{E}∣ψ⟩∈E与一个右矢⟨ϕ∣∈E∗\langle \phi | \in \mathcal{E}^*⟨ϕ∣∈E∗，要计算
⟨ϕ∣f(A^)∣ψ⟩\langle \phi | f(\hat A) | \psi \rangle⟨ϕ∣f(A^)∣ψ⟩

可以用恒等算符代换：
f(A^)=f(A^)1^={离散：f(A^)∑n∣an⟩⟨an∣离散且有重数：f(A^)∑n∑i=1gn∣ani⟩⟨ani∣连续：f(A^)∫−∞+∞da∣a⟩⟨a∣f(\hat A) = f(\hat A) \hat 1 = \begin{cases} 离散： f(\hat A) \sum_{n} | a_n \rangle \langle a_n | \\ 离散且有重数：f(\hat A) \sum_n \sum_{i=1}^{g_n} |a_n^i \rangle \langle a_n^i| \\ 连续：f(\hat A) \int_{-\infty}^{+\infty} da |a \rangle \langle a | \end{cases}f(A^)=f(A^)1^=⎩⎪⎨⎪⎧离散：f(A^)∑n∣an⟩⟨an∣离散且有重数：f(A^)∑n∑i=1gn∣ani⟩⟨ani∣连续：f(A^)∫−∞+∞da∣a⟩⟨a∣

例1 证明位置算符是厄尔米特算符
⟨x∣X^†∣ψ⟩=(⟨ψ∣X^∣x⟩)∗=(⟨ψ∣1^X^∣x⟩)∗=(∫−∞+∞dx′⟨ψ∣x′⟩⟨x′∣X^∣x⟩)∗=(∫−∞+∞dx′ψ∗(x′)x′⟨x′∣x⟩)∗=(∫−∞+∞dx′ψ∗(x′)x′δ(x′−x))∗=xψ(x)=⟨x∣X^∣ψ⟩\begin{aligned}\langle x|\hat X^{\dag}|\psi\rangle & = \left(\langle \psi|\hat X|x\rangle \right)^* \\ & =\left(\langle \psi| \hat 1\hat X|x\rangle \right)^* \\ &= \left(\int_{-\infty}^{+\infty}dx'\langle \psi| x' \rangle \langle x' |\hat X|x\rangle \right)^* \\& =\left(\int_{-\infty}^{+\infty}dx' \psi^*(x') x'\langle x' |x\rangle \right)^* \\ &=\left(\int_{-\infty}^{+\infty}dx' \psi^*(x') x'\delta(x'-x) \right)^* \\ & = x \psi(x) = \langle x|\hat X|\psi\rangle \end{aligned}⟨x∣X^†∣ψ⟩=(⟨ψ∣X^∣x⟩)∗=(⟨ψ∣1^X^∣x⟩)∗=(∫−∞+∞dx′⟨ψ∣x′⟩⟨x′∣X^∣x⟩)∗=(∫−∞+∞dx′ψ∗(x′)x′⟨x′∣x⟩)∗=(∫−∞+∞dx′ψ∗(x′)x′δ(x′−x))∗=xψ(x)=⟨x∣X^∣ψ⟩

从第二个等号到第三个等号用了恒等算符代换，
X^=1^X^=∫−∞+∞dx′∣x′⟩⟨x′∣X^\hat X = \hat 1 \hat X = \int_{-\infty}^{+\infty} dx' |x' \rangle \langle x' | \hat XX^=1^X^=∫−∞+∞dx′∣x′⟩⟨x′∣X^

