[1] x ^ \hat x x ^ p ^ \hat p p ^ a ^ , a ^ † \hat a,\hat a^\dagger a ^ , a ^ †
(2) は、a ^ , a ^ † \hat a,\hat a^\dagger a ^ , a ^ † x ^ , p ^ \hat x,\hat p x ^ , p ^ x ^ , p ^ \hat x,\hat p x ^ , p ^ x ^ \hat x x ^ p ^ \hat p p ^
係数の確認に次元を使う。ℏ / ( m ω ) \hbar/(m\omega) ℏ/ ( mω ) m ℏ ω m\hbar\omega m ℏ ω a ^ , a ^ † \hat a,\hat a^\dagger a ^ , a ^ † [ x ^ , p ^ ] [\hat x,\hat p] [ x ^ , p ^ ] [ x ^ , p ^ ] = − i ℏ 2 [ a ^ + a ^ † , a ^ − a ^ † ] = − i ℏ 2 ( − 1 − 1 ) = i ℏ
[\hat x,\hat p]=-i\frac\hbar2\,[\hat a+\hat a^\dagger,\ \hat a-\hat a^\dagger]=-i\frac\hbar2\,(-1-1)=i\hbar
[ x ^ , p ^ ] = − i 2 ℏ [ a ^ + a ^ † , a ^ − a ^ † ] = − i 2 ℏ ( − 1 − 1 ) = i ℏ [ a ^ , a ^ † ] = 1 [\hat a,\hat a^\dagger]=1 [ a ^ , a ^ † ] = 1 [ a ^ , a ^ ] = [ a ^ † , a ^ † ] = 0 [\hat a,\hat a]=[\hat a^\dagger,\hat a^\dagger]=0 [ a ^ , a ^ ] = [ a ^ † , a ^ † ] = 0 [ x ^ , p ^ ] = i ℏ [\hat x,\hat p]=i\hbar [ x ^ , p ^ ] = i ℏ
誤りは、p ^ \hat p p ^ p ^ = + i m ℏ ω / 2 ( a ^ − a ^ † ) \hat p=+i\sqrt{m\hbar\omega/2}\,(\hat a-\hat a^\dagger) p ^ = + i m ℏ ω /2 ( a ^ − a ^ † ) [ x ^ , p ^ ] = − i ℏ [\hat x,\hat p]=-i\hbar [ x ^ , p ^ ] = − i ℏ
[2] 交換関係 [ H ^ , a ^ † a ^ ] [\hat H,\hat a^\dagger\hat a] [ H ^ , a ^ † a ^ ]
交換子 [ A ^ , B ^ ] = A ^ B ^ − B ^ A ^ [\hat A,\hat B]=\hat A\hat B-\hat B\hat A [ A ^ , B ^ ] = A ^ B ^ − B ^ A ^ [ A ^ , A ^ ] = 0 [\hat A,\hat A]=0 [ A ^ , A ^ ] = 0 c c c [ A ^ + c , B ^ ] = [ A ^ , B ^ ] [\hat A+c,\hat B]=[\hat A,\hat B] [ A ^ + c , B ^ ] = [ A ^ , B ^ ] H ^ \hat H H ^ a ^ † a ^ \hat a^\dagger\hat a a ^ † a ^ a ^ † a ^ \hat a^\dagger\hat a a ^ † a ^ H ^ \hat H H ^ a ^ † a ^ \hat a^\dagger\hat a a ^ † a ^ H ^ \hat H H ^ a ^ , a ^ † \hat a,\hat a^\dagger a ^ , a ^ †
別解として、[ H ^ , a ^ ] = − ℏ ω a ^ [\hat H,\hat a]=-\hbar\omega\hat a [ H ^ , a ^ ] = − ℏ ω a ^ [ H ^ , a ^ † ] = ℏ ω a ^ † [\hat H,\hat a^\dagger]=\hbar\omega\hat a^\dagger [ H ^ , a ^ † ] = ℏ ω a ^ † [ A ^ , B ^ C ^ ] = [ A ^ , B ^ ] C ^ + B ^ [ A ^ , C ^ ] [\hat A,\hat B\hat C]=[\hat A,\hat B]\hat C+\hat B[\hat A,\hat C] [ A ^ , B ^ C ^ ] = [ A ^ , B ^ ] C ^ + B ^ [ A ^ , C ^ ] [ H ^ , a ^ † a ^ ] = [ H ^ , a ^ † ] a ^ + a ^ † [ H ^ , a ^ ] = ℏ ω a ^ † a ^ − ℏ ω a ^ † a ^ = 0
[\hat H,\hat a^\dagger\hat a]=[\hat H,\hat a^\dagger]\hat a+\hat a^\dagger[\hat H,\hat a]=\hbar\omega\hat a^\dagger\hat a-\hbar\omega\hat a^\dagger\hat a=0
[ H ^ , a ^ † a ^ ] = [ H ^ , a ^ † ] a ^ + a ^ † [ H ^ , a ^ ] = ℏ ω a ^ † a ^ − ℏ ω a ^ † a ^ = 0 H ^ \hat H H ^ a ^ † a ^ \hat a^\dagger\hat a a ^ † a ^ ∣ n ⟩ \ket n ∣ n ⟩ H ^ ∣ n ⟩ = ℏ ω ( n + 1 2 ) ∣ n ⟩ \hat H\ket n=\hbar\omega(n+\frac12)\ket n H ^ ∣ n ⟩ = ℏ ω ( n + 2 1 ) ∣ n ⟩
