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ϕ4模型真空态间的转换算符

刘晖 郭恒源 周遥 JarahEvslin

刘晖, 郭恒源, 周遥, JarahEvslin. ϕ4模型真空态间的转换算符[J]. 原子核物理评论, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
引用本文: 刘晖, 郭恒源, 周遥, JarahEvslin. ϕ4模型真空态间的转换算符[J]. 原子核物理评论, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
Hui LIU, Hengyuan GUO, Yao ZHOU, Evslin Jarah. Operator Changing Between Two Vacua of the ϕ4 Quantum Field Theory[J]. Nuclear Physics Review, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
Citation: Hui LIU, Hengyuan GUO, Yao ZHOU, Evslin Jarah. Operator Changing Between Two Vacua of the ϕ4 Quantum Field Theory[J]. Nuclear Physics Review, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018

ϕ4模型真空态间的转换算符

doi: 10.11804/NuclPhysRev.37.2020018
基金项目: 国家自然科学基金资助项目(11875296)
详细信息
    作者简介:

    刘晖(1994-),男,江苏盐城人,硕士,从事粒子物理与原子核物理研究,E-mail: liuhui@impcas.ac.cn

  • 中图分类号: O571

Operator Changing Between Two Vacua of the ϕ4 Quantum Field Theory

Funds: National Natural Science Foundation of China(111875296)
  • 摘要: 本工作在正则量子化的基础上,对$\phi^4$模型的哈密顿量采取正规序来正规化真空零点能,然后微扰计算了至次领头阶的真空态修正。同时首次得到可以在$\phi^4$模型的两个真空态之间转换的算符,我们相信这个算符也是适当极限条件下产生$\phi^4$扭结(kink)的算符的形式。最后简要说明了真空能的应用。
  • [1] DASHEN R F, HASSLACHER B, NEVEU A. Phys Rev D, 1974, 10: 4130. doi:  10.1103/PhysRevD.10.4130
    [2] REBHAN A, VAN NIEUWENHUIZEN P. Nucl Phys B, 1997, 508: 449. doi:  10.1016/S0550-3213(97)00625-1,10.1016/S0550-3213(97)80021-1
    [3] RAJARAMAN R. Phys Rept, 1975, 21: 227. doi:  10.1016/0370-1573(75)90016-2
    [4] LIU H, ZHOU Y, JARAH E. Eur Phys J C, 2020, 80: 357. doi:  10.1140/epjc/s10052-020-7934-6
    [5] JAFFE A M. Commun Math Phys, 1965, 1: 127. doi:  10.1007/BF01646496
    [6] GLIMM J, JAFFE A M. Phys Rev, 1968, 176: 1945. doi:  10.1103/PhysRev.176.1945
    [7] TAYLOR J G. Annals Phys, 1978, 115: 153. doi:  0.1016/0003-4916(78)90179-3
    [8] GUO H Y, JARAH E. J High Energ Phys, 2020, 2020: 140. doi:  10.1007/JHEP02(2020)140
    [9] LUTHER A. Phys Rev B, 1976, 14: 2153. doi:  110.1103/PhysRevB.14.2153
    [10] JARAH E. J High Energ Phys, 2019, 2019: 161. doi:  10.1007/JHEP11(2019)161
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出版历程
  • 收稿日期:  2020-03-12
  • 修回日期:  2020-06-30
  • 刊出日期:  2020-12-30

ϕ4模型真空态间的转换算符

doi: 10.11804/NuclPhysRev.37.2020018
    基金项目:  国家自然科学基金资助项目(11875296)
    作者简介:

    刘晖(1994-),男,江苏盐城人,硕士,从事粒子物理与原子核物理研究,E-mail: liuhui@impcas.ac.cn

  • 中图分类号: O571

摘要: 本工作在正则量子化的基础上,对$\phi^4$模型的哈密顿量采取正规序来正规化真空零点能,然后微扰计算了至次领头阶的真空态修正。同时首次得到可以在$\phi^4$模型的两个真空态之间转换的算符,我们相信这个算符也是适当极限条件下产生$\phi^4$扭结(kink)的算符的形式。最后简要说明了真空能的应用。

English Abstract

刘晖, 郭恒源, 周遥, JarahEvslin. ϕ4模型真空态间的转换算符[J]. 原子核物理评论, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
引用本文: 刘晖, 郭恒源, 周遥, JarahEvslin. ϕ4模型真空态间的转换算符[J]. 原子核物理评论, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
Hui LIU, Hengyuan GUO, Yao ZHOU, Evslin Jarah. Operator Changing Between Two Vacua of the ϕ4 Quantum Field Theory[J]. Nuclear Physics Review, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
Citation: Hui LIU, Hengyuan GUO, Yao ZHOU, Evslin Jarah. Operator Changing Between Two Vacua of the ϕ4 Quantum Field Theory[J]. Nuclear Physics Review, 2020, 37(4): 833-841. doi: 10.11804/NuclPhysRev.37.2020018
    • 场论中的非微扰方法是对微扰方法的补充,微扰方法是在真空附近对场做微扰处理的计算方法,这样当理论的真空出现简并时,微扰方法会导致我们丢失一些关于真空对称性的信息。这些信息只能通过非微扰的方法来得到。例如1+1维中$ \phi^4 $理论的孤子解。而要知道孤子的质量首先需要知道对应的真空的能量,这二者都有高阶修正。我们工作的最终目标是希望理解杨-米尔斯理论中的禁闭现象,而直接研究3+1维量子色动力学(QCD)中的禁闭太过复杂,但是很多2维有效理论同样有禁闭现象,因此我们可以从它们入手来研究禁闭的一些性质。孤子和杨-米尔斯色禁闭的关系类似于超导理论中的对偶迈森纳效应,两个孤子之间有可能产生对偶迈森纳效应,我们希望借此来理解QCD中的禁闭现象。为了了解量子场论中孤子的产生算符,我们从最简单的孤子入手,即1+1维$ \phi^4 $模型中的扭结。但是一直以来对于扭结的量子质量修正都有争议,而要知道扭结的质量就必须要知道真空能量的修正。

