Calculations on Coherent States and Squeezed States (1)

作者: D.F. Walls,Gerard J. Milburn
出版社/メーカー: Springer
発売日: 2010/02/12
メディア: ペーパーバック
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This is intended to fill some gaps in calculation details in the book above.

Basic formulas and relationships

　 $\displaystyle e^ABe^{-A} = \sum_{n=0}^\infty \frac{1}{n!}({\rm C}_A)^n B$ 　where ${\rm C}_A B := [A,\,B]$ ----- (1.1)

　 $\displaystyle e^{A+B} = e^Ae^Be^{-\frac{1}{2}[A,\,B]}$ 　when $[A,\,B]$ is commutable with $A$ and $B$ ----- (1.2)

　 $[A,\,BC] = [A,\,B]C + B[A,\,C]$

　 $[a,\,a^\dagger]=1$

　 $[a,\,a^{\dagger n}] = na^{\dagger (n-1)}$

　 $a{\mid 0\rangle}=0$

　 $\displaystyle{\mid n\rangle} := \frac{1}{\sqrt{n!}}a^{\dagger n}{\mid 0 \rangle}$

　 $a^\dagger a{\mid n\rangle}=n{\mid n\rangle}$

2.3 Coherent States

Displacement operator $D(\alpha) := \exp(\alpha a^\dagger-\alpha^*a)$

　1. $D(-\alpha) = D^\dagger(\alpha)$
　2. $D(\alpha)$ is Unitary. i.e. $D^\dagger(\alpha)D(\alpha)=1$

1. is obvious. For 2., using anti-Hermite $A:=-(\alpha a^\dagger-\alpha^*a)$ (i.e $A^\dagger = -A$ ), $D(\alpha)=e^{-A}$ . Hence $D^\dagger(\alpha)D(\alpha) = e^Ae^{-A}=1$ .

　3. $D^\dagger(\alpha)a D(\alpha) = a+\alpha$
　4. $D^\dagger(\alpha)a^\dagger D(\alpha) = a^\dagger+\alpha^*$

For 3., $[A,\,a] = [-(\alpha a^\dagger-\alpha^*a),\,a] = \alpha$ . Hence, from (1.1), $D^\dagger(\alpha)a D(\alpha) = e^Aae^{-A} = a +\alpha$ . 4. is conjugate of 3.

　5. $D^\dagger(\alpha)a^n D(\alpha) = (a+\alpha)^n$
　6. $D^\dagger(\alpha)a^{\dagger n} D(\alpha) = (a^\dagger+\alpha^*)^n$

For 5., $D^\dagger(\alpha)a^n D(\alpha) = (D^\dagger(\alpha)a D(\alpha))^n$ since $D^\dagger(\alpha)D(\alpha)=1$ . 6. is conjugate of 5.

　7. $D^\dagger(\alpha)a^{\dagger n}a^m D(\alpha) = (a^\dagger+\alpha^*)^n(a+\alpha)^m$
　8. $D^\dagger(\alpha)a^{n}a^{\dagger m} D(\alpha) = (a+\alpha)^n(a^\dagger+\alpha^*)^m$

For 7., $D^\dagger(\alpha)a^{\dagger n}a^m D(\alpha)=(D^\dagger(\alpha)a^\dagger D(\alpha))^n(D^\dagger(\alpha)a D(\alpha))^m = (a^\dagger+\alpha^*)^n(a+\alpha)^m$ . 8. is conjugate of 7.

Coherent state $\mid \alpha\rangle := D(\alpha){\mid 0\rangle}$

　9. $a{\mid \alpha\rangle} = \alpha{\mid \alpha\rangle}$

From 3., $a D(\alpha) = D(\alpha)(a+\alpha)$ . Hence $a{\mid \alpha\rangle}=a D(\alpha){\mid 0\rangle} = D(\alpha)(a+\alpha){\mid 0\rangle} = \alpha D(\alpha){\mid 0\rangle}=\alpha{\mid \alpha\rangle}$

　10. $\displaystyle\mid \alpha\rangle = e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}{\mid n\rangle}$
　11. $\displaystyle P(n) = |{\langle n\mid \alpha\rangle}|^2 = \frac{1}{n!}|\alpha|^{2n}e^{-|\alpha|^2}$
　12. $\displaystyle\overline n = {\langle\alpha\mid}a^{\dagger}a{\mid\alpha\rangle}=|\alpha|^2$

