凝聚態(tài)場(chǎng)論常用公式（9）：Landau能級(jí)的對(duì)稱規(guī)范與代數(shù)解法

2023-03-26 19:51 作者:打電動(dòng)的阿偉嘻嘻嘻 0人讀過(guò) | 我要投稿

對(duì)無(wú)相互作用的二維電子氣，施加強(qiáng)磁場(chǎng)，同時(shí)限制對(duì)稱性為：旋轉(zhuǎn)對(duì)稱性.? ??

考慮對(duì)稱規(guī)范： $A_x%3D%5Cfrac%7BB%7D%7B2%7Dy%2C%5C%20A_y%3D-%5Cfrac%7BB%7D%7B2%7Dx%2C%5C%20A_z%3D0.$

可以驗(yàn)證其哈密頓量滿足旋轉(zhuǎn)不變性（證明略）：

$%5Chat%7BH%7D%3D%5Cfrac%7B1%7D%7B2m%7D%5B(%5Chat%7Bp%7D_x-%5Cfrac%7BeB%7D%7B2c%7Dy)%5E2%2B(%5Chat%7Bp%7D_y%2B%5Cfrac%7BeB%7D%7B2c%7Dx)%5E2%5D.$

我們使用代數(shù)解法求解，首先必須聲明的是，以下所有的計(jì)算必須化為? $x%2C%5Chat%7Bp%7D_x%2Cy%2C%5Chat%7Bp%7D_y$ ?的表象下才有意義，并利用? $%5Bx%2C%5Chat%7Bp%7D_x%5D%3D%5By%2C%5Chat%7Bp%7D_y%5D%3Di$ 才能得到算符量子化公式,但是一些計(jì)算也利用了復(fù)變函數(shù)偏導(dǎo)的技巧.

$a_0%3D(%5Cfrac%7B%5Chbar%20c%7D%7BeB%7D)%5E%7B%5Cfrac%7B1%7D%7B2%7D%7D%2Cz%3D%5Cfrac%7B1%7D%7B2a_0%7D(x%2Biy)%2C%5C%20z%5E%7B%5Cast%7D%3D%5Cfrac%7B1%7D%7B2a_0%7D(x-iy).$

利用復(fù)變函數(shù)導(dǎo)數(shù)定義（這里的計(jì)算需要仔細(xì)用定義去算，不是筆誤）：

$%5Cpartial_z%3Da_0%5B%5Cpartial_x-i%5Cpartial_y%5D%2C%5C%20%5Cpartial_%7Bz%5E%7B%5Cast%7D%7D%3Da_0%5B%5Cpartial_x%2Bi%5Cpartial_y%5D.$ (注意在位置動(dòng)量表象下 $(%5Cpartial_z)%5E%7B%5Cast%7D%3D-%5Cpartial_%7Bz%5E%7B%5Cast%7D%7D.$ )

定義能量升降算符 $a%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D(z%2B%5Cpartial_%7Bz%5E%7B%5Cast%7D%7D)%2Ca%5E%7B%5Cdagger%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D(z%5E%7B%5Cast%7D-%5Cpartial_z)%2C%5C%20%5Chat%7Bn%7D%3Da%5E%7B%5Cdagger%7Da.$

可以計(jì)算得到 $%5Chat%7BH%7D%3D%5Chbar%5Comega_c(%5Chat%7Bn%7D%2B%5Cfrac%7B1%7D%7B2%7D)%2C%5C%20%5Ba%2Ca%5E%7B%5Cdagger%7D%5D%3D1.$

類比能量升降算符，定義z方向角動(dòng)量升降算符 $b%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D(z%5E%7B%5Cast%7D%2B%5Cpartial_z)%2C%5C%20b%5E%7B%5Cdagger%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D(z-%5Cpartial_%7Bz%5E%7B%5Cast%7D%7D).$

可以計(jì)算得到 $%5Bb%2Cb%5E%7B%5Cdagger%7D%5D%3D1%2C%5Bb%2Ca%5E%7B%5Cdagger%7D%5D%3D%5Bb%2Ca%5D%3D0.$

