Coherent State Definition

In classical wave mechanics

A sinusoidal wave is the ideal coherent state. The first-order complex degree of temporal coherence is given by:

$$\gamma(\tau)=\frac{\braket{E(r,t)E^*(r,t + \tau)}}{\braket{|E(r,t)|^2}}$$

This measures the wave's similarity to its time-shifted self (ignoring the global phase). Thus, a perfect sinusoidal wave is perfectly coherent, yielding $|\gamma(\tau)| = 1$ for all $\tau$.

In quantum mechanics

The coherent state $\ket{\alpha}$ is defined as the eigenstates of the annihilation operator $\hat{a}$:

$$\hat{a}\ket{\alpha}=\alpha\ket{\alpha}$$

We will prove below that this definition is consistent with the classical definition. Intuitively, because $\hat{a}$ removes a photon without changing the state, measuring the field does not alter its quantum state. The wave remains identical before and after annihilation, making it fundamentally coherent.

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Expand $\ket{\alpha}$ in the Fock Basis

We can express the coherent state as a linear combination of number (Fock) states:

$$\ket{\alpha}=\sum_{n=0}^\infty c_n \ket{n}$$

Using the property $\hat{a}\ket{n} = \sqrt{n}\ket{n-1}$ alongside the eigenvalue equation, we obtain the recurrence relation . Referencing the Taylor expansion of , this yields:

$$\ket{\alpha}=c_0\sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}\ket{n}$$

By enforcing the normalization condition $\braket{\alpha|\alpha} = 1$, we find:

$$c_0 = e^{-\frac{1}{2}|\alpha|^2}$$

Therefore, the full expansion is:

$$\ket{\alpha} = e^{-\frac{1}{2}|\alpha|^2} \sum_{n=0}^\infty \frac{\alpha^n}{\sqrt{n!}}\ket{n}$$

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The Physical Meaning of $\alpha$

The eigenvalue $\alpha$ is generally a complex number because the annihilation operator $\hat{a}$ is non-Hermitian.

The expectation value of the photon number operator $\hat{n} = \hat{a}^\dagger\hat{a}$ is:

$$\braket{n}=\bra{\alpha}\hat{a}^\dagger\hat{a}\ket{\alpha}=|\alpha|^2$$

Thus, the magnitude $|\alpha| = \sqrt{\braket{n}}$ represents the square root of the average photon number. The phase of $\alpha$ depends on the choice of time origin, as shown below.

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Proving Consistency with Classical Mechanics

Using the Heisenberg picture for free evolution, the time dependence of the annihilation operator is:

$$\hat{a}(t)=\hat{a}_0 e^{-i\omega t}$$

(Note: The standard physical convention uses $e^{-i\omega t}$ so that the positive frequency component matches classical wave mechanics conventions).

Let $\alpha = |\alpha|e^{i\phi}$. The expectation value of the position operator $\hat{x}(t)$ evolves as:

$$\begin{array}{rl}{\displaystyle ⟨x(t)⟩}& {\displaystyle =\sqrt{\frac{\mathrm{\hslash }}{2m\omega }}⟨\alpha \mathrm{\mid }\widehat{a}(t)+{\widehat{a}}^{†}(t)\mathrm{\mid }\alpha ⟩}\\ {\displaystyle }& {\displaystyle =\sqrt{\frac{\mathrm{\hslash }}{2m\omega }}(\mathrm{\mid }\alpha \mathrm{\mid }{e}^{i\varphi }{e}^{-i\omega t}+\mathrm{\mid }\alpha \mathrm{\mid }{e}^{-i\varphi }{e}^{i\omega t})}\\ {\displaystyle }& {\displaystyle =\sqrt{\frac{\mathrm{\hslash }⟨n⟩}{2m\omega }}(e^{-i(\omega t-\varphi )}+e^{i(\omega t-\varphi )})}\\ {\displaystyle }\end{array}$$

