Time Evolution
Unitary Time Evolution
[Assumption] : The basic assumption of quantum dynamics is that , the time evolution of a closed system is unitary .
[Definition] : Time evolution operator : That means if we note the quantum state at time $t$ . : as $\ket{\psi(t)}$ . , then there is a Unitary operator
$\hat U(t',t)$ , one have:
$$\ket{\psi(t')} = \hat U(t',t)\ket{\psi(t)}$$
The evolution operator has the following property:
$$\hat U(t'',t')\hat U(t',t)=\hat U(t'',t) \ \ ; \ \hat U(t,t)=\hat 1$$
It is usually continuous and reversible
The infinitesimal time evolution:
$$\hat U(t+\td t,t) = \hat 1 -\frac {\ti} {\hbar} \hat H(t)\td t +O(dt^2)$$
Where the Hamiltonian is:
$$\hat H(t) = \ti \hbar \frac {\td} {\td t'} \hat U(t',t)\Big|_{t'=t}=\ti \hbar \frac {\td \hat U(t',t)} {\td t'} \hat U^{\dagger}(t',t)\Big|_{t'=t}$$
Then for $(\partial_1 \hat U(t',t))\hat U^{\dagger}(t',t)+\hat U(t',t)\partial_1\hat U^{\dagger}(t',t)=0$ , one can find that $\hat H$ is Hermitian
Moreover, one can prove that $\frac {\td \hat U(t',t)} {\td t'} \hat U^{\dagger}(t',t)$ is independent to $t$ . : :
Proof
$$\begin{aligned}[] \frac {\td \hat U(t,t_0)} {\td t} \hat U^{\dagger}(t,t_0) &= \frac {\td \hat U(t,t'_0)} {\td t} \hat U(t'_0,t_0) (\hat U(t,t'_0)\hat U(t'_0,t_0))^{\dagger} \\ &= \frac {\td \hat U(t,t'_0)} {\td t} \hat U(t'_0,t_0) \hat U^{\dagger}(t'_0,t_0) \hat U^{\dagger}(t,t'_0) \\ &= \frac {\td \hat U(t,t'_0)} {\td t} \hat U^{\dagger}(t,t'_0) \end{aligned}$$ Q.E.D.Then we define that:
[Definition] : Hamiltonian is
$$\hat H(t) = \ti \hbar \frac {\td \hat U(t,t_0)} {\td t} \hat U^{\dagger} (t,t_0)$$
which is independent to initial time $t_0$
By this definition, one can write down the equation that discribes the evolution of evolution operator:
$$\ti \hbar \frac {\partial} {\partial t} \hat U(t,t_0) = \hat H(t) \hat U(t,t_0)$$
Schrodinger Equation and Schrodinger Picture
Applying the evolution equation of $\hat U(t,t_0)$ ,, one can derive the evolution equation of quantum state:
$$\ti \hbar \frac{\td} {\td t} \ket{\psi(t)} = \hat H(t)\ket{\psi(t)}$$
When $\hat H$ is independent to $t$ . : , one can “solve” this as:
$$\ket{\psi(t)} = \exp((t-t_0) \hat H/\ti \hbar) \ket{\psi(t_0)} \ ; \ \ \hat U(t,t_0) = \exp\big(\frac {(t-t_0)} {\ti \hbar} \hat H\big)$$
In general, one can write that :
$$\hat U(t,t_0) = \mathcal{T}\Big(\exp\big(-\frac {\ti} {\hbar} \int_{t'=t_0}^t \hat H (t')\td t' \big) \Big)\equiv \sum_{n=0}^{\infty} \Big(\frac {1} {\ti \hbar}\Big)^n \frac 1 {n!} \int_{t_1,\cdots,t_n=t_0}^t \td t_1\cdots \td t_n \mathcal{T}\big(\prod_{i} \hat H (t_i) \big)$$
Where $\mathcal{T}$ means time-ordering:
- For
Bosonic$\hat A,\hat B$ , $\mathcal{T}(\hat A(t)\hat B(t')) = \begin{cases} \hat A(t)\hat B(t') & t\gt t' \\ \hat B(t')\hat A(t) & t'\gt t\end{cases}$ - For
Fermionic$\hat A,\hat B$ , $\mathcal{T}(\hat A(t)\hat B(t')) = \begin{cases} \hat A(t)\hat B(t') & t\gt t' \\ (-1)\hat B(t')\hat A(t) & t'\gt t\end{cases}$
And this is equivalent to the Dyson series :
$$\hat U(t,t_0) = \hat 1 + \frac {1} {\ti \hbar} \int_{t_1=t_0}^t \hat H(t_1)\td t_1 + \Big(\frac 1 {\ti\hbar} \Big)^2 \int_{t_2=t_0}^t \int_{t_1=t_0}^{t_2} \hat H(t_2)\hat H(t_1)\td t_2\td t_1 +\cdots$$
[Theorem] : eigenstates of Hamiltonian are Stationary states :
$$\ket{E,t} = e^{-\ti E(t-t_0)}\ket{E,t_0} \ \ \text{ where }\hat H\ket{E,t}=E\ket{E,t}$$
And expectation value of any operator does not change over time, density matrix does not change over time.