設(shè)其本征態(tài)為 $%5Cpsi_%7Bnm%7D%3A%5C%20%5Chat%7BH%7D%5Cpsi_%7Bnm%7D%3D%5Chbar%5Comega_c(n%2B%5Cfrac%7B1%7D%7B2%7D)%5Cpsi_%7Bnm%7D%2C%5C%20b%5E%7B%5Cdagger%7Db%5Cpsi_%7Bnm%7D%3Dm%5Cpsi_%7Bnm%7D.$

利用代數(shù)方法求其本征態(tài)的表達(dá)式：

$a%5Cpsi_%7B0m%7D%3D0%5CRightarrow(z%2B%5Cpartial_%7Bz%5E%7B%5Cast%7D%7D)%5Cpsi_%7B0m%7D%3D0%5CRightarrow%5Cpsi_%7B00%7D%3D%7B%5Crm%20Const%7D%5C%20e%5E%7B-zz%5E%7B%5Cast%7D%7D.$

$%5Cpsi_%7B0m%7D%3D%7B%5Crm%20Const%7D%5C%20(b%5E%7B%5Cdagger%7D)%5Em%5Cpsi_%7B00%7D%3D%7B%5Crm%20Const%7D%5C%20z%5Eme%5E%7B-zz%5E%7B%5Cast%7D%7D.$

考慮z方向的角動(dòng)量算符? $%5Chat%7BL%7D_z%3D-i%5Chbar%5Bx%5Cpartial_y-y%5Cpartial_x%5D%3D%5Chbar%5Bz%5Cpartial_z-z%5E%7B%5Cast%7D%5Cpartial_%7Bz%5E%7B%5Cast%7D%7D%5D.$

可以計(jì)算得到 $%5Chat%7BL%7D_z%5Cpsi_%7B0m%7D%3D%5Chbar%20m%5Cpsi_%7B0m%7D.$

設(shè)m的上限M,接下來(lái)求解m=M態(tài)，此過(guò)程實(shí)際上是確定m的上界，此物理圖像極其重要！此物理圖像極其重要！此物理圖像極其重要！

考慮m=M態(tài)的? $%5Cfrac%7B1%7D%7B4a_0%5E2%7D%3Cx%5E2%2By%5E2%3E%3D%5Cfrac%7B%5Cint(zz%5E%7B%5Cast%7D)%5EMe%5E%7B-2zz%5E%7B%5Cast%7D%7Dz%5E%7B%5Cast%7Dz%5C%20dz%5C%20dz%5E%7B%5Cast%7D%7D%7B%5Cint(zz%5E%7B%5Cast%7D)%5EMe%5E%7B-2zz%5E%7B%5Cast%7D%7D%5C%20dz%5C%20dz%5E%7B%5Cast%7D%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CGamma(M%2B2)%7D%7B%5CGamma(M%2B1)%7D%3D%5Cfrac%7BM%2B1%7D%7B2%7D%2C$

$%E5%85%B6%E4%B8%AD%E5%88%A9%E7%94%A8%E4%BA%86zz%5E%7B%5Cast%7D%3D(x%5E2%2By%5E2)%3Dr%5E2%5Cin(0%2C%2B%5Cinfty).$

$M%5Cgg1%E6%97%B6%2C%5C%20M%3D%5Cfrac%7BB%5Ccdot%20%7B%5Crm%20Area%7D%7D%7B%5Cfrac%7Bhc%7D%7Be%7D%7D%3D%5Cfrac%7B%5CPhi%7D%7B%5CPhi_0%7D.$

可以驗(yàn)證 $b%5E%7B%5Cdagger%7Db%3Da%5E%7B%5Cdagger%7Da%2B%5Cfrac%7B1%7D%7B%5Chbar%7D%5Chat%7BL%7D_z%2C%5C%20%5Chat%7BL_z%7D%5Cpsi_%7Bnm%7D%3D%5Chbar(m-n)%5Cpsi_%7Bnm%7D%2C%5C%20%5B%5Chat%7BL%7D_z%2Cb%5D%3D-%5Chbar%20b%2C%5C%20%5B%5Chat%7BL%7D_z%2Cb%5E%7B%5Cdagger%7D%5D%3D%5Chbar%20b%5E%7B%5Cdagger%7D.$