      非微扰场论的伟大成就之一是计算了$ \phi^4 $扭结质量的领头阶修正[1]。这个理论通过周期盒子以及固定有限数量的模态来实现正规化。很多人进行过这项计算,然而因为选用不同的重整化方案,他们得到了很多不同的结果。文献[2]中解释了计算结果的分歧来自于领头阶的修正依赖于重整化方案的选取。研究指出$ \phi^4 $模型的正规化和重整化不是必须的[3],我们采用正规序的哈密顿量[4],因为不涉及到重整化方案所以有希望能够无异议地计算出扭结的质量和扭结态本身。为实现这个目标,我们首先需要在相同阶微扰计算$ \phi^4 $的真空态以及它们的能量,本工作将对$ \phi^4 $的真空态展开详细的计算,同时得到从自由真空态到$ \phi^4 $ 模型的两个真空态各自的转换算符,并且在最后将这两个算符拼接得到$ \phi^4 $模型真空态间的转换算符。这个转换算符可以用来和我们后续工作中将要计算得到的中心在$ x_0 $点处的$ \phi^4 $扭结的产生算符做对比,我们希望后者在$ x_0\rightarrow \infty $时会约化到最后给出的算符$ \mathcal{O} $

      必须指出,在这种情况微扰理论不收敛[5],因此为了严格描述$ \phi^4 $模型,我们必须采用空间截断并且对哈密顿量进行卷积[6]。这里不考虑不收敛的问题,因为我们的最终目的是理解超对称理论,其核心问题中不会遇到发散的真空能,所以在下文的微扰计算中都不会采用空间截断。$ \phi^4 $扭结是一种十分简单的扭结,研究它能让我们知道如何处理扭结的高阶修正。对于后续的研究,我们关心的其实是BPS(Bogomol'nyi,Prasad和Sommerfield)扭结的质量。在超对称理论中,玻色子和费米子对于真空能的贡献是抵消的,所以不会出现圈修正。类似地,BPS扭结的质量也是完全由超对称代数对应的中心荷给出。

    • 首先我们给出克莱因-高登理论的拉氏量

      $$ \mathcal{L} = \frac{1}{2}\left(\rm\partial_\mu\phi\right)^2-\frac{1}{2}m^2\phi^2, $$ (1)

      这里的$ m $是质量;$ \phi $是场算符。克莱因-高登理论的真空态是自由真空,用$ |{0}\rangle $表示。

      现在我们考虑如下拉氏量

      $$ \begin{split} \mathcal{L} =& \frac{1}{2}\left(\partial_\mu\phi\right)^2+\frac{1}{4}m^2 \phi^2 -\frac{\lambda}{4}\phi^4-\frac{\lambda}{4} v^4\\ =& \frac{1}{2}\left(\partial_\mu\phi\right)^2-\frac{\lambda}{4}\left(\phi^2-v^2\right)^2, \end{split} $$ (2)

      这里$ m $$ \lambda $都是正数,$ m $与传统的质量参数相差一个$ \sqrt{2} $ 因子,后面我们将会看出$ m $的物理意义是平移后的场$ \phi\pm v $的质量。$ \lambda $$ \phi^4 $相互作用的耦合常数,表示相互作用的强弱。$ v $是常数$ \sqrt{\frac{m^2} {2\lambda}} $。因为我们讨论的是$ 1+1 $维情形,$ \phi $是无量纲的,$ \lambda $有量纲$ m^2 $$ v $也无量纲。接下来我们将以$ 1/v $作为微扰展开的系数。在这个$ \phi^4 $理论中有两个简并的真空态,它们分别位于$ \phi = v $$ \phi = -v $,我们分别记它们为$ |{+v}\rangle $$ |{-v}\rangle $

      虽然$ \phi $不是自由场,但是在薛定谔表象我们仍然可以对它进行傅立叶变换来定义湮灭算符$ a $和产生算符$ a^\dagger $

      $$ \begin{split} \phi(x) =& \int \frac{{\rm d} p}{2 \pi} \frac{1}{\sqrt{2 \omega_{p}}}\left(a_{p}+a_{-p}^{\dagger}\right) {\rm e}^{{\rm i} p x},\\ \pi(x) =& \int \frac{{\rm d} p}{2 \pi}(-i) \sqrt{\frac{\omega_{p}}{2}}\left(a_{p}-a_{-p}^{\dagger}\right) {\rm e}^{{\rm i} p x}, \end{split} $$ (3)

      其中:$ p $表示动量,且

      $$ \omega_p = \sqrt{m^2+p^2}{\text{。}} $$ (4)

      如此定义的$ a $作用在自由真空$ |{0}\rangle $上为0。从正则对易关系

      $$ [\phi(x),\pi(y)] = {\rm i}\delta(x-y) , $$ (5)

      我们可以得到

      $$ [a_p,a^\dagger_q] = 2\pi\delta(p-q){\text{。}} $$ (6)

      其中$ \delta $是狄拉克$ \delta $函数。

      现在我们希望找到可以将态$ |{0}\rangle $变换到态$ |{+v}\rangle $或者$ |{-v}\rangle $ 的算符。首先来看态$ |{+v}\rangle $

    • 平移算符$ \mathcal{D}_\alpha $的作用是把场$ \phi $的值平移$ \alpha $,它应当满足对易关系

      $$ \left[\phi,\mathcal{D}_\alpha\right] = \alpha \mathcal{D}_\alpha{\text{。}} $$ (7)

      只有当$ \mathcal{D}_\alpha $有如下形式时

      $$ \mathcal{D}_\alpha = {\rm e}^{{\rm -i}\alpha\int {\rm d}x \pi(x)} = {\rm e}^{\alpha \sqrt{m/2}\left(a_0^\dagger-a_0\right)}, $$ (8)

      对易关系才能满足,这里$ a_0 $$ p = 0 $$ a_p $的特殊情形。需要指出,这里的平移算符$ \mathcal{D} $既不是厄米算符也不是幺正算符。前面提到,我们从一开始就对哈密顿量做正规序(将在下面给出正规序的定义)。我们引入平移算符的目的是对哈密顿量做平移使其可以写成微扰展开的形式,理由如下:

