Condition of coherent state§
Coherent state have sinusoidal expectation value and static noise, which means
{ ⟨ x ⟩ = X 0 cos ( ω t − ϕ ) ⟨ p ⟩ = P 0 sin ( ω t − ϕ ) ∂ t σ x 2 = 0 ∂ t σ p 2 = 0 \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} ⎩ ⎨ ⎧ ⟨ x ⟩ ⟨ p ⟩ ∂ t σ x 2 ∂ t σ p 2 = X 0 cos ( ω t − ϕ ) = P 0 sin ( ω t − ϕ ) = 0 = 0 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
⟨ a ⟩ = α e − i ϕ \braket{a}=\alpha e^{-i\phi} ⟨ a ⟩ = α e − i ϕ Note, the derivation below show why should the magnitude be constant
The expectation value of the position,
⟨ x ⟩ = ℏ 2 m ω ⟨ ψ ∣ a ^ e i ω t + a ^ † e − i ω t ∣ ψ ⟩ = ℏ 2 m ω ( ⟨ a ⟩ e i ω t + ⟨ a ⟩ ∗ e − i ω t ) = 2 ℏ α 2 m ω cos ( ω t − ϕ ) \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*} ⟨ x ⟩ = 2 mω ℏ ⟨ ψ ∣ a ^ e iω t + a ^ † e − iω t ∣ ψ ⟩ = 2 mω ℏ ( ⟨ a ⟩ e iω t + ⟨ a ⟩ ∗ e − iω t ) = mω 2ℏ α 2 cos ( ω t − ϕ ) 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
⟨ p ⟩ = 2 ℏ m ω α 2 sin ( ω t − ϕ ) \braket{p}=\sqrt{2\hbar m \omega\alpha^2}\sin(\omega t - \phi) ⟨ p ⟩ = 2ℏ mω α 2 sin ( ω t − ϕ ) For future use, let's also calculate ⟨ x 2 ⟩ \braket{x^2} ⟨ x 2 ⟩
⟨ a 2 ⟩ = A 2 e − 2 i γ \braket{a^2}=A^2e^{-2i\gamma} ⟨ a 2 ⟩ = A 2 e − 2 iγ ⟨ x 2 ⟩ = ℏ 2 m ω ⟨ ψ ∣ a ^ 2 e 2 i ω t + ( a ^ † ) 2 e − 2 i ω t + { a ^ , a ^ † } ∣ ψ ⟩ = ℏ 2 m ω ( ⟨ a 2 ⟩ e 2 i ω t + ⟨ a 2 ⟩ ∗ e − 2 i ω t + 2 n + 1 ) = ℏ m ω ( A 2 cos ( 2 ω t − 2 γ ) + n + 1 2 ) \begin{align*}
\braket{x^2}&=\frac{\hbar}{2m\omega}\bra{\psi}\hat{a}^2e^{2i\omega t}+\left(\hat{a}^\dagger\right)^2 e^{-2i\omega t}+\{\hat{a},\hat{a}^\dagger\}\ket{\psi}\\
&=\frac{\hbar}{2m\omega}\left(\braket{a^2}e^{2i\omega t}+\braket{a^2}^* e^{-2i\omega t}+2n+1\right)\\
&=\frac{\hbar}{m\omega}\left(A^2\cos(2\omega t - 2\gamma)+n+\frac{1}{2}\right)\\
