有限温度量子场论-第二章 虚时路径积分（2.3 中性标量场）

从2.3节，正式进入场的计算。
利用虚时路径积分可以计算简单的无相互作用中性标量场。中性意味着没有守恒荷，哈密顿量没有项。标量意味着每个时空点的场的值用一个变量表示。
虽然对于无相互作用的简单情况，可以用二次量子化简单计算。但是，虚时路径积分在有相互作用情况下很有用，因此本章意在用简单情况展示这一有用方法。

2 Imaginary-time Path Integral

2.3 Neutral scalar field

First we can write a free neutral scalar field in Fourier form:

$$\hat{\phi}(\mathbf{x}, t)=\sum_{\mathbf{k} }\left[\hat{a}_{\mathbf{k} } e^{-i\left(E_{\mathbf{k} } t-\mathbf{k} \cdot \mathbf{x}\right)}+\hat{a}_{\mathbf{k} }^{\dagger} e^{i\left(E_{\mathbf{k} } t-\mathbf{k} \cdot \mathbf{x}\right)}\right]$$

The Fourier amplitudes are written in annihilation and creation operators. They have the commutation rules:

$$\left[\hat{a}_{\mathbf{k} }, \hat{a}_{\mathbf{k}^{\prime} }^{\dagger}\right]=\delta_{\mathbf{k}, \mathbf{k}^{\prime} }, \quad\left[\hat{a}_{\mathbf{k} }, \hat{a}_{\mathbf{k}^{\prime} }\right]=\left[\hat{a}_{\mathbf{k} }^{\dagger}, \hat{a}_{\mathbf{k}^{\prime} }^{\dagger}\right]=0$$

Then we can express the field under the second-quantized basis:

$$\left|n_{\mathbf{k}_{1} }, n_{\mathbf{k}_{2} }, \ldots, n_{\mathbf{k}_{i} }, \ldots\right\rangle \equiv \prod_{i=1}^{\infty}\left|n_{\mathbf{k}_{i} }\right\rangle$$

The Hamiltonian of the field is a harmonic oscillator:

$$\hat{H}=\sum_{\mathbf{k} } E_{\mathbf{k} }\left(\frac{1}{2}+\hat{a}_{\mathbf{k} }^{\dagger} \hat{a}_{\mathbf{k} }\right)$$

The partition function is:

$$\mathcal{Z}_{0}=\prod_{\mathbf{k} } \sum_{n_{\mathbf{k} }=0}^{\infty}\left\langle n_{\mathbf{k} }\left|e^{-\beta \hat{H}_{\mathbf{k} } }\right| n_{\mathbf{k} }\right\rangle=\prod_{\mathbf{k} } \frac{e^{-\beta E_{\mathbf{k} } / 2} }{1-e^{-\beta E_{\mathbf{k} } } }$$ $$\ln \mathcal{Z}_{0}=-\sum_{\mathbf{k} }[\frac{\beta E_{\mathbf{k} } }{2}+\ln \left(1-e^{-\beta E_{\mathbf{k} } }\right)]$$

Here the first term is infinite. This is the vacuum energy. The second term is the temperature effect.

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Next we use the path integral. This method is very useful in interacting systems. Now in simple non-interacting system we develop this useful tool.

Here without the second quantization, we write the field in spacial coordinates and time . Denote the field operator in Schordinger picture, at :

$$\hat{\phi}(x,0)|\phi_0\rangle \equiv \phi_0 (x)|\phi_0\rangle$$

The complete and orthogonal relations of :

$$\int \prod_{\mathbf{x} } d \phi_{0}(\mathbf{x})\left|\phi_{0}\right\rangle\left\langle\phi_{0}\right|=\hat{1}$$ $$\left\langle\phi_{a} \mid \phi_{b}\right\rangle=\prod_{\mathbf{x} } \delta\left(\phi_{a}(\mathbf{x})-\phi_{b}(\mathbf{x})\right) \equiv \delta\left[\phi_{a}-\phi_{b}\right]$$

This is for the “x-basis”. Now we define the “p-basis”. The field canonically conjugated to is:

$$\pi(\mathbf{x}, t)=\frac{\partial \mathcal{L} }{\partial\left[\partial_{t} \phi(\mathbf{x}, t)\right]}$$

For we have similar complete and orthogonal relations as .

