Condition of coherent state

Coherent state have sinusoidal expectation value and static noise, which means

$$\begin{cases} \braket{x}&=X_0\cos(\omega t - \phi)\\ \braket{p}&=P_0\sin(\omega t - \phi)\\ \partial_t\sigma^2_x &= 0\\ \partial_t\sigma^2_p &= 0\\ \end{cases}$$

let's check if this is the sufficient condition for minimum uncertainty

Check

Unpack everything

Expectation value of amplitude

Without conceptual connection to the coherent state, let's just call

$$\braket{a}=\alpha e^{-i\phi}$$

Note, the derivation below show why should the magnitude be constant

The expectation value of the position,

$$\begin{align*} \braket{x}&=\sqrt{\frac{\hbar}{2m\omega}}\bra{\psi}\hat{a}e^{i\omega t}+\hat{a}^\dagger e^{-i\omega t}\ket{\psi}\\ &=\sqrt{\frac{\hbar}{2m\omega}}\left(\braket{a}e^{i\omega t}+\braket{a}^* e^{-i\omega t}\right)\\ &=\sqrt{\frac{2\hbar \alpha^2}{m\omega}}\cos(\omega t - \phi)\\ \end{align*}$$

If the magnitude of the expectation value is non constant, then this will not be sinusoidal. As the expectation value of the annihilation operator is sufficient to calculate the expectation value of momentum. Let's just have the result as

$$\braket{p}=\sqrt{2\hbar m \omega\alpha^2}\sin(\omega t - \phi)$$

For future use, let's also calculate $\braket{x^2}$ And call

$$\braket{a^2}=A^2e^{-2i\gamma}$$ $$\begin{array}{rl}{\displaystyle ⟨x^2⟩}& {\displaystyle =\frac{\mathrm{\hslash }}{2m\omega }⟨\psi \mathrm{\mid }{\widehat{a}}^2{e}^{2i\omega t}+{\left({\widehat{a}}^{†}\right)}^2{e}^{-2i\omega t}+\{\widehat{a},{\widehat{a}}^{†}\}\mathrm{\mid }\psi ⟩}\\ {\displaystyle }& {\displaystyle =\frac{\mathrm{\hslash }}{2m\omega }(⟨a^2⟩e^{2i\omega t}+{⟨a^2⟩}^{\ast }{e}^{-2i\omega t}+2n+1)}\\ {\displaystyle }& {\displaystyle =\frac{\mathrm{\hslash }}{m\omega }(A^2\cos (2\omega t-}\end{array}$$

And

$$\braket{x}^2=\frac{\hbar \alpha^2}{m\omega}\left(1+\cos(2\omega t - 2\phi)\right)$$

Static noise

$$\partial_t\sigma_x^2=\partial_t\braket{x^2}-\partial_t\braket{x}^2$$

Take the time derivative of each term

$$\partial_t \braket{x^2}=-\frac{2\hbar A^2}{m}\sin(2\omega t -2\gamma)$$

And

$$\partial_t \braket{x}^2=-\frac{2\hbar\alpha^2}{m}\sin(2\omega t - 2\phi)$$

Thus

$$\begin{align*} \partial_t \sigma_x^2 &= \partial_t \braket{x^2} - \partial_t \braket{x}^2 \\[4pt] &= -\frac{2\hbar A^2}{m}\sin(2\omega t - 2\gamma) + \frac{2\hbar\alpha^2}{m}\sin(2\omega t - 2\phi) \\[4pt] &= \frac{2\hbar}{m}\Bigl[ \alpha^2 \sin(2\omega t - 2\phi) - A^2 \sin(2\omega t - 2\gamma) \Bigr]. \end{align*}$$

Momentum variance

$$\begin{array}{rl}{\displaystyle ⟨p^2⟩}& {\displaystyle =-\frac{\mathrm{\hslash }m\omega }{2}[⟨{\widehat{a}}^2⟩e^{2i\omega t}+{⟨{\widehat{a}}^2⟩}^{\ast }{e}^{-2i\omega t}-(2n+1)]}\\ {\displaystyle }& {\displaystyle =-\frac{\mathrm{\hslash }m\omega }{2}[2A^2\cos (2\omega t-2\gamma )-(2n+1)]}\\ {\displaystyle }& {\displaystyle =\mathrm{\hslash }m\omega (n+\frac{1}{2}-A^2\cos (2\omega t-2\gamma ))\mathrm{.}}\end{array}$$ $$\begin{align*} \braket{p}^2 &= 2\hbar m \omega\,\alpha^2 \sin^2(\omega t - \phi) \\[4pt] &= \hbar m \omega\,\alpha^2 \bigl[1 - \cos(2\omega t - 2\phi)\bigr]. \end{align*}$$

