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量子场论(一):简谐振子的正则量子化

2022-10-24 01:06 作者:我的世界-华汁  | 我要投稿


一维简谐振子的哈密顿量为:

H%3D%5Cfrac%7Bp%5E2%7D%7B2m%7D%2B%5Cfrac12m%5Comega%5E2x%5E2.%5Ctag%7B1.1%7D

其中m是质量,%5Comega是角频率。第一项是动能项,第二项是势能项。在量子力学中,把坐标与动量这对共轭量视为厄米算符,满足正则对易关系:

%5Bx%2C%5Chat%20p%5D%3Di.%5Ctag%7B1.2%7D

现在构造两个非厄米的无量纲算符:

%5Chat%20a%3D%5Cfrac1%7B%5Csqrt%7B2m%5Comega%7D%7D(m%5Comega%20x%2Bi%5Chat%20p)%2C%5Chat%20a%5E%5Cdagger%3D%5Cfrac1%7B%5Csqrt%7B2m%5Comega%7D%7D(m%5Comega%20x-i%5Chat%20p).%5Ctag%7B1.3%7D

%5Chat%20a称为湮灭算符,%5Chat%20a%5E%5Cdagger称为产生算符。两者互为对方的厄米共轭算符。两者的对易关系为:

%5B%5Chat%20a%2C%5Chat%20a%5E%5Cdagger%5D%3D%5Cfrac1%7B2m%5Comega%7D%5Bm%5Comega%20x%2Bi%5Chat%20p%2Cm%5Comega%20x-i%5Chat%20p%5D%3D%5Cfrac1%7B2m%5Comega%7D(%5Bm%5Comega%20x%2C-i%5Chat%20p%5D%2B%5Bi%5Chat%20p%2Cm%5Comega%20x%5D)%5C%5C%3D%5Cfrac12(-i%5Bx%2C%5Chat%20p%5D%2Bi%5B%5Chat%20p%2Cx%5D)%3D-i%5Bx%2C%5Chat%20p%5D%3D1%5Ctag%7B1.4%7D

用这两个算符来反表示坐标和动量算符:

x%3D%5Cfrac1%7B%5Csqrt%7B2m%5Comega%7D%7D(%5Chat%20a%2B%5Chat%20a%5E%5Cdagger)%2C%5Chat%20p%3D-i%5Csqrt%7B%5Cfrac%7Bm%5Comega%7D2%7D(%5Chat%20a-%5Chat%20a%5E%5Cdagger).%5Ctag%7B1.5%7D

从而,哈密顿算符就可以表达成:

%5Chat%20H%3D-%5Cfrac1%7B2m%7D%5Cfrac%7Bm%5Comega%7D2(%5Chat%20a-%5Chat%20a%5E%5Cdagger)%5E2%2B%5Cfrac12m%5Comega%5E2%5Cfrac1%7B2m%5Comega%7D(%5Chat%20a%2B%5Chat%20a%5E%5Cdagger)%5E2%5C%5C%3D-%5Cfrac%5Comega%204(%5Chat%20a%5Chat%20a-%5Chat%20a%5Chat%20a%5E%5Cdagger-%5Chat%20a%5E%5Cdagger%5Chat%20a%2B%5Chat%20a%5E%5Cdagger%5Chat%20a%5E%5Cdagger)%2B%5Cfrac%5Comega%204(%5Chat%20a%5Chat%20a%2B%5Chat%20a%5Chat%20a%5E%5Cdagger%2B%5Chat%20a%5E%5Cdagger%5Chat%20a%2B%5Chat%20a%5E%5Cdagger%5Chat%20a%5E%5Cdagger)%5C%5C%3D%5Cfrac%5Comega%202(%5Chat%20a%5Chat%20a%5E%5Cdagger%2B%5Chat%20a%5E%5Cdagger%5Chat%20a).%5Ctag%7B1.6%7D

由对易关系得知,%5Chat%20a%5Chat%20a%5E%5Cdagger%3D%5Chat%20a%5E%5Cdagger%5Chat%20a%2B1%2C于是:

%5Chat%20H%3D%5Cfrac%5Comega%202(2%5Chat%20a%5E%5Cdagger%5Chat%20a%2B1)%3D%5Comega(%5Chat%20a%5E%5Cdagger%5Chat%20a%2B%5Cfrac12)%3D%5Comega(%5Chat%20N%2B%5Cfrac12).%5Ctag%7B1.7%7D

其中%5Chat%20N%5Cequiv%5Chat%20a%5E%5Cdagger%5Chat%20a是厄米算符,称为粒子数算符。对于任意量子态%7C%5Cpsi%5Crangle%2C%5Chat%20N的期待值非负:

%5Clangle%5Cpsi%7C%5Chat%20N%7C%5Cpsi%5Crangle%3D%5Clangle%5Cpsi%7C%5Chat%20a%5E%5Cdagger%5Chat%20a%7C%5Cpsi%5Crangle%3D%5Clangle%5Chat%20a%5Cpsi%7C%5Chat%20a%5Cpsi%5Crangle%5Cge0.%5Ctag%7B1.8%7D

因此,哈密顿算符是正定算符,%5Clangle%5Cpsi%7C%5Chat%20H%7C%5Cpsi%5Crangle%3E0.