      容易验证,平移算符$ \mathcal{D} $和正规序的算符函数之间有如下关系

      $$ : F\big[\pi(x), \phi(x)\big]: \mathcal{D}_{f} = \mathcal{D}_{f}: F\big[\pi(x), \phi(x)+f(x)\big], $$ (9)

      其中$ F(\pi(x),\; \phi(x)) $是任意的算符函数。::是正规序符号,它表示将所有的$ a^\dagger $放置在$ a $左边。将上式用在哈密顿量$ H $上,我们就能得到平移后的哈密顿量$ H' $

      $$ H^{\prime} = \mathcal{D}_{-v} H \mathcal{D}_{-v}^{-1}, \quad {\text{其中}}\mathcal{D}_{-v}^{-1} = \mathcal{D}_{+v}{\text{。}} $$ (10)

      $$ \begin{split} H^{\prime} =& \int {\rm d} x: \frac{1}{2}\left(\phi'\right)^{2}+\frac{1}{2} \pi^{2} +\frac{1}{2} m^{2} \phi^{2}+ \\& \frac{1}{v} \frac{m^{2}}{2} \phi^{3}+\frac{1}{v^{2}} \frac{m^{2}}{8} \phi^{4}:\,, \end{split} $$ (11)

      这里以及后面出现的$ \phi' $都表示对空间的求导而非时间。

    • 利用$ \phi^4 $的拉氏量,我们可以写出其对应的哈密顿量

      $$ \begin{split} H_{\phi^4} =& \int {\rm d}x :\frac{1}{2}{\phi'}^2 + \frac{1}{2}\pi^2 +\frac{\lambda}{4}\left(\phi^2-v^2\right)^2 :\\ = &\int {\rm d}x : \frac{1}{2}{\phi'}^2 + \frac{1}{2}\pi^2 +\frac{1}{2}m^2(\phi-v)^2+\\& \frac{1}{v}\frac{m^2}{2}(\phi-v)^3+\frac{1}{v^2}\frac{m^2}{8}(\phi-v)^4 :{\text{。}}\end{split} $$ (12)

      这里我们已经采用了正规序。现在看另一个哈密顿量

      $$ H' = \int {\rm d}x : \frac{1}{2}{\phi'}^2 + \frac{1}{2}\pi^2 +\frac{1}{2}m^2\phi^2+\frac{1}{v}\frac{m^2}{2}\phi^3+\frac{1}{v^2}\frac{m^2}{8}\phi^4 :{\text{。}} $$ (13)

      这两个哈密顿量的区别在于$ \phi $的场值平移了$ v $。我们把式(13)的基态记作$ |{g_1}\rangle $。现在我们有如下关系

      $$ |{+v}\rangle = \mathcal{D}_{+v}|{g_1}\rangle{\text{。}} $$ (14)

      我们已经知道了$ \mathcal{D}_\alpha $,现在的任务就变为找到一个把$ |{0}\rangle $转换成$ |{g_1}\rangle $的算符。下面开始微扰计算。

      首先我们改写式(13),

      $$ \begin{split} H' =& \int {\rm d}x {:\frac{1}{2}{\phi'}^2+\frac{1}{2}\pi^2+\frac{1}{2}m^2\phi^2:}+ \\& {\frac{1}{v}\int {\rm d}x :\frac{m^2}{2}\phi^3:}+{\frac{1}{v^2}\int {\rm d}x :\frac{m^2}{8}\phi^4:}\\ =& H_0+\frac{1}{v}H_1+\frac{1}{v^2}H_2{\text{。}}\end{split} $$ (15)

      $ H_0 $是克莱因-高登场的哈密顿量,后两项分别是一阶微 扰项$ O\left(\frac{1}{v}\right) $和二阶 微扰项$ O\left(\frac{1}{v^2}\right) $。因为我们现在只研究到第二阶,我们微扰地写出如下的薛定谔方程

      $$ \begin{split} & \left[H_0+\frac{1}{v}H_1 + \left(\frac{1}{v}\right)^2 H_2\right]\Bigg\{|{0}\rangle+\frac{1}{v}|{1}\rangle+ \left(\frac{1}{v}\right)^2|{2}\rangle+\mathcal{O}\left[\left(\frac{1}{v}\right)^3\right]\Bigg\}\\ =& \left\{E_0+\frac{1}{v}E_1 + \left(\frac{1}{v}\right)^2 E_2+\mathcal{O}\left[\left(\frac{1}{v}\right)^3\right]\right\}\left\{|{0}\rangle+\frac{1}{v}|{1}\rangle+ \left(\frac{1}{v}\right)^2|{2}\rangle+\mathcal{O}\left[\left(\frac{1}{v}\right)^3\right]\right\}{\text{。}}\end{split} $$ (16)

      这里$ \left(|{0}\rangle+\frac{1}{v}|{1}\rangle+\frac{1}{v^2}|{2}\rangle \right) $就是直到第二阶的$ |{g_1}\rangle $。需要注意这里微扰的适用条件是$ \frac{1}{v}\ll 1 $,即$ v\gg 1 $

      现在我们的任务就是用$ |{0}\rangle $来表示$ |{1}\rangle $$ |{2}\rangle $。应当注意的是,为了表达式的简洁,我们并没有对$ |{g_1}\rangle $采取归一化,我们的归一化方案是

      $$\left\langle {0|0} \right\rangle = 1,\;\;{\rm{ }}\left\langle {0|1} \right\rangle = \left\langle {0|2} \right\rangle = ... = \left\langle {0|n} \right\rangle = 0,$$ (17)

      这里$ |{n}\rangle $是第$ n $阶修正。

      第0阶

      首先我们看式(16)中只涉及到第$ 0 $阶项,我们得到的

      $$ H_0 |{0}\rangle = E_0 |{0}\rangle{\text{。}} $$ (18)

      因为$ H_0 = \int \frac{{\rm d}p}{2\pi}\omega_p a_p^\dagger a_p $,而其中$ a_p $作用到$ |{0}\rangle $得到0,所以$ E_0 = 0 $

