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The coherent state was introduced previously for the harmonic oscillator here and it is the state that describe more closely the classical oscillating electromagnetic field:
| \Psi_{\lambda,N} \rangle = \sum_{n=0}^\infty c_{\lambda,n} e^{-i\left(n_\lambda + \frac{1}{2}\right)\omega_\lambda t} | n_{\lambda} \rangle
with:
c_{\lambda,n} = \sqrt{\frac{N^{n_\lambda} e^{-N}}{n_\lambda!}}
where N is the expected number of photons in the mode, |c_{\lambda,n}|^2 is a Poisson distribution from statistics, with a mean N and a standard deviation \sqrt N, so the number of photons in this mode is not determined.
The coherent state is an eigenstate of the annihilation operator \mathbf a_\lambda. It is possible to prove it; writing the eigenvalue equation and checking if is satisfied:
\mathbf a_\lambda | \Psi_{\lambda,N} \rangle = \alpha_\lambda | \Psi_{\lambda,N} \rangle
Applying the annihilation operator:
\begin{aligned} \mathbf a_\lambda | \Psi_{\lambda,N} \rangle & =\mathbf a_\lambda \sum_{n=0}^\infty c_{\lambda,n} e^{-i\left(n_{\lambda} + \frac{1}{2}\right)\omega_\lambda t} a_\lambda | n_{\lambda} \rangle \\ & = \sum_{n=0}^\infty c_{\lambda,n} e^{-i\left(n_{\lambda} + \frac{1}{2}\right)\omega_\lambda t} \sqrt{n_\lambda} | n_{\lambda} - 1 \rangle \\ & = 0 + \sum_{n=1}^\infty c_{\lambda,n} e^{-i\left(n_{\lambda} + \frac{1}{2}\right)\omega_\lambda t} \sqrt{n_\lambda} | n_{\lambda} - 1 \rangle \\ & = \sum_{n=0}^\infty c_{\lambda,n+1} e^{-i\left((n_{\lambda}+1) + \frac{1}{2}\right)\omega_\lambda t} \sqrt{n_\lambda + 1} | n_{\lambda} \rangle \\ & = e^{-i\omega_\lambda t}\sum_{n=0}^\infty c_{\lambda,n+1} e^{-i\left(n_{\lambda} + \frac{1}{2}\right)\omega_\lambda t} \sqrt{n_\lambda + 1} | n_{\lambda} \rangle \end{aligned}
Apply the relationship:
c_{\lambda,n+1} = c_{\lambda,n}\sqrt{\frac{N}{n_\lambda + 1}}
gives:
\mathbf a_\lambda | \Psi_{\lambda,N} \rangle = \sqrt{N} e^{-i\omega_\lambda t}\sum_{n=0}^\infty c_{\lambda,n} e^{-i\left(n_{\lambda} + \frac{1}{2}\right)\omega_\lambda t} | n_{\lambda} \rangle = \sqrt{N} e^{-i\omega_\lambda t} | \Psi_{\lambda,N} \rangle
so it verified that the coherent state is an eigenstate of the annihilator operator with an eigenenergy:
\alpha = \sqrt{N} e^{-i\omega_\lambda t}
using these results and taking the Hermitian conjugate:
\langle \Psi_{\lambda,N} | \mathbf{a}_\lambda^\dagger = \bar \alpha_\lambda \langle \Psi_{\lambda,N} |
It is possible to find the expectation for the quantity \xi_\lambda and \pi_\lambda using these results:
\begin{aligned} & \xi_\lambda = \frac{1}{\sqrt 2} \left(\mathbf a^\dag_\lambda + \mathbf a_\lambda\right) \\ & \pi_\lambda = \frac{i}{\sqrt 2} \left(\mathbf a^\dag_\lambda - \mathbf a^\dag\right) \end{aligned}
The expectations are:
\begin{aligned} \langle \xi_\lambda \rangle & = \frac{1}{\sqrt 2} \langle \Psi_{\lambda,N} | \left(\mathbf a^\dag_\lambda + \mathbf a_\lambda\right) | \Psi_{\lambda,N} \rangle = \frac{1}{\sqrt 2} \langle \Psi_{\lambda,N} | \mathbf a^\dag_\lambda | \Psi_{\lambda,N} \rangle +\frac{1}{\sqrt 2} \langle \Psi_{\lambda,N} |\mathbf a_\lambda | \Psi_{\lambda,N} \rangle \\ & = \frac{1}{\sqrt 2} \left(\langle \Psi_{\lambda,N} | \mathbf a^\dag_\lambda\right) | \Psi_{\lambda,N} \rangle + \frac{1}{\sqrt 2} \langle \Psi_{\lambda,N} \left(|\mathbf a_\lambda | \Psi_{\lambda,N} \rangle\right) \\ & = \frac{\bar \alpha}{\sqrt 2} \langle \Psi_{\lambda,N} | \Psi_{\lambda,N} \rangle + \frac{\alpha}{\sqrt 2} \langle \Psi_{\lambda,N} | \Psi_{\lambda,N} \rangle\\ & = \frac{\bar \alpha}{\sqrt 2} + \frac{\alpha}{\sqrt 2} \\ & \frac{\sqrt{N} e^{i\omega_\lambda t}}{\sqrt 2} + \frac{\sqrt{N} e^{-i\omega_\lambda