%7Cn%5Crangle%5Chat%20N的本征态,归一化为%5Clangle%20n%7Cn%5Crangle%3D1.它满足本征方程:

%5Chat%20N%7Cn%5Crangle%3Dn%7Cn%5Crangle.%5Ctag%7B1.9%7D

由于n%3D%5Clangle%20n%7Cn%7Cn%5Crangle%3D%5Clangle%20n%7C%5Chat%20N%7Cn%5Crangle%5Cge0%2C所以本征值n非负。利用对易子公式:

%5BAB%2CC%5D%3DA%5BB%2CC%5D%2B%5BA%2CC%5DB.%5Ctag%7B1.10%7D

%5BA%2CBC%5D%3D%5BA%2CB%5DC%2BB%5BA%2CC%5D.%5Ctag%7B1.11%7D

推导出:

%5B%5Chat%20N%2C%5Chat%20a%5E%5Cdagger%5D%3D%5B%5Chat%20a%5E%5Cdagger%5Chat%20a%2C%5Chat%20a%5E%5Cdagger%5D%3D%5Chat%20a%5E%5Cdagger%5B%5Chat%20a%2C%5Chat%20a%5E%5Cdagger%5D%3D%5Chat%20a%5E%5Cdagger%2C%5B%5Chat%20N%2C%5Chat%20a%5D%3D%5B%5Chat%20a%5E%5Cdagger%5Chat%20a%2C%5Chat%20a%5D%3D%5B%5Chat%20a%5E%5Cdagger%2C%5Chat%20a%5D%5Chat%20a%3D-%5Chat%20a.%5Ctag%7B1.12%7D

因此,有:

%5Chat%20N%5Chat%20a%5E%5Cdagger%3D%5Chat%20a%5E%5Cdagger%5Chat%20N%2B%5Chat%20a%5E%5Cdagger%2C%5Chat%20N%5Chat%20a%3D%5Chat%20a%5Chat%20N-%5Chat%20a.%5Ctag%7B1.13%7D

这样可以推导出:

%5Chat%20N%5Chat%20a%5E%5Cdagger%7Cn%5Crangle%3D(%5Chat%20a%5E%5Cdagger%5Chat%20N%2B%5Chat%20a%5E%5Cdagger)%7Cn%5Crangle%3D(n%2B1)%5Chat%20a%5E%5Cdagger%7Cn%5Crangle.%5Ctag%7B1.14%7D

%5Chat%20N%5Chat%20a%7Cn%5Crangle%3D(%5Chat%20a%5Chat%20N-%5Chat%20a)%7Cn%5Crangle%3D(n-1)%5Chat%20a%7Cn%5Crangle.%5Ctag%7B1.15%7D

可见,%5Chat%20a%5E%5Cdagger%7Cn%5Crangle%5Chat%20a%7Cn%5Crangle都是%5Chat%20N的本征态,本征值分别为n%2B1n-1.也就是说:

%5Chat%20a%5E%5Cdagger%7Cn%5Crangle%3Dc_1%7Cn%2B1%5Crangle%2C%5Chat%20a%7Cn%5Crangle%3Dc_2%7Cn-1%5Crangle.%5Ctag%7B1.16%7D

其中c_1c_2是归一化常数。产生算符%5Chat%20a%5E%5Cdagger能把本征值为n的态变成本征值为n%2B1的态,因此也称为升算符。湮灭算符%5Chat%20a能把本征值为n的态变成本征值为n-1的态,因此也称为降算符。为了确定归一化常数,进行以下推导:

n%2B1%3D%5Clangle%20n%7C%5Chat%20N%2B1%7Cn%5Crangle%3D%5Clangle%20n%7C%5Chat%20a%5E%5Cdagger%5Chat%20a%2B1%7Cn%5Crangle%3D%5Clangle%20n%7C%5Chat%20a%5Chat%20a%5E%5Cdagger%7Cn%5Crangle%3D%7Cc_1%7C%5E2%5Clangle%20n%2B1%7Cn%2B1%5Crangle%3D%7Cc_1%7C%5E2.%5Ctag%7B1.17%7D

n%3D%5Clangle%20n%7C%5Chat%20N%7Cn%5Crangle%3D%5Clangle%20n%7C%5Chat%20a%5E%5Cdagger%5Chat%20a%7Cn%5Crangle%3D%7Cc_2%7C%5E2%5Clangle%20n-1%7Cn-1%5Crangle%3D%7Cc_2%7C%5E2.%5Ctag%7B1.18%7D

将归一化常数取为正实数,我们就得到:

%5Chat%20a%5E%5Cdagger%7Cn%5Crangle%3D%5Csqrt%7Bn%2B1%7D%7Cn%2B1%5Crangle%2C%5Chat%20a%7Cn%5Crangle%3D%5Csqrt%20n%7Cn-1%5Crangle.%5Ctag%7B1.19%7D

%5Chat%20N的某个本征态%7Cn%5Crangle出发,用降算符接连作用,得到本征值逐步减小的一系列本征态:

%5Chat%20a%7Cn%5Crangle%2C%5Chat%20a%5E2%7Cn%5Crangle%2C%5Chat%20a%5E3%7Cn%5Crangle%2C%E2%80%A6.%5Ctag%7B1.20%7D

本征值分别为:

n-1%2Cn-2%2Cn-3%2C%E2%80%A6.%5Ctag%7B1.21%7D

由于n%5Cge0%2C 所以必定存在一个最小本征值n_0%2C它的本征态满足:

%5Chat%20a%7Cn_0%5Crangle%3D0.%5Ctag%7B1.22%7D

于是:

%5Chat%20N%7Cn_0%5Crangle%3D%5Chat%20a%5E%5Cdagger%5Chat%20a%7Cn_0%5Crangle%3D0%3D0%7Cn_0%5Crangle.%5Ctag%7B1.23%7D

可见n_0%3D0%2C即:

%7Cn_0%5Crangle%3D%7C0%5Crangle.%5Ctag%7B1.24%7D

反过来,从%7C0%5Crangle出发,用升算符接连作用,得到本征值逐步增加的一系列本征态:

%5Chat%20a%5E%5Cdagger%7C0%5Crangle%2C(%5Chat%20a%5E%5Cdagger)%5E2%7C0%5Crangle%2C(%5Chat%20a%5E%5Cdagger)%5E3%7C0%5Crangle%2C%E2%80%A6.%5Ctag%7B1.25%7D

它们的本征值分别为:

1%2C2%2C3%2C%E2%80%A6.%5Ctag%7B1.26%7D

所以,本征值n是非负整数,是量子化的。本征态%7Cn%5Crangle可以表示为:

%7Cn%5Crangle%3Dc_3(%5Chat%20a%5E%5Cdagger)%5En%7C0%5Crangle.%5Ctag%7B1.27%7D

为了确定归一化常数c_3%2C进行下面的运算:

%0A%5Clangle%20n%7Cn%5Crangle%3D%7Cc_3%7C%5E2%5Clangle0%7C%5Chat%20a%5En(%5Chat%20a%5E%5Cdagger)%5En%7C0%5Crangle%3D1%5Ccdot%7Cc_3%7C%5E2%5Clangle1%7C%5Chat%20a%5E%7Bn-1%7D(%5Chat%20a%5E%5Cdagger)%5E%7Bn-1%7D%7C1%5Crangle%3D1%5Ccdot2%5Ccdot%7Cc_3%7C%5E2%5Clangle2%7C%5Chat%20a%5E%7Bn-2%7D(%5Chat%20a%5E%5Cdagger)%5E%7Bn-2%7D%7C2%5Crangle%3D%E2%80%A6%5C%5C%3D(n-1)!%7Cc_3%7C%5E2%5Clangle%20n-1%7C%5Chat%20a(%5Chat%20a%5E%5Cdagger)%7Cn-1%5Crangle%3Dn!%7Cc_3%7C%5E2%5Clangle%20n%7Cn%5Crangle.%5Ctag%7B1.28%7D

取归一化常数为正实数,则有c_3%3D%5Cfrac1%7B%5Csqrt%7Bn!%7D%7D%2C于是:

%0A%7Cn%5Crangle%3D%5Cfrac1%7B%5Csqrt%7Bn!%7D%7D(%5Chat%20a%5E%5Cdagger)%5En%7C0%5Crangle.%5Ctag%7B1.29%7D

从(1.7)也可以看出,%7Cn%5Crangle也是哈密顿算符的本征态:

%5Chat%20H%7Cn%5Crangle%3D%5Comega(%5Chat%20N%2B%5Cfrac12)%7Cn%5Crangle%3D%5Comega(n%2B%5Cfrac12)%7Cn%5Crangle%3DE_n%7Cn%5Crangle.%5Ctag%7B1.30%7D

相应的能量本征值为:

E_n%3D%5Comega(n%2B%5Cfrac12).%5Ctag%7B1.31%7D

基态%7C0%5Crangle的能量本征值并不为零,而是E_0%3D%5Cfrac%5Comega2%2C称为零点能。我们可以把%7C0%5Crangle看做真空态,把n%3E0%7Cn%5Crangle看做含有n个声子的激发态。粒子数算符描述声子数。产生算符的作用是产生一个声子,湮灭算符的作用是湮灭一个声子。

量子场论(一):简谐振子的正则量子化的评论 (共 条)

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