      第1阶

      然后我们看式(16)涉及到$ O\left(\frac{1}{v}\right) $ 这一阶的项,

      $$ H_0|{1}\rangle+H_1|{0}\rangle = E_1|{0}\rangle{\text{。}} $$ (19)

      这里我们用到$ E_0 = 0 $。为解出$ |{1}\rangle $首先我们要去掉$ H_1 $的正规序。

      $$ \begin{split} H_1 = \frac{m^2}{2}\!\int {\rm d}x :\phi^3: =& \frac{m^2}{2}\!\int {\rm d}x \!\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\!\sqrt{2 \omega_{p_2}}\!\sqrt{2 \omega_{p_3}}}{\rm e}^{{\rm i}(p_1+p_2+p_3)x}\Big({a_{p_1}a_{p_2}a_{p_3}\!+\!a_{-p_2}^\dagger a_{p_1}a_{p_3}\!+\!a_{-p_1}^\dagger a_{p_2}a_{p_3}}\!+ \\ &a_{-p_1}^\dagger a_{-p_2}^\dagger a_{p_3}+a_{-p_3}^\dagger a_{p_1}a_{p_2}+a_{-p_3}^\dagger a_{-p_2}^\dagger a_{p_1}+a_{-p_3}^\dagger a_{-p_1}^\dagger a_{p_2}+a_{-p_3}^\dagger a_{-p_1}^\dagger a_{-p_2}^\dagger \Big){\text{。}} \end{split} $$ (20)

      然后式(19)变为

      $$\begin{split} H_0|{1}\rangle = E_1|{0}\rangle-\frac{m^2}{2}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger|{0}\rangle{\text{。}}\end{split}$$ (21)

      从式(21)我们可以知道

      $$ E_1 = 0, $$ (22)

      并且$ |{1}\rangle $有如下形式

      $$ |{1}\rangle = \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi} f(p_1,p_2,p_3) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger |{0}\rangle{\text{。}} $$ (23)

      这里$ f(p_1,p_2,p_3) $是一个待定函数,一旦知道$ f(p_1,p_2,p_3) $,我们就知道了$ |{1}\rangle $。因为$ H_0 = \int \frac{{\rm d}p}{2\pi}\omega_p a_p^\dagger a_p $,所以我们有

      $$ \begin{split}& \int \frac{{\rm d}p}{2\pi}\omega_p a_p^\dagger a_p \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi} f(p_1,p_2,p_3) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger |{0}\rangle = \\&\quad -\frac{m^2}{2}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger|{0}\rangle{\text{。}} \end{split} $$ (24)

      利用式(24)可以解出$ f(p_1,p_2,p_3) $

      $$ \begin{split} LHS =& \int \frac{{\rm d}p}{2\pi} \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\omega_p f(p_1,p_2,p_3)a_p^\dagger a_p a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger |{0}\rangle\\ =& \int\frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\left(\omega_{p_1}+\omega_{p_2}+\omega_{p_3}\right)f(p_1,p_2,p_3) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger |{0}\rangle = RHS{\text{。}} \end{split} $$ (25)

      计算过程中反复用到了式(6)的对易关系,把$ a_p $移到最右边后使其可以作用在$ |{0}\rangle $上得到0。所以,

      $$ \left(\omega_{p_1}+\omega_{p_2}+\omega_{p_3}\right) f(p_1,p_2,p_3) = -\frac{m^2}{2}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}} 2\pi \delta(p_1+p_2+p_3){\text{。}} $$ (26)

      然后

      $$ f(p_1,p_2,p_3) = -\frac{m^2}{2}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}} \frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3){\text{。}}$$ (27)

      $$ |{1}\rangle = -\frac{m^2}{2}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}} \frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger|{0}\rangle{\text{。}} $$ (28)

      第2阶

      现在我们继续往下一阶进行微扰计算。我们现在看式(16)中涉及到$ O\left(\frac{1}{v^2}\right) $这一阶的项,

      $$ H_0|{2}\rangle+H_1|{1}\rangle+H_2|{0}\rangle = E_2|{0}\rangle, $$ (29)

      改变项的顺序,

      $$ H_0|{2}\rangle = E_2|{0}\rangle-H_1|{1}\rangle-H_2|{0}\rangle{\text{。}} $$ (30)

      其中

      $$ H_2 = \frac{m^2}{8}\int {\rm d}x {:\phi^4:{\text{。}}} $$ (31)

      根据我们一阶计算的经验,我们只需要用到$ H_2 $里的一项,因为作用到$ |{0}\rangle $上后,大部分项都变为0.

      $$ \begin{split} H_2|{0}\rangle =& \frac{m^2}{8}\int {\rm d}x \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}p_4}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}} {\rm e}^{{\rm i}(p_1+p_2+p_3+p_4)x} a_{-p_1}^\dagger a_{-p_2}^\dagger a_{-p_3}^\dagger a_{-p_4}^\dagger|{0}\rangle \\ =& \frac{m^2}{8}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}p_4}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}}2\pi \delta(p_1+p_2+p_3+p_4) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger|{0}\rangle{\text{。}} \end{split} $$ (32)

      现在我们处理最为繁杂的项$ H_1|{1}\rangle $。它里面不含$ a^\dagger $的项有1个,含有2个$ a^\dagger $的项有3个,含有4个$ a^\dagger $的项有3个,含有6个$ a^\dagger $的项有1个。利用不含$ a^\dagger $的这一项我们可以算出二阶能量修正$ E_2 $,利用含有$ a^\dagger $的这些项我们可以算出$ |{2}\rangle $