For 10., using (1.2), $\displaystyle D(\alpha) = e^{-\frac{1}{2}|\alpha|^2}e^{\alpha a^\dagger}e^{-\alpha^*a}$ since $[\alpha a^\dagger,\,-\alpha^*a] = -|\alpha|^2$ . Then $\displaystyle D(\alpha){\mid 0\rangle} = e^{-\frac{1}{2}|\alpha|^2} e^{\alpha a^{\dagger}} e^{-\alpha^*a}{\mid 0\rangle} = e^{-\frac{1}{2}|\alpha|^2} e^{\alpha a^\dagger} {\mid 0\rangle} = e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^\infty\frac{\alpha^n}{n!} a^{\dagger n}{\mid 0\rangle} = e^{-\frac{1}{2}|\alpha|^2}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}} {\mid n\rangle}$ .

11. directly follows from 10. For 12., using 7. ${\langle\alpha\mid}a^{\dagger}a{\mid\alpha\rangle}={\langle 0\mid}D^\dagger(\alpha)a^{\dagger}aD(\alpha){\mid 0\rangle} ={\langle 0\mid}(a^\dagger+\alpha^*)(a+\alpha){\mid 0\rangle} = |\alpha|^2$ .

From 11. and 12., you can see that the distribution of photon numbers $P(n)$ follows the Poisson distribution, and the average is $\overline n = |\alpha|^2$

2.4 Squeezed States

Quadrature operators $X_1 := a + a^\dagger,\ X_2 := -i(a-a^\dagger)$ , equivalently, $\displaystyle a = \frac{1}{2}(X_1+iX_2),\ a^\dagger = \frac{1}{2}(X_1-iX_2)$

　20. ${\langle\alpha\mid}X_1{\mid\alpha\rangle} = \alpha + \alpha^*$
　21. ${\langle\alpha\mid}X_2{\mid\alpha\rangle} = -i(\alpha - \alpha^*)$

From 3. and 4., ${\langle\alpha\mid}a{\mid\alpha\rangle} = \alpha,\ {\langle\alpha\mid}a^\dagger{\mid\alpha\rangle} = \alpha^*$ .

　22. ${\langle\alpha\mid}X_1^2{\mid\alpha\rangle} = \alpha^2+\alpha^{*2}+2|\alpha|^2 + 1$
　23. ${\langle\alpha\mid}X_2^2{\mid\alpha\rangle} = -\alpha^2-\alpha^{*2}+2|\alpha|^2 + 1$

For 22., using 5. to 8.,
${\langle\alpha\mid}X_1^2{\mid\alpha\rangle}={\langle\alpha\mid}(a^2+a^{\dagger 2}+aa^\dagger+a^\dagger a){\mid\alpha\rangle}$
　 $={\langle 0\mid}\left\{(a+\alpha)^2+(a^\dagger+\alpha^*)^2+(a+\alpha)(a^\dagger+\alpha^*)+(a^\dagger+\alpha^*)(a+\alpha)\right\}{\mid 0\rangle}$
　 $=\alpha^2+\alpha^{\dagger 2} + {\langle 0\mid}aa^\dagger{\mid 0\rangle}+2|\alpha|^2$
　 $=\alpha^2+\alpha^{\dagger 2} +2|\alpha|^2 + 1$
23. is the same as 22.

　24. $(\Delta X_1)^2 := {\langle\alpha\mid}X_1^2{\mid\alpha\rangle} - ({\langle\alpha\mid}X_1{\mid\alpha\rangle})^2 = 1$
　25. $(\Delta X_2)^2 := {\langle\alpha\mid}X_2^2{\mid\alpha\rangle} - ({\langle\alpha\mid}X_2{\mid\alpha\rangle})^2 = 1$

Directly follows from 20. to 23.