This is the Schorodinger picture (subscript $_S$ hereafter) : time evolution is implemented on the state
The Heisenberg Picture
The Heisenberg picture (subscript $_H$ hereafter) : time evolution is encoded in operators, while the states have no evolution
[Definition] : the time-dependent operator $\hat O_H(t)$ in the Heisenberg picture is:
$$\hat O_H(t) = \hat U^{\dagger} (t) \hat O_S \hat U(t)$$
Note
Then for matrix elements in Heisenberg picture, one have: $\bra{\psi}\hat O_H(t)\ket{\phi} = \bra{\psi(t)}\hat O_S \ket{\phi(t)}$The Heisenberg equation of motion:
$$\hbar \frac {\td} {\td t} \hat O_H(t) = \ti [\hat H_H(t),\hat O_H(t)]$$
Where $\hat H_H(t) = \hat U^{\dagger}(t)\hat H_S(t)\hat U(t) =\ti \hbar \hat U^{\dagger}(t)\frac {\td} {\td t} \hat U(t) $
When $\hat H_S(t)$ is independent of time, then $\hat H$ is commutative with $\hat U$ , and $\hat H_H = \hat H_S$
Some Examples About Dynamics
Liouville Equation
In Schrodinger picture, For $\hat \rho(t) = \sum_i \lambda_i \ket{\psi_i(t)}\bra{\psi_i(t)}$ , one can write down:
$$\ti \hbar \frac {\td \hat \rho(t)} {\td t} = [\hat H_S(t),\hat \rho(t)]$$
But in Heisenberg picture, the density matrix will never change by time
Single Particle Dynamics
Consider time independent Hamiltonian:
$$\hat H = \frac {\hat {\bm{p}}^2} {2m} +V(\hat {\bm{x}})$$
- In Schrodinger picture, one can obtain a continuity equation:
$$\frac {\partial} {\partial t} \rho(\bm{x},t) + \nabla \cdot \bm{J}(\bm{x},t)=0$$
where probability density $\rho(\bm{x},t) = \psi^*(\bm{x},t)\psi(\bm{x},t) = \bra{\bm{x}}\hat \rho\ket{\bm{x}}$ <br>
and probability current $\bm{J}(\bm{x},t) = \frac {\hbar} m \text{Im}\big(\psi^*(\bm{x},t)\nabla\psi(\bm{x},t) \big)$
- In Heisenberg picture, one can obtain the Ehrenfest theorem:
the equation of motion of position $\hat{\bm{x}}_{H} $ and momentum $\hat{\bm{p}}_H$ are:
$$\begin{aligned} \frac {\td} {\td t} \hat{\bm{x}}_H(t) &= \frac {\ti} {\hbar} [\hat H,\hat{\bm{x}}_H] = \frac {\hat{\bm{p}}_H} {m} \\ \frac {\td} {\td t} \hat{\bm{p}}_H(t) &= \frac {\ti} {\hbar} [\hat H,\hat{\bm{p}}_H] = -\nabla V(\hat{\bm{x}}) \end{aligned}$$
Then one can write down a equation which is similar to the classical equation of motion :
$$m \frac {\td^2} {\td t^2} \langle \hat{\bm{x}}\rangle = - \langle \nabla V(\hat{\bm{x}})\rangle$$
1D Harmonic Oscillator
Having time-independent Hamiltonian:
$$\hat H = \frac {\hat p^2} {2m} +\frac {m \omega^2} 2 \hat x^2 = \hbar \omega(\hat a^{\dagger}\hat a + \frac 1 2)$$
Where $\hat a =\sqrt{\frac {m\omega} {2\hbar}}(\hat x +\frac {\ti} {m\omega} \hat p)$ and satisfies: $[\hat a,\hat a^{\dagger}] = 1$ ..
One can solve the equation of motion: $\frac {\td} {td t} \hat a(t) = \frac {\ti} {\hbar} [\hat H,\hat a(t)] = -\ti \omega \hat a(t)$ and obtain $\hat a(t) = e^{-\ti \omega t} \hat a(0)$ .. Then one can obtain the:
$$\begin{aligned} \hat x(t) &= \cos(\omega t) \hat x(0)+\frac 1 {m\omega} \sin(\omega t) \hat p(0) \\ \hat p(t) &= -m\omega \sin(\omega t) \hat x(0) + \cos(\omega t) \hat p(0) \end{aligned}$$
[Definition] : The coherent state :
$$e^{-|z|^2/2}e^{z\hat a^{\dagger}} \ket{0} \ \ \text{with } \langle \hat a(0)\rangle = z$$
Then one have :
$$\langle \hat x(0)\rangle = \sqrt{\frac {2\hbar} {m\omega}} \text{Re} z \ \ \langle\hat p(0)\rangle = \sqrt{2\hbar m \omega} \text{Im}(z)$$
The coherent satisfies the minimal uncertainty relation for $\hat x$ and $\hat p$ < :
$$\Delta x^2 = \langle \hat x^2\rangle - (\langle \hat x \rangle)^2 = \frac {\hbar} {2m\omega} \ \ \Delta p^2 = \langle \hat p^2\rangle - (\langle \hat p\rangle)^2 = \frac {\hbar m \omega} 2 $$
Then $\Delta x^2 \Delta p^2 = \hbar^2/4$
Adiabatic Theorem
[Theorem] : When system is instantaneous ground state(s) and the Hamiltonian changes slowly with time, then the system will remain to be (one of) the instantaneous ground state(s) at later times.
