東京大学 新領域創成科学研究科 複雑理工学専攻 2018年8月実施 専門基礎科目 第8問 Author Miyake
Description Consider a one-dimensional quantum mechanical harmonic oscillator described by the Hamiltonian
H ^ = p ^ 2 2 m + m ω 2 x ^ 2 2 , \hat H=\frac{\hat p^2}{2m}+\frac{m\omega^2\hat x^2}{2}, H ^ = 2 m p ^ 2 + 2 m ω 2 x ^ 2 , where m m m ω \omega ω x ^ \hat x x ^ p ^ \hat p p ^ a ^ † \hat a^\dagger a ^ † a ^ \hat a a ^
a ^ † = m ω 2 ℏ ( x ^ − i m ω p ^ ) , a ^ = m ω 2 ℏ ( x ^ + i m ω p ^ ) . \begin{gathered}
\hat a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x-\frac{i}{m\omega}\hat p\right),\\
\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i}{m\omega}\hat p\right).
\end{gathered} a ^ † = 2ℏ mω ( x ^ − mω i p ^ ) , a ^ = 2ℏ mω ( x ^ + mω i p ^ ) . Here, i i i ℏ \hbar ℏ h h h 2 π 2\pi 2 π [ a ^ , a ^ † ] = 1 [\hat a,\hat a^\dagger]=1 [ a ^ , a ^ † ] = 1 A A A A ^ \hat A A ^ ⟨ A ^ ⟩ \langle\hat A\rangle ⟨ A ^ ⟩
Δ A ≡ ⟨ ( A ^ − ⟨ A ^ ⟩ ) 2 ⟩ . \Delta A\equiv\sqrt{\langle(\hat A-\langle\hat A\rangle)^2\rangle}. Δ A ≡ ⟨( A ^ − ⟨ A ^ ⟩ ) 2 ⟩ . (Q.1) Express the position operator x ^ \hat x x ^ p ^ \hat p p ^ a ^ † \hat a^\dagger a ^ † a ^ \hat a a ^
(Q.2) The ground state ∣ 0 ⟩ |0\rangle ∣0 ⟩ a ^ ∣ 0 ⟩ = 0 \hat a|0\rangle=0 a ^ ∣0 ⟩ = 0
(1) Find the expectation values ⟨ x ^ ⟩ \langle\hat x\rangle ⟨ x ^ ⟩ ⟨ x ^ 2 ⟩ \langle\hat x^2\rangle ⟨ x ^ 2 ⟩ ⟨ p ^ ⟩ \langle\hat p\rangle ⟨ p ^ ⟩ ⟨ p ^ 2 ⟩ \langle\hat p^2\rangle ⟨ p ^ 2 ⟩
(2) Find the product of the standard deviations of the position and the momentum ( Δ x Δ p ) (\Delta x\Delta p) ( Δ x Δ p )
(Q.3) Rewrite the Hamiltonian H ^ \hat H H ^ n ^ = a ^ † a ^ \hat n=\hat a^\dagger\hat a n ^ = a ^ † a ^
(Q.4) Express the commutation relation [ a ^ , ( a ^ † ) n ] [\hat a,(\hat a^\dagger)^n] [ a ^ , ( a ^ † ) n ] n n n a ^ † \hat a^\dagger a ^ † n n n
(Q.5) The n n n n n n
∣ n ⟩ = 1 n ! ( a ^ † ) n ∣ 0 ⟩ . |n\rangle=\frac1{\sqrt{n!}}(\hat a^\dagger)^n|0\rangle. ∣ n ⟩ = n ! 1 ( a ^ † ) n ∣0 ⟩ . Show that ∣ n ⟩ |n\rangle ∣ n ⟩ n ^ \hat n n ^ ∣ n ⟩ |n\rangle ∣ n ⟩ H ^ \hat H H ^
(Q.6) Consider a state expressed in terms of a complex number α \alpha α
∣ α ⟩ = e − ∣ α ∣ 2 / 2 e α a ^ † ∣ 0 ⟩ = e − ∣ α ∣ 2 / 2 ∑ n = 0 ∞ α n n ! ( a ^ † ) n ∣ 0 ⟩ . \begin{gathered}
|\alpha\rangle=e^{-|\alpha|^2/2}e^{\alpha\hat a^\dagger}|0\rangle\\
=e^{-|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{n!}(\hat a^\dagger)^n|0\rangle.
