物理代写|量子场论代写Quantum field theory代考|Vacuum matrix elements

Doug I. Jones

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物理代写|量子场论代写Quantum field theory代考|Vacuum matrix elements

In deriving LSZ we used that the vacuum state $|\Omega\rangle$ was annihilated by the operators $a_p(t)$ in the interacting theory at a time $t=-\infty$. To relate this to a state for which we know how the free-field creation and annihilation operators act, we need to evolve it to the reference time $t_0$ where the free and interacting pictures are taken equal. This is straightforward: states evolve (in the Schrödinger picture) with $S\left(t, t_0\right)$, and thus $S\left(t, t_0\right)|\Omega\rangle$ is annihilated by $a_p\left(t_0\right)$ at $t=-\infty$. Equivalently (in the Heisenberg picture) the operator $a_p(t)=$ $S\left(t, t_0\right)^{\dagger} a_p\left(t_0\right) S\left(t, t_0\right)$ annihilates $|\Omega\rangle$ at $t=-\infty$.

In the free theory, there is a state $|0\rangle$, which is annihilated by the $a_p$. Since the $a_p$ evolve with a simple phase rotation, the same state $|0\rangle$ is annihilated by the (free theory) $a_p$ at any time. More precisely, even if we do not assume $|0\rangle$ has zero energy, then $a_p\left(t_0\right) e^{i H_0\left(t-t_0\right)}|0\rangle=0$ at $t=-\infty$. Since at the time $t_0$ the free and interacting theory creation and annihilation operators are equal, the $a_p$ in both theories annihilate $e^{i H_0\left(t-t_0\right)}|0\rangle$ and $S\left(t, t_0\right)|\Omega\rangle$. Thus, the two states must be proportional. Therefore
$$|\Omega\rangle=\mathcal{N}i \lim {t \rightarrow-\infty} S^{\dagger}\left(t, t_0\right) e^{i H_0\left(t-t_0\right)}|0\rangle=\mathcal{N}i U{0-\infty}|0\rangle$$
for some number $\mathcal{N}i$. Similarly, $\langle\Omega|=\mathcal{N}_f\langle 0| U{\infty 0}$ for some number $\mathcal{N}f$. Now let us see what happens when we rewrite correlation functions in the interaction picture. We are interested in time-ordered products $\left\langle\Omega\left|T\left{\phi\left(x_1\right) \cdots \phi\left(x_n\right)\right}\right| \Omega\right\rangle$. Since all the $\phi\left(x_i\right)$ are within a time-ordered product, we can write them in any order we want. So let us put them in time order, or equivalently we assume $t_1>\cdots>t_n$ without loss of generality. Then, \begin{aligned} & \left\langle\Omega\left|T\left{\phi\left(x_1\right) \cdots \phi\left(x_n\right)\right}\right| \Omega\right\rangle=\left\langle\Omega\left|\phi\left(x_1\right) \cdots \phi\left(x_n\right)\right| \Omega\right\rangle \ & \quad=\mathcal{N}_i \mathcal{N}_f\left\langle 0\left|U{\infty 0} U_{01} \phi_0\left(x_1\right) U_{10} U_{02} \phi_0\left(x_2\right) U_{20} \cdots U_{0 n} \phi_0\left(x_n\right) U_{n 0} U_{0-\infty}\right| 0\right\rangle \ & \quad=\mathcal{N}i \mathcal{N}_f\left\langle 0\left|U{\infty 1} \phi_0\left(x_1\right) U_{12} \phi_0\left(x_2\right) U_{23} \cdots U_{(n-1) n} \phi_0\left(x_n\right) U_{n-\infty}\right| 0\right\rangle \end{aligned}

物理代写|量子场论代写Quantum field theory代考|Interaction potential

The only thing left to understand is what $V_I(t)$ is. We have defined the time $t_0$ as when the interacting fields are the same as the free fields. For example, a cubic interaction would be
$$V\left(t_0\right)=\int d^3 x \frac{g}{3 !} \phi\left(\vec{x}, t_0\right)^3=\int d^3 x \frac{g}{3 !} \phi_0\left(\vec{x}, t_0\right)^3=\int d^3 x \frac{g}{3 !} \phi(\vec{x})^3,$$

Recall that the time dependence of the free fields is determined by the free Hamiltonian,
$$\phi_0(\vec{x}, t)=e^{i H_0\left(t-t_0\right)} \phi_0(\vec{x}) e^{-i H_0\left(t-t_0\right)},$$
and therefore
$$V_I=e^{i H_0\left(t-t_0\right)}\left[\int d^3 x \frac{g}{3 !} \phi_0(\vec{x})^3\right] e^{-i H_0\left(t-t_0\right)}=\int d^3 x \frac{g}{3 !} \phi_0(\vec{x}, t)^3 .$$
So the interaction picture potential is expressed in terms of the free fields at all times.
Now we will make our final transition away from non-Lorentz-invariant Hamiltonians to Lorentz-invariant Lagrangians, leaving old-fashioned perturbation theory for good. Recall that the potential is related to the Lagrangian by $V_I=-\int d^3 x \mathcal{L}{\text {int }}\left[\phi_0\right]$, where $\mathcal{L}{\text {int }}$ is the interacting part of the Lagrangian density. Then,
$$U_{\infty,-\infty}=\exp \left[-i \int_{-\infty}^{\infty} d t V_I(t)\right]=\exp \left[i \int_{-\infty}^{\infty} d^4 x \mathcal{L}{\text {int }}\left[\phi_0\right]\right]$$ The $\int{-\infty}^{\infty} d t$ combines with the $\int d^3 x$ to give a Lorentz-invariant integral.
In summary, matrix elements of interacting fields in the interacting vacuum are given by
$$\left\langle\Omega\left|\phi\left(x_1\right) \cdots \phi\left(x_n\right)\right| \Omega\right\rangle=\frac{\left\langle 0\left|U_{\infty 1} \phi_0\left(x_1\right) U_{12} \phi_0\left(x_2\right) U_{23} \cdots \phi_0\left(x_n\right) U_{n,-\infty}\right| 0\right\rangle}{\left\langle 0\left|U_{\infty,-\infty}\right| 0\right\rangle},$$
where $|\Omega\rangle$ is the ground state in the interacting theory and
$$U_{i j}=T\left{\exp \left[i \int_{t_j}^{t_i} d^4 x \mathcal{L}{\text {int }}\left[\phi_0\right]\right]\right}$$ with $\mathcal{L}{\text {int }}[\phi]=\mathcal{L}[\phi]-\mathcal{L}0[\phi]$, where $\mathcal{L}_0[\phi]$ is the free Lagrangian. The free Lagrangian is defined as whatever goes into the free-field evolution, usually taken to be just kinetic terms. For the special case of time-ordered products, such as what we need for $S$-matrix elements, this simplifies to $$\left\langle\Omega\left|T\left{\phi\left(x_1\right) \cdots \phi\left(x_n\right)\right}\right| \Omega\right\rangle=\frac{\left\langle 0\left|T\left{\phi_0\left(x_1\right) \cdots \phi_0\left(x_n\right) e^{i \int d^4 x \mathcal{L}{\operatorname{inn}[}\left[\phi_0\right]}\right}\right| 0\right\rangle}{\left\langle 0\left|T\left{e^{i \int d^4 x \mathcal{L}_{\text {int }}\left[\phi_0\right]}\right}\right| 0\right\rangle},$$
which is a remarkably simple and manifestly Lorentz-invariant result.

量子场论代考

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