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The Influence of Coherence on the First Law of Quantum Thermodynamics
DOI: 10.12677/MP.2022.126017, PDF, HTML, XML, 下载: 362  浏览: 698

Abstract: Quantum thermodynamics and classical thermodynamics are very different in essence, but they are related to each other. Quantum coherence plays an important role in the study of quantum thermodynamics. In this paper, the peculiar behavior of quantum coherence in thermodynamics is dis-covered by studying the process of atomic spontaneous radiation. Select different initial states of the system by studying a specific system, it is found that the heat generated by the external work of the system and coherent dynamics contribute to the internal energy of the system. The results are of great significance to the further improvement of the first law of quantum thermodynamics.

1. 引言

2. 基本原理(理论模型)

2.1. Kraus算符及其表示

$\epsilon \left[\stackrel{^}{\rho }\left(0\right)\right]=\underset{i}{\sum }{\stackrel{^}{E}}_{i}\stackrel{^}{\rho }\left(0\right){\stackrel{^}{E}}_{i}^{†}=\stackrel{^}{\rho }\left(t\right)$ (1)

$\underset{i}{\sum }{\stackrel{^}{E}}_{i}^{†}{\stackrel{^}{E}}_{i}=\stackrel{^}{I}$ (2)

2.2. 量子热力学第一定律

$\text{d}U=\text{d}W+\text{d}Q+\text{d}C$ (3)

$\Delta U\left(t\right)=\underset{n}{\sum }\underset{k}{\sum }\underset{0}{\overset{t}{\int }}\frac{\text{d}}{\text{d}{t}^{\prime }}\left({E}_{n}{\rho }_{k}{|{c}_{n,k}|}^{2}\right)\text{d}{t}^{\prime }$ (4)

${\stackrel{^}{H}}_{s}=\underset{n}{\sum }{E}_{n}|n〉〈n|$ (5)

$W\left(t\right)=\underset{n}{\sum }\underset{k}{\sum }\underset{0}{\overset{t}{\int }}{\rho }_{k}{|{c}_{n,k}|}^{2}\frac{\text{d}{E}_{n}}{\text{d}{t}^{\prime }}\text{d}{t}^{\prime }$ (6)

(7)

$C\left(t\right)=\underset{n}{\sum }\underset{k}{\sum }\underset{0}{\overset{t}{\int }}\left({E}_{n}{\rho }_{k}\right)\frac{\text{d}}{\text{d}{t}^{\prime }}{|{c}_{n,k}|}^{2}\text{d}{t}^{\prime }$ (8)

3. 相干性对量子热力学的影响

${E}_{1}=\sqrt{p}\left(\begin{array}{cc}1& 0\\ 0& \sqrt{1-\gamma }\end{array}\right),\text{\hspace{0.17em}}{E}_{2}=\sqrt{p}\left(\begin{array}{cc}0& \sqrt{\gamma }\\ 0& 0\end{array}\right),\text{\hspace{0.17em}}{E}_{3}=\sqrt{1-p}\left(\begin{array}{cc}\sqrt{1-\gamma }& 0\\ 0& 1\end{array}\right),\text{\hspace{0.17em}}{E}_{4}=\sqrt{1-p}\left(\begin{array}{cc}0& 0\\ \sqrt{\gamma }& 0\end{array}\right)$ (9)

$\rho \left(t\right)=\left(\begin{array}{cc}\left(1-\gamma +\gamma p\right){\rho }_{00}+\gamma {\rho }_{11}& \sqrt{1-\gamma }{\rho }_{01}\\ \sqrt{1-\gamma }{\rho }_{10}& \gamma \left(1-p\right){\rho }_{00}+\left(1-\gamma p\right){\rho }_{11}\end{array}\right)$ (10)

$\rho \left(t\right)=\left(\begin{array}{cc}{\rho }_{00}+\gamma {\rho }_{11}& \sqrt{1-\gamma }{\rho }_{01}\\ \sqrt{1-\gamma }{\rho }_{10}& \left(1-\gamma \right){\rho }_{11}\end{array}\right)$ (11)