誤りは、a ^ a ^ † \hat a\hat a^\dagger a ^ a ^ † a ^ † a ^ \hat a^\dagger\hat a a ^ † a ^ H ^ = ℏ ω a ^ † a ^ \hat H=\hbar\omega\hat a^\dagger\hat a H ^ = ℏ ω a ^ † a ^ 0 0 0 ℏ ω / 2 \hbar\omega/2 ℏ ω /2 Δ x Δ p ≥ ℏ / 2 \Delta x\Delta p\ge\hbar/2 Δ x Δ p ≥ ℏ/2 ⟨ H ^ ⟩ ≥ ℏ ω / 2 \langle\hat H\rangle\ge\hbar\omega/2 ⟨ H ^ ⟩ ≥ ℏ ω /2 a ^ a ^ † − a ^ † a ^ = 1 \hat a\hat a^\dagger-\hat a^\dagger\hat a=1 a ^ a ^ † − a ^ † a ^ = 1
[3] 基底状態の位置表示 ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩
位置の固有ケット ∣ x ⟩ \ket x ∣ x ⟩ x ^ ∣ x ⟩ = x ∣ x ⟩ \hat x\ket x=x\ket x x ^ ∣ x ⟩ = x ∣ x ⟩ x ^ \hat x x ^ ⟨ x ∣ x ^ = x ⟨ x ∣ \bra x\hat x=x\bra x ⟨ x ∣ x ^ = x ⟨ x ∣ p ^ \hat p p ^ − i ℏ d / d x -i\hbar\,d/dx − i ℏ d / d x a ^ ∣ 0 ⟩ = 0 \hat a\ket0=0 a ^ ∣ 0 ⟩ = 0 ψ 0 \psi_0 ψ 0
規格化条件 ∫ ∣ ψ 0 ∣ 2 d x = 1 \int|\psi_0|^2dx=1 ∫ ∣ ψ 0 ∣ 2 d x = 1 1 1 1 m ω / ℏ m\omega/\hbar mω /ℏ C 2 C^2 C 2 m ω / ( π ℏ ) \sqrt{m\omega/(\pi\hbar)} mω / ( π ℏ ) C C C 1 / 4 1/4 1/4
確かめとして、H ^ ψ 0 = 1 2 ℏ ω ψ 0 \hat H\psi_0=\frac12\hbar\omega\psi_0 H ^ ψ 0 = 2 1 ℏ ω ψ 0 ψ 0 ′ = − ( m ω / ℏ ) x ψ 0 \psi_0'=-(m\omega/\hbar)x\psi_0 ψ 0 ′ = − ( mω /ℏ ) x ψ 0 ψ 0 ′ ′ = ( − m ω / ℏ + m 2 ω 2 x 2 / ℏ 2 ) ψ 0 \psi_0''=\bigl(-m\omega/\hbar+m^2\omega^2x^2/\hbar^2\bigr)\psi_0 ψ 0 ′′ = ( − mω /ℏ + m 2 ω 2 x 2 / ℏ 2 ) ψ 0 − ℏ 2 2 m ψ 0 ′ ′ + 1 2 m ω 2 x 2 ψ 0 = ( ℏ ω 2 − 1 2 m ω 2 x 2 + 1 2 m ω 2 x 2 ) ψ 0 = ℏ ω 2 ψ 0
-\frac{\hbar^2}{2m}\psi_0''+\frac12m\omega^2x^2\psi_0=\Bigl(\frac{\hbar\omega}2-\frac12m\omega^2x^2+\frac12m\omega^2x^2\Bigr)\psi_0=\frac{\hbar\omega}2\psi_0
− 2 m ℏ 2 ψ 0 ′′ + 2 1 m ω 2 x 2 ψ 0 = ( 2 ℏ ω − 2 1 m ω 2 x 2 + 2 1 m ω 2 x 2 ) ψ 0 = 2 ℏ ω ψ 0 1 2 ℏ ω \frac12\hbar\omega 2 1 ℏ ω ∣ ψ 0 ∣ 2 ∝ e − x 2 / ℓ 2 |\psi_0|^2\propto e^{-x^2/\ell^2} ∣ ψ 0 ∣ 2 ∝ e − x 2 / ℓ 2 ℓ = ℏ / ( m ω ) \ell=\sqrt{\hbar/(m\omega)} ℓ = ℏ/ ( mω ) x ^ \hat x x ^ ℏ / ( 2 m ω ) = ℓ / 2 \sqrt{\hbar/(2m\omega)}=\ell/\sqrt2 ℏ/ ( 2 mω ) = ℓ / 2 2 \sqrt2 2
誤りは、規格化定数を ( m ω / π ℏ ) 1 / 2 (m\omega/\pi\hbar)^{1/2} ( mω / π ℏ ) 1/2 ∫ ∣ ψ 0 ∣ 2 d x = ( m ω / π ℏ ) 1 / 2 \int|\psi_0|^2dx=(m\omega/\pi\hbar)^{1/2} ∫ ∣ ψ 0 ∣ 2 d x = ( mω / π ℏ ) 1/2 m = ω = ℏ = 1 m=\omega=\hbar=1 m = ω = ℏ = 1 π − 1 / 2 ≃ 0.564 \pi^{-1/2}\simeq0.564 π − 1/2 ≃ 0.564 1 1 1 1 1 1
[4] 運動量の固有ケットの位置表示 ⟨ x ∣ p ⟩ \braket{x|p} ⟨ x ∣ p ⟩
運動量の固有値 p p p ⟨ p ∣ p ′ ⟩ = δ ( p − p ′ ) \braket{p|p'}=\delta(p-p') ⟨ p ∣ p ′ ⟩ = δ ( p − p ′ ) ∫ ∣ x ⟩ ⟨ x ∣ d x = I ^ \int|x\rangle\langle x|dx=\hat I ∫ ∣ x ⟩ ⟨ x ∣ d x = I ^ ⟨ p ∣ p ′ ⟩ \braket{p|p'} ⟨ p ∣ p ′ ⟩ x x x δ ( k ) = 1 2 π ∫ e i k x d x \delta(k)=\frac1{2\pi}\int e^{ikx}dx δ ( k ) = 2 π 1 ∫ e ik x d x
デルタ関数の性質 δ ( α k ) = δ ( k ) / ∣ α ∣ \delta(\alpha k)=\delta(k)/|\alpha| δ ( α k ) = δ ( k ) /∣ α ∣ δ ( ( p ′ − p ) / ℏ ) = ℏ δ ( p ′ − p ) \delta\bigl((p'-p)/\hbar\bigr)=\hbar\,\delta(p'-p) δ ( ( p ′ − p ) /ℏ ) = ℏ δ ( p ′ − p ) ℏ \hbar ℏ ⟨ x ∣ p ⟩ \braket{x|p} ⟨ x ∣ p ⟩ p p p 2 π ℏ / p 2\pi\hbar/p 2 π ℏ/ p