\end{gathered} ∣ α ⟩ = e − ∣ α ∣ 2 /2 e α a ^ † ∣0 ⟩ = e − ∣ α ∣ 2 /2 n = 0 ∑ ∞ n ! α n ( a ^ † ) n ∣0 ⟩ . Answer the following questions on this state. e e e
(1) Show that this state is an eigenstate of the annihilation operator a ^ \hat a a ^
(2) Obtain the expectation value ⟨ n ^ ⟩ \langle\hat n\rangle ⟨ n ^ ⟩ Δ n \Delta n Δ n n ^ = a ^ † a ^ \hat n=\hat a^\dagger\hat a n ^ = a ^ † a ^
题目描述
考虑由下列 Hamiltonian 描述的一维量子谐振子：
H ^ = p ^ 2 2 m + m ω 2 x ^ 2 2 \hat H=\frac{\hat p^2}{2m}+\frac{m\omega^2\hat x^2}{2} H ^ = 2 m p ^ 2 + 2 m ω 2 x ^ 2 其中 m m m ω \omega ω x ^ \hat x x ^ p ^ \hat p p ^ a ^ † \hat a^\dagger a ^ † a ^ \hat a a ^
a ^ † = m ω 2 ℏ ( x ^ − i m ω p ^ ) , a ^ = m ω 2 ℏ ( x ^ + i m ω p ^ ) 。 \begin{gathered}
\hat a^\dagger=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x-\frac{i}{m\omega}\hat p\right),\\
\hat a=\sqrt{\frac{m\omega}{2\hbar}}\left(\hat x+\frac{i}{m\omega}\hat p\right)。
\end{gathered} a ^ † = 2ℏ mω ( x ^ − mω i p ^ ) , a ^ = 2ℏ mω ( x ^ + mω i p ^ ) 。 i i i ℏ \hbar ℏ h h h 2 π 2\pi 2 π [ a ^ , a ^ † ] = 1 [\hat a,\hat a^\dagger]=1 [ a ^ , a ^ † ] = 1 A ^ \hat A A ^ A A A ⟨ A ^ ⟩ \langle\hat A\rangle ⟨ A ^ ⟩
Δ A ≡ ⟨ ( A ^ − ⟨ A ^ ⟩ ) 2 ⟩ \Delta A\equiv\sqrt{\langle(\hat A-\langle\hat A\rangle)^2\rangle} Δ A ≡ ⟨( A ^ − ⟨ A ^ ⟩ ) 2 ⟩ (Q.1) 用产生算符 a ^ † \hat a^\dagger a ^ † a ^ \hat a a ^ x ^ \hat x x ^ p ^ \hat p p ^
(Q.2) 基态 ∣ 0 ⟩ |0\rangle ∣0 ⟩ a ^ ∣ 0 ⟩ = 0 \hat a|0\rangle=0 a ^ ∣0 ⟩ = 0
(1) 求期望值 ⟨ x ^ ⟩ \langle\hat x\rangle ⟨ x ^ ⟩ ⟨ x ^ 2 ⟩ \langle\hat x^2\rangle ⟨ x ^ 2 ⟩ ⟨ p ^ ⟩ \langle\hat p\rangle ⟨ p ^ ⟩ ⟨ p ^ 2 ⟩ \langle\hat p^2\rangle ⟨ p ^ 2 ⟩
(2) 求位置与动量的标准差乘积 Δ x Δ p \Delta x\Delta p Δ x Δ p
(Q.3) 用粒子数算符 n ^ = a ^ † a ^ \hat n=\hat a^\dagger\hat a n ^ = a ^ † a ^ H ^ \hat H H ^
(Q.4) 用 a ^ † \hat a^\dagger a ^ † n n n [ a ^ , ( a ^ † ) n ] [\hat a,(\hat a^\dagger)^n] [ a ^ , ( a ^ † ) n ] n n n
(Q.5) 第 n n n n n n
