東北大学 工学研究科 電気・情報系 2016年8月実施 基礎科目 問題6 物理基礎2 Author 祭音Myyura (co-authored with GPT 5.6 SOL)
Description 日本語版
行列の交換関係に関して，次の問に答えよ。交換関係とは，[ A , B ] = A B − B A [A,B]=AB-BA [ A , B ] = A B − B A A A A B B B i i i O O O f ( t ) = exp ( t A ) B exp ( − t A ) , g ( λ ) = exp ( λ A ) exp ( λ B ) f(t)=\exp(tA)B\exp(-tA),\ g(\lambda)=\exp(\lambda A)\exp(\lambda B) f ( t ) = exp ( t A ) B exp ( − t A ) , g ( λ ) = exp ( λ A ) exp ( λ B ) A A A B B B t t t λ \lambda λ
exp ( X ) = ∑ n = 0 ∞ 1 n ! X n . \exp(X)=\sum_{n=0}^{\infty}\frac1{n!}X^n. exp ( X ) = n = 0 ∑ ∞ n ! 1 X n . ここで，X X X X 0 X^0 X 0 I I I
(1)
α = 1 2 ( 0 1 0 1 0 1 0 1 0 ) , β = 1 2 ( 0 − i 0 i 0 − i 0 i 0 ) , γ = ( 1 0 0 0 0 0 0 0 − 1 ) \alpha=\frac1{\sqrt2}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix},\quad
\beta=\frac1{\sqrt2}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix},\quad
\gamma=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix} α = 2 1 0 1 0 1 0 1 0 1 0 , β = 2 1 0 i 0 − i 0 i 0 − i 0 , γ = 1 0 0 0 0 0 0 0 − 1 とする。交換関係 [ α , β ] [\alpha,\beta] [ α , β ] [ α , ( α 2 + β 2 + γ 2 ) ] [\alpha,(\alpha^2+\beta^2+\gamma^2)] [ α , ( α 2 + β 2 + γ 2 )]
(2) f ( 0 ) f(0) f ( 0 )
(3) d d t exp ( t A ) = A exp ( t A ) = exp ( t A ) A \frac d{dt}\exp(tA)=A\exp(tA)=\exp(tA)A d t d exp ( t A ) = A exp ( t A ) = exp ( t A ) A
(4) 下記二つの関係式を示せ：
f ′ ( t ) = d d t f ( t ) = exp ( t A ) [ A , B ] exp ( − t A ) , f'(t)=\frac d{dt}f(t)=\exp(tA)[A,B]\exp(-tA), f ′ ( t ) = d t d f ( t ) = exp ( t A ) [ A , B ] exp ( − t A ) , f ′ ′ ( t ) = d d t f ′ ( t ) = exp ( t A ) [ A , [ A , B ] ] exp ( − t A ) . f''(t)=\frac d{dt}f'(t)=\exp(tA)[A,[A,B]]\exp(-tA). f ′′ ( t ) = d t d f ′ ( t ) = exp ( t A ) [ A , [ A , B ]] exp ( − t A ) . (5) [ A , [ A , B ] ] = [ B , [ A , B ] ] = O [A,[A,B]]=[B,[A,B]]=O [ A , [ A , B ]] = [ B , [ A , B ]] = O exp ( t A ) B exp ( − t A ) \exp(tA)B\exp(-tA) exp ( t A ) B exp ( − t A ) f ( t ) f(t) f ( t )