So in the partition function is:

$$\mathcal{Z} \equiv \operatorname{Tr} e^{-\beta \hat{H} }=\int \prod_{\mathbf{x} \in V} d \phi_{0}(\mathbf{x})\left\langle\phi_{0}\left|e^{-\beta \hat{H} }\right| \phi_{0}\right\rangle$$

is the volume of the system. Compare with the transition amplitude , replace .

After some calculation, we finally get the partition function as:

$$\begin{aligned} &\mathcal{Z}=\mathfrak{R} \int_{\text {periodic} } \mathcal{D} \pi(\mathbf{x}, \tau) \int_{\text {periodic} } \mathcal{D} \phi(\mathbf{x}, \tau) \\ &\exp \left\{\int_{0}^{\beta} d \tau \int_{V} d \mathbf{x}\left[i\pi(\mathbf{x}, \tau) \frac{\partial \phi(\mathbf{x}, \tau)}{\partial \tau}-\mathcal{H}(\phi, \pi, \nabla \phi)\right]\right\} \end{aligned}\tag{31,56}$$

Here the periodic means the periodic boundary condition for the integral interval. Note that for similar reson, for we can also use the periodic restriction.

There is a in as a mormalization constant, which ensures that is dimensionless. This is necessary because and have dimension.

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With some careful calculation, is finally given as:

$$\Re_{\pi}=\left(2 \pi \varepsilon a^{3}\right)^{N M^{3} / 2}, \quad \Re_{\phi}=\left(\frac{a^{3} }{2 \pi \varepsilon}\right)^{N M^{3} / 2}$$ $$\mathfrak{R} = \Re_{\pi} \Re_{\phi} = \left(2 \pi \varepsilon a^{3} \frac{a^{3} }{2 \pi \varepsilon}\right)^{N M^{3} / 2}=\left(a^{3}\right)^{N M^{3} } =\left(\frac{L^{3} }{M^{3} }\right)^{N M^{3} }=V^{N M^{3} } M^{-3 N M^{3} }$$

are the number of small interval we took for the integral for and coordinates. And the number of all space-time points is .

So the last part is irrelavant thermodynamically, can be seen as a function of : .

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Now we compute the partiton function .

Non-interacting neutral scalar field Hamiltonian:

$$\mathcal{H}=\mathcal{H}_{0}=\frac{1}{2} \pi^{2}+\frac{1}{2}(\nabla \phi)^{2}+\frac{m^{2} }{2} \phi^{2}$$

Making a shift of : ,

$$\begin{aligned} i \pi \partial_{\tau} \phi-\mathcal{H}_{0} &=-\frac{1}{2} \pi^{\prime 2}+\frac{1}{2}\left[\left(i \partial_{\tau} \phi\right)^{2}-(\nabla \phi)^{2}-m^{2} \phi^{2}\right] \\ &=-\frac{1}{2} \pi^{2}-\frac{1}{2}\left[\left(\partial_{\tau} \phi\right)^{2}+(\nabla \phi)^{2}+m^{2} \phi^{2}\right] \end{aligned}$$

We see that the later term is the Euclidean form:

$$\mathcal{L}_{E}=\frac{1}{2}\left[\left(\partial_{\tau} \phi\right)^{2}+(\nabla \phi)^{2}+m^{2} \phi^{2}\right]$$

Thus the partiton function is:

$$\mathcal{Z}=\mathfrak{R} \int_{\text {periodic } } \mathcal{D} \pi \exp \left(-\frac{1}{2} \int_{X} \pi^{2}\right) \int_{\text {periodic } } \mathcal{D} \phi \exp \left(-\int_{X} \mathcal{L}_{E}\right)\tag{60}$$