∣ n ⟩ = 1 n ! ( a ^ † ) n ∣ 0 ⟩ |n\rangle=\frac1{\sqrt{n!}}(\hat a^\dagger)^n|0\rangle ∣ n ⟩ = n ! 1 ( a ^ † ) n ∣0 ⟩ 证明 ∣ n ⟩ |n\rangle ∣ n ⟩ n ^ \hat n n ^ ∣ n ⟩ |n\rangle ∣ n ⟩ H ^ \hat H H ^
(Q.6) 考虑用复数 α \alpha α
∣ α ⟩ = e − ∣ α ∣ 2 / 2 e α a ^ † ∣ 0 ⟩ = e − ∣ α ∣ 2 / 2 ∑ n = 0 ∞ α n n ! ( a ^ † ) n ∣ 0 ⟩ 。 \begin{gathered}
|\alpha\rangle=e^{-|\alpha|^2/2}e^{\alpha\hat a^\dagger}|0\rangle\\
=e^{-|\alpha|^2/2}\sum_{n=0}^\infty\frac{\alpha^n}{n!}(\hat a^\dagger)^n|0\rangle。
\end{gathered} ∣ α ⟩ = e − ∣ α ∣ 2 /2 e α a ^ † ∣0 ⟩ = e − ∣ α ∣ 2 /2 n = 0 ∑ ∞ n ! α n ( a ^ † ) n ∣0 ⟩ 。 就这一状态回答下列问题。e e e
(1) 证明该状态是湮灭算符 a ^ \hat a a ^
(2) 求粒子数算符 n ^ = a ^ † a ^ \hat n=\hat a^\dagger\hat a n ^ = a ^ † a ^ ⟨ n ^ ⟩ \langle\hat n\rangle ⟨ n ^ ⟩ Δ n \Delta n Δ n
Kai （問1）
x ^ = ℏ 2 m ω ( a ^ † + a ^ ) p ^ = i ℏ m ω 2 ( a ^ † − a ^ ) \begin{aligned}
\hat{x}
&=
\sqrt{\frac{\hbar}{2 m \omega}}
\left( \hat{a}^\dagger + \hat{a} \right)
\\
\hat{p}
&=
i \sqrt{\frac{\hbar m \omega}{2}}
\left( \hat{a}^\dagger - \hat{a} \right)
\end{aligned} x ^ p ^ = 2 mω ℏ ( a ^ † + a ^ ) = i 2 ℏ mω ( a ^ † − a ^ ) （問2）
(1)
a ^ ∣ 0 ⟩ = 0 \hat{a} | 0 \rangle = 0 a ^ ∣0 ⟩ = 0 ⟨ 0 ∣ a ^ † = 0 \langle 0 | \hat{a}^\dagger = 0 ⟨ 0∣ a ^ † = 0
⟨ x ^ ⟩ = ℏ 2 m ω ⟨ 0 ∣ ( a ^ † + a ^ ) ∣ 0 ⟩ = 0 ⟨ x ^ 2 ⟩ = ℏ 2 m ω ⟨ 0 ∣ ( ( a ^ † ) 2 + ( a ^ ) 2 + a ^ † a ^ + a ^ a ^ † ) ∣ 0 ⟩ = ℏ 2 m ω ⟨ 0 ∣ ( ( a ^ † ) 2 + ( a ^ ) 2 + 2 a ^ † a ^ + 1 ) ∣ 0 ⟩ = ℏ 2 m ω ⟨ p ^ ⟩ = i ℏ m ω 2 ⟨ 0 ∣ ( a ^ † − a ^ ) ∣ 0 ⟩ = 0 ⟨ p ^ 2 ⟩ = − ℏ m ω 2 ⟨ 0 ∣ ( ( a ^ † ) 2 + ( a ^ ) 2 − a ^ † a ^ − a ^ a ^ † ) ∣ 0 ⟩ = − ℏ m ω 2 ⟨ 0 ∣ ( ( a ^ † ) 2 + ( a ^ ) 2 − 2 a ^ † a ^ − 1 ) ∣ 0 ⟩ = ℏ m ω 2 \begin{aligned}
\langle \hat{x} \rangle
&=
\sqrt{\frac{\hbar}{2 m \omega}}
\langle 0 | \left( \hat{a}^\dagger + \hat{a} \right) | 0 \rangle\\
&=
0\\
\langle \hat{x}^2 \rangle
&=
\frac{\hbar}{2 m \omega}
\langle 0 | \left(