次元の確認をする。⟨ p ∣ p ′ ⟩ \braket{p|p'} ⟨ p ∣ p ′ ⟩ ( 運動量 ) − 1 {\rm (運動量)^{-1}} ( 運動量 ) − 1 ∫ d x \int dx ∫ d x ∣ c ∣ 2 |c|^2 ∣ c ∣ 2 ( 長さ ⋅ 運動量 ) − 1 {\rm (長さ{\cdot}運動量)^{-1}} ( 長さ ⋅ 運動量 ) − 1 ℏ \hbar ℏ 1 / ( 2 π ℏ ) 1/(2\pi\hbar) 1/ ( 2 π ℏ )
誤りは、ℏ \hbar ℏ c = 1 / 2 π c=1/\sqrt{2\pi} c = 1/ 2 π ⟨ p ∣ p ′ ⟩ = ℏ δ ( p − p ′ ) \braket{p|p'}=\hbar\,\delta(p-p') ⟨ p ∣ p ′ ⟩ = ℏ δ ( p − p ′ ) ℏ \hbar ℏ ∣ c ∣ 2 = 1 / ( 2 π ) |c|^2=1/(2\pi) ∣ c ∣ 2 = 1/ ( 2 π ) ℏ \hbar ℏ
[5] 運動量表示 ⟨ p ∣ 0 ⟩ \braket{p|0} ⟨ p ∣0 ⟩ σ \sigma σ
x x x p p p ⟨ p ∣ ψ ⟩ = ∫ ⟨ p ∣ x ⟩ ⟨ x ∣ ψ ⟩ d x \braket{p|\psi}=\int\braket{p|x}\braket{x|\psi}dx ⟨ p ∣ ψ ⟩ = ∫ ⟨ p ∣ x ⟩ ⟨ x ∣ ψ ⟩ d x ⟨ p ∣ x ⟩ = ⟨ x ∣ p ⟩ ∗ \braket{p|x}=\braket{x|p}^* ⟨ p ∣ x ⟩ = ⟨ x ∣ p ⟩ ∗ γ \gamma γ ∫ e − ( x − i γ ) 2 d x = π \int e^{-(x-i\gamma)^2}dx=\sqrt\pi ∫ e − ( x − iγ ) 2 d x = π e − z 2 e^{-z^2} e − z 2 x x x i p / ( m ω ) ip/(m\omega) i p / ( mω )
確かめは3つある。1つ目は規格化で、∫ ∣ ⟨ p ∣ 0 ⟩ ∣ 2 d p = ( π m ℏ ω ) − 1 / 2 ∫ e − p 2 / ( m ℏ ω ) d p = 1 \int|\braket{p|0}|^2dp=(\pi m\hbar\omega)^{-1/2}\int e^{-p^2/(m\hbar\omega)}dp=1 ∫ ∣ ⟨ p ∣0 ⟩ ∣ 2 d p = ( π m ℏ ω ) − 1/2 ∫ e − p 2 / ( m ℏ ω ) d p = 1 ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩ ⟨ x 2 ⟩ = ℏ / ( 2 m ω ) \langle x^2\rangle=\hbar/(2m\omega) ⟨ x 2 ⟩ = ℏ/ ( 2 mω ) ⟨ p ∣ 0 ⟩ \braket{p|0} ⟨ p ∣0 ⟩ ⟨ p 2 ⟩ = m ℏ ω / 2 \langle p^2\rangle=m\hbar\omega/2 ⟨ p 2 ⟩ = m ℏ ω /2 Δ x Δ p = ℏ / 2 \Delta x\Delta p=\hbar/2 Δ x Δ p = ℏ/2 H ^ \hat H H ^ H ^ = m ω 2 2 [ x ^ 2 + ( p ^ / m ω ) 2 ] \hat H=\frac{m\omega^2}2\bigl[\hat x^2+(\hat p/m\omega)^2\bigr] H ^ = 2 m ω 2 [ x ^ 2 + ( p ^ / mω ) 2 ] x ↔ p / ( m ω ) x\leftrightarrow p/(m\omega) x ↔ p / ( mω ) ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩ x 2 x^2 x 2 p 2 / ( m ω ) 2 p^2/(m\omega)^2 p 2 / ( mω ) 2 exp ( − p 2 / ( 2 m ℏ ω ) ) \exp(-p^2/(2m\hbar\omega)) exp ( − p 2 / ( 2 m ℏ ω ))
誤りは、σ 2 = m ℏ ω \sigma^2=m\hbar\omega σ 2 = m ℏ ω 2 2 2 ∣ ⟨ p ∣ 0 ⟩ ∣ 2 ∝ e − 2 p 2 / ( m ℏ ω ) |\braket{p|0}|^2\propto e^{-2p^2/(m\hbar\omega)} ∣ ⟨ p ∣0 ⟩ ∣ 2 ∝ e − 2 p 2 / ( m ℏ ω ) ⟨ p 2 ⟩ = m ℏ ω / 4 \langle p^2\rangle=m\hbar\omega/4 ⟨ p 2 ⟩ = m ℏ ω /4 Δ x Δ p = ℏ / ( 2 2 ) ≃ 0.35 ℏ \Delta x\Delta p=\hbar/(2\sqrt2)\simeq0.35\hbar Δ x Δ p = ℏ/ ( 2 2 ) ≃ 0.35ℏ 0.5 ℏ 0.5\hbar 0.5ℏ
[6] 第1励起状態の位置表示 ⟨ x ∣ 1 ⟩ \braket{x|1} ⟨ x ∣1 ⟩
a ^ † \hat a^\dagger a ^ † a ^ † \hat a^\dagger a ^ † m ω / 2 ℏ ( x − ℏ m ω d d x ) \sqrt{m\omega/2\hbar}\,(x-\frac\hbar{m\omega}\frac{d}{dx}) mω /2ℏ ( x − mω ℏ d x d ) ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩ d d x ⟨ x ∣ 0 ⟩ \frac d{dx}\braket{x|0} d x d ⟨ x ∣0 ⟩