The same calculation can be carried out for the momentum expectation value $\braket{p(t)}$, utilizing the definition $\hat{p}(t) = -i\sqrt{\frac{m\hbar\omega}{2}}\left(\hat{a}(t) - \hat{a}^\dagger(t)\right)$:

$$\begin{array}{rl}{\displaystyle ⟨p(t)⟩}& {\displaystyle =-i\sqrt{\frac{m\mathrm{\hslash }\omega }{2}}⟨\alpha \mathrm{\mid }\widehat{a}(t)-{\widehat{a}}^{†}(t)\mathrm{\mid }\alpha ⟩}\\ {\displaystyle }& {\displaystyle =-i\sqrt{\frac{m\mathrm{\hslash }\omega }{2}}(\mathrm{\mid }\alpha \mathrm{\mid }{e}^{i\varphi }{e}^{-i\omega t}-\mathrm{\mid }\alpha \mathrm{\mid }{e}^{-i\varphi }{e}^{i\omega t})}\\ {\displaystyle }& {\displaystyle =-i\sqrt{\frac{m\mathrm{\hslash }\omega ⟨n⟩}{2}}(e^{-i(\omega t-\varphi )}-e^{i(\omega t-\varphi )}}\end{array}$$

Here is the complete, finalized section for your note. I have integrated your derivation smoothly, cleaned up the final algebraic typos from your draft section, and written out the step-by-step calculations for $\braket{x^2(t)}$, $\braket{p^2(t)}$, and the final uncertainty product $\sigma_x \sigma_p$ using the displacement operator frame.

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The Displacement Operator

To have further discussion of the coherent state with clean math, we can utilize the Lie algebra view of quantum mechanics (we can proceed without it, but the math is exceptionally heavy).

The Aim

We want a unitary operator $\hat{D}(\alpha)$ that constructs our coherent state directly from the vacuum:

$$\ket{\alpha}=\hat{D}(\alpha)\ket{0}$$

Why can we leverage Lie theory here? Because the displacement parameter $\alpha$ is a continuous, smooth complex coordinate on phase space. Since our physical ladder operators satisfy the canonical commutation relation $[\hat{a}, \hat{a}^\dagger] = \hat{I}$, they span the 3-dimensional Heisenberg-Weyl Lie algebra . Any continuous transformation preserving this structure must be generated by a linear combination of these basis elements:

$$\hat{D} = e^{\lambda_1\hat{a}+\lambda_2\hat{a}^\dagger+\lambda_3\hat{I}}$$

For $\hat{D}$ to be a physical, probability-preserving unitary operator, its generator must be anti-Hermitian ($\hat{G}^\dagger = -\hat{G}$):

$$\lambda_1^*\hat{a}^\dagger+\lambda_2^*\hat{a}+\lambda_3^*\hat{I}=-\lambda_1\hat{a}-\lambda_2\hat{a}^\dagger-\lambda_3\hat{I}$$

Matching operator coefficients yields $\lambda_2 = -\lambda_1^*$ and $\lambda_3 = i\theta$ (where ). Because the identity operator commutes with all elements of the algebra, it factors out as a global phase that we can physically neglect:

$$\hat{D}=e^{\lambda_1\hat{a}-\lambda_1^*\hat{a}^\dagger}$$

To determine the unknown geometric parameter $\lambda_1$ in terms of our physical eigenvalue $\alpha$, we enforce the definition of the coherent state:

$$\hat{a}\hat{D}(\lambda_1)\ket{0}=\alpha \hat{D}(\lambda_1)\ket{0}$$

We evaluate how $\hat{a}$ transforms under this group action by inserting an identity operator $\hat{I} = \hat{D}\hat{D}^\dagger$:

$$\hat{a}\hat{D}\ket{0} = \hat{D} \left[ \hat{D}^\dagger \hat{a} \hat{D} \right] \ket{0} = \alpha \hat{D}\ket{0}$$