t}}{\sqrt 2} = \frac{\sqrt{N}}{\sqrt 2} \left[cos\left(\omega_\lambda t\right) + sin\left(\omega_\lambda t\right) + cos\left(\omega_\lambda t\right) - sin\left(\omega_\lambda t\right) \right]\\ & = \frac{\sqrt{N}}{\sqrt 2}2cos\left(\omega_\lambda t\right) = \sqrt{2N} cos\left(\omega_\lambda t\right) \\ \langle \pi_\lambda \rangle & = \frac{i}{\sqrt 2} \langle \Psi_{\lambda,N} | \left(\mathbf a^\dag_\lambda - \mathbf a_\lambda\right) | \Psi_{\lambda,N} \rangle = \frac{i}{\sqrt 2} \langle \Psi_{\lambda,N} | \mathbf a^\dag_\lambda | \Psi_{\lambda,N} \rangle -\frac{i}{\sqrt 2} \langle \Psi_{\lambda,N} |\mathbf a_\lambda | \Psi_{\lambda,N} \rangle \\ & = \frac{i}{\sqrt 2} \left(\langle \Psi_{\lambda,N} | \mathbf a^\dag_\lambda\right) | \Psi_{\lambda,N} \rangle - \frac{1}{\sqrt 2} \langle \Psi_{\lambda,N} \left(|\mathbf a_\lambda | \Psi_{\lambda,N} \rangle\right) \\ & = \frac{i\bar \alpha}{\sqrt 2} \langle \Psi_{\lambda,N} | \Psi_{\lambda,N} \rangle - \frac{i\alpha}{\sqrt 2} \langle \Psi_{\lambda,N} | \Psi_{\lambda,N} \rangle\\ & = \frac{i\bar \alpha}{\sqrt 2} - \frac{i\alpha}{\sqrt 2} \\ & \frac{i\sqrt{N} e^{i\omega_\lambda t}}{\sqrt 2} - \frac{i\sqrt{N} e^{-i\omega_\lambda t}}{\sqrt 2} = \frac{i\sqrt{N}}{\sqrt 2} \left[cos\left(\omega_\lambda t\right) + sin\left(\omega_\lambda t\right) - cos\left(\omega_\lambda t\right) + sin\left(\omega_\lambda t\right) \right] \\ & = \frac{i\sqrt{N}}{\sqrt 2}2i \sin\left(\omega_\lambda t\right) = -\sqrt{2N} sin\left(\omega_\lambda t\right) \\ \end{aligned}
To find the expectation value of the square of the dimensionless position operator \langle \xi^2 \rangle:
\xi^2 = \left(\frac{1}{\sqrt{2}} (\mathbf a + \mathbf a^\dagger)\right)^2 = \frac{1}{2} (\mathbf a + \mathbf a^\dagger)^2
Expanding this:
\xi^2 = \frac{1}{2} (\mathbf a^2 + \mathbf a \mathbf a^\dagger + \mathbf a^\dagger \mathbf a + (\mathbf a^\dagger)^2)
Using the commutation relation [\mathbf a, \mathbf a^\dagger] = 1, where \mathbf a \mathbf a^\dagger = 1 + \mathbf a^\dagger \mathbf a:
\xi^2 = \frac{1}{2} (\mathbf a^2 + 1 + 2 \mathbf a^\dagger \mathbf a + (\mathbf a^\dagger)^2) = \frac{1}{2} (\mathbf a^2 + (\mathbf a^\dagger)^2 + 2 \mathbf a^\dagger\mathbf a + 1)
Now, we compute the expectation value of each term:
\begin{aligned} & \langle \Psi_{\lambda,N} | \mathbf a^2 | \Psi_{\lambda,N} \rangle = \alpha^2 \\ & \langle \Psi_{\lambda,N} | (\mathbf a^\dagger)^2 | \Psi_{\lambda,N} \rangle = \overline{\alpha}^2 \\ & \langle \Psi_{\lambda,N} | \mathbf a^\dagger \mathbf a | \Psi_{\lambda,N} \rangle = |\alpha|^2 = N \end{aligned}
Therefore:
\langle \xi^2 \rangle = \frac{1}{2} \left(\alpha^2 + \overline{\alpha}^2 + 2N + 1\right)
Using \alpha = \sqrt{N} e^{-i \omega_\lambda t} and \overline{\alpha} = \sqrt{N} e^{i \omega_\lambda t}:
\begin{aligned} \langle \xi^2 \rangle & = \frac{1}{2} \left(N e^{-2i \omega_\lambda t} + N e^{2i \omega_\lambda t} + 2N + 1\right) \\ & = \frac{1}{2} \left(2N \cos(2 \omega_\lambda t) + 2N + 1\right) \\ &= N \cos(2 \omega_\lambda t) + N + \frac{1}{2} \end{aligned}
Since:
\langle \xi \rangle = \sqrt{2N} \cos(\omega_\lambda t)
Therefore, the variance \sigma_\xi^2 is:
\sigma_\xi^2 = \langle \xi^2 \rangle - \langle \xi \rangle^2 = \left(N \cos(2 \omega_\lambda t) + N + \frac{1}{2}\right) - 2N \cos^2(\omega_\lambda t)
Using the double-angle formula \cos(2 \theta) = 2 \cos^2(\theta) - 1, we find:
\sigma_\xi^2 = N (2 \cos^2(\omega_\lambda t) - 1) + N + \frac{1}{2} - 2N \cos^2(\omega_\lambda t) = \frac{1}{2}
This result indicates that the position spread (or uncertainty) is constant and minimal for a coherent state, reflecting its quantum ground state-like behavior.