With the integal . To calculate the Gaussian integral, the and are written in Fourier term:

$$\begin{aligned} \phi(\mathbf{x}, \tau) &=\sqrt{\frac{\beta}{V} } \sum_{n=-\infty}^{\infty} \sum_{\mathbf{k} } e^{i \omega_{n} \tau+i \mathbf{k} \cdot \mathbf{x} } \tilde{\phi}\left(\omega_{n}, \mathbf{k}\right) \\ \pi(\mathbf{x}, \tau) &=\sqrt{\frac{1}{\beta V} } \sum_{n=-\infty}^{\infty} \sum_{\mathbf{k} } e^{i \omega_{n} \tau+i \mathbf{k} \cdot \mathbf{x} } \tilde{\pi}\left(\omega_{n}, \mathbf{k}\right) \end{aligned}\tag{61}$$

We note that the normalization factor:

$$\sqrt{\frac{\beta}{V} }=\sqrt{\frac{N \varepsilon}{M^{3} a^{3} } }, \quad \sqrt{\frac{1}{\beta V} }=\sqrt{\frac{1}{N \varepsilon M^{3} a^{3} } }$$

Therefore we have a term, which cancels the .

With the periodic boundary condition, the must be:
. These discrete values are called boson Matsubara frequency.

For Fermions, .

We assume the system is a finite volume . And for convinience, we write the and in 4-vectors:

$$\begin{array}{ll} K^{\mu}=\left(k_{0}, \mathbf{k}\right), & k_{0} \equiv-i \omega_{n} \\ X^{\mu}=\left(x_{0}, \mathbf{x}\right), & x_{0} \equiv-i \tau \end{array}$$

Then we have $\omega{n} \tau+\mathbf{k} \cdot \mathbf{x}=-k{0} x{0}+\mathbf{k} \cdot \mathbf{x}=-K{\mu} X^{\mu} \equiv-K \cdot X(61)$ is:

$$\begin{aligned} \phi(X) &=\sqrt{\frac{\beta}{V} } \sum_{K} e^{-i K \cdot X} \phi_{K} \\ \pi(X) &=\sqrt{\frac{1}{\beta V} } \sum_{K} e^{-i K \cdot X} \pi_{K} \end{aligned}\tag{72}$$

The Fourier basis are orthogonal:

$$\int_X e^{-i(K+Q)}\cdot X = \beta V\delta^{(4)}_{K,Q}$$

With the orthogonal relation, the first Gaussian part in is evaluated:

$$\begin{aligned} -\frac{1}{2} \int_{X} \pi^{2}(X) &=-\frac{1}{2 \beta V} \sum_{K} \sum_{Q} \pi_{K} \pi_{Q} \int_{X} e^{-i(K+Q) \cdot X} \\ &=-\frac{1}{2 \beta V} \sum_{K} \sum_{Q} \pi_{K} \pi_{Q} \beta V \delta_{K,-Q}^{(4)} \\ &=-\frac{1}{2} \sum_{K} \pi_{K} \pi_{-K} . \end{aligned}$$

And the second term:

$$\begin{aligned} -\frac{1}{2} \int_{X} \mathcal{L}_{E} &=-\frac{\beta}{2 V} \sum_{K} \sum_{Q}\left(K \cdot Q+m^{2}\right) \phi_{K} \phi_{Q} \int_{X} e^{-i(K+Q) \cdot X}\\ &=-\frac{1}{2} \sum_{K} \beta^{2}\left(m^{2}-K^{2}\right) \phi_{K} \phi_{-K}\\ &=-\frac{1}{2} \sum_{K}\left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right] \phi_{K} \phi_{-K} \end{aligned}$$

Here a propergator called Thermal propergator is defined:

$$\mathcal{D}_{0}^{-1}(K) \equiv m^{2}-K^{2}=m^{2}+\mathbf{k}^{2}+\omega_{n}^{2}=\omega_{n}^{2}+E_{\mathbf{k} }^{2}, E_k = \sqrt{\mathbf{k}^2+m^2}$$

From the fact that and are real, the Fourier basis are not all independent. We can partite all basis into 3 parts: , and .

