Calculations on Coherent States and Squeezed States (2)

Quantum Optics

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.

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 $r$ s.

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.