      现在看$ H_1|{1}\rangle $中含有2个$ a^\dagger $ 的这3 项,这三项实际上是相等的,只是积分的哑指标不同,所以我们只需要看其中一项,首先我们计算$ a_{-p_2}^\dagger a_{p_1} a_{p_3} a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle $

      $$ \begin{split} a_{-p_2}^\dagger a_{p_1} a_{p_3} a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle = & \big[2\pi \delta(p_3-q_1) 2\pi \delta(p_1-q_2)a_{-p_2}^\dagger a_{q_3}^\dagger +2\pi \delta(p_3-q_1) 2\pi \delta(p_1-q_3)a_{-p_2}^\dagger a_{q_2}^\dagger+\\& 2\pi \delta(p_3-q_2) 2\pi \delta(p_1-q_1)a_{-p_2}^\dagger a_{q_3}^\dagger +2\pi \delta(p_3-q_2) 2\pi \delta(p_1-q_3)a_{-p_2}^\dagger a_{q_1}^\dagger+\\& 2\pi \delta(p_3-q_3) 2\pi \delta(p_1-q_1)a_{-p_2}^\dagger a_{q_2}^\dagger +2\pi \delta(p_3-q_3) 2\pi \delta(p_1-q_2)a_{-p_2}^\dagger a_{q_1}^\dagger \big]|{0}\rangle{\text{。}} \end{split} $$ (33)

      所以在$ H_1|{1}\rangle $中含有2个$ a^\dagger $ 的项是

      $$ \begin{split} &-3\times \frac{m^4}{4}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{q_3}}}\times \\& \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+q_3) a_{-p_2}^\dagger a_{p_1}a_{p_3} a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger|{0}\rangle \\=& -\frac{3}{4}m^4 \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{q_3}}}\times\\ & \frac{1}{\omega_{p_3}+\omega_{p_1}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_3+p_1+q_3) a_{-p_2}^\dagger a_{q_3}^\dagger|{0}\rangle-\\& \frac{3}{4}m^4 \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{p_1}}}\times\\ & \frac{1}{\omega_{p_3}+\omega_{q_2}+\omega_{p_1}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_3+q_2+p_1) a_{-p_2}^\dagger a_{q_2}^\dagger|{0}\rangle-\\ & \frac{3}{4}m^4 \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{q_3}}}\times \\& \frac{1}{\omega_{p_1}+\omega_{p_3}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_1+p_3+q_3) a_{-p_2}^\dagger a_{q_3}^\dagger|{0}\rangle-\end{split} $$
      $$ \begin{split} & \frac{3}{4}m^4 \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_1}}}\times \\ & \frac{1}{\omega_{q_1}+\omega_{p_3}+\omega_{p_1}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+p_3+p_1) a_{-p_2}^\dagger a_{q_1}^\dagger|{0}\rangle-\\& \frac{3}{4}m^4 \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{p_3}}}\times \\& \frac{1}{\omega_{p_1}+\omega_{q_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_1+q_2+p_3) a_{-p_2}^\dagger a_{q_3}^\dagger|{0}\rangle-\\& \frac{3}{4}m^4 \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_3}}} \times\\ & \frac{1}{\omega_{q_1}+\omega_{p_1}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+p_1+p_3) a_{-p_2}^\dagger a_{q_1}^\dagger|{0}\rangle-\\ =& \frac{9}{2}m^4\int \frac{{\rm d}p}{2\pi}\frac{{\rm d}q}{2\pi}\frac{1}{8\omega_p \omega_q \omega_{p+q}}\frac{1}{\omega_p +\omega_q +\omega_{p+q}}a_{-q}^\dagger a_q|{0}\rangle{\text{。}} \end{split} $$ (34)

      现在我们看$ H_1|{1}\rangle $中含有4个$ a^\dagger $的那3项。同样,这三项是相等的,我们只需计算$ a_{-p_1}^\dagger a_{-p_2}^\dagger a_{p_3} a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle $

      $$ \begin{split} &a_{-p_1}^\dagger a_{-p_2}^\dagger a_{p_3} a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle\\ =& \Big[2\pi \delta(p_3-q_1)a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger+2\pi \delta(p_3-q_2)a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_1}^\dagger a_{q_3}^\dagger+2\pi \delta(p_3-q_3)a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_1}^\dagger a_{q_2}^\dagger\Big]|{0}\rangle{\text{。}} \end{split}$$ (35)

      所以在$ H_1|{1}\rangle $中含有4个$ a^\dagger $ 的项是

      $$ \begin{split}& -3\times \frac{m^4}{4}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{q_3}}} \times\\ & \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+q_3) a_{-p_1}^\dagger a_{-p_2}^\dagger a_{p_3} a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle\\ =& -\frac{3}{4}m^4\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{q_3}}} \times \\& \frac{1}{\omega_{p_3}+\omega_{q_2}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_3+q_2+q_3) a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle-\\ &\frac{3}{4}m^4\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{q_3}}}\times \\& \frac{1}{\omega_{q_1}+\omega_{p_3}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+p_3+q_3) a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_1}^\dagger a_{q_3}^\dagger |{0}\rangle- \\& \frac{3}{4}m^4\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{p_3}}}\times \\ & \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+p_3) a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_1}^\dagger a_{q_2}^\dagger |{0}\rangle-\\ = &\frac{9}{4}m^4\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{p_3}}}\times \\ & \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+p_3) a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_1}^\dagger a_{q_2}^\dagger |{0}\rangle{\text{。}} \end{split} $$ (36)

      $ H_1|{1}\rangle $中含有6个$ a^\dagger $的项我们可以直接写出:

      $$ \begin{split}& -\frac{m^4}{4}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{dq_2}{2\pi}\frac{{\rm d}q_3}{2\pi} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{q_3}}}\times \\ & \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+q_3) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle{\text{。}} \end{split} $$ (37)

      现在我们可以改写式(30),

      $$ \begin{split} H_0|{2}\rangle =& \frac{9}{2}m^4\int \frac{{\rm d}p}{2\pi}\frac{{\rm d}q}{2\pi}\frac{1}{8\omega_p \omega_q \omega_{p+q}}\frac{1}{\omega_p +\omega_q +\omega_{p+q}}a_{-q}^\dagger a_q|{0}\rangle+\\& \frac{9}{4}m^4\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{p_3}}} \times\\ & \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+p_3) a_{-p_1}^\dagger a_{-p_2}^\dagger a_{q_1}^\dagger a_{q_2}^\dagger |{0}\rangle+\\& \frac{m^4}{4}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}q_1}{2\pi}\frac{{\rm d}q_2}{2\pi}\frac{{\rm d}q_3}{2\pi}\frac{1} {\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\sqrt{2 \omega_{q_1}}\sqrt{2 \omega_{q_2}}\sqrt{2 \omega_{q_3}}}\times \\ & \frac{1}{\omega_{q_1}+\omega_{q_2}+\omega_{q_3}} 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(q_1+q_2+q_3) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{q_1}^\dagger a_{q_2}^\dagger a_{q_3}^\dagger |{0}\rangle-\\& \frac{m^2}{8}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}p_4}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}} 2\pi \delta(p_1+p_2+p_3+p_4) a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger|{0}\rangle{\text{。}} \end{split} $$ (38)

