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Calculations on Coherent States and Squeezed States (2)

Quantum Optics

Quantum Optics

This is intended to fill some gaps in calculation details in the book above.

2.6 Variance in the Electric Field

Hamiltonian H = \hbar\omega\left(a^\dagger a+\frac{1}{2}\right)

Electric field operator E = i E_0 (ae^{i\mathbf k\cdot\mathbf r}-a^\dagger e^{-i\mathbf k\cdot\mathbf r}) = -E_0X_2

Time evolution (in the Heisenberg picture)

　1. a(t) = ae^{-i\omega t}

　2. a^\dagger(t) = a^\dagger e^{i\omega t}

　3. E(t) = i E_0 \left\{ae^{-i(\omega t-\mathbf k\cdot\mathbf r)}-a^\dagger e^{i(\omega t-\mathbf k\cdot\mathbf r)}\right\}

For 1., \displaystyle \left[i\frac{H}{\hbar}t,\, a\right] = -i\omega t\cdot a, hence \displaystyle a(t) = e^{i\frac{H}{\hbar}t}ae^{-i\frac{H}{\hbar}t}=\sum_{n=0}^\infty\frac{1}{n!}({\rm C_{i\frac{H}{\hbar}t}})^n a = ae^{-i\omega t}.
2. is conjugate of 1. 3. follows directly from 1. and 2.

　4. E(t) = E_0\left\{X_1\sin (\omega t-\mathbf k\cdot\mathbf r)-X_2\cos(\omega t-\mathbf k\cdot\mathbf r)\right\}

Rewriting 3. with \displaystyle a = \frac{1}{2}(X_1+iX_2),\,a^\dagger = \frac{1}{2}(X_1-iX_2)

Time dependent variance of E(t)

　5. V(E(t)) = E_0^2\left\{V(X_1)\sin^2(\omega t-\mathbf k\cdot\mathbf r)+V(X_2)\cos^2(\omega t-\mathbf k\cdot\mathbf r)-2V(X_1,\,X_2)\sin(\omega t-\mathbf k\cdot\mathbf r)\cos(\omega t-\mathbf k\cdot\mathbf r)\right\}

\langle E^2(t)\rangle = E_0^2\left[\sin^2(\omega t-\mathbf k\cdot\mathbf r)\langle X_1^2\rangle+cos^2(\omega t-\mathbf k\cdot\mathbf r)\langle X_2^2\rangle -\langle\{X_1,\,X_2\}\rangle\sin(\omega t-\mathbf k\cdot\mathbf r)\cos(\omega t -\mathbf k\cdot\mathbf r)\right]

\langle E(t)\rangle^2 = E_0^2\left[\sin^2(\omega t-\mathbf k\cdot\mathbf r)\langle X_1\rangle^2+\cos^2(\omega t-\mathbf k\cdot\mathbf r)\langle X_2\rangle^2-2\langle X_1\rangle\langle X_2\rangle\sin(\omega t-\mathbf k\cdot\mathbf r)\cos(\omega t-\mathbf k\cdot\mathbf r)\right]

Hence, V(E(t)) = \langle E^2(t)\rangle - \langle E(t)\rangle^2 results in 5.
where V(X_1) = \langle X_1^2\rangle - \langle X_1\rangle^2,\,\langle X_2^2\rangle - \langle X_2\rangle^2,\,V(X_1,\,X_2) = \frac{1}{2}\langle\{X_1,\,X_2\}\rangle-2\langle X_1\rangle\langle X_2\rangle

　6. \displaystyle V(X_1,\,X_2) = 0　when \displaystyle V(X_1)V(X_2) = \left|\frac{1}{2}\langle [X_1,\,X_2]\rangle\right|^2

Apply the Schroedinger's uncertainty relation \displaystyle V(X_1)V(X_2)\ge |V(X_1,\,X_2)|^2 + \left|\frac{1}{2}\langle [X_1,\,X_2]\rangle\right|^2

Especially when V(X_1) = e^{-2r},\,V(X_2) = e^{2r}, under the condition that V(X_1,\,X_2)=0, you get the next result.