確かめを3つ行う。規格化は ∫ ∣ ⟨ x ∣ 1 ⟩ ∣ 2 d x = 2 m ω ℏ ⟨ x 2 ⟩ 0 = 2 m ω ℏ ⋅ ℏ 2 m ω = 1 \int|\braket{x|1}|^2dx=\frac{2m\omega}\hbar\langle x^2\rangle_0=\frac{2m\omega}\hbar\cdot\frac\hbar{2m\omega}=1 ∫ ∣ ⟨ x ∣1 ⟩ ∣ 2 d x = ℏ 2 mω ⟨ x 2 ⟩ 0 = ℏ 2 mω ⋅ 2 mω ℏ = 1 ⟨ x ∣ 1 ⟩ \braket{x|1} ⟨ x ∣1 ⟩ x x x ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩ ⟨ 0 ∣ 1 ⟩ = 0 \braket{0|1}=0 ⟨ 0∣1 ⟩ = 0 a ^ ∣ 1 ⟩ = ∣ 0 ⟩ \hat a\ket1=\ket0 a ^ ∣ 1 ⟩ = ∣ 0 ⟩ H ^ ∣ 1 ⟩ = 3 2 ℏ ω ∣ 1 ⟩ \hat H\ket1=\frac32\hbar\omega\ket1 H ^ ∣ 1 ⟩ = 2 3 ℏ ω ∣ 1 ⟩ x = 0 x=0 x = 0
誤りは2つある。1つ目は a ^ † \hat a^\dagger a ^ † x + ℏ m ω d d x x+\frac\hbar{m\omega}\frac d{dx} x + mω ℏ d x d a ^ \hat a a ^ ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩ a ^ ∣ 0 ⟩ = 0 \hat a\ket0=0 a ^ ∣ 0 ⟩ = 0 2 2 2 m ω / 2 ℏ x ⟨ x ∣ 0 ⟩ \sqrt{m\omega/2\hbar}\,x\braket{x|0} mω /2ℏ x ⟨ x ∣0 ⟩ m ω 2 ℏ ⋅ ℏ 2 m ω = 1 4 \frac{m\omega}{2\hbar}\cdot\frac\hbar{2m\omega}=\frac14 2ℏ mω ⋅ 2 mω ℏ = 4 1 1 1 1
解答
[1] x ^ \hat x x ^ p ^ \hat p p ^ a ^ , a ^ † \hat a,\hat a^\dagger a ^ , a ^ †
式 (2) の和と差をとる。
a ^ + a ^ † = m ω 2 ℏ ⋅ 2 x ^ , a ^ − a ^ † = m ω 2 ℏ ⋅ 2 i m ω p ^ .
\hat a+\hat a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\cdot2\hat x,\qquad
\hat a-\hat a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\cdot\frac{2i}{m\omega}\hat p .
a ^ + a ^ † = 2ℏ mω ⋅ 2 x ^ , a ^ − a ^ † = 2ℏ mω ⋅ mω 2 i p ^ . x ^ \hat x x ^ p ^ \hat p p ^ x ^ = ℏ 2 m ω ( a ^ + a ^ † ) , p ^ = m ω 2 i 2 ℏ m ω ( a ^ − a ^ † ) = − i m ℏ ω 2 ( a ^ − a ^ † ) .
\hat x=\sqrt{\frac{\hbar}{2m\omega}}\,(\hat a+\hat a^\dagger),\qquad
\hat p=\frac{m\omega}{2i}\sqrt{\frac{2\hbar}{m\omega}}\,(\hat a-\hat a^\dagger)=-i\sqrt{\frac{m\hbar\omega}{2}}\,(\hat a-\hat a^\dagger).
x ^ = 2 mω ℏ ( a ^ + a ^ † ) , p ^ = 2 i mω mω 2ℏ ( a ^ − a ^ † ) = − i 2 m ℏ ω ( a ^ − a ^ † ) .
[2] 交換関係 [ H ^ , a ^ † a ^ ] [\hat H,\hat a^\dagger\hat a] [ H ^ , a ^ † a ^ ]
[1] の結果から
x ^ 2 = ℏ 2 m ω ( a ^ + a ^ † ) 2 , p ^ 2 = − m ℏ ω 2 ( a ^ − a ^ † ) 2 .
\hat x^2=\frac{\hbar}{2m\omega}(\hat a+\hat a^\dagger)^2,\qquad
\hat p^2=-\frac{m\hbar\omega}{2}(\hat a-\hat a^\dagger)^2 .
x ^ 2 = 2 mω ℏ ( a ^ + a ^ † ) 2 , p ^ 2 = − 2 m ℏ ω ( a ^ − a ^ † ) 2 . H ^ \hat H H ^ H ^ = 1 2 m ( − m ℏ ω 2 ) ( a ^ − a ^ † ) 2 + 1 2 m ω 2 ⋅ ℏ 2 m ω ( a ^ + a ^ † ) 2 = ℏ ω 4 [ ( a ^ + a ^ † ) 2 − ( a ^ − a ^ † ) 2 ] = ℏ ω 2 ( a ^ a ^ † + a ^ † a ^ ) . \begin{align*}
\hat H&=\frac1{2m}\Bigl(-\frac{m\hbar\omega}2\Bigr)(\hat a-\hat a^\dagger)^2+\frac12m\omega^2\cdot\frac\hbar{2m\omega}(\hat a+\hat a^\dagger)^2\\
&=\frac{\hbar\omega}4\Bigl[(\hat a+\hat a^\dagger)^2-(\hat a-\hat a^\dagger)^2\Bigr]
=\frac{\hbar\omega}2\bigl(\hat a\hat a^\dagger+\hat a^\dagger\hat a\bigr).