例2 计算动量算符的作用
⟨x∣P^∣ψ⟩=⟨x∣P^1^∣ψ⟩=∫−∞+∞⟨x∣P^∣p⟩⟨p∣ψ⟩dp=∫−∞+∞p⟨x∣p⟩⟨p∣ψ⟩dp=12πℏ∫−∞+∞pψˉ(p)eixp/ℏdp=iℏ∂∂x12πℏ∫−∞+∞ψˉ(p)eixp/ℏdp=−iℏ∂∂xψ(x)\begin{aligned} \langle x|\hat P|\psi\rangle & =\langle x|\hat P \hat 1|\psi\rangle \\ & = \int_{-\infty}^{+\infty} \langle x|\hat P |p \rangle \langle p|\psi\rangle dp \\ & = \int_{-\infty}^{+\infty} p\langle x |p \rangle \langle p|\psi\rangle dp \\ &= \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{+\infty} p\bar \psi(p)e^{ixp/\hbar} dp \\ & = \frac{i}{\hbar} \frac{\partial }{\partial x} \frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{+\infty} \bar \psi(p)e^{ixp/\hbar} dp \\ & = -i \hbar \frac{\partial }{\partial x} \psi(x)\end{aligned}⟨x∣P^∣ψ⟩=⟨x∣P^1^∣ψ⟩=∫−∞+∞⟨x∣P^∣p⟩⟨p∣ψ⟩dp=∫−∞+∞p⟨x∣p⟩⟨p∣ψ⟩dp=2πℏ1∫−∞+∞pψˉ(p)eixp/ℏdp=ℏi∂x∂2πℏ1∫−∞+∞ψˉ(p)eixp/ℏdp=−iℏ∂x∂ψ(x)

从第一个等号到第二个等号用了恒等算符代换，
P^=P^1^=∫−∞+∞dpP^∣p⟩⟨p∣\hat P = \hat P \hat 1 = \int_{-\infty}^{+\infty} dp \hat P |p \rangle \langle p | P^=P^1^=∫−∞+∞dpP^∣p⟩⟨p∣

例3 计算S^(x0)\hat S(x_0)S^(x0)的作用，其中S^(x0)=e−ix0P^/ℏ,x0∈R\hat S(x_0)=e^{-ix_0\hat P/\hbar},x_0 \in \mathbb{R}S^(x0)=e−ix0P^/ℏ,x0∈R，
⟨x∣S^(x0)∣ψ⟩=⟨x∣S^(x0)1^∣ψ⟩=∫−∞+∞dp⟨x∣e−ix0P^/ℏ∣p⟩⟨p∣ψ⟩=∫−∞+∞dpe−ix0p/ℏ⟨x∣p⟩⟨p∣ψ⟩=∫−∞+∞dpe−ix0p/ℏ⟨x∣p⟩⟨p∣ψ⟩=∫−∞+∞dpe−ix0p/ℏ12πℏeixp/ℏψˉ(p)=∫−∞+∞dp12πℏeip(x−x0)/ℏψˉ(p)=FT−1[ψˉ(p)](x−x0)=ψ(x−x0)\begin{aligned} \langle x |\hat S(x_0) | \psi \rangle & = \langle x |\hat S(x_0) \hat 1 | \psi \rangle \\ & = \int_{-\infty}^{+\infty}dp \langle x|e^{-ix_0\hat P/\hbar}|p \rangle \langle p | \psi \rangle \\ & = \int_{-\infty}^{+\infty} dp e^{-ix_0p/\hbar} \langle x | p\rangle \langle p|\psi \rangle \\ & = \int_{-\infty}^{+\infty} dp e^{-ix_0p/\hbar} \langle x | p\rangle \langle p|\psi \rangle \\ & = \int_{-\infty}^{+\infty}dpe^{-ix_0p/\hbar} \frac{1}{\sqrt{2\pi \hbar}} e^{ixp/\hbar} \bar \psi(p) \\ & = \int_{-\infty}^{+\infty} dp \frac{1}{\sqrt{2\pi \hbar}} e^{ip(x-x_0)/\hbar}\bar \psi(p) \\ & = FT^{-1}[\bar \psi(p)](x-x_0) = \psi(x-x_0)\end{aligned}⟨x∣S^(x0)∣ψ⟩=⟨x∣S^(x0)1^∣ψ⟩=∫−∞+∞dp⟨x∣e−ix0P^/ℏ∣p⟩⟨p∣ψ⟩=∫−∞+∞dpe−ix0p/ℏ⟨x∣p⟩⟨p∣ψ⟩=∫−∞+∞dpe−ix0p/ℏ⟨x∣p⟩⟨p∣ψ⟩=∫−∞+∞dpe−ix0p/ℏ2πℏ1eixp/ℏψˉ(p)=∫−∞+∞dp2πℏ1eip(x−x0)/ℏψˉ(p)=FT−1[ψˉ(p)](x−x0)=ψ(x−x0)