Using the Baker-Campbell-Hausdorff (BCH) expansion, the core commutator evaluates to a simple scalar:

$$[\lambda_1 \hat{a} - \lambda_1^* \hat{a}^\dagger, \hat{a}] = -\lambda_1^* [\hat{a}^\dagger, \hat{a}] = \lambda_1^*$$

Because this commutator is a scalar, all higher-order nested brackets vanish. The similarity transformation shifts the operator linearly:

$$\hat{D}^\dagger \hat{a} \hat{D} = \hat{a} + \lambda_1^*$$

Substituting this back into our primary expression:

$$\begin{align*} \hat{D}\left( \hat{a} + \lambda_1^* \right)\ket{0} &= \alpha\hat{D}\ket{0} \\ \hat{D}\hat{a}\ket{0} + \lambda_1^*\hat{D}\ket{0} &= \alpha\hat{D}\ket{0} \end{align*}$$

Since the vacuum state cannot be annihilated further ($\hat{a}\ket{0}=0$), the first term vanishes, leaving:

$$\lambda_1^* = \alpha \implies \lambda_1 = \alpha^*$$

Substituting $\lambda_1 = \alpha^*$ back into our generator yields the standard form of the Displacement Operator:

$$\hat{D}(\alpha)=e^{\alpha\hat{a}^\dagger-\alpha^*\hat{a}}$$

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The Uncertainty Relation

Instead of tracking dynamic states, we can stand firmly on the vacuum state $\ket{0}$ and compute our time-dependent variances by shifting the operators themselves via the Heisenberg picture:

$$\hat{a}(t) = \hat{a}_0 e^{-i\omega t}, \quad \hat{a}^\dagger(t) = \hat{a}_0^\dagger e^{i\omega t}$$

From our BCH framework, we know the initial displacement operator shifts the boundary operators at $t=0$ by a classical constant:

$$\hat{D}^\dagger(\alpha_0)\hat{a}_0\hat{D}(\alpha_0) = \hat{a}_0 + \alpha_0, \quad \hat{D}^\dagger(\alpha_0)\hat{a}_0^\dagger\hat{D}(\alpha_0) = \hat{a}_0^\dagger + \alpha_0^*$$

1. Calculating $\braket{x^2(t)}$

The square of the time-dependent position operator is given by:

$$\hat{x}^2(t) = \frac{\hbar}{2m\omega}\left( \hat{a}_0^2 e^{-2i\omega t} + (\hat{a}_0^\dagger)^2 e^{2i\omega t} + 2\hat{a}_0^\dagger\hat{a}_0 + 1 \right)$$

Evaluating the expectation value $\bra{0}\hat{D}^\dagger \hat{x}^2(t) \hat{D}\ket{0}$ amounts to shifting the internal operators:

$$\braket{x^2(t)} = \frac{\hbar}{2m\omega} \bra{0} \left[ (\hat{a}_0 + \alpha_0)^2 e^{-2i\omega t} + (\hat{a}_0^\dagger + \alpha_0^*)^2 e^{2i\omega t} + 2(\hat{a}_0^\dagger + \alpha_0^*)(\hat{a}_0 + \alpha_0) + 1 \right] \ket{0}$$

Expanding this product out, any term containing a standalone quantum operator $\hat{a}_0$ or $\hat{a}_0^\dagger$ drops out against the vacuum boundaries ($$ and ). Only the classical parameters and the identity survive:

$$\braket{x^2(t)} = \frac{\hbar}{2m\omega} \left( \alpha_0^2 e^{-2i\omega t} + (\alpha_0^*)^2 e^{2i\omega t} + 2|\alpha_0|^2 + 1 \right)$$

Substituting the polar form $\alpha_0 = |\alpha_0|e^{i\phi}$:

$$\begin{align*} \braket{x^2(t)} &= \frac{\hbar}{2m\omega} \left( |\alpha_0|^2 e^{-2i(\omega t - \phi)} + |\alpha_0|^2 e^{2i(\omega t - \phi)} + 2|\alpha_0|^2 + 1 \right) \\ &= \frac{\hbar}{2m\omega} \left( 2|\alpha_0|^2 \cos(2\omega t - 2\phi) + 2|\alpha_0|^2 + 1 \right) \end{align*}$$