\end{align*} H ^ = 2 m 1 ( − 2 m ℏ ω ) ( a ^ − a ^ † ) 2 + 2 1 m ω 2 ⋅ 2 mω ℏ ( a ^ + a ^ † ) 2 = 4 ℏ ω [ ( a ^ + a ^ † ) 2 − ( a ^ − a ^ † ) 2 ] = 2 ℏ ω ( a ^ a ^ † + a ^ † a ^ ) . a ^ 2 \hat a^2 a ^ 2 ( a ^ † ) 2 (\hat a^\dagger)^2 ( a ^ † ) 2 a ^ a ^ † + a ^ † a ^ \hat a\hat a^\dagger+\hat a^\dagger\hat a a ^ a ^ † + a ^ † a ^ [ a ^ , a ^ † ] = 1 [\hat a,\hat a^\dagger]=1 [ a ^ , a ^ † ] = 1 a ^ a ^ † = a ^ † a ^ + 1 \hat a\hat a^\dagger=\hat a^\dagger\hat a+1 a ^ a ^ † = a ^ † a ^ + 1 H ^ = ℏ ω ( a ^ † a ^ + 1 2 ) .
\hat H=\hbar\omega\Bigl(\hat a^\dagger\hat a+\frac12\Bigr).
H ^ = ℏ ω ( a ^ † a ^ + 2 1 ) . 1 2 \frac12 2 1 [ a ^ † a ^ , a ^ † a ^ ] = 0 [\hat a^\dagger\hat a,\hat a^\dagger\hat a]=0 [ a ^ † a ^ , a ^ † a ^ ] = 0 [ H ^ , a ^ † a ^ ] = ℏ ω [ a ^ † a ^ , a ^ † a ^ ] = 0.
[\hat H,\hat a^\dagger\hat a]=\hbar\omega\,[\hat a^\dagger\hat a,\hat a^\dagger\hat a]=0 .
[ H ^ , a ^ † a ^ ] = ℏ ω [ a ^ † a ^ , a ^ † a ^ ] = 0.
[3] 基底状態の位置表示 ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩
⟨ x ∣ a ^ ∣ 0 ⟩ = 0 \braket{x|\hat a|0}=0 ⟨ x ∣ a ^ ∣0 ⟩ = 0 a ^ \hat a a ^ m ω 2 ℏ ( ⟨ x ∣ x ^ ∣ 0 ⟩ + i m ω ⟨ x ∣ p ^ ∣ 0 ⟩ ) = 0.
\sqrt{\frac{m\omega}{2\hbar}}\Bigl(\braket{x|\hat x|0}+\frac{i}{m\omega}\braket{x|\hat p|0}\Bigr)=0 .
2ℏ mω ( ⟨ x ∣ x ^ ∣0 ⟩ + mω i ⟨ x ∣ p ^ ∣0 ⟩ ) = 0. ⟨ x ∣ \bra x ⟨ x ∣ x ^ \hat x x ^ ⟨ x ∣ x ^ = x ⟨ x ∣ \bra x\hat x=x\bra x ⟨ x ∣ x ^ = x ⟨ x ∣ ⟨ x ∣ x ^ ∣ 0 ⟩ = x ⟨ x ∣ 0 ⟩ \braket{x|\hat x|0}=x\braket{x|0} ⟨ x ∣ x ^ ∣0 ⟩ = x ⟨ x ∣0 ⟩ ∣ ξ ⟩ = ∣ 0 ⟩ \ket\xi=\ket0 ∣ ξ ⟩ = ∣ 0 ⟩ ⟨ x ∣ p ^ ∣ 0 ⟩ = − i ℏ d d x ⟨ x ∣ 0 ⟩ \braket{x|\hat p|0}=-i\hbar\,\dfrac{d}{dx}\braket{x|0} ⟨ x ∣ p ^ ∣0 ⟩ = − i ℏ d x d ⟨ x ∣0 ⟩ i ⋅ ( − i ) = 1 i\cdot(-i)=1 i ⋅ ( − i ) = 1 x ⟨ x ∣ 0 ⟩ + ℏ m ω d d x ⟨ x ∣ 0 ⟩ = 0.
x\braket{x|0}+\frac\hbar{m\omega}\frac{d}{dx}\braket{x|0}=0 .
x ⟨ x ∣0 ⟩ + mω ℏ d x d ⟨ x ∣0 ⟩ = 0. ψ 0 ( x ) = ⟨ x ∣ 0 ⟩ \psi_0(x)=\braket{x|0} ψ 0 ( x ) = ⟨ x ∣0 ⟩ ψ 0 ′ = − m ω ℏ x ψ 0 \psi_0'=-\dfrac{m\omega}{\hbar}x\,\psi_0 ψ 0 ′ = − ℏ mω x ψ 0 d ψ 0 ψ 0 = − m ω ℏ x d x ⟹ ln ∣ ψ 0 ∣ = − m ω 2 ℏ x 2 + c o n s t , ψ 0 ( x ) = C e − m ω x 2 / 2 ℏ .
\frac{d\psi_0}{\psi_0}=-\frac{m\omega}\hbar x\,dx\ \Longrightarrow\ \ln|\psi_0|=-\frac{m\omega}{2\hbar}x^2+{\rm const},\qquad
\psi_0(x)=C\,e^{-m\omega x^2/2\hbar}.
ψ 0 d ψ 0 = − ℏ mω x d x ⟹ ln ∣ ψ 0 ∣ = − 2ℏ mω x 2 + const , ψ 0 ( x ) = C e − mω x 2 /2ℏ . C C C u = m ω / ℏ x u=\sqrt{m\omega/\hbar}\,x u = mω /ℏ x ∫ − ∞ ∞ e − u 2 d u = π \int_{-\infty}^\infty e^{-u^2}du=\sqrt\pi ∫ − ∞ ∞ e − u 2 d u = π 1 = C 2 ∫ − ∞ ∞ e − m ω x 2 / ℏ d x = C 2 ℏ m ω π ⟹ C = ( m ω π ℏ ) 1 / 4 .
1=C^2\int_{-\infty}^\infty e^{-m\omega x^2/\hbar}dx=C^2\sqrt{\frac\hbar{m\omega}}\sqrt\pi
\ \Longrightarrow\ C=\Bigl(\frac{m\omega}{\pi\hbar}\Bigr)^{1/4}.