\left( \hat{a}^\dagger \right)^2 + \left( \hat{a} \right)^2 +
\hat{a}^\dagger \hat{a} + \hat{a} \hat{a}^\dagger
\right) | 0 \rangle\\
&=
\frac{\hbar}{2 m \omega}
\langle 0 | \left(
\left( \hat{a}^\dagger \right)^2 + \left( \hat{a} \right)^2 +
2 \hat{a}^\dagger \hat{a} + 1
\right) | 0 \rangle\\
&=
\frac{\hbar}{2 m \omega}\\
\langle \hat{p} \rangle
&=
i \sqrt{\frac{\hbar m \omega}{2}}
\langle 0 | \left( \hat{a}^\dagger - \hat{a} \right) | 0 \rangle\\
&=
0\\
\langle \hat{p}^2 \rangle
&=
- \frac{\hbar m \omega}{2}
\langle 0 | \left(
\left( \hat{a}^\dagger \right)^2 + \left( \hat{a} \right)^2 -
\hat{a}^\dagger \hat{a} - \hat{a} \hat{a}^\dagger
\right) | 0 \rangle\\
&=
- \frac{\hbar m \omega}{2}
\langle 0 | \left(
\left( \hat{a}^\dagger \right)^2 + \left( \hat{a} \right)^2 -
2 \hat{a}^\dagger \hat{a} - 1
\right) | 0 \rangle\\
&=
\frac{\hbar m \omega}{2}
\end{aligned} ⟨ x ^ ⟩ ⟨ x ^ 2 ⟩ ⟨ p ^ ⟩ ⟨ p ^ 2 ⟩ = 2 mω ℏ ⟨ 0∣ ( a ^ † + a ^ ) ∣0 ⟩ = 0 = 2 mω ℏ ⟨ 0∣ ( ( a ^ † ) 2 + ( a ^ ) 2 + a ^ † a ^ + a ^ a ^ † ) ∣0 ⟩ = 2 mω ℏ ⟨ 0∣ ( ( a ^ † ) 2 + ( a ^ ) 2 + 2 a ^ † a ^ + 1 ) ∣0 ⟩ = 2 mω ℏ = i 2 ℏ mω ⟨ 0∣ ( a ^ † − a ^ ) ∣0 ⟩ = 0 = − 2 ℏ mω ⟨ 0∣ ( ( a ^ † ) 2 + ( a ^ ) 2 − a ^ † a ^ − a ^ a ^ † ) ∣0 ⟩ = − 2 ℏ mω ⟨ 0∣ ( ( a ^ † ) 2 + ( a ^ ) 2 − 2 a ^ † a ^ − 1 ) ∣0 ⟩ = 2 ℏ mω を得る。（ ⟨ 0 ∣ 0 ⟩ = 1 \langle 0 | 0 \rangle = 1 ⟨ 0∣0 ⟩ = 1
(2)
Δ A = ⟨ ( A ^ − ⟨ A ^ ⟩ ) 2 ⟩ = ⟨ A ^ 2 ⟩ − ⟨ A ^ ⟩ 2 \begin{aligned}
\Delta A
&=
\sqrt{ \langle ( \hat{A} - \langle \hat{A} \rangle )^2 \rangle }
\\
&=
\sqrt{ \langle \hat{A}^2 \rangle - \langle \hat{A} \rangle^2 }
\end{aligned} Δ A = ⟨( A ^ − ⟨ A ^ ⟩ ) 2 ⟩ = ⟨ A ^ 2 ⟩ − ⟨ A ^ ⟩ 2 であるから、 (1) より、
Δ x Δ p = ℏ 2 m ω ⋅ ℏ m ω 2 = ℏ 2 \begin{aligned}
\Delta x \Delta p
&=
\sqrt{ \frac{\hbar}{2 m \omega} } \cdot
\sqrt{ \frac{\hbar m \omega}{2} }
\\
&=
\frac{\hbar}{2}
\end{aligned} Δ x Δ p = 2 mω ℏ ⋅ 2 ℏ mω = 2 ℏ を得る。
（問3）
H ^ = ℏ ω ( n ^ + 1 2 ) \begin{aligned}