Squeeze operator $S(\epsilon) := \exp\frac{1}{2}(\epsilon^*a^2-\epsilon a^{\dagger 2})$ 　where $\epsilon = r e^{2i\phi}$

　26. $S(-\epsilon) =- S^\dagger(\epsilon)$
　27. $S(\epsilon)$ is Unitary. i.e. $S^\dagger(\epsilon)S(\epsilon)=1$

26. is obvious. For 27., using anti-Hermie $\displaystyle A:= -\frac{1}{2}(\epsilon^*a^2-\epsilon a^{\dagger 2}),\,S(\epsilon)=e^{-A}$ . Hence $S^\dagger(\epsilon)S(\epsilon)=e^Ae^{-A}=1$

　28. $S^\dagger(\epsilon)aS(\epsilon) = a\cosh r-a^\dagger e^{2i\phi}\sinh r$
　29. $S^\dagger(\epsilon)a^\dagger S(\epsilon) = a^\dagger\cosh r-a e^{-2i\phi}\sinh r$

For 28., we use (1.1) with $\displaystyle A:= -\frac{1}{2}(\epsilon^*a^2-\epsilon a^{\dagger 2})$ and $B=a$ .
$\displaystyle {\rm C}_A a = -\frac{1}{2}[\epsilon^*a^2-\epsilon a^{\dagger 2},\,a] = -\frac{1}{2}\epsilon[a,\,a^{\dagger 2}] = -\epsilon a^\dagger$
$\displaystyle {\rm C}^2_A a = -\frac{1}{2}[\epsilon^*a^2-\epsilon a^{\dagger 2},\,-\epsilon a^\dagger] = \frac{1}{2}|\epsilon|^2 [a^2,\,a^\dagger] =r^2 a$
By mathematical induction, in general,
$\displaystyle {\rm C}^n_A a =\begin{cases}-\epsilon r^{n-1}a^\dagger=-e^{2i\phi} r^{n}a^\dagger & (n=1,3,5,\cdots)\\r^{n}a & (n=0, 2,4,\cdots)\end{cases}$
Hence,
$\displaystyle S^\dagger(\epsilon)aS(\epsilon) = \sum_{n=0}^\infty \frac{1}{n!}{\rm C}_A^n a = -e^{2i\phi}a^\dagger \sum_{k=0}^\infty \frac{r^{2k+1}}{(2k+1)!} + a\sum_{k=0}^\infty \frac{r^{2k}}{(2k)!}=a\cosh r-a^\dagger e^{2i\phi}\sinh r$

29. is conjugate of 28.

　30. $S^\dagger(\epsilon)a^nS(\epsilon) = (a\cosh r-a^\dagger e^{2i\phi}\sinh r)^n$
　31. $S^\dagger(\epsilon)a^{\dagger n}S(\epsilon) = (a^\dagger\cosh r-a e^{-2i\phi}\sinh r)^n$
　32. $S^\dagger(\epsilon)a^{\dagger m}a^nS(\epsilon) = (a^\dagger\cosh r-a e^{-2i\phi}\sinh r)^m(a\cosh r-a^\dagger e^{2i\phi}\sinh r)^n$

The same as 5., 6., 7.

　33. ${\langle 0\mid}S^\dagger(\epsilon)a^2S(\epsilon) {\mid 0\rangle} = -e^{2i\phi}\cosh r\sinh r$
　34. ${\langle 0\mid}S^\dagger(\epsilon)a^{\dagger 2}S(\epsilon) {\mid 0\rangle} = -e^{-2i\phi}\cosh r\sinh r$
　35. ${\langle 0\mid}S^\dagger(\epsilon)a^{\dagger}aS(\epsilon) {\mid 0\rangle} = \sinh^2 r$
　36. ${\langle 0\mid}S^\dagger(\epsilon)aa^{\dagger}S(\epsilon) {\mid 0\rangle} = \cosh^2 r$

For 33., using 30.,
${\langle 0\mid}S^\dagger(\epsilon)a^2S(\epsilon) {\mid 0\rangle} = {\langle 0\mid}(a\cosh r-a^\dagger e^{2i\phi}\sinh r)^2{\mid 0\rangle}$
　 $=-e^{2i\phi}\cosh r\sinh r{\langle 0\mid}aa^\dagger{\mid 0\rangle}=-e^{2i\phi}\cosh r\sinh r$
34. is conjugate of 33.