1 = C 2 ∫ − ∞ ∞ e − mω x 2 /ℏ d x = C 2 mω ℏ π ⟹ C = ( π ℏ mω ) 1/4 . C C C
[4] 運動量の固有ケットの位置表示 ⟨ x ∣ p ⟩ \braket{x|p} ⟨ x ∣ p ⟩
p ^ ∣ p ⟩ = p ∣ p ⟩ \hat p\ket p=p\ket p p ^ ∣ p ⟩ = p ∣ p ⟩ ∣ ξ ⟩ = ∣ p ⟩ \ket\xi=\ket p ∣ ξ ⟩ = ∣ p ⟩ p ⟨ x ∣ p ⟩ = ⟨ x ∣ p ^ ∣ p ⟩ = − i ℏ d d x ⟨ x ∣ p ⟩ ⟹ d d x ⟨ x ∣ p ⟩ = i p ℏ ⟨ x ∣ p ⟩ ⟹ ⟨ x ∣ p ⟩ = c e i p x / ℏ .
p\braket{x|p}=\braket{x|\hat p|p}=-i\hbar\frac{d}{dx}\braket{x|p}
\ \Longrightarrow\ \frac{d}{dx}\braket{x|p}=\frac{ip}\hbar\braket{x|p}
\ \Longrightarrow\ \braket{x|p}=c\,e^{ipx/\hbar}.
p ⟨ x ∣ p ⟩ = ⟨ x ∣ p ^ ∣ p ⟩ = − i ℏ d x d ⟨ x ∣ p ⟩ ⟹ d x d ⟨ x ∣ p ⟩ = ℏ i p ⟨ x ∣ p ⟩ ⟹ ⟨ x ∣ p ⟩ = c e i p x /ℏ . c c c x x x ⟨ p ∣ p ′ ⟩ = δ ( p − p ′ ) \braket{p|p'}=\delta(p-p') ⟨ p ∣ p ′ ⟩ = δ ( p − p ′ ) c c c ∫ ∣ x ⟩ ⟨ x ∣ d x = I ^ \int|x\rangle\langle x|dx=\hat I ∫ ∣ x ⟩ ⟨ x ∣ d x = I ^ ⟨ p ∣ x ⟩ = ⟨ x ∣ p ⟩ ∗ \braket{p|x}=\braket{x|p}^* ⟨ p ∣ x ⟩ = ⟨ x ∣ p ⟩ ∗ ⟨ p ∣ p ′ ⟩ = ∫ − ∞ ∞ ⟨ p ∣ x ⟩ ⟨ x ∣ p ′ ⟩ d x = ∣ c ∣ 2 ∫ − ∞ ∞ e i ( p ′ − p ) x / ℏ d x = ∣ c ∣ 2 2 π δ ( p ′ − p ℏ ) = ∣ c ∣ 2 2 π ℏ δ ( p ′ − p ) .
\braket{p|p'}=\int_{-\infty}^\infty\braket{p|x}\braket{x|p'}dx=|c|^2\int_{-\infty}^\infty e^{i(p'-p)x/\hbar}dx
=|c|^2\,2\pi\,\delta\Bigl(\frac{p'-p}\hbar\Bigr)=|c|^2\,2\pi\hbar\,\delta(p'-p).
⟨ p ∣ p ′ ⟩ = ∫ − ∞ ∞ ⟨ p ∣ x ⟩ ⟨ x ∣ p ′ ⟩ d x = ∣ c ∣ 2 ∫ − ∞ ∞ e i ( p ′ − p ) x /ℏ d x = ∣ c ∣ 2 2 π δ ( ℏ p ′ − p ) = ∣ c ∣ 2 2 π ℏ δ ( p ′ − p ) . δ ( k / ℏ ) = ℏ δ ( k ) \delta(k/\hbar)=\hbar\,\delta(k) δ ( k /ℏ ) = ℏ δ ( k ) ⟨ p ∣ p ′ ⟩ = δ ( p − p ′ ) \braket{p|p'}=\delta(p-p') ⟨ p ∣ p ′ ⟩ = δ ( p − p ′ ) ∣ c ∣ 2 = 1 / ( 2 π ℏ ) |c|^2=1/(2\pi\hbar) ∣ c ∣ 2 = 1/ ( 2 π ℏ ) c = 1 / 2 π ℏ c=1/\sqrt{2\pi\hbar} c = 1/ 2 π ℏ
[5] 運動量表示 ⟨ p ∣ 0 ⟩ \braket{p|0} ⟨ p ∣0 ⟩ σ \sigma σ
完全性 ∫ ∣ x ⟩ ⟨ x ∣ d x = I ^ \int|x\rangle\langle x|dx=\hat I ∫ ∣ x ⟩ ⟨ x ∣ d x = I ^ ⟨ p ∣ x ⟩ = ⟨ x ∣ p ⟩ ∗ = e − i p x / ℏ / 2 π ℏ \braket{p|x}=\braket{x|p}^*=e^{-ipx/\hbar}/\sqrt{2\pi\hbar} ⟨ p ∣ x ⟩ = ⟨ x ∣ p ⟩ ∗ = e − i p x /ℏ / 2 π ℏ ⟨ p ∣ 0 ⟩ = ∫ − ∞ ∞ ⟨ p ∣ x ⟩ ⟨ x ∣ 0 ⟩ d x = 1 2 π ℏ ( m ω π ℏ ) 1 / 4 ∫ − ∞ ∞ exp ( − m ω 2 ℏ x 2 − i p ℏ x ) d x .
\braket{p|0}=\int_{-\infty}^\infty\braket{p|x}\braket{x|0}dx
=\frac1{\sqrt{2\pi\hbar}}\Bigl(\frac{m\omega}{\pi\hbar}\Bigr)^{1/4}\int_{-\infty}^\infty\exp\Bigl(-\frac{m\omega}{2\hbar}x^2-\frac{ip}\hbar x\Bigr)dx .
⟨ p ∣0 ⟩ = ∫ − ∞ ∞ ⟨ p ∣ x ⟩ ⟨ x ∣0 ⟩ d x = 2 π ℏ 1 ( π ℏ mω ) 1/4 ∫ − ∞ ∞ exp ( − 2ℏ mω x 2 − ℏ i p x ) d x . − m ω 2 ℏ x 2 − i p ℏ x = − m ω 2 ℏ ( x + i p m ω ) 2 − p 2 2 m ℏ ω .