\end{align*} ⟨ x 2 ⟩ = 2 mω ℏ ⟨ ψ ∣ a ^ 2 e 2 iω t + ( a ^ † ) 2 e − 2 iω t + { a ^ , a ^ † } ∣ ψ ⟩ = 2 mω ℏ ( ⟨ a 2 ⟩ e 2 iω t + ⟨ a 2 ⟩ ∗ e − 2 iω t + 2 n + 1 ) = mω ℏ ( A 2 cos ( 2 ω t − 2 γ ) + n + 2 1 ) And
⟨ x ⟩ 2 = ℏ α 2 m ω ( 1 + cos ( 2 ω t − 2 ϕ ) ) \braket{x}^2=\frac{\hbar \alpha^2}{m\omega}\left(1+\cos(2\omega t - 2\phi)\right) ⟨ x ⟩ 2 = mω ℏ α 2 ( 1 + cos ( 2 ω t − 2 ϕ ) ) Static noise§
∂ t σ x 2 = ∂ t ⟨ x 2 ⟩ − ∂ t ⟨ x ⟩ 2 \partial_t\sigma_x^2=\partial_t\braket{x^2}-\partial_t\braket{x}^2 ∂ t σ x 2 = ∂ t ⟨ x 2 ⟩ − ∂ t ⟨ x ⟩ 2 Take the time derivative of each term
∂ t ⟨ x 2 ⟩ = − 2 ℏ A 2 m sin ( 2 ω t − 2 γ ) \partial_t \braket{x^2}=-\frac{2\hbar A^2}{m}\sin(2\omega t -2\gamma) ∂ t ⟨ x 2 ⟩ = − m 2ℏ A 2 sin ( 2 ω t − 2 γ ) And
∂ t ⟨ x ⟩ 2 = − 2 ℏ α 2 m sin ( 2 ω t − 2 ϕ ) \partial_t \braket{x}^2=-\frac{2\hbar\alpha^2}{m}\sin(2\omega t - 2\phi) ∂ t ⟨ x ⟩ 2 = − m 2ℏ α 2 sin ( 2 ω t − 2 ϕ ) Thus
∂ t σ x 2 = ∂ t ⟨ x 2 ⟩ − ∂ t ⟨ x ⟩ 2 = − 2 ℏ A 2 m sin ( 2 ω t − 2 γ ) + 2 ℏ α 2 m sin ( 2 ω t − 2 ϕ ) = 2 ℏ m [ α 2 sin ( 2 ω t − 2 ϕ ) − A 2 sin ( 2 ω t − 2 γ ) ] . \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*} ∂ t σ x 2 = ∂ t ⟨ x 2 ⟩ − ∂ t ⟨ x ⟩ 2 = − m 2ℏ A 2 sin ( 2 ω t − 2 γ ) + m 2ℏ α 2 sin ( 2 ω t − 2 ϕ ) = m 2ℏ [ α 2 sin ( 2 ω t − 2 ϕ ) − A 2 sin ( 2 ω t − 2 γ ) ] . Momentum variance§
⟨ p 2 ⟩ = − ℏ m ω 2 [ ⟨ a ^ 2 ⟩ e 2 i ω t + ⟨ a ^ 2 ⟩ ∗ e − 2 i ω t − ( 2 n + 1 ) ] = − ℏ m ω 2 [ 2 A 2 cos ( 2 ω t − 2 γ ) − ( 2 n + 1 ) ] = ℏ m ω ( n + 1 2 − A 2 cos ( 2 ω t − 2 γ ) ) . \begin{align*}
\braket{p^2}
&= -\frac{\hbar m \omega}{2}
\Bigl[ \braket{\hat{a}^2} e^{2i\omega t} + \braket{\hat{a}^2}^* e^{-2i\omega t} - (2n+1) \Bigr] \\[4pt]
&= -\frac{\hbar m \omega}{2}
\Bigl[ 2A^2\cos(2\omega t - 2\gamma) - (2n+1) \Bigr] \\[4pt]
&= \hbar m \omega \Bigl( n + \frac12 - A^2\cos(2\omega t - 2\gamma) \Bigr).