For 35., using 32.,
${\langle 0\mid}S^\dagger(\epsilon)a^{\dagger}aS(\epsilon) {\mid 0\rangle} ={\langle 0\mid} (a^\dagger\cosh r-a e^{-2i\phi}\sinh r)(a\cosh r-a^\dagger e^{2i\phi}\sinh r){\mid 0\rangle}$
　 $= \sinh^2 r {\langle 0\mid}aa^\dagger{\mid 0\rangle}=\sinh^2 r$
For 36., ${\langle 0\mid}S^\dagger(\epsilon)a^{\dagger}aS(\epsilon) {\mid 0\rangle} = {\langle 0\mid}S^\dagger(\epsilon)(aa^{\dagger}+1)S(\epsilon) {\mid 0\rangle}=\sinh^2 r+1=\cosh^2 r$

Squeezed state ${\mid\alpha,\,\epsilon\rangle} := D(\alpha)S(\epsilon){\mid 0\rangle}$

Rotated amplitude $Y_1 = e^{-i\phi}a + e^{i\phi}a^\dagger,\, Y_2=-i(e^{-i\phi}a-e^{i\phi}a^\dagger)$ , equivalently, $Y_1+iY_2 = e^{-i\phi}(X_1+iX_2)$

　37. ${\langle\alpha,\,\epsilon\mid}Y_1{\mid\alpha,\,\epsilon\rangle} = e^{-i\phi}\alpha+e^{i\phi}\alpha^*$
　38. ${\langle\alpha,\,\epsilon\mid}Y_2{\mid\alpha,\,\epsilon\rangle} = -i(e^{-i\phi}\alpha-e^{i\phi}\alpha^*)$

Since $Y_1+iY_2 = e^{-i\phi}(X_1+iX_2) = 2e^{-i\phi}a$ ,
${\langle\alpha,\,\epsilon\mid}(Y_1+iY_2){\mid\alpha,\,\epsilon\rangle}=2e^{-i\phi}{\langle 0\mid}S^\dagger(\epsilon)D^\dagger(\alpha)aD(\alpha)S(\epsilon){\mid 0\rangle}$
　 $= 2e^{-i\phi}{\langle 0\mid}S^\dagger(\epsilon)(a+\alpha)S(\epsilon){\mid 0\rangle}$ (from 3.)
　 $= 2e^{-i\phi}\alpha$ (from 28., ${\langle 0\mid}S^\dagger(\epsilon)aS(\epsilon){\mid 0\rangle}=0$ .)
Hence, ${\langle\alpha,\,\epsilon\mid}Y_1{\mid\alpha,\,\epsilon\rangle} = {\rm Re}\,(2e^{-i\phi}\alpha) = e^{-i\phi}\alpha+e^{i\phi}\alpha^*$ , and ${\langle\alpha,\,\epsilon\mid}Y_2{\mid\alpha,\,\epsilon\rangle}={\rm Im}\,(2e^{-i\phi}\alpha) = -i(e^{-i\phi}\alpha-e^{i\phi}\alpha^*)$ .

　39. ${\langle\alpha,\,\epsilon\mid}Y_1^2{\mid\alpha,\,\epsilon\rangle} = e^{-2i\phi}\alpha^2+e^{2i\phi}\alpha^{*2}+2|\alpha|^2 + e^{-2r}$
　40. ${\langle\alpha,\,\epsilon\mid}Y_2^2{\mid\alpha,\,\epsilon\rangle} = -(e^{-2i\phi}\alpha^2+e^{2i\phi}\alpha^{*2})+2|\alpha|^2 + e^{2r}$