&
\hat{H}
=
\hbar \omega \left( \hat{n} + \frac{1}{2} \right)
\end{aligned} H ^ = ℏ ω ( n ^ + 2 1 ) （問4）
[ a ^ , a ^ † ] = 1 [ a ^ , ( a ^ † ) 2 ] = a ^ † [ a ^ , a ^ † ] + [ a ^ , a ^ † ] a ^ † = 2 a ^ † [ a ^ , ( a ^ † ) 3 ] = a ^ † [ a ^ , ( a ^ † ) 2 ] + [ a ^ , a ^ † ] ( a ^ † ) 2 = 3 ( a ^ † ) 2 [ a ^ , ( a ^ † ) 4 ] = a ^ † [ a ^ , ( a ^ † ) 3 ] + [ a ^ , a ^ † ] ( a ^ † ) 3 = 4 ( a ^ † ) 3 \begin{aligned}
\left[ \hat{a}, \hat{a}^\dagger \right]
&=
1
\\
\left[ \hat{a}, \left( \hat{a}^\dagger \right)^2 \right]
&=
\hat{a}^\dagger \left[ \hat{a}, \hat{a}^\dagger \right] +
\left[ \hat{a}, \hat{a}^\dagger \right] \hat{a}^\dagger
\\
&=
2 \hat{a}^\dagger
\\
\left[ \hat{a}, \left( \hat{a}^\dagger \right)^3 \right]
&=
\hat{a}^\dagger
\left[ \hat{a}, \left( \hat{a}^\dagger \right)^2 \right] +
\left[ \hat{a}, \hat{a}^\dagger \right]
\left( \hat{a}^\dagger \right)^2
\\
&=
3 \left( \hat{a}^\dagger \right)^2
\\
\left[ \hat{a}, \left( \hat{a}^\dagger \right)^4 \right]
&=
\hat{a}^\dagger
\left[ \hat{a}, \left( \hat{a}^\dagger \right)^3 \right] +
\left[ \hat{a}, \hat{a}^\dagger \right]
\left( \hat{a}^\dagger \right)^3
\\
&=
4 \left( \hat{a}^\dagger \right)^3
\end{aligned} [ a ^ , a ^ † ] [ a ^ , ( a ^ † ) 2 ] [ a ^ , ( a ^ † ) 3 ] [ a ^ , ( a ^ † ) 4 ] = 1 = a ^ † [ a ^ , a ^ † ] + [ a ^ , a ^ † ] a ^ † = 2 a ^ † = a ^ † [ a ^ , ( a ^ † ) 2 ] + [ a ^ , a ^ † ] ( a ^ † ) 2 = 3 ( a ^ † ) 2 = a ^ † [ a ^ , ( a ^ † ) 3 ] + [ a ^ , a ^ † ] ( a ^ † ) 3 = 4 ( a ^ † ) 3 一般に [ A , B C ] = [ A , B ] C + B [ A , C ] [A,BC]=[A,B]C+B[A,C] [ A , BC ] = [ A , B ] C + B [ A , C ]
[ a ^ , ( a ^ † ) n ] = n ( a ^ † ) n − 1 \begin{aligned}
\left[ \hat{a}, \left( \hat{a}^\dagger \right)^n \right]
&=
n \left( \hat{a}^\dagger \right)^{n-1}
\end{aligned} [ a ^ , ( a ^ † ) n ] = n ( a ^ † ) n − 1 である。
（問5）
n ^ ∣ n ⟩ = 1 n ! a ^ † a ^ ( a ^ † ) n ∣ 0 ⟩ = 1 n ! a ^ † { ( a ^ † ) n a ^ + n ( a ^ † ) n − 1 } ∣ 0 ⟩ = n ∣ n ⟩ \begin{aligned}