${\rho }_{0}\left(t\right)=\frac{1}{2}{\text{e}}^{-t\Gamma }\left({\rho }_{00}{\text{e}}^{t\Gamma }+{\rho }_{11}{\text{e}}^{t\Gamma }+m\right)$ (12)

${\rho }_{1}\left(t\right)=\frac{1}{2}{\text{e}}^{-t\Gamma }\left({\rho }_{00}{\text{e}}^{t\Gamma }+{\rho }_{11}{\text{e}}^{t\Gamma }-m\right)$ (13)

$|{k}_{0}\left(t\right)〉=\frac{\left[A|g〉+|e〉\right]}{2{\rho }_{10}\sqrt{D+1}}$ (14)

$|{k}_{1}\left(t\right)〉=\frac{\left[B|g〉+|e〉\right]}{2{\rho }_{10}\sqrt{F+1}}$ (15)

$Q\left(t\right)=\underset{n}{\sum }\underset{k}{\sum }\underset{0}{\overset{t}{\int }}{E}_{n}{|{c}_{n,k}|}^{2}\frac{\text{d}{\rho }_{k}}{\text{d}{t}^{\prime }}\text{d}{t}^{\prime }=a{E}_{g}+b{E}_{e}$ (16)

$a=\underset{0}{\overset{t}{\int }}{|〈g|{k}_{0}\left({t}^{\prime }\right)〉|}^{2}\frac{\text{d}}{\text{d}t}{\rho }_{0}\left({t}^{\prime }\right)\text{d}{t}^{\prime }+\underset{0}{\overset{t}{\int }}{|〈g|{k}_{1}\left({t}^{\prime }\right)〉|}^{2}\frac{\text{d}}{\text{d}t}{\rho }_{1}\left({t}^{\prime }\right)\text{d}{t}^{\prime }$

$b=\underset{0}{\overset{t}{\int }}{|〈e|{k}_{0}\left({t}^{\prime }\right)〉|}^{2}\frac{\text{d}}{\text{d}t}{\rho }_{0}\left({t}^{\prime }\right)\text{d}{t}^{\prime }+\underset{0}{\overset{t}{\int }}{|〈e|{k}_{1}\left({t}^{\prime }\right)〉|}^{2}\frac{\text{d}}{\text{d}t}{\rho }_{1}\left({t}^{\prime }\right)\text{d}{t}^{\prime }$

$C\left(t\right)=\underset{n}{\sum }\underset{k}{\sum }\underset{0}{\overset{t}{\int }}\left({E}_{n}{\rho }_{k}\right)\frac{\text{d}}{\text{d}{t}^{\prime }}{|{c}_{n,k}|}^{2}\text{d}{t}^{\prime }=d{E}_{g}+f{E}_{e}$ (17)

$d=\underset{0}{\overset{t}{\int }}{\rho }_{0}\left({t}^{\prime }\right)\frac{\text{d}}{\text{d}t}{|〈g|{k}_{0}\left({t}^{\prime }\right)〉|}^{2}\text{d}{t}^{\prime }+\underset{0}{\overset{t}{\int }}{\rho }_{1}\left({t}^{\prime }\right)\frac{\text{d}}{\text{d}t}{|〈g|{k}_{1}\left({t}^{\prime }\right)〉|}^{2}\text{d}{t}^{\prime }$

$f=\underset{0}{\overset{t}{\int }}{\rho }_{0}\left({t}^{\prime }\right)\frac{\text{d}}{\text{d}t}{|〈e|{k}_{0}\left({t}^{\prime }\right)〉|}^{2}\text{d}{t}^{\prime }+\underset{0}{\overset{t}{\int }}{\rho }_{1}\left({t}^{\prime }\right)\frac{\text{d}}{\text{d}t}{|〈e|{k}_{0}\left({t}^{\prime }\right)〉|}^{2}\text{d}{t}^{\prime }$