-\frac{m\omega}{2\hbar}x^2-\frac{ip}\hbar x=-\frac{m\omega}{2\hbar}\Bigl(x+\frac{ip}{m\omega}\Bigr)^2-\frac{p^2}{2m\hbar\omega}.
− 2ℏ mω x 2 − ℏ i p x = − 2ℏ mω ( x + mω i p ) 2 − 2 m ℏ ω p 2 . u = m ω / 2 ℏ x u=\sqrt{m\omega/2\hbar}\,x u = mω /2ℏ x γ = p / 2 m ℏ ω \gamma=p/\sqrt{2m\hbar\omega} γ = p / 2 m ℏ ω m ω / 2 ℏ ( x + i p / m ω ) = u + i γ \sqrt{m\omega/2\hbar}\,(x+ip/m\omega)=u+i\gamma mω /2ℏ ( x + i p / mω ) = u + iγ d x = 2 ℏ / m ω d u dx=\sqrt{2\hbar/m\omega}\,du d x = 2ℏ/ mω d u γ \gamma γ ∫ e − ( u − i γ ) 2 d u = π \int e^{-(u-i\gamma)^2}du=\sqrt\pi ∫ e − ( u − iγ ) 2 d u = π γ \gamma γ ∫ − ∞ ∞ exp ( − m ω 2 ℏ x 2 − i p ℏ x ) d x = 2 ℏ m ω e − p 2 / ( 2 m ℏ ω ) ∫ e − ( u + i γ ) 2 d u = 2 π ℏ m ω e − p 2 / ( 2 m ℏ ω ) .
\int_{-\infty}^\infty\exp\Bigl(-\frac{m\omega}{2\hbar}x^2-\frac{ip}\hbar x\Bigr)dx=\sqrt{\frac{2\hbar}{m\omega}}\,e^{-p^2/(2m\hbar\omega)}\int e^{-(u+i\gamma)^2}du=\sqrt{\frac{2\pi\hbar}{m\omega}}\,e^{-p^2/(2m\hbar\omega)} .
∫ − ∞ ∞ exp ( − 2ℏ mω x 2 − ℏ i p x ) d x = mω 2ℏ e − p 2 / ( 2 m ℏ ω ) ∫ e − ( u + iγ ) 2 d u = mω 2 π ℏ e − p 2 / ( 2 m ℏ ω ) . 1 2 π ℏ 2 π ℏ m ω ( m ω π ℏ ) 1 / 4 = 1 m ω ( m ω π ℏ ) 1 / 4 = ( π m ℏ ω ) − 1 / 4 \dfrac1{\sqrt{2\pi\hbar}}\sqrt{\dfrac{2\pi\hbar}{m\omega}}\Bigl(\dfrac{m\omega}{\pi\hbar}\Bigr)^{1/4}=\dfrac1{\sqrt{m\omega}}\Bigl(\dfrac{m\omega}{\pi\hbar}\Bigr)^{1/4}=(\pi m\hbar\omega)^{-1/4} 2 π ℏ 1 mω 2 π ℏ ( π ℏ mω ) 1/4 = mω 1 ( π ℏ mω ) 1/4 = ( π m ℏ ω ) − 1/4 ⟨ p ∣ 0 ⟩ = ( π m ℏ ω ) − 1 / 4 exp ( − p 2 2 m ℏ ω ) .
\braket{p|0}=(\pi m\hbar\omega)^{-1/4}\exp\Bigl(-\frac{p^2}{2m\hbar\omega}\Bigr).
⟨ p ∣0 ⟩ = ( π m ℏ ω ) − 1/4 exp ( − 2 m ℏ ω p 2 ) . exp ( − p 2 / σ 2 ) \exp(-p^2/\sigma^2) exp ( − p 2 / σ 2 ) σ 2 = 2 m ℏ ω \sigma^2=2m\hbar\omega σ 2 = 2 m ℏ ω
[6] 第1励起状態の位置表示 ⟨ x ∣ 1 ⟩ \braket{x|1} ⟨ x ∣1 ⟩
a ^ † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ \hat a^\dagger\ket n=\sqrt{n+1}\ket{n+1} a ^ † ∣ n ⟩ = n + 1 ∣ n + 1 ⟩ n = 0 n=0 n = 0 ∣ 1 ⟩ = a ^ † ∣ 0 ⟩ \ket1=\hat a^\dagger\ket0 ∣ 1 ⟩ = a ^ † ∣ 0 ⟩ a ^ † \hat a^\dagger a ^ † ⟨ x ∣ x ^ = x ⟨ x ∣ \bra x\hat x=x\bra x ⟨ x ∣ x ^ = x ⟨ x ∣ ∣ ξ ⟩ = ∣ 0 ⟩ \ket\xi=\ket0 ∣ ξ ⟩ = ∣ 0 ⟩ ⟨ x ∣ 1 ⟩ = m ω 2 ℏ ( ⟨ x ∣ x ^ ∣ 0 ⟩ − i m ω ⟨ x ∣ p ^ ∣ 0 ⟩ ) = m ω 2 ℏ ( x ⟨ x ∣ 0 ⟩ − i m ω ( − i ℏ ) d d x ⟨ x ∣ 0 ⟩ ) .
\braket{x|1}=\sqrt{\frac{m\omega}{2\hbar}}\Bigl(\braket{x|\hat x|0}-\frac{i}{m\omega}\braket{x|\hat p|0}\Bigr)
=\sqrt{\frac{m\omega}{2\hbar}}\Bigl(x\braket{x|0}-\frac{i}{m\omega}(-i\hbar)\frac{d}{dx}\braket{x|0}\Bigr).