Similar steps can be done for \langle \pi^2 \rangle:
\pi^2 = \left(\frac{i}{\sqrt{2}} (\mathbf a^\dagger - \mathbf a)\right)^2 = -\frac{1}{2} (\mathbf a^\dagger - \mathbf a)^2
Expanding this:
\pi^2 = -\frac{1}{2} ((\mathbf a^\dagger)^2 - \mathbf a^\dagger \mathbf a - \mathbf a \mathbf a^\dagger + \mathbf a^2)
Using the commutation relation [\mathbf a, \mathbf a^\dagger] = 1, where \mathbf a \mathbf a^\dagger = 1 + \mathbf a^\dagger \mathbf a:
\pi^2 = -\frac{1}{2} ((\mathbf a^\dagger)^2 - 2\mathbf a^\dagger \mathbf a - 1 + \mathbf a^2)
reusing the value previously computed:
\langle \pi^2 \rangle = -\frac{1}{2} (\overline{\alpha}^2 - 2N - 1 + \alpha^2)
Using \alpha = \sqrt{N} e^{-i \omega_\lambda t} and \overline{\alpha} = \sqrt{N} e^{i \omega_\lambda t}:
\begin{aligned} \langle \pi^2 \rangle & = -\frac{1}{2} (N e^{2i \omega_\lambda t} - 2N - 1 + N e^{-2i \omega_\lambda t}) \\ & = -\frac{1}{2} (2N \cos(2 \omega_\lambda t) - 2N - 1)\\ & = N \cos(2 \omega_\lambda t) - N - \frac{1}{2} \end{aligned}
The variance \sigma_\pi^2 is:
\sigma_\pi^2 = \langle \pi^2 \rangle - \langle \pi \rangle^2 = \left(N \cos(2 \omega_\lambda t) - N - \frac{1}{2}\right) - 2N \sin^2(\omega_\lambda t)
Using the identity \cos(2 \theta) = 1 - 2 \sin^2(\theta), we find:
\sigma_\pi^2 = N (1 - 2 \sin^2(\omega_\lambda t)) - N - \frac{1}{2} - 2N \sin^2(\omega_\lambda t) = \frac{1}{2}
The variance \sigma_\pi^2 of \pi in the coherent state |\Psi_{\lambda,N} \rangle is also \frac{1}{2}, and thus the standard deviation \sigma_\pi is:
\sigma_\pi = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}
This result reflects the symmetric quantum mechanical uncertainties for position and momentum in the ground state-like coherent state, maintaining the minimal Heisenberg uncertainty principle limit.
The Heisenberg Uncertainty Principle for position x and momentum p can be expressed as:
\sigma_x \sigma_p \geq \frac{\hbar}{2}
where \sigma_x and \sigma_p are the standard deviations of position and momentum, respectively, and \hbar is the reduced Planck constant.
For the dimensionless position \xi and momentum \pi defined in a harmonic oscillator, the uncertainty relationship translates to:
\sigma_\xi \sigma_\pi \geq \frac{1}{2}
In a coherent state we found that the uncertainties in dimensionless position and momentum are:
\sigma_\xi = \sigma_\pi = \frac{1}{\sqrt{2}}
Multiplying these gives:
\sigma_\xi \sigma_\pi = \frac{1}{2}
This is exactly the equality condition in the uncertainty principle for these dimensionless variables. It shows that coherent states are minimum uncertainty states, where the product of the uncertainties reaches the theoretical lower bound established by quantum mechanics.