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Calculations on Coherent States and Squeezed States (1)

Quantum Optics

Quantum Optics

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)

Hence D(\alpha)=S(\epsilon)D(\beta)S^\dagger(\epsilon), and by multiplying S(\epsilon) from the right side, you get 50.