For system’s Hamiltonian is dependent of time by a parameter $R$ :
$$\hat H(t) = \hat H(R(t))$$
The instantaneous eigenstates of Hamiltonian are $\hat H(t)\ket{E_n(t)} = E_n(t) \ket{E_n(t)}$ ..
The Adiabatic theorem showd that:
$$\ket{\psi(t=0)} = \ket{E_0(t=0)} \Rightarrow \ket{\psi(t)} = \ket{E_0(t)}$$
The theorem holds if
$$ \Big|\frac{1} {E_n(t)-E_m(t)} \bra{E_n(t)}\frac {\td \hat H(t)} {\td t} \ket{E_m(t)}\Big| \ll 1. $$
Proof(non-degenerate cases)
For the evolution can be discriped by Schrodinger equation: $$\ti \hbar \frac {\td} {\td t} \ket{\psi(t)} = \hat H(R(t)) \ket{\psi(t)}$$ One can expand the $\ket{\psi(t)}$ . by the eigenstates $\ket{E_n(t)}$ ,, as: $$\ket{\psi(t)} = \sum_i c_i(t)e^{\ti \theta_n(t)} \ket{E_n(t)}$$ Where $\theta_n(t) = \frac {-1} {\hbar} \int_0^t \td t' E_n(t')$Then one have the equation: $$\ti \hbar \frac {\td} {\td t} c_n(t) = - \sum_m c_m(t) \bra{E_n(t)}\frac {\td} {\td t} \ket{E_m(t)} e^{\ti(\theta_m(t)-\theta_n(t))}$$ For differentiating the both side of the eigenstates function: $$\hat H(t) \frac {\td} {\td t}\ket{E_n(t)} + \frac {\td \hat H(t)} {\td t} \ket{E_n(t)} = \frac {\td E_n(t)} {\td t} \ket{E_n(t)} + E_n(t)\frac {\td } {\td t} \ket{E_n(t)}$$ One have when $m\neq n$ :: $$\bra{E_n(t)}\frac {\td} {\td t} \ket{E_m(t)} = \frac {1} {E_m(t)-E_n(t)} \bra{E_n(t)}\frac {\td \hat H(t)} {\td t} \ket{E_m(t)} $$ Then we have: $$\ti \hbar \frac {\td } {\td t} c_n(t) = -c_n(t) \bra{E_n(t)}\frac {\td} {\td t} \ket{E_n(t)} - \sum_{m\neq n} \frac {c_m(t)} {E_n(t)-E_m(t)} \bra{E_n(t)}\frac {\td \hat H(t)} {\td t} \ket{E_m(t)} e^{\ti(\theta_m(t)-\theta_n(t))}$$ With the initial condition: $c_n(t=0) = \delta_{0,n}$ , and the Hamiltonian changes very slow. One can solve the equation: $$c_n(t) = \delta_{0,n} \exp\big(\int_0^t \td t' \bra{E_n(t')}\frac {\td} {\td t'} \ket{E_n(t')} \big) $$ Which means that $\ket{\psi(t)}$ . is the simultaneous ground state . On the other hand, we have the condition for adiabatic theorem: $$ \Big|\frac{1} {E_n(t)-E_m(t)} \bra{E_n(t)}\frac {\td \hat H(t)} {\td t} \ket{E_m(t)}\Big| \ll 1. $$
Proof(non-degenerate cases, with another adiabatic condition)
Consider the adiabatic process with $$ \hat H(t) \ket{E_n(t)} = E_n(t) \ket{E_n(t)}. $$ Thus, the Schrodinger equation of arbitrary state $\ket{\psi(t)} = \sum_n a_n(t) e^{-\ti \int^t E_n(t)\td t} \ket{E_n(t)}$ reads (with $\hbar = 1$ )) $$ \ti \sum_n \Big(\frac {\td a_n} {\td t} \ket{E_n(t)} + a_n(-\ti E_n)\ket{E_n(t)} +a_n \partial_t \ket{E_n(t)}\Big)e^{-\ti \int^t E_n(t)\td t} = \sum_n a_n e^{-\ti \int^t E_n(t)\td t} E_n(t) \ket{E_n(t)}. $$ Or $$ \ti \frac {\td a_n} {\td t} + \ti \sum_m a_m e^{-\ti \theta_{nm}(t)} \braket{n|\partial_t|m} = 0. $$ Given matrix $$ (\bm{A})_{nm} = -e^{\ti \theta_{nm}(t)} \braket{n|\partial_t|m} \equiv - \braket{E_n(t)|\partial_t|E_m(t)} e^{\ti \int_0^t (E_n - E_m)\td t}, $$ the equation of motion can be write as matrix differential equation (with $(\bm{x})_n = a_n$ )), like $$ \frac {\td \bm{x}} {\td t} = \bm{A}(t) \bm{x} \Leftrightarrow \frac {\td a_n} {\td t} = A_{nm} a_m. $$ Consider the diagonal term contribution, let $$ b_n(t) = a_n(t) \exp\Big(-\int_0^t A_{nn}(t)\td t\Big) = a_n(t) \exp\Big(-\ti \int_0^t (\ti \braket{n|\partial_t|n})\td t\Big) = a_n(t)e^{-\ti \gamma_n(t)}. $$ Then $$ \frac {\td b_n} {\td t} = \frac {\td a_n} {\td t} e^{-\ti \gamma_n} - A_{nn} a_n e^{-\ti \gamma_n} = (A_{nm} - A_{nn}\delta_{nm}) b_n. $$ Or $$ \frac {\td \bm{y}} {\td t} = \bm{B} \bm{y} $$ with $\bm{y} = (b_1 ,\cdots)$ .. Note that $\bm{y}(0) = \bm{x}(0)$ ,, and when $\|\bm{y}(t) - \bm{y}(0)\|\ll 1$ ,, the off-diagonal terms could be neglected. One can write down $$ \begin{aligned} \|\bm{y}(t) - \bm{y}(0)\| &= \Big\|\int_0^t \bm{B}(s) (\bm{y}(s) - \bm{y}(0) + \bm{y}(0))\td s \Big\| \\ &\leq \Big\|\int_0^t \bm{B}(s) (\bm{y}(s) - \bm{y}(0))\td s \Big\| + \Big\|\int_0^t \bm{B}(s) \bm{y}(0) \td s \Big\| \\ &\leq \int_0^t \|\bm{B}(s)\|\cdot \|\bm{y}(s) - \bm{y}(0)\|\td s + \Big\|\int_0^t \bm{B}(s) \td s \Big\| \cdot \|\bm{y}(0)\| \end{aligned} $$ Denoting $z(t) = \|\bm{y}(t) - \bm{y}(0)\|$ ,, $M(s) = \|\bm{B}(s)\|$ (being a metrix-norm, as a positive real number), and note $\|\bm{y}(0)\| = 1$ as the wavefunction is normalized. Thus, the inequality reads $$ \begin{aligned} z(t) &\leq \int_0^t M(s) z(s)\td s + \Big\|\int_0^t \bm{B}(s) \td s \Big\| \\ &\leq \int_0^t M(s) z(s)\td s + \sum_{n, m} \Big|\int_0^t B_{nm}(s) \td s \Big|. \end{aligned} $$ Note $$ \begin{aligned} \int_0^t B_{nm}(s) \td s &= \int_0^t (1- \delta_{nm}) A_{nm}(s) \td s = -(1- \delta_{nm}) \int_0^t e^{\ti \theta_{nm}(s)} \braket{n|\partial_t|m} \td s \\ &=-(1- \delta_{nm}) \int_0^t e^{\ti \theta_{nm}(s)} \frac {\braket{n|\partial_t|m}} {E_n - E_m} \td \theta_{nm} \\ &= -(1- \delta_{nm}) \int \frac {\braket{n|\partial_t|m}} {\ti(E_n - E_m)} \td e^{\ti \theta_{nm}} \end{aligned} $$ Thus $$ \begin{aligned} \Big|\int_0^t B_{nm}(s) \td s \Big| &= (1-\delta_{nm})\Bigg|\int \frac {\braket{n|\partial_t|m}} {E_n - E_m} \td e^{\ti \theta_{nm}} \Bigg|\\ &\leq (1-\delta_{nm}) \sup_{0\leq s \leq t} \Big|\frac {\braket{n|\partial_t|m}} {E_n - E_m}\Big| \cdot \Big|\int \td e^{\ti \theta_{nm}}\Big| \\ &\leq 2(1-\delta_{nm}) \sup_{0\leq s \leq t} \Big|\frac {\braket{n|\partial_t|m}} {E_n - E_m}\Big| \end{aligned} $$ Finally, the inequality about $z(t)$ becomes $$ z(t) \leq \int_0^t M(s)z(s)\td s + 2 \sum_{n\neq m} \sup_{0\leq s \leq t} \Big|\frac {\braket{n|\partial_t|m}} {E_n - E_m}\Big| \equiv \int_0^t M(s)z(s)\td s + \Xi(t). $$ Note $\Xi(t)$ is non-decrease and $\Xi(0) \geq 0$ .. Consider function (The following proof is known as the integral form of [Grönwall's inequality](https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality)) $$ w(t) = \exp\Big(- \int_0^t M(s)\td s \Big) \int_0^t M(s) z(s)\td s. $$ One has $$ \begin{aligned} w'(t) &= -M(t) e^{-\int_0^t M(s)\td s}\int_0^t M(s) z(s)\td s + (M(t)z(t))e^{-\int_0^t M(s)\td s} \\ &= M(t)e^{-\int_0^t M(s)\td s} \Big(z(t) - \int_0^t M(s)z(s) \td s\Big) \\ &\leq M(t)e^{-\int_0^t M(s)\td s} \Xi(t) \end{aligned} $$ Thus $$ w(t) - w(0) \leq \int_0^t \td s \ M(s)e^{-\int_0^s M(x)\td x} \Xi(s) \leq \Xi(t)\int_0^t \td s \ M(s)e^{-\int_0^s M(x)\td x} = \Xi(t) (1 - e^{-\int_0^t M(s)\td s}) $$ Note $w(0) = 0$ ,, then we have $$ \exp\Big(- \int_0^t M(s)\td s \Big) \int_0^t M(s) z(s)\td s \leq \Xi(t)(1 - e^{-\int_0^t M(s)\td s}). $$ And $$ z(t) \leq \int_0^t M(s) z(s)\td s + \Xi(t) \leq \Xi(t)e^{\int_0^t M(s)\td s} \leq \Xi(t) e^{t \sup M}. $$ That means $$ \|\bm{y}(t) - \bm{y}(0)\| \leq e^{K t} \sum_{n\neq m} \sup_{0\leq s \leq t} \Big|\frac {\braket{n|\partial_t|m}} {E_n - E_m}\Big|. $$ There is another adiabatic condition, which reads $$ \sup_{0\leq s \leq t} \Big|\frac {\braket{n|\partial_t|m}} {E_n - E_m}\Big| \ll 1. $$Propagator and Path Integral
Propagator
[Definition] : The propagator is the time-evolution operator represented in coordinate basis:
$$K(x',t';x,t):=\langle x'|\hat U(t',t)|x\rangle$$
And $K(x',t;x,t)=\delta(x'-x)$ . By this definition we have: $\psi(x',t') = \int K(x',t';x,t)\psi(x,t)\td x$ ..
