### 物理代写|量子场论代写Quantum field theory代考|PHYS3101

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## 物理代写|量子场论代写Quantum field theory代考|Quantum Mechanics in Imaginary Time

The unitary time evolution operator has the spectral representation
$$\hat{K}(t)=\mathrm{e}^{-\mathrm{i} \hat{H} t}=\int \mathrm{e}^{-\mathrm{i} E t} \mathrm{~d} \hat{P}{\mathrm{E}},$$ where $\hat{P}{\mathrm{E}}$ is the spectral family of the Hamiltonian. If $\hat{H}$ has discrete spectrum, then $\hat{P}{\mathrm{E}}$ is the orthogonal projector onto the subspace of $\mathscr{H}$ spanned by all eigenfunctions with energies less than $E$. In the following we assume that the Hamiltonian operator is bounded from below. Then we can subtract its ground state energy to obtain a non-negative $\hat{H}$ for which the integration limits in (2.35) are 0 and $\infty$. With the substitution $t \rightarrow t-\mathrm{i} \tau$, we obtain $$\mathrm{e}^{-(\tau+\mathrm{i} t) \hat{H}}=\int{0}^{\infty} \mathrm{e}^{-E(\tau+\mathrm{i} t)} \mathrm{d} \hat{P}_{\mathrm{E}}$$
This defines a holomorphic semigroup in the lower complex half-plane
$${z=t-\mathrm{i} \tau \in \mathbb{C}, \tau \geq 0}$$
If the operator $(2.36)$ is known on the negative imaginary axis $(t=0, \tau \geq 0)$, one can perform an analytic continuation to the real axis $(t, \tau=0)$. The analytic continuation to complex time $t \rightarrow-\mathrm{i} \tau$ corresponds to a transition from the Minkowski metric $\mathrm{d} s^{2}=d t^{2}-\mathrm{d} x^{2}-\mathrm{d} y^{2}-\mathrm{d} z^{2}$ to a metric with Euclidean signature. Hence a theory with imaginary time is called Euclidean theory.

The time evolution operator $\hat{K}(t)$ exists for real time and defines a oneparametric unitary group. It fulfills the Schrödinger equation
$$\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} t} \hat{K}(t)=\hat{H} \hat{K}(t)$$
with a complex and oscillating kernel $K\left(t, q^{\prime}, q\right)=\left\langle q^{\prime}|\hat{K}(t)| q\right\rangle$. For imaginary time we have a Hermitian (and not unitary) evolution operator
$$\hat{K}(\tau)=\mathrm{e}^{-\tau \hat{H}}$$
with positive spectrum. The $\hat{K}(\tau)$ exist for positive $\tau$ and form a semi-group only. For almost all initial data, evolution back into the “imaginary past” is impossible.
The evolution operator for imaginary time satisfies the heat equation
$$\frac{\mathrm{d}}{\mathrm{d} \tau} \hat{K}(\tau)=-\hat{H} \hat{K}(\tau)$$
instead of the Schrödinger equation and has kernel
$$K\left(\tau, q^{\prime}, q\right)=\left\langle q^{\prime}\left|\mathrm{e}^{-\tau \hat{H}}\right| q\right\rangle, \quad K\left(0, q^{\prime}, q\right)=\delta\left(q^{\prime}, q\right)$$
This kernel is real ${ }^{1}$ for a real Hamiltonian. Furthermore it is strictly positive:

## 物理代写|量子场论代写Quantum field theory代考|Imaginary Time Path Integral

To formulate the path integral for imaginary time, we employ the product formula $(2.28)$, which follows from the product formula (2.27) through the substitution of it by $\tau$. For such systems the analog of $(2.31)$ for Euclidean time $\tau$ is obtained by the substitution of $i \varepsilon$ by $\varepsilon$. Thus we tind
\begin{aligned} K\left(\tau, q^{\prime}, q\right) &=\left\langle\hat{q}^{\prime}\left|\mathrm{e}^{-\tau \hat{H} / \hbar}\right| \hat{q}\right\rangle \ &=\lim {n \rightarrow \infty} \int \mathrm{d} q{1} \cdots \mathrm{d} q_{n-1}\left(\frac{m}{2 \pi \hbar \varepsilon}\right)^{n / 2} \mathrm{e}^{-S_{\mathrm{E}}\left(q_{0}, q_{1}, \ldots, q_{n}\right) / \hbar} \ S_{\mathrm{E}}(\ldots) &=\varepsilon \sum_{j=0}^{n-1}\left{\frac{m}{2}\left(\frac{q_{j+1}-q_{j}}{\varepsilon}\right)^{2}+V\left(q_{j}\right)\right} \end{aligned}
where $q_{0}=q$ and $q_{n}=q^{\prime}$. The multidimensional integral represents the sum over all broken-line paths from $q$ to $q^{\prime}$. Interpreting $S_{\mathrm{E}}$ as Hamiltonian of a classical lattice model and $\hbar$ as temperature, it is (up to the fixed endpoints) the partition function of a one-dimensional lattice model on a lattice with $n+1$ sites. The realvalued variable $q_{j}$ defined on site $j$ enters the action $S_{\mathrm{E}}$ which contains interactions between the variables $q_{j}$ and $q_{j+1}$ at neighboring sites. The values of the lattice field
$${0,1, \ldots, n-1, n} \rightarrow\left{q_{0}, q_{1}, \ldots, q_{n-1}, q_{n}\right}$$
are prescribed at the endpoints $q_{0}=q$ and $q_{n}=q^{\prime}$. Note that the classical limit $\hbar \rightarrow 0$ corresponds to the low-temperature limit of the lattice system.