　7. \displaystyle V(E(t)) = E_0^2\left\{\cosh 2r+\sinh 2r\cdot\cos 2(\omega t-\mathbf k\cdot\mathbf r)\right\}

\displaystyle V(E(t)) = E_0^2\left\{e^{-2r}\left(\frac{1-\cos(2(\omega t-\mathbf k\cdot\mathbf r))}{2}\right) + e^{2r}\left(\frac{1+\cos2(\omega t-\mathbf k\cdot\mathbf r)}{2}\right)\right\}
　\displaystyle=E_0^2\left\{\frac{e^{2r}+e^{-2r}}{2}+\frac{e^{2r}-e^{-2r}}{2}\cos2(\omega t-\mathbf k\cdot\mathbf r)\right\}
　\displaystyle=E_0^2\left\{\cosh 2r+\sinh 2r\cdot\cos 2(\omega t-\mathbf k\cdot\mathbf r)\right\}

Hence, for r=0 (unsqueezed coherent state), the variance becomes constant whereas for r\ne 0 (squeezed coherent state), the variance flactuates with the angular velocity 2\omega.

　8. {\langle \alpha,\,\epsilon\mid}E(t){\mid \alpha,\,\epsilon\rangle} = i E_0 \left\{\alpha e^{-i(\omega t-\mathbf k\cdot\mathbf r)}-\alpha^* e^{i(\omega t-\mathbf k\cdot\mathbf r)}\right\}

This follows from 3. and {\langle \alpha,\,\epsilon\mid}a{\mid \alpha,\,\epsilon\rangle}=\alpha.

Especially for \displaystyle\alpha = \frac{1}{2}, it results in \langle E(t) \rangle = E_0\sin(\omega t-\mathbf k\cdot\mathbf r)

The following shows graphs of \langle E(t)\rangle \pm \sqrt{V(t)} for various rs.


Time evolution of squeezed states.

Schematic view of oscillation

Electric field operator E = iE_0(a-a^\dagger)

Time evolution E(t) = iE_0(ae^{-i\omega t}-a^\dagger e^{i\omega t})

Expected value \langle E(t)\rangle = iE_0(\alpha e^{-i\omega t}-\alpha^* e^{i\omega t})= -2E_0{\rm Im}\,(\alpha e^{-i\omega t})

In terms of X_1,\,X_2, it corresponds to:
　\langle X_1(t)\rangle = (\alpha e^{-i\omega t}+\alpha^* e^{i\omega t}) = 2{\rm Re}\,(\alpha e^{-i\omega t})
　\displaystyle\langle X_2(t)\rangle = -i(\alpha e^{-i\omega t}-\alpha^* e^{i\omega t})=2{\rm Im}\,(\alpha e^{-i\omega t})=\frac{-1}{E_0}\langle E(t)\rangle

Define the rotated coordinates Y_1+iY_2 : = (X_1+iX_2)e^{i\omega t_0} = 2ae^{i\omega t_0} for some fixed t_0.
Then at t=t_0, Y_1(t=t_0) +iY_2(t=t_0) = 2 a e^{-i\omega t_0}e^{i\omega t_0} = 2a = X_1+iX_2. Hence,
　\langle Y_1(t_0)\rangle = \langle X_1 \rangle = 2 {\rm Re}\,\alpha,\ V(Y_1(t_0)) = V(X_1)
　\langle Y_2(t_0)\rangle = \langle X_2 \rangle = 2 {\rm Im}\,\alpha,\ V(Y_2(t_0)) = V(X_2)

These results can be summarized in the following diagram.

3.7.1 Squeezed State

Photon number fluctuation for the squeezed state {\mid \alpha,\,r\rangle}.