Now replace the variables of int in to the Fourier amplitudes, we have:

$$\mathfrak{R D} \pi(X) \mathcal{D} \phi(X)=\mathfrak{R} \mathcal{J} d \pi_{0} d \phi_{0} \prod_{K>0} d \operatorname{Re} \pi_{K} d \operatorname{Im} \pi_{K} d \operatorname{Re} \phi_{K} d \operatorname{Im} \phi_{K}$$

Here the Jacobian is a constant. And further we can have a dimensionless :

$$\mathcal{J}=\left(\sqrt{\frac{1}{\beta V} }\right)^{\mathcal{F} }\left(\sqrt{\frac{\beta}{V} }\right)^{\mathcal{F} } \mathcal{J}^{\prime}=\frac{1}{V^{\mathcal{F} } } \mathcal{J}^{\prime}$$

Then multiply the , we obtain which is very divergent.

• The functional integral over :

$$\begin{aligned} & \int d \pi_{0} \prod_{K>0} d \operatorname{Re} \pi_{K} d \operatorname{Im} \pi_{K} e^{-\frac{1}{2} \sum_{K}\left|\pi_{K}\right|^{2} } \\ =& \int d \pi_{0} e^{-\pi_{0}^{2} / 2} \int \prod_{K>0} d \operatorname{Re} \pi_{K} d \operatorname{Im} \pi_{K} e^{-\sum_{K>0}\left[\left(\operatorname{Re} \pi_{K}\right)^{2}+\left(\operatorname{Im} \pi_{K}\right)^{2}\right]} \\ =& \int d \pi_{0} e^{-\pi_{0}^{2} / 2} \prod_{K>0} \int d \operatorname{Re} \pi_{K} d \operatorname{Im} \pi_{K} e^{-\left[\left(\operatorname{Re} \pi_{K}\right)^{2}+\left(\operatorname{Im} \pi_{K}\right)^{2}\right]} \\ =& \sqrt{\frac{\pi}{2} } \prod_{K>0} \pi \equiv \mathcal{Q} . \end{aligned}$$

This is a dimensionless but divergent number.

• The functional integral over $\phiK\mathcal{D}{0}^{-1}(-K)=\mathcal{D}_{0}^{-1}(K)$:

$$\begin{aligned} & \int d \phi_{0} \prod_{K>0} d \operatorname{Re} \phi_{K} d \operatorname{Im} \phi_{K} e^{-\frac{1}{2} \sum_{K} \beta^{2} \mathcal{D}_{0}^{-1}(K)\left|\phi_{K}\right|^{2} } \\ =& \int d \phi_{0} e^{-\frac{1}{2} \beta^{2} m^{2} \phi_{0}^{2} } \prod_{K>0} \int d \operatorname{Re} \phi_{K} d \operatorname{Im} \phi_{K} \exp \left\{-\beta^{2} \mathcal{D}_{0}^{-1}(K)\left[\left(\operatorname{Re} \phi_{K}\right)^{2}+\left(\operatorname{Im} \phi_{K}\right)^{2}\right]\right\} \\ =& \frac{1}{\beta m} \sqrt{\frac{\pi}{2} } \prod_{K>0} \pi\left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right]^{-1} \\ =& \mathcal{Q} \prod_{K}\left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right]^{-1 / 2} . \end{aligned}$$

This result could be interpreted as a determinant of matrix in momentum space with all the eigenvalues :

$$\prod_{K}\left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right]^{-1 / 2} \equiv\left[\operatorname{det}_{K}\left(\beta^{2} \mathcal{D}_{0}^{-1}\right)\right]^{-1 / 2}$$

So not very precisely, the ln partition function is:

$$\begin{aligned} \ln \mathcal{Z}_{0} &=-\frac{1}{2} \ln \operatorname{det}_{K}\left(\beta^{2} \mathcal{D}_{0}^{-1}\right) \\ &=-\frac{1}{2} \operatorname{Tr}_{K} \ln \left(\beta^{2} \mathcal{D}_{0}^{-1}\right) \\ &=-\frac{1}{2} \sum_{K} \ln \left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right] \end{aligned}$$