\hat{n} | n \rangle
&=
\frac{1}{\sqrt{n!}} \hat{a}^\dagger \hat{a}
\left( \hat{a}^\dagger \right)^n | 0 \rangle
\\
&=
\frac{1}{\sqrt{n!}} \hat{a}^\dagger
\left\{
\left( \hat{a}^\dagger \right)^n \hat{a}
+ n \left( \hat{a}^\dagger \right)^{n-1}
\right\} | 0 \rangle
\\
&=
n | n \rangle
\end{aligned} n ^ ∣ n ⟩ = n ! 1 a ^ † a ^ ( a ^ † ) n ∣0 ⟩ = n ! 1 a ^ † { ( a ^ † ) n a ^ + n ( a ^ † ) n − 1 } ∣0 ⟩ = n ∣ n ⟩ であるから、
∣ n ⟩ | n \rangle ∣ n ⟩ n ^ \hat{n} n ^ n n n ∣ n ⟩ | n \rangle ∣ n ⟩ H ^ \hat{H} H ^
ℏ ω ( n + 1 2 ) \begin{aligned}
&
\hbar \omega \left( n + \frac{1}{2} \right)
\end{aligned} ℏ ω ( n + 2 1 ) に属する固有状態である。
（問6）
(1)
a ^ ∣ α ⟩ = e − ∣ α ∣ 2 2 ∑ n = 0 ∞ α n n ! a ^ ( a ^ † ) n ∣ 0 ⟩ = e − ∣ α ∣ 2 2 ∑ n = 1 ∞ α n n ! a ^ ( a ^ † ) n ∣ 0 ⟩ ( ∵ a ^ ∣ 0 ⟩ = 0 ) = e − ∣ α ∣ 2 2 ∑ n = 1 ∞ α n n ! { ( a ^ † ) n a ^ + n ( a ^ † ) n − 1 } ∣ 0 ⟩ ( ∵ （問4） ) = e − ∣ α ∣ 2 2 ∑ n = 1 ∞ α n ( n − 1 ) ! ( a ^ † ) n − 1 ∣ 0 ⟩ ( ∵ a ^ ∣ 0 ⟩ = 0 ) = α ∣ α ⟩ \begin{aligned}
\hat{a} | \alpha \rangle
&=
e^{ - \frac{|\alpha|^2}{2} } \sum_{n=0}^\infty
\frac{\alpha^n}{n!} \hat{a} \left( \hat{a}^\dagger \right)^n
| 0 \rangle\\
&=
e^{ - \frac{|\alpha|^2}{2} } \sum_{n=1}^\infty
\frac{\alpha^n}{n!} \hat{a} \left( \hat{a}^\dagger \right)^n
| 0 \rangle
\ \ \ \ \ \ \ \
( \because \hat{a} | 0 \rangle = 0 )\\
&=
e^{ - \frac{|\alpha|^2}{2} } \sum_{n=1}^\infty
\frac{\alpha^n}{n!}
\left\{ \left( \hat{a}^\dagger \right)^n \hat{a}
+ n \left( \hat{a}^\dagger \right)^{n-1}
\right\}
| 0 \rangle
\ \ \ \ \ \ \ \
( \because \text{（問4）} )\\
&=
e^{ - \frac{|\alpha|^2}{2} } \sum_{n=1}^\infty
\frac{\alpha^n}{(n-1)!} \left( \hat{a}^\dagger \right)^{n-1}
| 0 \rangle
\ \ \ \ \ \ \ \
( \because \hat{a} | 0 \rangle = 0 )\\
&=
\alpha | \alpha \rangle