　1. \langle n \rangle = \sinh^2 r + |\alpha|^2
　2. V(n) = \langle n^2\rangle - \langle n\rangle^2 = \langle n\rangle + |\alpha|^2(\cosh 2r-\sinh 2r\cos 2\theta-1)+\sinh^2r\cosh 2r 　where \alpha = |\alpha|e^{i\phi}

For 1., \langle n \rangle = {\langle 0 \mid}S^\dagger(r)D^\dagger(\alpha)a^\dagger aD(\alpha)S(r){\mid 0\rangle}={\langle 0 \mid}S^\dagger(r)(a^\dagger+\alpha^*)(a+\alpha)S(r){\mid 0\rangle}
　={\langle 0 \mid}S^\dagger(r)(a^\dagger a+|\alpha|^2)S(r){\mid 0\rangle}=\sinh^2r+|\alpha|^2

For 2., \langle n^2\rangle = {\langle 0 \mid}S^\dagger(r)D^\dagger(\alpha)a^\dagger aa^\dagger aD(\alpha)S(r){\mid 0\rangle}
　={\langle 0 \mid}S^\dagger(r)D^\dagger(\alpha)a^{\dagger 2} a^2D(\alpha)S(r){\mid 0\rangle}+\langle n\rangle　since a\dagger a a^\dagger a = a^\dagger(a^\dagger a + 1)a = a^{\dagger 2}a^2 + a^\dagger a

Hence, \langle n^2\rangle-\langle n\rangle={\langle 0 \mid}S^\dagger(r)(a^\dagger+\alpha^*)^2(a+\alpha)^2S(r){\mid 0\rangle}
　={\langle 0 \mid}S^\dagger(r)\left[a^{\dagger 2}a^2+2\alpha a^{\dagger 2}a+2\alpha^*a^{2}a^\dagger + \alpha^2a^{\dagger 2}+\alpha^{*2}a^2+4|\alpha|^2a^\dagger a+2\alpha^*\alpha^2a^\dagger + 2\alpha^2\alpha^{*2}a+|\alpha|^4\right]S(r){\mid 0\rangle}
　={\langle 0 \mid}\left[(a^\dagger\cosh r-a\sinh r)^2(a\cosh r-a^\dagger\sinh r)^2+\left\{\alpha^2(a^\dagger\cosh r-a\sinh r)^2 + {\rm (Conjugate)}\right\}+4|\alpha|^2(a^\dagger\cosh r-a\sinh r)(a\cosh r-a^\dagger\sinh r)+|\alpha|^4\right]{\mid 0\rangle}
　={\langle 0 \mid}\left[a^2a^{\dagger 2}\sinh^4r+aa^\dagger aa^\dagger\sinh^2r\cosh^2r-\left\{\alpha^2aa^\dagger\sinh r\cosh r+ {\rm (Conjugate)}\right\}+4|\alpha|^2aa^\dagger\sinh^2r+|\alpha|^4\right]{\mid 0\rangle}
　=2\sinh^4r+\sinh^2r\cosh^2r-2|\alpha|^2\cos2\phi\sinh r\cosh r+4|\alpha|^2\sinh^2r+|\alpha|^4
　since {\langle 0 \mid}a^2a^{\dagger 2}{\mid 0\rangle}=2,\ {\langle 0 \mid}aa^{\dagger}aa^{\dagger}{\mid 0\rangle}={\langle 0 \mid}aa^{\dagger}{\mid 0\rangle}=1

As a result, V(n)-\langle n\rangle =2|\alpha|^2(\sinh^2 r-\cos2\phi\sinh r\cosh r)+\sinh^4r+sinh^2r\cosh^2r
　= |\alpha|^2(\cosh 2r-\sinh 2r\cos 2\phi-1)+\sinh^2r\cosh 2r
　since 2\sinh^2r=\sinh^2r+\cosh^2r-1=\cosh 2r-1,\ 2\sinh r\cosh r = \sinh 2r

From these results, you can see that V(n)=\langle n\rangle for r=0 (non-squeezed coherent state) since it obeys the Poisson distribution.