      所以$ |{2}\rangle $应当具有如下形式

      $$ \begin{split} |{2}\rangle =& \bigg(\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}g(p_1,p_2)a_{p_1}^\dagger a_{p_2}^\dagger+\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}p_4}{2\pi} h(p_1,p_2,p_3,p_4)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger+\\& \int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{{\rm d}p_4}{2\pi}\frac{{\rm d}p_5}{2\pi}\frac{{\rm d}p_6}{2\pi} l(p_1,p_2,p_3,p_4,p_5,p_6)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger a_{p_5}^\dagger a_{p_6}^\dagger \bigg) |{0}\rangle{\text{。}}\end{split} $$ (39)

      将式(39)代回式(38),我们可以解出$ g(p_1,p_2) $$ h(p_1,p_2,p_3,p_4) $$ l(p_1, p_2,p_3,p_4,p_5,p_6) $。然后得到

      $$ \begin{split} |{2}\rangle =& \bigg \{{\frac{9}{4}m^4\int \frac{{\rm d}p}{2\pi}\frac{{\rm d}q}{2\pi} \frac{1}{\omega_p} \frac{1}{8\omega_{p+q} \omega_p \omega_q} \frac{1}{\omega_{p+q}+\omega_p+\omega_q}a_p^\dagger a_{-p}^\dagger}+\\& \int \frac{{\rm d}p_1}{2\pi} \frac{{\rm d}p_2}{2\pi} \frac{{\rm d}p_3}{2\pi} \frac{{\rm d}p_4}{2\pi} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}+\omega_{p_4}}\times\\ & \left[{\frac{9}{4}m^4\int \frac{{\rm d}q}{2\pi}\frac{1}{2\omega_q} \frac{1}{\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_q} 2\pi \delta(-p_1-p_2+q) 2\pi \delta(p_3+p_4+q)}- {\frac{m^2}{8} 2\pi \delta(p_1\!+\!p_2\!+\!p_3\!+\!p_4)} \right]a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger +\\& \int \frac{{\rm d}p_1 ... {\rm d}p_6}{2\pi^6} \frac{m^4}{4}\frac{1}{\omega_{p_1}\!+\!\omega_{p_2}\!+\!\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_{p_5}\!+\!\omega_{p_6}} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}\sqrt{2 \omega_{p_5}}\sqrt{2 \omega_{p_6}}} \frac{1}{\omega_{p_4}+\omega_{p_5}+\omega_{p_6}}\times\\ & 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_4+p_5+p_6)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger a_{p_5}^\dagger a_{p_6}^\dagger\bigg\} |{0}\rangle{\text{。}} \end{split} $$ (40)

      现在我们得到了直到二阶的$ |{g_1}\rangle $

      $$ \begin{split} |{g_1}\rangle = &\Bigg \{ 1-\frac{1}{v}\frac{m^2}{2}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger+\\& \frac{1}{v^2}\bigg \{{\frac{9}{4}m^4\int \frac{{\rm d}p}{2\pi}\frac{{\rm d}q}{2\pi} \frac{1}{\omega_p} \frac{1}{8\omega_{p+q} \omega_p \omega_q} \frac{1}{\omega_{p+q}+\omega_p+\omega_q}a_p^\dagger a_{-p}^\dagger}+\\& \int \frac{{\rm d}p_1}{2\pi} \frac{{\rm d}p_2}{2\pi} \frac{{\rm d}p_3}{2\pi} \frac{{\rm d}p_4}{2\pi} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}+\omega_{p_4}}\times\\& \left[{\frac{9}{4}m^4\int \frac{{\rm d}q}{2\pi}\frac{1}{2\omega_q} \frac{1}{\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_q} 2\pi \delta(-p_1-p_2+q) 2\pi \delta(p_3+p_4+q)}- {\frac{m^2}{8} 2\pi \delta(p_1\!+\!p_2\!+\!p_3\!+\!p_4)} \right]a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger +\\& \int \frac{{\rm d}p_1 ... {\rm d}p_6}{2\pi^6} \frac{m^4}{4}\frac{1}{\omega_{p_1}\!+\!\omega_{p_2}\!+\!\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_{p_5}\!+\!\omega_{p_6}}\times \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}\sqrt{2 \omega_{p_5}}\sqrt{2 \omega_{p_6}}} \frac{1}{\omega_{p_4}\!+\!\omega_{p_5}\!+\!\omega_{p_6}}\times\\& 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_4+p_5+p_6)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger a_{p_5}^\dagger a_{p_6}^\dagger\bigg\} \Bigg \} |{0}\rangle{\text{。}} \end{split} $$ (41)

      我们把所有大括号里面的项看作一个算符并记作$ \mathcal{O}' $。把我们需要找的算符,即把自由真空$ |{0}\rangle $变换为$ \phi^4 $真空$ |{+v}\rangle $的算符记作$ \mathcal{O}_1 $,我们便算出了$ \mathcal{O}_1 $$ \mathcal{O}_1 = \mathcal{D}_{+v} \mathcal{O}' $