\end{align*} ⟨ p 2 ⟩ = − 2 ℏ mω [ ⟨ a ^ 2 ⟩ e 2 iω t + ⟨ a ^ 2 ⟩ ∗ e − 2 iω t − ( 2 n + 1 ) ] = − 2 ℏ mω [ 2 A 2 cos ( 2 ω t − 2 γ ) − ( 2 n + 1 ) ] = ℏ mω ( n + 2 1 − A 2 cos ( 2 ω t − 2 γ ) ) . ⟨ p ⟩ 2 = 2 ℏ m ω α 2 sin 2 ( ω t − ϕ ) = ℏ m ω α 2 [ 1 − cos ( 2 ω t − 2 ϕ ) ] . \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*} ⟨ p ⟩ 2 = 2ℏ mω α 2 sin 2 ( ω t − ϕ ) = ℏ mω α 2 [ 1 − cos ( 2 ω t − 2 ϕ ) ] . Hence
σ p 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ℏ m ω [ n + 1 2 − α 2 + α 2 cos ( 2 ω t − 2 ϕ ) − A 2 cos ( 2 ω t − 2 γ ) ] . \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]. σ p 2 = ⟨ p 2 ⟩ − ⟨ p ⟩ 2 = ℏ mω [ n + 2 1 − α 2 + α 2 cos ( 2 ω t − 2 ϕ ) − A 2 cos ( 2 ω t − 2 γ ) ] . Static noise for momentum§
∂ t σ p 2 = 2 ℏ m ω 2 [ A 2 sin ( 2 ω t − 2 γ ) − α 2 sin ( 2 ω t − 2 ϕ ) ] . \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]. ∂ t σ p 2 = 2ℏ m ω 2 [ A 2 sin ( 2 ω t − 2 γ ) − α 2 sin ( 2 ω t − 2 ϕ ) ] . Imposing ∂ t σ p 2 = 0 \partial_t \sigma_p^2 = 0 ∂ t σ p 2 = 0 ∂ t σ x 2 = 0 \partial_t \sigma_x^2 = 0 ∂ t σ x 2 = 0 t t t
A 2 sin ( 2 ω t − 2 γ ) = α 2 sin ( 2 ω t − 2 ϕ ) . A^2 \sin(2\omega t - 2\gamma) = \alpha^2 \sin(2\omega t - 2\phi). A 2 sin ( 2 ω t − 2 γ ) = α 2 sin ( 2 ω t − 2 ϕ ) . Equality for all times forces
A 2 = α 2 and γ = ϕ ( m o d π ) . \boxed{A^2 = \alpha^2 \qquad\text{and}\qquad \gamma = \phi \pmod{\pi}}. A 2 = α 2 and γ = ϕ ( mod π ) . Uncertainty product§
Substituting A = α A = \alpha A = α γ = ϕ \gamma = \phi γ = ϕ
σ x 2 = ℏ m ω ( n + 1 2 − α 2 ) , σ p 2 = ℏ m ω ( n + 1 2 − α 2 ) , \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), σ x 2 = mω ℏ ( n + 2 1 − α 2 ) , σ p 2 = ℏ mω ( n + 2 1 − α 2 ) , σ x σ p = ℏ ( n + 1 2 − α 2 ) . \sigma_x \sigma_p = \hbar\left(n + \frac12 - \alpha^2\right). σ x σ p = ℏ ( n + 2 1 − α 2 ) . Since n = ⟨ a ^ † a ^ ⟩ ≥ ∣ ⟨ a ^ ⟩ ∣ 2 = α 2 n = \braket{\hat{a}^\dagger \hat{a}} \ge |\braket{\hat{a}}|^2 = \alpha^2 n = ⟨ a ^ † a ^ ⟩ ≥ ∣ ⟨ a ^ ⟩ ∣ 2 = α 2
σ x σ p ≥ ℏ 2 , \sigma_x \sigma_p \ge \frac{\hbar}{2}, σ x σ p ≥ 2 ℏ , Where we can see, n ≥ α 2 n \ge \alpha^2 n ≥ α 2 n = α 2 n = \alpha^2 n = α 2
We already show that the ground state in this system ( note that ground state is not defined by n = 0 n=0 n = 0
One more condition§
So, one additional condition needed:
n = α 2 ⟨ a ^ † a ^ ⟩ = ⟨ a ^ ⟩ ⟨ a ^ † ⟩ ⟨ ( a ^ − ⟨ a ^ ⟩ ) † ( a ^ − ⟨ a ^ ⟩ ) ⟩ = 0. \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*} n ⟨ a ^ † a ^ ⟩ ⟨ ( a ^ − ⟨ a ^ ⟩ ) † ( a ^ − ⟨ a ^ ⟩ ) ⟩ = α 2 = ⟨ a ^ ⟩ ⟨ a ^ † ⟩ = 0. For a pure state this means ( a ^ − ⟨ a ^ ⟩ ) ∣ ψ ⟩ = 0 (\hat{a} - \braket{\hat{a}}) |\psi\rangle = 0 ( a ^ − ⟨ a ^ ⟩ ) ∣ ψ ⟩ = 0 ∣ ψ ⟩ |\psi\rangle ∣ ψ ⟩ a ^ \hat{a} a ^
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 = α 2 n=\alpha^2 n = α 2
σ x 2 = ℏ m ω ( A 2 cos ( 2 ω t − 2 γ ) + n + 1 2 − α 2 − α 2 cos ( 2 ω t − 2 ϕ ) ) \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) σ x 2 = mω ℏ ( A 2 cos ( 2 ω t − 2 γ ) + n + 2 1 − α 2 − α 2 cos ( 2 ω t − 2 ϕ ) ) σ p 2 = ℏ m ω ( n + 1 2 − α 2 + α 2 cos ( 2 ω t − 2 ϕ ) − A 2 cos ( 2 ω t − 2 γ ) ) \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) σ p 2 = ℏ mω ( n + 2 1 − α 2 + α 2 cos ( 2 ω t − 2 ϕ ) − A 2 cos ( 2 ω t − 2 γ ) ) This simplifies to
m ω σ x 2 = ℏ [ sinh 2 r + 1 2 + R cos ( 2 ω t − δ ) ] m\omega\sigma_x^2 = \hbar\Bigl[ \sinh^2{r} + \tfrac12 + R \cos(2\omega t - \delta) \Bigr] mω σ x 2 = ℏ [ sinh 2 r + 2 1 + R cos ( 2 ω t − δ ) ] 1 m ω σ p 2 = ℏ [ sinh 2 r + 1 2 − R cos ( 2 ω t − δ ) ] \frac{1}{m\omega}\sigma_p^2 = \hbar \Bigl[ \sinh^2{r} + \tfrac12 - R \cos(2\omega t - \delta)\Bigr] mω 1 σ p 2 = ℏ [ sinh 2 r + 2 1 − R cos ( 2 ω t − δ ) ] where
R = A 4 + α 4 − 2 A 2 α 2 cos ( 2 γ − 2 ϕ ) , δ = arctan 2 ( A 2 sin 2 γ − α 2 sin 2 ϕ , A 2 cos 2 γ − α 2 cos 2 ϕ ) . 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). R = A 4 + α 4 − 2 A 2 α 2 cos ( 2 γ − 2 ϕ ) , δ = arctan 2 ( A 2 sin 2 γ − α 2 sin 2 ϕ , A 2 cos 2 γ − α 2 cos 2 ϕ ) . And
n − α 2 = sinh 2 r n-\alpha^2 = \sinh^2{r} n − α 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 x x p p p
This gives
σ x 2 σ p 2 = ℏ 2 [ ( sinh 2 r + 1 2 ) 2 − R 2 cos 2 ( 2 ω t − δ ) ] \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] σ x 2 σ p 2 = ℏ 2 [ ( sinh 2 r + 2 1 ) 2 − R 2 cos 2 ( 2 ω t − δ ) ] Hence the uncertainty product is
σ x σ p = ℏ ( sinh 2 r + 1 2 ) 2 − R 2 cos 2 ( 2 ω t − δ ) . \sigma_x \sigma_p = \hbar \, \sqrt{ \left( \sinh^2{r} + \frac{1}{2} \right)^2 - R^2 \cos^2(2\omega t - \delta) }. σ x σ p = ℏ ( sinh 2 r + 2 1 ) 2 − R 2 cos 2 ( 2 ω t − δ ) . Which is also oscillating. And periodically dipped to minimum uncertainty.