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量子場論(一):簡諧振子的正則量子化

2022-10-24 01:06 作者:我的世界-華汁  | 我要投稿


一維簡諧振子的哈密頓量為:

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

其中m是質(zhì)量,%5Comega是角頻率。第一項是動能項,第二項是勢能項。在量子力學中,把坐標與動量這對共軛量視為厄米算符,滿足正則對易關系:

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

現(xiàn)在構造兩個非厄米的無量綱算符:

%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稱為湮滅算符,%5Chat%20a%5E%5Cdagger稱為產(chǎn)生算符。兩者互為對方的厄米共軛算符。兩者的對易關系為:

%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

用這兩個算符來反表示坐標和動量算符:

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

從而,哈密頓算符就可以表達成:

%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

由對易關系得知,%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,稱為粒子數(shù)算符。對于任意量子態(tài)%7C%5Cpsi%5Crangle%2C%5Chat%20N的期待值非負:

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

%7Cn%5Crangle%5Chat%20N的本征態(tài),歸一化為%5Clangle%20n%7Cn%5Crangle%3D1.它滿足本征方程:

%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非負。利用對易子公式:

%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

推導出:

%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

這樣可以推導出:

%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

可見,%5Chat%20a%5E%5Cdagger%7Cn%5Crangle%5Chat%20a%7Cn%5Crangle都是%5Chat%20N的本征態(tài),本征值分別為n%2B1n-1.也就是說:

%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是歸一化常數(shù)。產(chǎn)生算符%5Chat%20a%5E%5Cdagger能把本征值為n的態(tài)變成本征值為n%2B1的態(tài),因此也稱為升算符。湮滅算符%5Chat%20a能把本征值為n的態(tài)變成本征值為n-1的態(tài),因此也稱為降算符。為了確定歸一化常數(shù),進行以下推導:

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

將歸一化常數(shù)取為正實數(shù),我們就得到:

%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的某個本征態(tài)%7Cn%5Crangle出發(fā),用降算符接連作用,得到本征值逐步減小的一系列本征態(tài):

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

本征值分別為:

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

由于n%5Cge0%2C?所以必定存在一個最小本征值n_0%2C它的本征態(tài)滿足:

%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

可見n_0%3D0%2C即:

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

反過來,從%7C0%5Crangle出發(fā),用升算符接連作用,得到本征值逐步增加的一系列本征態(tài):

%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

它們的本征值分別為:

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

所以,本征值n是非負整數(shù),是量子化的。本征態(tài)%7Cn%5Crangle可以表示為:

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

為了確定歸一化常數(shù)c_3%2C進行下面的運算:

%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

取歸一化常數(shù)為正實數(shù),則有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

從(1.7)也可以看出,%7Cn%5Crangle也是哈密頓算符的本征態(tài):

%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

相應的能量本征值為:

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

基態(tài)%7C0%5Crangle的能量本征值并不為零,而是E_0%3D%5Cfrac%5Comega2%2C稱為零點能。我們可以把%7C0%5Crangle看做真空態(tài),把n%3E0%7Cn%5Crangle看做含有n個聲子的激發(fā)態(tài)。粒子數(shù)算符描述聲子數(shù)。產(chǎn)生算符的作用是產(chǎn)生一個聲子,湮滅算符的作用是湮滅一個聲子。

量子場論(一):簡諧振子的正則量子化的評論 (共 條)

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