The multidimensional integral (2.52) corresponds to the summation over all path on the time lattice. What happens to the finite-dimensional integral when we take the continuum limit $n \rightarrow \infty$ ? Then we obtain the Euclidean path integral representation for the positive kernel
$$K\left(\tau, q^{\prime}, q\right)=\left\langle q^{\prime}\left|\mathrm{e}^{-\tau \hat{H} / h}\right| q\right\rangle=C \int_{q(0)=q}^{q(\tau)=q^{\prime}} \mathscr{D} q \mathrm{e}^{-S_{\mathrm{E}}[q] / h}$$
The integrand contains the Euclidean action
$$S_{\mathrm{E}}[q]=\int_{0}^{\tau} d \sigma\left{\frac{m}{2} \dot{q}^{2}+V(q(\sigma))\right}$$
which for many physical systems is bounded from below.

## 物理代写|量子场论代写Quantum field theory代考|Path Integral in Quantum Statistics

The Euclidean path integral formulation immediately leads to an interesting connection between quantum statistical mechanics and classical statistical physics. Indeed, if we set $\tau / \hbar \equiv \beta$ and integrate over $q=q^{\prime}$ in (2.53), then we end up with the path integral representation for the canonical partition function of a quantum system with Hamiltonian $\hat{H}$ at inverse temperature $\beta=1 / k_{B} T$. More precisely, setting $q=q^{\prime}$ and $\tau=\hbar \beta$ in the left-hand side of this formula, then the integral over $q$ yields the trace of $\exp (-\beta \hat{H})$, which is just the canonical partition function,
$$\int \mathrm{d} q K(\hbar \beta, q, q)=\operatorname{tr} \mathrm{e}^{-\beta \hat{H}}=Z(\beta)=\sum \mathrm{e}^{-\beta E_{n}} \quad \text { with } \quad \beta=\frac{1}{k_{B} T}$$
Setting $q=q^{\prime}$ in the Euclidean path integral in (2.53) means that we integrate over paths beginning and ending at $q$ during the imaginary time interval $[0, \hbar \beta]$. The final integral over $q$ leads to the path integral over all periodic paths with period $\hbar \beta$
$$Z(\beta)=C \oint \mathscr{D} q \mathrm{e}^{-S_{\mathrm{E}}[q] / \hbar}, \quad q(\hbar \beta)=q(0)$$
For example, the kernell of the harmonic oscillator in $(2.43)$ on the diagonal is
$$K_{\omega}(\beta, q, q)=\sqrt{\frac{m \omega}{2 \pi \sinh (\omega \beta)}} \exp \left{-m \omega \tanh (\omega \beta / 2) q^{2}\right}$$
where we used units with $\hbar=1$. The integral over $q$ yields the partition function
\begin{aligned} Z(\beta) &=\sqrt{\frac{m \omega}{2 \pi \sinh (\omega \beta)}} \int \mathrm{d} q \exp \left{-m \omega \tanh (\omega \beta / 2) q^{2}\right} \ &=\frac{1}{2 \sinh (\omega \beta / 2)}=\frac{\mathrm{e}^{-\omega \beta / 2}}{1-\mathrm{e}^{-\omega \beta}}=\mathrm{e}^{-\omega \beta / 2} \sum_{n=0}^{\infty} \mathrm{e}^{-n \omega \beta} \end{aligned}
where we used $\sinh x=2 \sinh x / 2 \cosh x / 2$. A comparison with the spectral sum over all energies in (2.55) yields the energies of the oscillator with (angular) frequency $\omega$,
$$E_{n}=\omega\left(n+\frac{1}{2}\right), \quad n=0,1,2, \ldots$$
For large values of $\omega \beta$, i.e., for very low temperature, the spectral sum is dominated by the contribution of the ground state energy. Thus for cold systems, the free energy converges to the ground state energy
$$F(\beta) \equiv-\frac{1}{\beta} \log Z(\beta) \stackrel{\omega \beta \rightarrow \infty}{\longrightarrow} E_{0}$$
One often is interested in the energies and wave functions of excited states. We now discuss an elegant method to extract this information from the path integral.