    • 为了得到$ |{-v}\rangle $(另一个$ \phi^4 $真空态)需要把$ H_{\phi^4} $写成另一种形式

      $$ \begin{split} H_{\phi^4} =& \int {\rm d}x :\frac{1}{2}{\phi'}^2 + \frac{1}{2}\pi^2 +\frac{\lambda}{4}\left(\phi^2-v^2\right) :\\ =& \int {\rm d}x : \frac{1}{2}{\phi'}^2 + \frac{1}{2}\pi^2 +\frac{1}{2}m^2(\phi+v)^2 -\\&\frac{1}{v}\frac{m^2}{2}(\phi+v)^3+\frac{1}{v^2}\frac{m^2}{8}(\phi+v)^4 :{\text{。}} \end{split} $$ (42)

      类似的需要得到

      $$\begin{split} H =& \int {\rm d}x {:\frac{1}{2}{\phi'}^2+\frac{1}{2}\pi^2+\frac{1}{2}m^2\phi^2:}-\\&{\frac{1}{v}\int {\rm d}x :\frac{m^2}{2}\phi^3:}+\\&\frac{1}{v^2}\int {\rm d}x :\frac{m^2}{8}\phi^4: \end{split} $$ (43)

      对应的基态。我们把这个基态记作$ |{g_2}\rangle $,然后有$ |{-v}\rangle = \mathcal{D}_{-v}|{g_2}\rangle $。经过几乎与上面相同的计算,直到第2阶$ |{g_2}\rangle $有着与$ |{g_1}\rangle $几乎相同的表达式(只是$ \frac{1}{v} $这一项有 正负号的变化),

      $$ \begin{split} |{g_2}\rangle =& \Bigg \{ 1+\frac{1}{v}\frac{m^2}{2}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger+\\& \frac{1}{v^2}\bigg \{{\frac{9}{4}m^4\int \frac{{\rm d}p}{2\pi}\frac{{\rm d}q}{2\pi} \frac{1}{\omega_p} \frac{1}{8\omega_{p+q} \omega_p \omega_q} \frac{1}{\omega_{p+q}+\omega_p+\omega_q}a_p^\dagger a_{-p}^\dagger}+\\& \int \frac{{\rm d}p_1}{2\pi} \frac{{\rm d}p_2}{2\pi} \frac{{\rm d}p_3}{2\pi} \frac{{\rm d}p_4}{2\pi} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}+\omega_{p_4}}\times\\ & \left[{\frac{9}{4}m^4\int \frac{{\rm d}q}{2\pi}\frac{1}{2\omega_q} \frac{1}{\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_q} 2\pi \delta(-p_1-p_2+q) 2\pi \delta(p_3+p_4+q)}- {\frac{m^2}{8} 2\pi \delta(p_1\!+\!p_2\!+\!p_3\!+\!p_4)} \right]a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger +\\& \int \frac{{\rm d}p_1 ... {\rm d}p_6}{2\pi^6} \frac{m^4}{4}\frac{1}{\omega_{p_1}\!+\!\omega_{p_2}\!+\!\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_{p_5}\!+\!\omega_{p_6}} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}\sqrt{2 \omega_{p_5}}\sqrt{2 \omega_{p_6}}} \frac{1}{\omega_{p_4}+\omega_{p_5}+\omega_{p_6}}\times\\ & 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_4+p_5+p_6)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger a_{p_5}^\dagger a_{p_6}^\dagger\bigg\} \Bigg \} |{0}\rangle{\text{。}} \end{split} $$ (44)

      我们把大括号中的算符记作$ \mathcal{O}'' $,记把$ |{0}\rangle $变换到$ |{-v}\rangle $的算符为$ \mathcal{O}_2 $,于是,$ \mathcal{O}_2 = \mathcal{D}_{-v} \mathcal{O}'' $

      现在我们便得到了把$ \phi^4 $的一个真空态$ |{+v}\rangle $变换到另一个真空态$ |{-v}\rangle $ 的算符$ \mathcal{O} $ (即$ |{-v}\rangle = \mathcal{O}|{+v}\rangle $)。

      $$ \mathcal{O} = \mathcal{O}_2\mathcal{O}_1^{-1} = \mathcal{D}_{-v} \mathcal{O}'' \left(\mathcal{D}_{+v} \mathcal{O}'\right)^{-1} = \mathcal{D}_{-v} \mathcal{O}''\mathcal{O}'^{-1} \mathcal{D}_{-v}{\text{。}} $$ (45)

      为了简洁我们可以简写$ \mathcal{O}' $$ \mathcal{O}'' $

      $$ \mathcal{O}' = 1-\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\mathcal{B}, $$ (46)
      $$ \mathcal{O}'' = 1+\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\mathcal{B}{\text{。}} $$ (47)

      这里

      $$ \mathcal{A} = \frac{m^2}{2}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger, $$ (48)
      $$ \begin{split} \mathcal{B} =& {\frac{9}{4}m^4\int \frac{{\rm d}p}{2\pi}\frac{{\rm d}q}{2\pi} \frac{1}{\omega_p} \frac{1}{8\omega_{p+q} \omega_p \omega_q} \frac{1}{\omega_{p+q}+\omega_p+\omega_q}a_p^\dagger a_{-p}^\dagger}+\\ &\int \frac{{\rm d}p_1}{2\pi} \frac{{\rm d}p_2}{2\pi} \frac{{\rm d}p_3}{2\pi} \frac{{\rm d}p_4}{2\pi} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}+\omega_{p_4}}\times\\& \left[{\frac{9}{4}m^4\int \frac{{\rm d}q}{2\pi}\frac{1}{2\omega_q} \frac{1}{\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_q} 2\pi \delta(-p_1-p_2+q) 2\pi \delta(p_3+p_4+q)}- {\frac{m^2}{8} 2\pi \delta(p_1\!+\!p_2\!+\!p_3\!+\!p_4)} \right]a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger + \end{split} $$
      $$ \begin{split} & \int \frac{{\rm d}p_1 ... {\rm d}p_6}{2\pi^6} \frac{m^4}{4}\frac{1}{\omega_{p_1}\!+\!\omega_{p_2}\!+\!\omega_{p_3}\!+\!\omega_{p_4}\!+\!\omega_{p_5}\!+\!\omega_{p_6}} \frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}\sqrt{2 \omega_{p_4}}\sqrt{2 \omega_{p_5}}\sqrt{2 \omega_{p_6}}} \frac{1}{\omega_{p_4}+\omega_{p_5}+\omega_{p_6}}\times\\& 2\pi \delta(p_1+p_2+p_3) 2\pi \delta(p_4+p_5+p_6)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger a_{p_4}^\dagger a_{p_5}^\dagger a_{p_6}^\dagger{\text{。}} \end{split} $$ (49)