Propagator is the transition probability amplitude for the particle(system) to state at $x$ at time $t$ . : and end up at $x',t'$ ..
Customarily, when $t'\lt t$ , define that $K(x',t';x,t)\equiv 0$ . And $K$ is the Green function satisfying:
$$\Big\{H\big(x',-\ti \hbar \frac {\partial} {\partial x'},t\big)-\ti \hbar \frac {\partial} {\partial t}\Big\} K(x',t';x,t) = -\ti \hbar \delta(x'-x)\delta(t'-t)$$
Where $H(x,p,t)$ is the Hamiltonian.
Proof
When $t'\lt t$ ,, the equation is hold.For ( $t'\gt t$ ),, for the equation of motion of evolution operator: $$\ti \hbar \frac {\td} {\td t'} \hat U(t',t) = \hat H \hat U(t',t)$$ one can write down: $$\bra{x'}\hat H \hat U(t',t) - \ti \hbar \frac {\td} {\td t'} \hat U(t',t) \ket{x} = (H(x',-\ti\hbar \partial_{x'},t') - \ti \hbar \frac {\td} {\td t'})K(x',t';x,t)=0$$ Then we consider the situation of $t'=t$ ,, then $K(x',t';x,t)=\delta(x'-x)$ , And: $$\int_{t-\epsilon}^{t+\epsilon} (H(x',-\ti \hbar \partial_{x'},t')-\ti \hbar \frac {\td} {\td t'}) K(x',t';x,t) \td t' = -\ti \hbar (K(x',t+\epsilon;x,t)-K(x',t-\epsilon;x,t)) \approx -\ti\hbar \delta(x'-x)$$ For the first term is "continuous" of $t$ . : . Then the equation proved.
More over, when $\hat H$ is independent of time, we can write down the expansion of propagator on energy-eigenstates:
$$K(x',t';x,t) = \sum_E e^{-\ti E(t'-t)/\hbar} \langle x'|E\rangle\langle E|x\rangle \ \ \text{ for } t'\gt t$$
For free particle: $\hat H = \hat p^2/2m$ , one can compute:
$$K(x',t';x,t) = \sqrt{\frac {m} {2\pi\hbar (t'-t)\ti}} \exp\big[\frac {\ti m (x'-x)^2} {2\hbar (t'-t)} \big]$$
Causal Functions
The propagator and $G(t',t):=\text{Tr}\hat U(t',t) = \int \td x K(x,t';x,t)$ are Causal functions : nonzero only for later( ( $t'\gt t$ ),) times.
The fourier transform (with a little difference from the one in mathematical) of causal function $G(t)$ with $G(t\lt 0)=0$ is:
$$\tilde{G}(\omega) = -\ti \int G(t) e^{\ti \omega t}\td t$$
For $G(t\lt 0)=0$ , $\tilde{G}(\omega'+\ti \omega'')$ will be non-singular(analytic) for all $\omega''\in\mathbb{R}^+$ ,, and tends to zero fast enough at infinity with $\omega''\gt 0$
[Theorem] : Kramer-Kronig relation(Hilbert transform). For such functions $\tilde{G}(\omega)$ , one have:
$$\begin{aligned} \text{Re} \tilde{G}(\omega) &= \frac 1 {\pi} \text{P} \int_{-\infty}^{\infty} \frac {\text{Im}\tilde{G}(\omega')} {\omega'-\omega} \td \omega' \\ \text{Im} \tilde{G}(\omega) &= -\frac 1 {\pi} \text{P} \int_{-\infty}^{\infty} \frac {\text{Re}\tilde{G}(\omega')} {\omega'-\omega} \td \omega' \end{aligned}$$
Where $\text{P}$ means the Cauchy principle value of integral.