## 物理代写|量子场论代写Quantum field theory代考|Quantum Mechanics in Imaginary Time

ķ^(吨)=和−一世H^吨=∫和−一世和吨 d磷^和,在哪里磷^和是哈密顿量的谱族。如果H^有离散谱，则磷^和是在子空间上的正交投影H由能量小于的所有特征函数跨越和. 在下文中，我们假设哈密顿算子是从下面有界的。然后我们可以减去它的基态能量得到一个非负的H^(2.35) 中的积分限制为 0 并且∞. 随着替换吨→吨−一世τ， 我们获得

ķ^(τ)=和−τH^

ddτķ^(τ)=−H^ķ^(τ)

ķ(τ,q′,q)=⟨q′|和−τH^|q⟩,ķ(0,q′,q)=d(q′,q)

## 物理代写|量子场论代写Quantum field theory代考|Imaginary Time Path Integral

\begin{aligned} 左(\tau, q^{\prime}, q\right) &=\left\langle\hat{q}^{\prime}\left|\mathrm{e}^{- \tau \hat{H} / \hbar}\right| \hat{q}\right\rangle\&=\lim{n\rightarrow\infty}\int \mathrm{d}q{1}\cdots\mathrm{d}q_{n-1}\left(\frac {m}{2 \pi \hbar \value psilon}\right)^{n/2}\mathrm{e}^{-S_{\mathrm{E}}\left(q_{0}, q_{1} , \ldots, q_{n}\right) / \hbar}\S_{\mathrm{E}}(\ldots) &=\varepsilon \sum_{j=0}^{n-1}\left{\frac { m}{2}\left(\frac{q_{j+1}-q_{j}}{\valuepsilon}\right)^{2}+V\left(q_{j}\right)\right} \结束{对齐}\begin{aligned} 左(\tau, q^{\prime}, q\right) &=\left\langle\hat{q}^{\prime}\left|\mathrm{e}^{- \tau \hat{H} / \hbar}\right| \hat{q}\right\rangle\&=\lim{n\rightarrow\infty}\int \mathrm{d}q{1}\cdots\mathrm{d}q_{n-1}\left(\frac {m}{2 \pi \hbar \value psilon}\right)^{n/2}\mathrm{e}^{-S_{\mathrm{E}}\left(q_{0}, q_{1} , \ldots, q_{n}\right) / \hbar}\S_{\mathrm{E}}(\ldots) &=\varepsilon \sum_{j=0}^{n-1}\left{\frac { m}{2}\left(\frac{q_{j+1}-q_{j}}{\valuepsilon}\right)^{2}+V\left(q_{j}\right)\right} \结束{对齐}

{0,1, \ldots, n-1, n} \rightarrow\left{q_{0}, q_{1}, \ldots, q_{n-1}, q_{n}\right}{0,1, \ldots, n-1, n} \rightarrow\left{q_{0}, q_{1}, \ldots, q_{n-1}, q_{n}\right}

ķ(τ,q′,q)=⟨q′|和−τH^/H|q⟩=C∫q(0)=qq(τ)=q′Dq和−小号和[q]/H

S_{\mathrm{E}}[q]=\int_{0}^{\tau} d \sigma\left{\frac{m}{2} \dot{q}^{2}+V(q( \sigma))\对}S_{\mathrm{E}}[q]=\int_{0}^{\tau} d \sigma\left{\frac{m}{2} \dot{q}^{2}+V(q( \sigma))\对}

## 物理代写|量子场论代写Quantum field theory代考|Path Integral in Quantum Statistics

∫dqķ(ℏb,q,q)=tr⁡和−bH^=从(b)=∑和−b和n 和 b=1ķ乙吨

K_{\omega}(\beta, q, q)=\sqrt{\frac{m \omega}{2 \pi \sinh (\omega \beta)}} \exp \left{-m \omega \tanh ( \omega \beta / 2) q^{2}\right}K_{\omega}(\beta, q, q)=\sqrt{\frac{m \omega}{2 \pi \sinh (\omega \beta)}} \exp \left{-m \omega \tanh ( \omega \beta / 2) q^{2}\right}

\begin{对齐}Z(\beta)&=\sqrt{\frac{m\omega}{2\pi\sinh(\omega\beta)}}\int\mathrm{d}q\exp\left{- m\omega\tanh(\omega\beta/2)q^{2}\right}\&=\frac{1}2\sinh(\omega\beta/2)}=\mathrm{e}^{- \omega\beta/2}{1-\mathrm{e}^{-\omega\beta}}=\mathrm{e}^{-\omega\beta/2}\sum_{n=0}^{\ infty} \ mathrm {e} ^ {- n \ omega \ beta} \ end {对齐}\begin{对齐}Z(\beta)&=\sqrt{\frac{m\omega}{2\pi\sinh(\omega\beta)}}\int\mathrm{d}q\exp\left{- m\omega\tanh(\omega\beta/2)q^{2}\right}\&=\frac{1}2\sinh(\omega\beta/2)}=\mathrm{e}^{- \omega\beta/2}{1-\mathrm{e}^{-\omega\beta}}=\mathrm{e}^{-\omega\beta/2}\sum_{n=0}^{\ infty} \ mathrm {e} ^ {- n \ omega \ beta} \ end {对齐}

F(b)≡−1b日志⁡从(b)⟶ωb→∞和0

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