Using 37. and 38. (double sign in same order),
${\langle\alpha,\,\epsilon\mid}Y_1^2{\mid\alpha,\,\epsilon\rangle},\,{\langle\alpha,\,\epsilon\mid}Y_2^2{\mid\alpha,\,\epsilon\rangle} = \pm{\langle 0\mid}S^\dagger(\epsilon)D^\dagger(\alpha)(e^{-i\phi}\alpha\pm e^{i\phi}\alpha^*)^2D(\alpha)S(\epsilon){\mid 0\rangle}$
　 $=\pm{\langle 0\mid}S^\dagger(\epsilon)D^\dagger(\alpha)\left\{e^{-2i\phi}a^2+e^{2i\phi}a^{\dagger 2}\pm (aa^\dagger + a^\dagger a)\right\}D(\alpha)S(\epsilon){\mid 0\rangle}$
　 $=\pm{\langle 0\mid}S^\dagger(\epsilon)\left\{e^{-2i\phi}(a+\alpha)^2+e^{2i\phi}(a^{\dagger}+\alpha^*)^2\pm \left((a+\alpha)(a^\dagger+\alpha^*)+(a^\dagger+\alpha^*)(a+\alpha)\right)\right\}S(\epsilon){\mid 0\rangle}$ 　(Using 3. to 8.)
　 $=\pm\left[e^{-2i\phi}(-e^{2i\phi}\cosh r\sinh r+\alpha^2)+e^{2i\phi}(-e^{-2i\phi}\cosh r\sinh r+\alpha^{*2})\pm(\cosh^2r+|\alpha|^2+\sinh^2r+|\alpha|^2)\right]$ 　(Using 33. to 36. and ${\langle 0\mid}S^\dagger(\epsilon)aS(\epsilon){\mid 0\rangle}={\langle 0\mid}S^\dagger(\epsilon)a^\dagger S(\epsilon){\mid 0\rangle}=0$ )
　 $=\pm(e^{-2i\phi}\alpha^2+e^{2i\phi}\alpha^{*2})+2|\alpha|^2 + \cosh^2r+\sinh^2\mp 2\cosh r\sinh r$
　 $=\pm(e^{-2i\phi}\alpha^2+e^{2i\phi}\alpha^{*2})+2|\alpha|^2 + (\cosh r\mp\sinh r)^2$
　 $=\pm(e^{-2i\phi}\alpha^2+e^{2i\phi}\alpha^{*2})+2|\alpha|^2 + e^{\mp 2r}$

　41. $(\Delta Y_1)^2 := {\langle\alpha,\,\epsilon\mid}Y_1^2{\mid\alpha,\,\epsilon\rangle} - ({\langle\alpha,\,\epsilon\mid}Y_1{\mid\alpha,\,\epsilon\rangle})^2 = e^{-2r}$
　42. $(\Delta Y_2)^2 := {\langle\alpha,\,\epsilon\mid}Y_2^2{\mid\alpha,\,\epsilon\rangle} - ({\langle\alpha,\,\epsilon\mid}Y_2{\mid\alpha,\,\epsilon\rangle})^2 = e^{2r}$

Directly from 37. to 40.

2.5 Two-Photon Coherent States

　50. $D(\alpha)S(\epsilon) = S(\epsilon)D(\beta)$ 　where $\alpha = \beta\cosh r-\beta^* e^{2i\phi}\sinh r$

$b := S(\epsilon)aS^\dagger(\epsilon) = S^\dagger(-\epsilon)aS(-\epsilon) = a\cosh r + a^\dagger e^{2i\phi}\sinh r$

$D_g(\beta) := \exp(\beta b^\dagger-\beta^*b)$
　 $=\exp\left\{\beta(a^\dagger\cosh r+ae^{-2i\phi}\sinh r)-\beta^*(a\cosh r+a^\dagger e^{2i\phi}\sinh r\right\}$
　 $=\exp\left\{(\beta\cosh r-\beta^* e^{2i\phi}\sinh r)a^\dagger-(\beta^*\cosh r-\beta e^{-2i\phi}\sinh r)a\right\}$
　 $=D(\alpha)$ 　(where $\alpha = \beta\cosh r-\beta^* e^{2i\phi}\sinh r$ )

On the other hand,
$\displaystyle D_g(\beta)=\sum_{n=0}^\infty \frac{1}{n!}(\beta b^\dagger-\beta^*b)^n=\sum_{n=0}^\infty \frac{1}{n!}\left\{S(\epsilon)(\beta a^\dagger-\beta^*a)S^\dagger(\epsilon)\right\}^n$
　 $\displaystyle=S(\epsilon)\sum_{n=0}^\infty \frac{1}{n!}(\beta a^\dagger-\beta^*a)^nS^\dagger(\epsilon)$ 　(since $S^\dagger(\epsilon)S(\epsilon)=1$ .)
　 $=S(\epsilon)D(\beta)S^\dagger(\epsilon)$