The sum of contains two parts: $\omegan{E\mathbf{k} }\omega_n$ is called Matsubara sum. This term could be divided into T-dependent part and a thermodynamically irrelavant divergent part.

$$\begin{aligned} \frac{1}{2} \sum_{n} \ln \left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right] &=\frac{1}{2} \sum_{n} \ln \left[\beta^{2}\left(\omega_{n}^{2}+E_{\mathbf{k} }^{2}\right)\right] \\ &=\frac{1}{2} \sum_{n} \ln \left[(2 \pi n)^{2}+\beta^{2} E_{\mathbf{k} }^{2}\right] \end{aligned}$$

Trick: write the logarithm into integral.

$$\ln \left[(2 \pi n)^{2}+\beta^{2} E_{\mathbf{k} }^{2}\right]=2 \int_{1}^{\beta E_{\mathbf{k} } } d x \frac{x}{(2 \pi n)^{2}+x^{2} }+\ln \left[(2 \pi n)^{2}+1\right]$$

With the help of the math handbook, the sum could be evaluated:

$$\sum_{n=-\infty}^\infty \frac{1}{(2\pi n)^2+x^2} = \frac{1}{x}(\frac{1}{2}+\frac{1}{e^x-1})$$

Now the divergent sum is devided into a vacuum energy and a boson statistical term.

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Next: Use residue theorem（留数定理）to calculate again.

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Finally write down all the terms of ln partition function:

$$\begin{aligned} -\frac{1}{2} \sum_{K} \ln \left[\beta^{2} \mathcal{D}_{0}^{-1}(K)\right]=&-\sum_{\mathbf{k}}\left[\frac{\beta E_{\mathbf{k}}}{2}+\ln \left(1-e^{-\beta E_{\mathbf{k}}}\right)\right] \\ &-\frac{1}{2} \sum_{\mathbf{k}} \sum_{n} \ln \left[(2 \pi n)^{2}+1\right] \\ &+\sum_{\mathbf{k}}\left[\frac{1}{2}+\ln \left(1-e^{-1}\right)\right] \end{aligned}$$

The first term is about , and the second and tird term are very divergent. Sometimes they are directly ignored because they are not thermal. Now use and to represent these two terms, we get the partition func as:

$$\mathcal{Z}_{0}=\mathfrak{\Re} \mathcal{J} \mathcal{Q}^{2} e^{\mathcal{A}+\mathcal{B}} \prod_{\mathbf{k}} \exp \left\{-\left[\frac{\beta E_{\mathbf{k}}}{2}+\ln \left(1-e^{-\beta E_{\mathbf{k}}}\right)\right]\right\}$$

If the prefactor is 1, the divergent terms could be ignored. Recall that in HO case, the divergent terms were canceled, thus we guess that the rdivergent terms here could also be canceled using the similar method.

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Conclusion:

• The calculation of neutral scalar field without interaction, the imaginary-time path integral is much harder than the partical-number representation.
• For interacting systems insome approximation (such as mean-field approximation), the partical-number representation is also convenient.
• However for interacting cases, the functional path integral formalism is powerful to perform the perturbative expansion and develop some non-perturbative methods.
• For perturbative calculation the divergence is not a probleme. For example we have an interactive action ,

$$\begin{aligned} \mathcal{Z} &=\sum_{n=0}^{\infty} \frac{1}{n !} \int \mathcal{D} \phi S_{I}^{n}[\phi] e^{S_{0}[\phi]} \\ &=\mathcal{Z}_{0} \sum_{n=0}^{\infty} \frac{1}{n !} \frac{\int \mathcal{D} \phi S_{I}^{n}[\phi] e^{S_{0}[\phi]}}{\int \mathcal{D} \phi e^{S_{0}[\phi]}} \end{aligned}$$

• Since the path integral over the conjugated field only contributes a constant, and it is irrelevant in most cases, we may simply drop it and write the imaginary-time path integral representation of the partition function as a int of .