例4 计算⟨x∣U^(t)∣x′⟩\langle x|\hat U(t)|x' \rangle⟨x∣U^(t)∣x′⟩，其中U^(t)=e−iP^2t/2mℏ\hat U(t)=e^{-i\hat P^2 t/2m\hbar}U^(t)=e−iP^2t/2mℏ，
⟨x∣U^(t)∣x′⟩=⟨x∣e−iP^2t/2mℏ∣x′⟩=∫−∞+∞dp⟨x∣e−iP^2t/2mℏ∣p⟩⟨p∣x′⟩=∫−∞+∞dpe−ip2t2mℏ⟨x∣p⟩⟨p∣x′⟩=∫−∞+∞dpe−ip2t2mℏ12πℏeixpℏ12πℏe−ipx′ℏ=12πℏ∫−∞+∞exp⁡(−i(t2mℏp2−x−x′ℏp))dp=2mℏtexp⁡(im(x−x′)22ℏt)2πℏ∫−∞+∞exp⁡(−i(t2mℏp−x−x′ℏ2t2mℏ)2)dp=exp⁡(im(x−x′)22ℏt)2πℏπexp⁡(−iπ/4)\begin{aligned} \langle x|\hat U(t)|x' \rangle &= \langle x|e^{-i\hat P^2 t/2m\hbar}|x' \rangle \\ & =\int_{-\infty}^{+\infty}dp \langle x|e^{-i\hat P^2 t/2m\hbar}|p \rangle \langle p|x' \rangle \\ & = \int_{-\infty}^{+\infty} dp e^{-\frac{ip^2t}{2m\hbar}} \langle x | p \rangle \langle p | x' \rangle \\ & = \int_{-\infty}^{+\infty} dp e^{-\frac{ip^2t}{2m\hbar}} \frac{1}{\sqrt{2\pi\hbar}}e^{\frac{ixp}{\hbar}} \frac{1}{\sqrt{2\pi\hbar}} e^{-\frac{ipx'}{\hbar}} \\ & = \frac{1}{2\pi \hbar} \int_{-\infty}^{+\infty} \exp \left( -i\left( \frac{t}{2m\hbar}p^2-\frac{x-x'}{\hbar}p \right) \right)dp \\ & =\sqrt{\frac{2m\hbar}{t}} \frac{\exp \left( \frac{im(x-x')^2}{2\hbar t} \right)}{2\pi \hbar} \int_{-\infty}^{+\infty} \exp \left( - i \left( \sqrt{\frac{t}{2m\hbar}}p-\frac{\frac{x-x'}{\hbar}}{2 \sqrt{\frac{t}{2m\hbar}}} \right)^2 \right)dp \\ & = \frac{\exp \left( \frac{im(x-x')^2}{2\hbar t} \right)}{2\pi\hbar} \sqrt{\pi} \exp \left( -i\pi/4 \right)\end{aligned}⟨x∣U^(t)∣x′⟩=⟨x∣e−iP^2t/2mℏ∣x′⟩=∫−∞+∞dp⟨x∣e−iP^2t/2mℏ∣p⟩⟨p∣x′⟩=∫−∞+∞dpe−2mℏip2t⟨x∣p⟩⟨p∣x′⟩=∫−∞+∞dpe−2mℏip2t2πℏ1eℏixp2πℏ1e−ℏipx′=2πℏ1∫−∞+∞exp(−i(2mℏtp2−ℏx−x′p))dp=t2mℏ2πℏexp(2ℏtim(x−x′)2)∫−∞+∞exp⎝⎜⎛−i⎝⎛2mℏtp−22mℏtℏx−x′⎠⎞2⎠⎟⎞dp=2πℏexp(2ℏtim(x−x′)2)πexp(−iπ/4)