f ( t ) = f ( 0 ) + t 1 ! f ′ ( 0 ) + t 2 2 ! f ′ ′ ( 0 ) + t 3 3 ! f ( 3 ) ( 0 ) + ⋯ . f(t)=f(0)+\frac t{1!}f'(0)+\frac{t^2}{2!}f''(0)+\frac{t^3}{3!}f^{(3)}(0)+\cdots. f ( t ) = f ( 0 ) + 1 ! t f ′ ( 0 ) + 2 ! t 2 f ′′ ( 0 ) + 3 ! t 3 f ( 3 ) ( 0 ) + ⋯ . (6) [ A , [ A , B ] ] = [ B , [ A , B ] ] = O [A,[A,B]]=[B,[A,B]]=O [ A , [ A , B ]] = [ B , [ A , B ]] = O [ exp ( λ A ) , B ] = − λ [ B , A ] exp ( λ A ) [\exp(\lambda A),B]=-\lambda[B,A]\exp(\lambda A) [ exp ( λ A ) , B ] = − λ [ B , A ] exp ( λ A ) d d λ g ( λ ) = ( A + B + λ [ A , B ] ) g ( λ ) \frac d{d\lambda}g(\lambda)=(A+B+\lambda[A,B])g(\lambda) d λ d g ( λ ) = ( A + B + λ [ A , B ]) g ( λ )
(7) [ A , [ A , B ] ] = [ B , [ A , B ] ] = O [A,[A,B]]=[B,[A,B]]=O [ A , [ A , B ]] = [ B , [ A , B ]] = O exp ( A ) exp ( B ) = exp ( A + B + 1 2 [ A , B ] ) \exp(A)\exp(B)=\exp\left(A+B+\frac12[A,B]\right) exp ( A ) exp ( B ) = exp ( A + B + 2 1 [ A , B ] )
题目描述
矩阵交换子定义为 [ A , B ] = A B − B A [A,B]=AB-BA [ A , B ] = A B − B A e X = ∑ n ≥ 0 X n / n ! e^X=\sum_{n\ge0}X^n/n! e X = ∑ n ≥ 0 X n / n !
f ( t ) = e t A B e − t A , g ( λ ) = e λ A e λ B , f(t)=e^{tA}Be^{-tA},\qquad g(\lambda)=e^{\lambda A}e^{\lambda B}, f ( t ) = e t A B e − t A , g ( λ ) = e λ A e λ B , A , B A,B A , B t , λ t,\lambda t , λ O O O
给定
α = 1 2 ( 0 1 0 1 0 1 0 1 0 ) , β = 1 2 ( 0 − i 0 i 0 − i 0 i 0 ) , γ = ( 1 0 0 0 0 0 0 0 − 1 ) , \alpha=\frac1{\sqrt2}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix},\quad
\beta=\frac1{\sqrt2}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix},\quad
\gamma=\begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}, α = 2 1 0 1 0 1 0 1 0 1 0 , β = 2 1 0 i 0 − i 0 i 0 − i 0 , γ = 1 0 0 0 0 0 0 0 − 1 , 求 [ α , β ] [\alpha,\beta] [ α , β ] [ α , α 2 + β 2 + γ 2 ] [\alpha,\alpha^2+\beta^2+\gamma^2] [ α , α 2 + β 2 + γ 2 ] f ( 0 ) f(0) f ( 0 ) d d t e t A = A e t A = e t A A \frac d{dt}e^{tA}=Ae^{tA}=e^{tA}A d t d e t A = A e t A = e t A A f ′ ( t ) = e t A [ A , B ] e − t A f'(t)=e^{tA}[A,B]e^{-tA} f ′ ( t ) = e t A [ A , B ] e − t A f ′ ′ ( t ) = e t A [ A , [ A , B ] ] e − t A f''(t)=e^{tA}[A,[A,B]]e^{-tA} f ′′ ( t ) = e t A [ A , [ A , B ]] e − t A [ A , [ A , B ] ] = [ B , [ A , B ] ] = O [A,[A,B]]=[B,[A,B]]=O [ A , [ A , B ]] = [ B , [ A , B ]] = O e A B e − A e^ABe^{-A} e A B e − A [ e λ A , B ] = − λ [ B , A ] e λ A [e^{\lambda A},B]=-\lambda[B,A]e^{\lambda A} [ e λ A , B ] = − λ [ B , A ] e λ A g ′ ( λ ) = ( A + B + λ [ A , B ] ) g ( λ ) g'(\lambda)=(A+B+\lambda[A,B])g(\lambda) g ′ ( λ ) = ( A + B + λ [ A , B ]) g ( λ ) e A e B = e A + B + [ A , B ] / 2 e^Ae^B=e^{A+B+[A,B]/2} e A e B = e A + B + [ A , B ] /2