$Q\left(\tau \right)=\frac{3}{4}-\frac{3}{4}{\text{e}}^{-\tau }-\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{3}{\sqrt{7}}-\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{3-6{\text{e}}^{-\tau }}{\sqrt{7}}+\frac{1}{16}\mathrm{log}\left(1+\frac{9}{4}{\text{e}}^{-2\tau }-\frac{9}{4}{\text{e}}^{-\tau }\right)$ (18)

$C\left(\tau \right)=\frac{\tau }{8}+\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{1}{3\sqrt{7}}+\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{8{\text{e}}^{\tau }-9}{3\sqrt{7}}+\frac{1}{16}\mathrm{log}\left(4\right)-\frac{1}{16}\mathrm{log}\left(9+4{\text{e}}^{2\tau }-9{\text{e}}^{\tau }\right)$ (19)

$\begin{array}{c}\Delta U\left(\tau \right)=\frac{3}{4}-\frac{3}{4}{\text{e}}^{-\tau }-\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{3}{\sqrt{7}}-\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{3-6{\text{e}}^{-\tau }}{\sqrt{7}}+\frac{1}{16}\mathrm{log}\left(1+\frac{9}{4}{\text{e}}^{-2\tau }-\frac{9}{4}{\text{e}}^{-\tau }\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{\tau }{8}+\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{1}{3\sqrt{7}}+\frac{\sqrt{7}}{8}{\mathrm{tan}}^{-1}\frac{8{\text{e}}^{\tau }-9}{3\sqrt{7}}+\frac{1}{16}\mathrm{log}\left(4\right)-\frac{1}{16}\mathrm{log}\left(9+4{\text{e}}^{2\tau }-9{\text{e}}^{\tau }\right)\end{array}$ (20)

Figure 1. The first law of atomic spontaneous emission process, heat and coherent dynamics are equal to the change of internal energy, where the parameter is θ = 30˚

$Q\left(\tau \right)=\frac{1}{4}-\frac{1}{4}{\text{e}}^{\tau }-\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{1}{\sqrt{15}}+\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{2{\text{e}}^{-\tau }-1}{\sqrt{15}}-\frac{\mathrm{log}\left(8\right)}{8}+\frac{3}{16}\mathrm{log}\left(4+{\text{e}}^{-2\tau }-{\text{e}}^{-\tau }\right)$ (21)

$C\left(\tau \right)=\frac{3\tau }{8}-\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{7}{15}+\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{8{\text{e}}^{\tau }-1}{\sqrt{15}}+\frac{1}{16}\mathrm{log}\left(64\right)-\frac{3}{16}\mathrm{log}\left(1+4{\text{e}}^{2\tau }-{\text{e}}^{\tau }\right)$ (22)

$\begin{array}{c}\Delta U\left(\tau \right)=\frac{1}{4}-\frac{1}{4}{\text{e}}^{\tau }-\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{1}{\sqrt{15}}+\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{2{\text{e}}^{-\tau }-1}{\sqrt{15}}-\frac{\mathrm{log}\left(8\right)}{8}+\frac{3}{16}\mathrm{log}\left(4+{\text{e}}^{-2\tau }-{\text{e}}^{-\tau }\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{3\tau }{8}-\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{7}{15}+\frac{\sqrt{15}}{8}{\mathrm{tan}}^{-1}\frac{8{\text{e}}^{\tau }-1}{\sqrt{15}}+\frac{1}{16}\mathrm{log}\left(64\right)-\frac{3}{16}\mathrm{log}\left(1+4{\text{e}}^{2\tau }-{\text{e}}^{\tau }\right)\end{array}$ (23)

Figure 2. According to the first law of atomic spontaneous emission process, the energy generated by heat and coherent dynamics is equal to the change of internal energy. Blue represents thermal energy, parameter is θ = 60˚

Figure 3. First-law description of the nonunitaryspontaneous emission process. Both and dynamics of coherence contribute to the change in the internal energy, $\Delta U\left(t\right)=Q\left(t\right)+C\left(t\right)$. Where the parameter is θ = 45˚

4. 结论

NOTES

*通讯作者。

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