Hence

$$\sigma_p^2 = \braket{p^2} - \braket{p}^2 = \hbar m \omega\Bigl[ n + \tfrac12 - \alpha^2 + \alpha^2\cos(2\omega t - 2\phi) - A^2\cos(2\omega t - 2\gamma) \Bigr].$$

Static noise for momentum

$$\partial_t \sigma_p^2 = 2\hbar m \omega^2 \Bigl[ A^2 \sin(2\omega t - 2\gamma) - \alpha^2 \sin(2\omega t - 2\phi) \Bigr].$$

Imposing $\partial_t \sigma_p^2 = 0$ and $\partial_t \sigma_x^2 = 0$ gives, for all ,

$$A^2 \sin(2\omega t - 2\gamma) = \alpha^2 \sin(2\omega t - 2\phi).$$

Equality for all times forces

$$\boxed{A^2 = \alpha^2 \qquad\text{and}\qquad \gamma = \phi \pmod{\pi}}.$$

Uncertainty product

Substituting $A = \alpha$, $\gamma = \phi$,

$$\sigma_x^2 = \frac{\hbar}{m\omega}\bigl(n + \tfrac12 - \alpha^2\bigr), \qquad \sigma_p^2 = \hbar m\omega \bigl(n + \tfrac12 - \alpha^2\bigr),$$ $$\sigma_x \sigma_p = \hbar\left(n + \frac12 - \alpha^2\right).$$

Since $n = \braket{\hat{a}^\dagger \hat{a}} \ge |\braket{\hat{a}}|^2 = \alpha^2$,

$$\sigma_x \sigma_p \ge \frac{\hbar}{2},$$

Where we can see, $n \ge \alpha^2$ with equality iff $n = \alpha^2$.

We already show that the ground state in this system ( note that ground state is not defined by $n=0$ but rather the lowest energy state possible ) have been shifted. Or even better, all state is shifted uniformly

One more condition

So, one additional condition needed:

$$\begin{align*} n &= \alpha^2 \\ \braket{\hat{a}^\dagger \hat{a}} &= \braket{\hat{a}}\braket{\hat{a}^\dagger} \\ \braket{(\hat{a} - \braket{\hat{a}})^\dagger (\hat{a} - \braket{\hat{a}})} &= 0 . \end{align*}$$

For a pure state this means $(\hat{a} - \braket{\hat{a}}) |\psi\rangle = 0$, i.e. $|\psi\rangle$ is an eigenstate of $\stackrel{}{a}$. Or so called, the coherent state.

Non-coherent state

From this point of view, well maybe we can drop the condition that the noise must be stable, which since we know coherent state isn't that, $n=\alpha^2$ must also be droped. Thus, now

$$\sigma^2_x = \frac{\hbar}{m\omega}\left(A^2\cos(2\omega t - 2\gamma)+n+\frac{1}{2}-\alpha^2-\alpha^2\cos(2\omega t - 2\phi)\right)$$ $$\sigma^2_p = \hbar m \omega\left( n + \frac{1}{2} - \alpha^2 + \alpha^2\cos(2\omega t - 2\phi) - A^2\cos(2\omega t - 2\gamma) \right)$$

This simplifies to

$$m\omega\sigma_x^2 = \hbar\Bigl[ \sinh^2{r} + \tfrac12 + R \cos(2\omega t - \delta) \Bigr]$$ $$\frac{1}{m\omega}\sigma_p^2 = \hbar \Bigl[ \sinh^2{r} + \tfrac12 - R \cos(2\omega t - \delta)\Bigr]$$

where

$$R = \sqrt{ A^4 + \alpha^4 - 2A^2\alpha^2\cos(2\gamma - 2\phi) }, \qquad \delta = \arctan2\!\bigl( A^2\sin 2\gamma - \alpha^2\sin 2\phi,\; A^2\cos 2\gamma - \alpha^2\cos 2\phi \bigr).$$

And

$$n-\alpha^2 = \sinh^2{r}$$

( Which indeed is quite arbitrary currently, we will derive this later )

This is the squeezed state under free harmonic oscillating force, where the uncertainty of $x$ and $p$ oscillate and squeezed into each others periodically.

This gives

$$\sigma_x^2 \sigma_p^2 = \hbar^2 \Bigl[ \bigl( \sinh^2{r} + \tfrac12 \bigr)^2 - R^2 \cos^2(2\omega t - \delta) \Bigr]$$

Hence the uncertainty product is

$$\sigma_x \sigma_p = \hbar \, \sqrt{ \left( \sinh^2{r} + \frac{1}{2} \right)^2 - R^2 \cos^2(2\omega t - \delta) }.$$

Which is also oscillating. And periodically dipped to minimum uncertainty.