User:EverettYou/Second Quantization

Base on the Lecture note.

Minimal Uncertainty States
Heisenberg uncertainty principle: for any Hermitian operator $$\hat{A}$$ and $$\hat{B}$$ and any state $$|\psi\rangle$$, the following inequality holds
 * $$\sqrt{\langle(\delta\hat{A})^2\rangle\langle(\delta\hat{B})^2\rangle}\geq \frac{1}{2}|\langle[\hat{A},\hat{B}]\rangle|$$,

where $$\delta\hat{A}=\hat{A}-\langle\hat{A}\rangle$$, $$\delta\hat{B}=\hat{B}-\langle\hat{B}\rangle$$, and $$\langle\cdots\rangle\equiv\langle\psi|\cdots|\psi\rangle$$.

The equality is achieved if and only if $$|\psi\rangle$$ is a solution of the minimal uncertainty equation
 * $$(\cos\theta\;\hat{A}+i\sin\theta\;\hat{B})|\psi\rangle = 0$$,

for any $$\theta\in[0,2\pi)$$. There is an one-to-one correspondence between the angle θ and the state $$|\psi\rangle$$ that minimize the uncertainty between $$\hat{A}$$ and $$\hat{B}$$.

Displacement operator
Definition: for $$\alpha\in\mathbb{C}$$,
 * $$\hat{D}(\alpha) = e^{(\alpha\hat{a}^\dagger-\alpha^*\hat{a})}$$.

Unitarity: $$\hat{D}^\dagger(\alpha)=\hat{D}^{-1}(\alpha)=\hat{D}(-\alpha)$$.

Action of displacement operator performs translation in the phase space
 * $$\hat{D}(\alpha)\hat{a}\hat{D}^\dagger(\alpha)= \hat{a}-\alpha$$,
 * $$\hat{D}(\alpha)\hat{a}^\dagger\hat{D}^\dagger(\alpha)= \hat{a}^\dagger-\alpha^*$$.

Applying to the vacuum state leads to the coherent state $$|\alpha\rangle=\hat{D}(\alpha)|0\rangle$$, such that
 * $$\hat{a}|\alpha\rangle = \alpha|\alpha\rangle$$.

Properties of Coherent State
Expansion in particle number representation
 * $$|\alpha\rangle=e^{-|\alpha|^2/2}\sum_{n=0}^{\infty}\frac{\alpha^n}{\sqrt{n!}}|n\rangle$$

Overlap:
 * $$\langle\alpha|\alpha'\rangle=e^{-\frac{1}{2}|\alpha|^2-\frac{1}{2}|\alpha'|^2+\alpha^*\alpha'}$$.

Completeness:
 * $$\int\frac{\mathrm{d}[\alpha]}{\pi}|\alpha\rangle\langle\alpha|=\hat{1}$$.

Squeezing operator
Definition: for $$\xi\in\mathbb{C}$$,
 * $$\hat{U}(\xi) = e^{\xi(\hat{a}^\dagger \hat{a}^\dagger-\hat{a}\hat{a})/2}$$.

Unitarity: $$\hat{U}^\dagger(\alpha)=\hat{U}^{-1}(\alpha)=\hat{U}(-\alpha)$$.

Action of squeezing operator performs the Bogoliubov transform
 * $$\hat{U}(\xi)\hat{a}\hat{U}^\dagger(\xi)= \cosh\xi\;\hat{a} + \sinh\xi\;\hat{a}^\dagger$$,
 * $$\hat{U}(\xi)\hat{a}^\dagger\hat{U}^\dagger(\xi)= \cosh \xi\;\hat{a}^\dagger + \sinh\xi\;\hat{a}$$.

Applying to the vacuum state leads to the squeezed state $$|\xi\rangle=\hat{U}(\xi)|0\rangle$$, such that
 * $$(\cosh\xi\;\hat{a} - \sinh\xi\;\hat{a}^\dagger)|\xi\rangle =0$$.