On the other hand, when r\ne 0:

for \phi = 0, V(n)-\langle n\rangle=|\alpha|^2(e^{-2r}-1)+\sinh^2r\cosh 2r < 0\ (|\alpha|^2 \gg \sinh^2r\cosh 2r). It becomes the sub-Poissson.

for \displaystyle\phi = \frac{\pi}{2}, V(n)-\langle n\rangle=|\alpha|^2(e^{2r}-1)+\sinh^2r\cosh 2r > 0\ (|\alpha|^2 \gg \sinh^2r\cosh 2r). It becomes the super-Poissson.

Especially, for \alpha = 0 (Squeezed vacuum)], V(n)-\langle n\rangle=\sinh^2r\cosh 2r \ge 0. It's always the super-Poission.

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.

多体系の密度行列の非対角成分が0になる理由を系に対するノイズ（ランダムな摂動）の観点で説明する

何の話かというと

Statistical Physics: Volume 5

Statistical Physics: Volume 5

Landau の Statistical Physics の §6 では、密度行列の非対角成分が平衡状態では無視できる理由として、「時間依存性がないことが平衡状態の定義であり、したがって、時間依存性を持つ非対角成分が0の場合を考える」となかなかトートロジカルな説明がなされています。一応、脚注において、「系に含まれる粒子数が増加するなどして、相互作用の重要性が低下することにより非対角成分が0になる」と説明されていますが、この部分がイマイチ直感的に理解できません。というわけで、この点について、もう少し定量的な説明を試みたのが以下の内容です。

まず、他の粒子からの相互作用を1粒子のハミルトニアンに対する摂動パラメーター q と考えて、1粒子のエネルギー固有値が q に依存した関数になると考えます。

　\hat H=\hat H_0 + \hat V(q)\ \Rightarrow\ \omega_i = \omega_i(q)（\omega_i=\frac{E_i}{\hbar}）

したがって、1粒子のエネルギー固有状態は、時間依存を含めて、

　\mid \phi_i\rangle e^{-i\omega_i(q)t}

となります。（厳密には、\mid \phi_i\rangle も q に依存していますが、この点については後述します。）

この時、一般の重ね合わせ状態

　\mid\psi\rangle=\sum_i c_i{\mid\phi_i\rangle} e^{-i\omega_i(q)t}

について、Pure state としての密度行列を計算すると次が得られます。

　\rho = {\mid\psi\rangle}{\langle\psi\mid} = \sum_i |c_i|^2{\mid\phi_i\rangle}{\langle\phi_i\mid}+\sum_{i\ne j}c_ic_j^*{\mid\phi_i\rangle}{\langle\phi_j\mid} e^{-i\left\{\omega_i(q)-\omega_j(q)\right\}t} ---- (1)

一般に、\omega が大きい古典極限では非対角成分は時間変動が大きく、マクロな時間レベルで平均して0になると考えられます。ただし、これは、統計力学とは無関係の古典極限の話です。ここでは、非古典的な量子状態に対する統計的性質を考える必要があります。

今、（粒子が多数存在するなどの理由で）粒子間の相互作用が乱雑で、摂動パラメーター自体が時間に対して激しく変動すると仮定すると、\omega_i(q(t)) という時間依存により、\omega がそれほど大きくない領域においても、位相 -i\omega_i(q(t)) は激しく変動するため、古典極限と同様の近似として、非対角成分を0とみなすことができます。一般に、マクロな時間内に \Delta \omega(q) \sim 2\pi、すなわち、\Delta E(q) \sim h 程度の変動があれば十分です。

この話を一般化すると、系に対するランダムな摂動（ノイズ）がある場合は、非古典領域においても密度行列の非対角成分を0とする、（平衡系の）統計力学的な取り扱いが正当化されることになります。

なお、(1) でマクロな時間平均を取ると、上述の理由で非対角成分は0になりますが、同時に、\mid \phi_i\rangle の q 依存性についても平均が行われます。これは、他の粒子からの影響を平均場で近似して、平均場のもとに得られたエネルギー固有状態 \mid\phi_i\rangle に置き換える操作と理解できます。平均場が0になる場合（つまり、摂動項の平均が0になる場合）は、近似として、もとのハミルトニアンの固有状態をそのまま利用することができます。