\end{aligned} a ^ ∣ α ⟩ = e − 2 ∣ α ∣ 2 n = 0 ∑ ∞ n ! α n a ^ ( a ^ † ) n ∣0 ⟩ = e − 2 ∣ α ∣ 2 n = 1 ∑ ∞ n ! α n a ^ ( a ^ † ) n ∣0 ⟩ ( ∵ a ^ ∣0 ⟩ = 0 ) = e − 2 ∣ α ∣ 2 n = 1 ∑ ∞ n ! α n { ( a ^ † ) n a ^ + n ( a ^ † ) n − 1 } ∣0 ⟩ ( ∵ （問 4 ） ) = e − 2 ∣ α ∣ 2 n = 1 ∑ ∞ ( n − 1 )! α n ( a ^ † ) n − 1 ∣0 ⟩ ( ∵ a ^ ∣0 ⟩ = 0 ) = α ∣ α ⟩ であるから、 ∣ α ⟩ | \alpha \rangle ∣ α ⟩ a ^ \hat{a} a ^ α \alpha α
(2)
⟨ n ^ ⟩ = ⟨ α ∣ a ^ † a ^ ∣ α ⟩ = ∣ α ∣ 2 ⟨ n ^ 2 ⟩ = ⟨ α ∣ a ^ † a ^ a ^ † a ^ ∣ α ⟩ = ∣ α ∣ 2 ⟨ α ∣ a ^ a ^ † ∣ α ⟩ = ∣ α ∣ 2 ⟨ α ∣ ( a ^ † a ^ + 1 ) ∣ α ⟩ = ∣ α ∣ 2 ( ∣ α ∣ 2 + 1 ) ∴ Δ n 2 = ⟨ n ^ 2 ⟩ − ⟨ n ^ ⟩ 2 = ∣ α ∣ 2 ( ∣ α ∣ 2 + 1 ) − ∣ α ∣ 4 = ∣ α ∣ 2 ∴ Δ n = ∣ α ∣ \begin{aligned}
\langle \hat{n} \rangle
&=
\langle \alpha | \hat{a}^\dagger \hat{a} | \alpha \rangle
\\
&=
| \alpha |^2
\\
\langle \hat{n}^2 \rangle
&=
\langle \alpha | \hat{a}^\dagger \hat{a} \hat{a}^\dagger \hat{a}
| \alpha \rangle
\\
&=
| \alpha |^2
\langle \alpha | \hat{a} \hat{a}^\dagger | \alpha \rangle
\\
&=
| \alpha |^2
\langle \alpha |
\left( \hat{a}^\dagger \hat{a} + 1 \right)
| \alpha \rangle
\\
&=
| \alpha |^2 \left( | \alpha |^2 + 1 \right)
\\
\therefore \ \ \ \
\Delta n^2
&=
\langle \hat{n}^2 \rangle - \langle \hat{n} \rangle^2
\\
&=
| \alpha |^2 \left( | \alpha |^2 + 1 \right) - | \alpha |^4
\\
&=
| \alpha |^2
\\
\therefore \ \ \ \
\Delta n
&=
| \alpha |
\end{aligned} ⟨ n ^ ⟩ ⟨ n ^ 2 ⟩ ∴ Δ n 2 ∴ Δ n = ⟨ α ∣ a ^ † a ^ ∣ α ⟩ = ∣ α ∣ 2 = ⟨ α ∣ a ^ † a ^ a ^ † a ^ ∣ α ⟩ = ∣ α ∣ 2 ⟨ α ∣ a ^ a ^ † ∣ α ⟩ = ∣ α ∣ 2 ⟨ α ∣ ( a ^ † a ^ + 1 ) ∣ α ⟩ = ∣ α ∣ 2 ( ∣ α ∣ 2 + 1 ) = ⟨ n ^ 2 ⟩ − ⟨ n ^ ⟩ 2 = ∣ α ∣ 2 ( ∣ α ∣ 2 + 1 ) − ∣ α ∣ 4 = ∣ α ∣ 2 = ∣ α ∣ なお、⟨ 0 ∣ 0 ⟩ = 1 \langle0|0\rangle=1 ⟨ 0∣0 ⟩ = 1 ⟨ m ∣ n ⟩ = δ m n \langle m|n\rangle=\delta_{mn} ⟨ m ∣ n ⟩ = δ mn
⟨ α ∣ α ⟩ = e − ∣ α ∣ 2 ∑ n = 0 ∞ ∣ α ∣ 2 n n ! = 1. \langle\alpha|\alpha\rangle
=e^{-|\alpha|^2}\sum_{n=0}^\infty\frac{|\alpha|^{2n}}{n!}=1. ⟨ α ∣ α ⟩ = e − ∣ α ∣ 2 n = 0 ∑ ∞ n ! ∣ α ∣ 2 n = 1. Reference