      然后,

      $$ \mathcal{O}'^{-1} = \left(1-\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\mathcal{B}\right)^{-1} = 1-\left(-\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\mathcal{B}\right)+\left(-\frac{1}{v} \mathcal{A}+\frac{1}{v^2}\mathcal{B}\right)^2 = 1+\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\left(\mathcal{A}^2-\mathcal{B}\right){\text{。}} $$ (50)

      最终得到

      $$ \begin{split} \mathcal{O} =& \mathcal{D}_{-v} \left(1+\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\mathcal{B}\right) \left[1+\frac{1}{v}\mathcal{A}+\frac{1}{v^2}\left(\mathcal{A}^2-\mathcal{B}\right)\right]\mathcal{D}_{-v}\\ =& \mathcal{D}_{-v}\left(1+\frac{2}{v}\mathcal{A}+\frac{2}{v^2}\mathcal{A}^2\right)\mathcal{D}_{-v}\\ =& e^{-v\alpha \sqrt{m/2}\left(a_0^\dagger-a_0\right)}\times \Bigg[1+\frac{m^2}{v}\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger+\\& \frac{m^4}{2v^2}\left(\int \frac{{\rm d}p_1}{2\pi}\frac{{\rm d}p_2}{2\pi}\frac{{\rm d}p_3}{2\pi}\frac{1}{\sqrt{2 \omega_{p_1}}\sqrt{2 \omega_{p_2}}\sqrt{2 \omega_{p_3}}}\frac{1}{\omega_{p_1}+\omega_{p_2}+\omega_{p_3}} 2\pi \delta(p_1+p_2+p_3)a_{p_1}^\dagger a_{p_2}^\dagger a_{p_3}^\dagger\right)^2\Bigg]\times e^{-v\alpha \sqrt{m/2}\left(a_0^\dagger-a_0\right)}{\text{。}} \end{split} $$ (51)

      本工作的主要动机是提供一个有效的公式来计算$ \phi^4 $扭结的高阶修正。如果扭结的中心在$ x_0 $点,并且扭结的形状因子[7]$ x\rightarrow -\infty $时的$ -v $ 平滑地过渡到$ x\rightarrow \infty $时的$ v $,那我们便可以期望在$ x_0\rightarrow \infty $时产生$ \phi^4 $扭结的算符会约化到$ \mathcal{O} $。这是因为中心在$ x_0 $处的扭结的产生算符会将$ x\ll x_0 $处场的真空期望值$ -v $变为$ x\gg x_0 $ 处场的真空期望值$ v $,而当$ x\rightarrow \infty $ 时,扭结的产生算符将从$ -\infty $$ \infty $ 取值的所有$ x $,即一维全空间的场的真空期望值从$ -v $变为$ v $,这正对应着本工作找出的$ \phi^4 $模型真空态之间的转换算符。

    • 在第2节我们看到真空能的第0阶和1阶修正都为0,因为本工作的主题是求出$ \phi^4 $真空态之间的转换算符,所以并未给出真空能的非零的2阶修正。不过对于后续的工作而言,知道真空能的大小是十分必要的,在这里我们简要地举例说明真空能的应用[8]。首先给出sine-Gordon哈密顿量

      $$ H = \int {\rm d} x \mathcal{H}(x), $$ (52)
      $$ \begin{split} \mathcal{H}(x) =& \frac{1}{2}: \pi(x) \pi(x):+\frac{1}{2}: \phi'(x) \phi'(x): \\& -\frac{m^{2}}{\lambda}:\big\{\cos [\sqrt{\lambda} \phi(x)]-1\big\}: \end{split} $$ (53)

      sine-Gordon模型有一系列简并真空,用$ |{0}\rangle_k $来表示真空期望值为$ \frac{2\pi}{\sqrt{\lambda}}k $ 的真空态,即

      $$ _k\langle 0|\phi| 0\rangle_{k} = \frac{2 \pi}{\sqrt{\lambda}} k, \; \; k \in \mathbb{Z}, $$ (54)

      $ E_{0,sG} $$ E_K $为该哈密顿量对应的真空态$ |{0}\rangle_k $和孤子基态$ |{K}\rangle $的本征值

      $$ H|0\rangle_{k} = E_{0,sG}|0\rangle_{k}, \; \; H|K\rangle = E_{K}|K\rangle, $$ (55)

      这里的$ E_{0,sG} $是sine-Gordon模型的真空能,$ K $是kink的缩写。孤子的质量由下式定义

      $$ M_{K} = E_{K}-E_{0,sG}{\text{。}} $$ (56)

      因此要知道sine-Gordon孤子的质量必须要首先知道sine-Gordon模型的真空能$ E_{0,sG} $. 而本工作中所用的微扰计算的方法正可以给出真空能。在我们考虑质量的一圈修正的情况下,$ E_{0,sG} = 0 $。后面的一圈修正的计算过程中使用了模态展开的方法,在这里我们直接展示其结果。对得到的质量的一圈修正的表达式数值求解,得出sine-Gordon孤子质量的一圈修正为$ -0.318\,25 $ m,这和利用sine-Gordon模型的可积性给出的结果[9]$ Q = -m/\pi $一致。

    • 本工作我们做了详细的场论微扰计算,得到了$ \phi^4 $模型的真空态以及不同真空态之间的转换算符。利用这个$ \phi^4 $真空态的微扰表达式,我们可以在领头阶量子修正的基础上[10]更进一步地计算出 扭结态本身以及它的质量。作为本工作可能推广的应用,我们注意到Higgs 场有着与$ \phi^4 $类似的形式,不同之处在于Higgs场是复数场并且是$ 3+1 $维的,因而本文用于计算$ \phi^4 $模型真空能和真空态的方法也可以用于Higgs物理中。

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