⟨ x ∣1 ⟩ = 2ℏ mω ( ⟨ x ∣ x ^ ∣0 ⟩ − mω i ⟨ x ∣ p ^ ∣0 ⟩ ) = 2ℏ mω ( x ⟨ x ∣0 ⟩ − mω i ( − i ℏ ) d x d ⟨ x ∣0 ⟩ ) . − i m ω ⋅ ( − i ℏ ) = i 2 ℏ m ω = − ℏ m ω -\dfrac{i}{m\omega}\cdot(-i\hbar)=\dfrac{i^2\hbar}{m\omega}=-\dfrac\hbar{m\omega} − mω i ⋅ ( − i ℏ ) = mω i 2 ℏ = − mω ℏ ⟨ x ∣ 1 ⟩ = m ω 2 ℏ ( x − ℏ m ω d d x ) ⟨ x ∣ 0 ⟩ .
\braket{x|1}=\sqrt{\frac{m\omega}{2\hbar}}\Bigl(x-\frac\hbar{m\omega}\frac{d}{dx}\Bigr)\braket{x|0}.
⟨ x ∣1 ⟩ = 2ℏ mω ( x − mω ℏ d x d ) ⟨ x ∣0 ⟩ . d d x ⟨ x ∣ 0 ⟩ = − m ω ℏ x ⟨ x ∣ 0 ⟩ \dfrac{d}{dx}\braket{x|0}=-\dfrac{m\omega}\hbar x\braket{x|0} d x d ⟨ x ∣0 ⟩ = − ℏ mω x ⟨ x ∣0 ⟩ 2 x ⟨ x ∣ 0 ⟩ 2x\braket{x|0} 2 x ⟨ x ∣0 ⟩ ⟨ x ∣ 1 ⟩ = 2 m ω ℏ x ⟨ x ∣ 0 ⟩ = 2 m ω ℏ ( m ω π ℏ ) 1 / 4 x e − m ω x 2 / 2 ℏ .
\braket{x|1}=\sqrt{\frac{2m\omega}\hbar}\,x\braket{x|0}=\sqrt{\frac{2m\omega}\hbar}\Bigl(\frac{m\omega}{\pi\hbar}\Bigr)^{1/4}x\,e^{-m\omega x^2/2\hbar}.
⟨ x ∣1 ⟩ = ℏ 2 mω x ⟨ x ∣0 ⟩ = ℏ 2 mω ( π ℏ mω ) 1/4 x e − mω x 2 /2ℏ .
最終答
[1] x ^ \hat x x ^ p ^ \hat p p ^ a ^ , a ^ † \hat a,\hat a^\dagger a ^ , a ^ †
x ^ = ℏ 2 m ω ( a ^ + a ^ † ) , p ^ = − i m ℏ ω 2 ( a ^ − a ^ † ) \displaystyle\hat x=\sqrt{\frac{\hbar}{2m\omega}}\,(\hat a+\hat a^\dagger),\qquad\hat p=-i\sqrt{\frac{m\hbar\omega}{2}}\,(\hat a-\hat a^\dagger) x ^ = 2 mω ℏ ( a ^ + a ^ † ) , p ^ = − i 2 m ℏ ω ( a ^ − a ^ † )
[2] 交換関係 [ H ^ , a ^ † a ^ ] [\hat H,\hat a^\dagger\hat a] [ H ^ , a ^ † a ^ ]
[ H ^ , a ^ † a ^ ] = 0 [\hat H,\hat a^\dagger\hat a]=0 [ H ^ , a ^ † a ^ ] = 0 H ^ = ℏ ω ( a ^ † a ^ + 1 2 ) \hat H=\hbar\omega(\hat a^\dagger\hat a+\tfrac12) H ^ = ℏ ω ( a ^ † a ^ + 2 1 )
[3] 基底状態の位置表示 ⟨ x ∣ 0 ⟩ \braket{x|0} ⟨ x ∣0 ⟩
⟨ x ∣ 0 ⟩ = ( m ω π ℏ ) 1 / 4 exp ( − m ω 2 ℏ x 2 ) \displaystyle\braket{x|0}=\Bigl(\frac{m\omega}{\pi\hbar}\Bigr)^{1/4}\exp\Bigl(-\frac{m\omega}{2\hbar}x^2\Bigr) ⟨ x ∣0 ⟩ = ( π ℏ mω ) 1/4 exp ( − 2ℏ mω x 2 )
[4] 運動量の固有ケットの位置表示 ⟨ x ∣ p ⟩ \braket{x|p} ⟨ x ∣ p ⟩
⟨ x ∣ p ⟩ = 1 2 π ℏ e i p x / ℏ \displaystyle\braket{x|p}=\frac1{\sqrt{2\pi\hbar}}\,e^{ipx/\hbar} ⟨ x ∣ p ⟩ = 2 π ℏ 1 e i p x /ℏ
[5] 運動量表示 ⟨ p ∣ 0 ⟩ \braket{p|0} ⟨ p ∣0 ⟩ σ \sigma σ
σ = 2 m ℏ ω \displaystyle\sigma=\sqrt{2m\hbar\omega} σ = 2 m ℏ ω ⟨ p ∣ 0 ⟩ = ( π m ℏ ω ) − 1 / 4 exp ( − p 2 2 m ℏ ω ) \displaystyle\braket{p|0}=(\pi m\hbar\omega)^{-1/4}\exp\Bigl(-\frac{p^2}{2m\hbar\omega}\Bigr) ⟨ p ∣0 ⟩ = ( π m ℏ ω ) − 1/4 exp ( − 2 m ℏ ω p 2 )
[6] 第1励起状態の位置表示 ⟨ x ∣ 1 ⟩ \braket{x|1} ⟨ x ∣1 ⟩
⟨ x ∣ 1 ⟩ = m ω 2 ℏ ( x − ℏ m ω d d x ) ⟨ x ∣ 0 ⟩ ( = 2 m ω ℏ x ⟨ x ∣ 0 ⟩ ) \displaystyle\braket{x|1}=\sqrt{\frac{m\omega}{2\hbar}}\Bigl(x-\frac\hbar{m\omega}\frac{d}{dx}\Bigr)\braket{x|0}\ \Bigl(=\sqrt{\frac{2m\omega}\hbar}\,x\braket{x|0}\Bigr) ⟨ x ∣1 ⟩ = 2ℏ mω ( x − mω ℏ d x d ) ⟨ x ∣0 ⟩ ( = ℏ 2 mω x ⟨ x ∣0 ⟩ )