Proof
Consider the contour integral of $\frac {\tilde{G}(\omega')} {\omega'-\omega}$ on the upper half plane(complex) , with $C_l$ along the real axis expect singular point $\omega$ (using a little circle $C_r$ to evade it) and $C_R$ as the infinity large circle.
Using Jordon Lemma, the integral on $C_R$ is zero. And the whole integral is zero for there is no singular point inner the contour. Then:
$$\text{P} \int_{-\infty}^{\infty} \frac {\tilde{G}(\omega')} {\omega'-\omega} \td \omega' + \int_{C_r} \frac {\tilde{G}(\omega')} {\omega'-\omega} \td \omega' = 0$$
The integral on $C_r$ can be compute with the expansion of integrated function near the point:
$$\int_{C_r}\frac {\tilde{G}(\omega')} {\omega'-\omega} \sim \int_{0}^{\pi} \frac {\tilde{G}(\omega + r e^{\ti \theta})} {r e^{\ti \theta}} r \ti e^{\ti\theta} \td \theta = \ti \pi \tilde{G}(\omega) $$
Then one have:
$$\text{P} \int_{-\infty}^{\infty} \frac {\tilde{G}(\omega')} {\omega'-\omega} \td \omega' = -\ti \pi \tilde{G}(\omega)$$
Example: $G(t\gt 0) = e^{-\ti E t}$ , then $\tilde{G}(\omega) = \frac 1 {\omega -E} - \ti \pi \delta(\omega-E)$
Path Integral in Quantum Mechanics
The goal of path integral is that try to describe the quantum dynamics from a "classical" point of view , as particle moving in coordinate space(or coordinate-momentum phase space).
- Path Integral Version #1
$$K(x',x,t) = \int \mathcal{D}[x(\tau)] \exp\big(\frac {\ti} {\hbar} \int_0^t L(x(\tau),\flo{x}(\tau))\td \tau \big) = \int \mathcal{D}[x]\exp\big[\frac {\ti} {\hbar} S \big]$$
- $\int \mathcal{D}[x]$ :
functional integralover allpath$x(\tau)$ with $x(0)=x ; x(t)=x'$ . The measure of path is very diffcult to define. - $L(x,\flo{x})$ :
Lagrangian - $\flo{x} \equiv \frac {\td x(\tau)} {\td \tau}$ : the
velocityon path $x(\tau)$ - $S[x(\tau)]\equiv \int_0^t L(x(\tau),\flo{x}(\tau))\td \tau$ : the
actionof path $x(\tau)$
- Path Integral Version #2:
$$K(x',x,t) = \int \mathcal{D}[x(\tau)]\mathcal{D}[p(\tau)] \exp \big(\frac {\ti} {\hbar} \int_0^t \{p \flo{x} - H(p,x) \} \big)$$
- $\int \mathcal{D}[p]$ :
functional integralover all path in momentum space, with proper measure. Path of $p$ has no boundary condition - $H(p,x)$ :
Hamiltonian
One can check these two version of path integral by single particle Hamiltonian: $\hat H = \hat p^2/2m + V(\hat x) = T(\hat p) +V(\hat x)$ , the propagator is $K(x',x,t) = \bra{x'}e^{-\ti t \hat H / \hbar} \ket{x}$ ::
Check
1. Check for version 1:Divide this propagation over time $t$ . : into $N$ steps, each of time $\epsilon=t/N$ : $$K(x',x,t) = \bra{x'} (e^{-\frac {\ti} {\hbar} \epsilon \hat H})^N \ket{x} = \int \td x_{N-1}\td x_{N-2}\cdots \td x_{1} \bra{x'}e^{-\ti \epsilon \hat H/\hbar}\ket{x_{N-1}}\cdots\bra{x_1}e^{-\ti \epsilon \hat H/\hbar}\ket{x}$$ With the `Trotter-Suzuki Approximation` : $e^{\eta (\hat A+\hat B)} = e^{\eta \hat A} e^{\eta \hat B} + O(\eta^2)$ , one have: $$\bra{x_{i+1}}e^{-\ti \epsilon \hat H/\hbar} \ket{x_i} \approx \bra{x_{i+1}}e^{-\frac {\ti} {\hbar} \epsilon \frac {\hat p^2} {2m}} \ket{x_i} e^{-\frac {\ti} {\hbar} \epsilon V(x_i)} = \sqrt{\frac {m} {2\pi\hbar \epsilon \ti}} e^{\frac {\ti} {\hbar} \epsilon \big(\frac {m} 2 (\frac {x_{i+1}-x_i} \epsilon)^2 - V(x_i) \big)} $$ Using this expression, one can write the propagator as: $$K(x',x,t) \approx \int \td x_{N-1}\cdots \td x_1 \big(\frac m {2\pi \hbar \epsilon \ti} \big)^{N/2} \exp \Big(\frac {\ti} {\hbar} \sum_{i=0}^{N-1} \epsilon \big[\frac {m} 2 (\frac {x_{i+1}-x_i} \epsilon)^2 - V(x_i) \big] \Big)$$ Where $x_0=x, x_N= x'$ . When ( $N\rightarrow \infty$ ) the summation at the index will be the integral.