Kai (1)
直接计算得
[ α , β ] = i γ . \boxed{[\alpha,\beta]=i\gamma.} [ α , β ] = iγ . 又 α 2 + β 2 + γ 2 = 2 I \alpha^2+\beta^2+\gamma^2=2I α 2 + β 2 + γ 2 = 2 I
[ α , α 2 + β 2 + γ 2 ] = O . \boxed{[\alpha,\alpha^2+\beta^2+\gamma^2]=O.} [ α , α 2 + β 2 + γ 2 ] = O . (2)–(3)
f ( 0 ) = B . \boxed{f(0)=B.} f ( 0 ) = B . 逐项求导并移指标：
d d t e t A = ∑ n = 1 ∞ t n − 1 A n ( n − 1 ) ! = A ∑ k = 0 ∞ t k A k k ! = A e t A = e t A A . \frac d{dt}e^{tA}=\sum_{n=1}^{\infty}\frac{t^{n-1}A^n}{(n-1)!}=A\sum_{k=0}^{\infty}\frac{t^kA^k}{k!}=Ae^{tA}=e^{tA}A. d t d e t A = n = 1 ∑ ∞ ( n − 1 )! t n − 1 A n = A k = 0 ∑ ∞ k ! t k A k = A e t A = e t A A . (4)
由乘积法则及 A A A e t A e^{tA} e t A
f ′ = A e t A B e − t A − e t A B e − t A A = e t A ( A B − B A ) e − t A . f'=Ae^{tA}Be^{-tA}-e^{tA}Be^{-tA}A=e^{tA}(AB-BA)e^{-tA}. f ′ = A e t A B e − t A − e t A B e − t A A = e t A ( A B − B A ) e − t A . 再求导一次即得
f ′ ′ = e t A [ A , [ A , B ] ] e − t A . \boxed{f''=e^{tA}[A,[A,B]]e^{-tA}.} f ′′ = e t A [ A , [ A , B ]] e − t A . (5)
记 C = [ A , B ] C=[A,B] C = [ A , B ] f ′ ′ = 0 f''=0 f ′′ = 0 f ( 0 ) = B , f ′ ( 0 ) = C f(0)=B,f'(0)=C f ( 0 ) = B , f ′ ( 0 ) = C
f ( t ) = B + t C , e A B e − A = B + [ A , B ] . f(t)=B+tC,\qquad\boxed{e^ABe^{-A}=B+[A,B].} f ( t ) = B + tC , e A B e − A = B + [ A , B ] . (6)
由 e λ A B = ( B + λ C ) e λ A e^{\lambda A}B=(B+\lambda C)e^{\lambda A} e λ A B = ( B + λ C ) e λ A
g ′ = A e λ A e λ B + e λ A B e λ B = ( A + B + λ C ) g . g'=Ae^{\lambda A}e^{\lambda B}+e^{\lambda A}Be^{\lambda B}
=\boxed{(A+B+\lambda C)g}. g ′ = A e λ A e λ B + e λ A B e λ B = ( A + B + λ C ) g . (7)
因 [ A , C ] = [ B , C ] = 0 [A,C]=[B,C]=0 [ A , C ] = [ B , C ] = 0 A + B A+B A + B C C C
h ( λ ) = exp [ λ ( A + B ) + λ 2 2 C ] h(\lambda)=\exp\left[\lambda(A+B)+\frac{\lambda^2}2C\right] h ( λ ) = exp [ λ ( A + B ) + 2 λ 2 C ] 满足 h ′ = ( A + B + λ C ) h h'=(A+B+\lambda C)h h ′ = ( A + B + λ C ) h h ( 0 ) = I h(0)=I h ( 0 ) = I g g g g = h g=h g = h λ = 1 \lambda=1 λ = 1
e A e B = exp ( A + B + 1 2 [ A , B ] ) . \boxed{e^Ae^B=\exp\left(A+B+\frac12[A,B]\right).} e A e B = exp ( A + B + 2 1 [ A , B ] ) .