UA OPTI570 量子力学 恒等算符在算符计算中的应用
UA OPTI570 量子力學 恒等算符在算符計算中的應用
恒等算符代換是量子力學中一個很有用的技巧：考慮算符A^:E→E\hat A:\mathcal{E} \to \mathcal{E}A^:E→E的函數f(A^)f(\hat A)f(A^)，假設存在一個左矢∣ψ?∈E|\psi \rangle \in \mathcal{E}∣ψ?∈E與一個右矢??∣∈E?\langle \phi | \in \mathcal{E}^*??∣∈E?，要計算
??∣f(A^)∣ψ?\langle \phi | f(\hat A) | \psi \rangle??∣f(A^)∣ψ?
可以用恒等算符代換：
f(A^)=f(A^)1^={離散：f(A^)∑n∣an??an∣離散且有重數：f(A^)∑n∑i=1gn∣ani??ani∣連續：f(A^)∫?∞+∞da∣a??a∣f(\hat A) = f(\hat A) \hat 1 = \begin{cases} 離散： f(\hat A) \sum_{n} | a_n \rangle \langle a_n | \\ 離散且有重數：f(\hat A) \sum_n \sum_{i=1}^{g_n} |a_n^i \rangle \langle a_n^i| \\ 連續：f(\hat A) \int_{-\infty}^{+\infty} da |a \rangle \langle a | \end{cases}f(A^)=f(A^)1^=??????離散：f(A^)∑n?∣an???an?∣離散且有重數：f(A^)∑n?∑i=1gn??∣ani???ani?∣連續：f(A^)∫?∞+∞?da∣a??a∣?
例1 證明位置算符是厄爾米特算符
?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??
從第二個等號到第三個等號用了恒等算符代換，
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 計算動量算符的作用
?x∣P^∣ψ?=?x∣P^1^∣ψ?=∫?∞+∞?x∣P^∣p??p∣ψ?dp=∫?∞+∞p?x∣p??p∣ψ?dp=12π?∫?∞+∞pψˉ(p)eixp/?dp=i???x12π?∫?∞+∞ψˉ(p)eixp/?dp=?i???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=2π??1?∫?∞+∞?pψˉ?(p)eixp/?dp=?i??x??2π??1?∫?∞+∞?ψˉ?(p)eixp/?dp=?i??x??ψ(x)?
從第一個等號到第二個等號用了恒等算符代換，
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 計算S^(x0)\hat S(x_0)S^(x0?)的作用，其中S^(x0)=e?ix0P^/?,x0∈R\hat S(x_0)=e^{-ix_0\hat P/\hbar},x_0 \in \mathbb{R}S^(x0?)=e?ix0?P^/?,x0?∈R，
?x∣S^(x0)∣ψ?=?x∣S^(x0)1^∣ψ?=∫?∞+∞dp?x∣e?ix0P^/?∣p??p∣ψ?=∫?∞+∞dpe?ix0p/??x∣p??p∣ψ?=∫?∞+∞dpe?ix0p/??x∣p??p∣ψ?=∫?∞+∞dpe?ix0p/?12π?eixp/?ψˉ(p)=∫?∞+∞dp12π?eip(x?x0)/?ψˉ(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?ix0?P^/?∣p??p∣ψ?=∫?∞+∞?dpe?ix0?p/??x∣p??p∣ψ?=∫?∞+∞?dpe?ix0?p/??x∣p??p∣ψ?=∫?∞+∞?dpe?ix0?p/?2π??1?eixp/?ψˉ?(p)=∫?∞+∞?dp2π??1?eip(x?x0?)/?ψˉ?(p)=FT?1[ψˉ?(p)](x?x0?)=ψ(x?x0?)?
例4 計算?x∣U^(t)∣x′?\langle x|\hat U(t)|x' \rangle?x∣U^(t)∣x′?，其中U^(t)=e?iP^2t/2m?\hat U(t)=e^{-i\hat P^2 t/2m\hbar}U^(t)=e?iP^2t/2m?，
?x∣U^(t)∣x′?=?x∣e?iP^2t/2m?∣x′?=∫?∞+∞dp?x∣e?iP^2t/2m?∣p??p∣x′?=∫?∞+∞dpe?ip2t2m??x∣p??p∣x′?=∫?∞+∞dpe?ip2t2m?12π?eixp?12π?e?ipx′?=12π?∫?∞+∞exp?(?i(t2m?p2?x?x′?p))dp=2m?texp?(im(x?x′)22?t)2π?∫?∞+∞exp?(?i(t2m?p?x?x′?2t2m?)2)dp=exp?(im(x?x′)22?t)2π?πexp?(?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/2m?∣x′?=∫?∞+∞?dp?x∣e?iP^2t/2m?∣p??p∣x′?=∫?∞+∞?dpe?2m?ip2t??x∣p??p∣x′?=∫?∞+∞?dpe?2m?ip2t?2π??1?e?ixp?2π??1?e??ipx′?=2π?1?∫?∞+∞?exp(?i(2m?t?p2??x?x′?p))dp=t2m???2π?exp(2?tim(x?x′)2?)?∫?∞+∞?exp?????i???2m?t??p?22m?t???x?x′?????2????dp=2π?exp(2?tim(x?x′)2?)?π?exp(?iπ/4)?
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