With the measure of path being: $$\mathcal{D}[x] = \lim_{N\rightarrow \infty} \prod_{i=1}^{N-1} \td x_i \big(\frac {m N} {2\pi\hbar t \ti} \big)^{N/2}$$ one can write that: $$K(x',x,t) = \int \mathcal{D}[x(\tau)] \exp \big(\frac {\ti} {\hbar} \int_0^t [\frac {m \flo{x}^2} 2 - V(x)]\td \tau \big) = \int \mathcal{D}[x] \exp(\frac {\ti} {\hbar} \int_0^t L(x,\flo{x}) \td \tau)$$ 2. Check for version 2:
With the same setting: $K(x',x,t) = \bra{x'}(e^{-\ti \epsilon \hat H / \hbar})^N \ket{x}$ and $e^{-\ti \epsilon \hat H/\hbar} = e^{-\ti \epsilon T(\hat p)/\hbar} e^{-\ti \epsilon V(\hat x)/\hbar} + O(\epsilon^2)$
Then insert $N-1$ resolution of identity in terms of $\hat x$ eigenstates, and $N$ resolution of identity in terms of $\hat p$ < eigenstates: $$\begin{aligned} K(x',x,t) &\approx \int \td p_{N-1}\cdots \td p_1 \td x_{N-1} \cdots \td x_1 \prod_{i=1}^{N-1} \bra{x_i} e^{-\ti \epsilon T(\hat p)/\hbar} \ket{p_{i-1}} \bra{p_{i-1}} e^{-\ti \epsilon V(\hat x)/\hbar} \ket{x_{i-1}} \\ &=\int \td p_{N-1}\cdots \td p_1 \td x_{N-1} \cdots \td x_1\exp \big(-\frac {\ti} {\hbar} \sum_{i=0}^{N-1} \epsilon(T(p_i)+V(x_i)) \big) \prod_{i=1}^{N} \langle x_i|p_{i-1}\rangle\langle p_{i-1}|x_{i-1}\rangle \\ &=\int (2\pi\hbar)^N \td p_{N-1}\cdots \td p_1 \td x_{N-1} \cdots \td x_1 \exp\big(\frac {\ti} {\hbar}\sum_{i=0}^{N-1} \epsilon (p_i \frac {x_{i+1}-x_i} {\epsilon} - T(p_i)-V(x_i)) \big) \\ &\approx \int \mathcal{D}[x]\mathcal{D}[p] \exp(\frac {\ti} {\hbar} \int_0^t [p\flo{x}-H(p,x)]\td \tau ) \end{aligned}$$ Where we used $\langle x | p \rangle = (2\pi\hbar)^{-1/2} \exp(\ti px/\hbar)$ ,, and where the measure of paths $\mathcal{D}[x]\mathcal{D}[p]$ contains the $(2\pi\hbar)^{-N}$ normalization factor
Stationary Phase Approximation
For the integral $\int e^{\ti k f(x)}\td x$ with large $k$ , most contribution comes from $x_s$ where $f$ is stationary , namely $f'(x_s)=0$ . Then using the Taylor expansion to $(x-x_s)^2$ , one can prove this approximation:
$$\int e^{\ti k f(x)} \td x \approx \sum_{x_s \in \{x:f'(x)=0\}} \sqrt{\frac {2\pi \ti} {kf''(x_s)}} e^{\ti k f(x_s)}$$
For n-dimensional integral:
$$\int e^{\ti k f(\bm{x}) } \td^n \bm{x} \approx \sum_{\bm{x}_s \in \{\bm{x}:\nabla f(\bm{x}) = \bm{0} \}} \Big(\frac {2\pi \ti} k \Big)^{n/2} \Big(\det \frac {\partial^2 f(\bm{x}_s)} {\partial x_i \partial x_j} \Big)^{-1/2} e^{\ti k f(\bm{x}_s)}$$
Geometric Phase
Consider an adiabatic periodic evolution of a Hamiltonian $\hat H(t)$ with $\hat H(T)=\hat H(0)$ . Suppose the Hamiltonian always has a unique ground state $\ket{E_0(t)}$ ..
We care for that after the periodic evolution, what is the phase acquired by the ground state.
The phase factor is $\bra{E_0(0)} \hat U(T)\ket{E_0(0)}$ where $\hat U(T) = \mathcal{T} \exp(-\frac {\ti} {\hbar} \int_{0}^T \hat H(t')\td t' )$ .
Note that $\hat U(t)\ket{E_0(0)}$ is not exactly $\ket{E_0(t)}$ . , but by the adiabatic theorem, they will only differ by a complex phase.
Divide $T$ into $N$ intervals of $\epsilon = T/N$ , define $t_n=n\epsilon$ . Then up to $O(\epsilon^2)$ error, we have the expansion:
$$\bra{E_0(0)}\hat U(T)\ket{E_0(0)} \approx \prod_{i=0}^{N-1} \bra{E_0(t_{i+1})} e^{-\frac {\ti} {\hbar } \epsilon \hat H(t_i)} \ket{E_0(t_i)} = e^{-\frac {\ti} {\hbar} \epsilon \sum_{i=0}^{N-1} E(t_i) } \prod_{i=0}^{N-1} \langle E_0(t_{i+1})| E_0(t_i)\rangle$$
Note that $\ket{E_0(T)}=\ket{E_0(0)}$ , and by adiabatic theorem, $\hat U(t) \ket{E_0(0)}$ will always at the ground state. Then $\ket{E_0(t_i)}\bra{E_0(t_i)}$ is always like $\hat 1$ when operated on it.