Using the trigonometric identity $2\cos(2\theta) + 2 = 4\cos^2(\theta)$:

$$\braket{x^2(t)} = \frac{2\hbar|\alpha_0|^2}{m\omega}\cos^2(\omega t - \phi) + \frac{\hbar}{2m\omega}$$

2. Calculating $\braket{p^2(t)}$

We repeat the exact same process for the squared momentum operator:

$$\hat{p}^2(t) = -\frac{m\hbar\omega}{2}\left( \hat{a}_0^2 e^{-2i\omega t} + (\hat{a}_0^\dagger)^2 e^{2i\omega t} - 2\hat{a}_0^\dagger\hat{a}_0 - 1 \right)$$

Shifting the operators into the vacuum frame yields:

$$\braket{p^2(t)} = -\frac{m\hbar\omega}{2} \left( \alpha_0^2 e^{-2i\omega t} + (\alpha_0^*)^2 e^{2i\omega t} - 2|\alpha_0|^2 - 1 \right)$$

Substituting $\alpha_0 = |\alpha_0|e^{i\phi}$:

$$\begin{align*} \braket{p^2(t)} &= -\frac{m\hbar\omega}{2} \left( 2|\alpha_0|^2 \cos(2\omega t - 2\phi) - 2|\alpha_0|^2 - 1 \right) \end{align*}$$

Using the identity $2\cos(2\theta) - 2 = -4\sin^2(\theta)$:

$$\braket{p^2(t)} = 2m\hbar\omega|\alpha_0|^2\sin^2(\omega t - \phi) + \frac{m\hbar\omega}{2}$$

3. Position and Momentum Variance

The variances $\sigma_x^2$ and $\sigma_p^2$ measure the statistical quantum fluctuations around the classical trajectories. Utilizing our previous derivations for the mean values $\braket{x(t)}$ and :

$$\sigma_x^2 = \braket{x^2(t)} - \braket{x(t)}^2 = \left[ \frac{2\hbar|\alpha_0|^2}{m\omega}\cos^2(\omega t - \phi) + \frac{\hbar}{2m\omega} \right] - \frac{2\hbar|\alpha_0|^2}{m\omega}\cos^2(\omega t - \phi) = \frac{\hbar}{2m\omega}$$ $$\sigma_p^2 = \braket{p^2(t)} - \braket{p(t)}^2 = \left[ 2m\hbar\omega|\alpha_0|^2\sin^2(\omega t - \phi) + \frac{m\hbar\omega}{2} \right] - 2m\hbar\omega|\alpha_0|^2\sin^2(\omega t - \phi) = \frac{m\hbar\omega}{2}$$

Taking the square root of both variances gives:

$$\sigma_x = \sqrt{\frac{\hbar}{2m\omega}}, \quad \sigma_p = \sqrt{\frac{m\hbar\omega}{2}}$$

4. The Uncertainty Product

Multiplying the two standard deviations together yields:

$$\sigma_x \sigma_p = \sqrt{\frac{\hbar}{2m\omega}} \cdot \sqrt{\frac{m\hbar\omega}{2}} = \frac{\hbar}{2}$$

This is minimum uncerteinty.

Put simply, why coherent state is minimum uncertainty? Because all state are shifted uniformly upward by the classical force from the harmonic occilation ( meaning the force have no quantum effect ). And the coherent state is at the minimum, ie, it is the shifted $\ket{0}$ state which is known to have minimum uncertainty.

Algebraicly, the uniform shift, is the operator we wrote down,

$$\hat{\tilde{a}}=\hat{D}^\dagger\hat{a}\hat{D}=\hat{a}+\alpha$$