When $\epsilon \rightarrow 0$ ( ( $N\rightarrow \infty$ ) ) limit, the first factor becomes $e^{-\frac {\ti} {\hbar} \int_0^T E_0(t)\td t}$ ,, which is the expected dynamic phase acquired from time-evolution
The second factor is:
$$\prod_{i=0}^{N-1} \langle E_0(t_{i+1})| E_0(t_i)\rangle \approx \prod_{i=1}^N \langle E_0(t_i)| 1-\epsilon \frac {\partial} {\partial t}\ket{E_0(t_i)} \approx \exp\big(\sum_{i=1}^N \epsilon \ti A(t_i) \big) \approx \exp(\ti \int_0^T A(\tau)\td \tau)$$
Where $A(t) = \ti \bra {E_0(t)}\partial_t \ket{E_0(t)}$
[Definition] : the Berry's phase :
$$\int A(t)\td t \equiv \int \ti \bra{\psi(t)}\partial_t \ket{\psi(t)} \td t$$
where $t$ . : parametrizes a periodic evolution $\ket{\psi(t)}$ . . And $\ket{\psi(t)}$ . is the instantaneous eigenstate of time-dependent Hamiltonian $\hat H(t)$ , we also assume they are non-degenerate.
[Definition] : the Berry connection :
$$A(t) := \ti \bra{\psi(t)}\partial_t\ket{\psi(t)}$$
Note
1. The periodicity requirement : $\ket{\psi(t_{\text{final}})} = \ket{\psi(t_{\text{initial}})} $ 2. Here $t$ . : is just a parameter describing the path in Hilbert space. NOT time-evolution 3. With note 2 in mind, one can write the Berry phase like(just in this format): $$\oint_{C\in \mathcal{H}} \ti \bra{\psi}\td \ket{\psi}$$ 4. Then one can find that Berry phase does NOT depend on the speed of "evolution" , just depend on the path (geometric) in Hilbert Space 5. The Berry phase does NOT depend on $\hbar$ , for it does not depend on speed. 6. `Gauge transformation` of the Berry connection: with the transformation: $\ket{\psi(t)}\rightarrow e^{\ti \theta(t)}\ket{\psi(t)}$ where $\theta(T)-\theta(0)\equiv 0 \text{ mod }2\pi$ :Berry connection will be: $$A(t)\rightarrow \ti \bra{\psi(t)}e^{-\ti \theta(t)} \partial_t e^{\ti \theta(t)}\ket{\psi(t)} = A(t)-\frac {\td \theta} {\td t}$$ But the Berry phase will be: $$\int A(t)\td t \rightarrow \int A(t) \td t + [\theta(T)-\theta(0)] \equiv \int A(t)\td t \text{ mod }2\pi$$
Gauge Invariance & Electromagnetic Field
We have known that adding a global phase factor $\ket{\psi}\rightarrow e^{\ti\theta}\ket{\psi}$ with real $\theta$ independent of $\bm{x},t$ will not change the Schorodinger equation and any expectation value of observables.
However, with the transformation:
$$\psi(\bm{x},t) \rightarrow e^{\ti \theta(\bm{x},t)} \psi(\bm{x},t)$$
we have:
$$\begin{aligned} \partial_t \psi &\rightarrow e^{\ti\theta}\partial_t \psi + (\ti\partial_t \theta) e^{\ti \theta} \psi \\ \partial_{\bm{x}} \psi &\rightarrow e^{\ti\theta}\partial_{\bm{x}}\psi + (\ti\partial_{\bm{x}}\theta)e^{\ti\theta} \psi \end{aligned}$$
Then the Schrodinger equation will not be preserved.
To make the theory formally gauge invariant under arbitary $\psi\rightarrow e^{\ti \theta}\psi$ , one need to absorb the differential of $\theta$ into the transformation of a Gauge Field
Define a 4-component space-time-dependent real-value Gauge Field $(a_0,\bm{a})$ , and the canonical momentum : $\hat P = \hat p - \hbar \bm{a}$
Demand the gauge transform to be : $\psi\rightarrow e^{\ti\theta}\psi , \bm{a}\rightarrow \bm{a}+\partial_{\bm{x}}\theta , a_0\rightarrow a_0+\partial_t \theta$ . Then one can check That $\hat P \psi \rightarrow e^{\ti\theta}\hat P\psi$
Modify the Schrodinger equation as:
$$\big[\frac {\hat P^2} {2m}+V(\bm{x})\big]\psi = \hbar (\ti\partial_t + a_0)\psi$$
This will be invariant under the above gauge transformation
MOdify the definition of probability current density as
$$\bm{J}(\bm{x},t) = \text{Re}\big[\psi^* \frac {\hat P} {m} \psi \big] = -\ti \frac {\hbar} {2m} (\psi^*\partial_{\bm{x}}\psi - \psi \partial_{\bm{x}}\psi^*) - \frac {\hbar} m \bm{a}\rho$$
This is invariant under the gauge transformation, so continuity equation is preserved.