Perturbation theory with an explicitly time-dependent Schrodinger picture hamiltonian

Question

There's something that's been bugging me about time-dependent Schrodinger-picture interaction Hamiltonians for a while - namely, the perturbation theory Peskin and Schrodinger derive in Chapter 4 doesn't seem to apply. I'm going to use the Hamiltonian introduced in problem 4.1 as a concrete example, which is also discussed in this Phys.SE post. This Hamiltonian is explicitly time-dependent, even in the Schrodinger picture, whereas the results derived in the chapter seem to rely importantly on the Hamiltonian being time-independent (via the Gell-Mann and Low theorem).

The source given in problem 4.1 $j(x,t)$ is explicitly dependent on time. Hence the interaction Hamiltonian is time-dependent even in the Schrödinger picture, $$H_{\mathrm{int},S} = -\int j(t,\boldsymbol x) \phi(\boldsymbol x),\tag{p.126}$$ where I've put the subscript $S$ to emphasize that I'm talking about the Schrödinger picture even though the Hamiltonian is time-dependent.

(In the Heisenberg picture it's $H_{\mathrm{int},H} = -\int j(t,\boldsymbol x) \phi(x)$ as the $c$-number function $j(t,\boldsymbol x)$ commutes with all the exponentials with the Hamiltonian.)

Because the Schrödinger Hamiltonian is time-dependent, we have to use instantaneous eigenfunctions that diagonalize the Hamiltonian at some particular time $t_0$ (later taken to be zero). However, we still have the Schrödinger equation, so we can find the states at later times by: \begin{equation} \left. | \psi(t) \right> = Te^{-i \int_{t_0}^t H_S(t^\prime) dt^\prime} \left. | \psi(t_0) \right>, \end{equation} cf. e.g. this Phys.SE post.

Because the Hamiltonian is time-dependent we have to use this even for energy eigenstates $\left. | n \right>$, and the resulting states won't generally be eigenstates at $t \neq t_0$.

Since we have the Schrödinger equation, though, most of the results in chapter 4 go right through with $e^{-i H (T-t_0)}$ replaced with $Te^{-i \int_{t_0}^T H_S(t^\prime) \, dt^\prime}$. The one that I'm having trouble with is the unnumbered equation under (4.26): \begin{equation} e^{-iHT} \left. | 0 \right> = \sum_n e^{-i E_n T} \left. | n \right> \left< n | 0 \right>.\tag{p.86} \end{equation}

As mentioned, the states $\left. |n \right>$ don't necessarily diagonize the Hamiltonian at some time other than $t_0$. So while I think: \begin{equation} e^{-iH_S(t_0)} \left. | 0 \right> = \sum_n e^{-i E_n(t_0)} \left. | n \right> \left< n | 0 \right> \end{equation} is true, we can't integrate from $t_0 = 0$ to $T$ in the integrals on each side. I think it is true that \begin{equation} e^{-iH_S(t_0)T} \left. | 0 \right> = \sum_n e^{-i E_n(t_0)T} \left. | n \right> \left< n | 0 \right> \end{equation}

Although unnumbered, this equation is a key step in deriving the Gell-Mann and Low Theorem (4.27), which is key to getting (4.31) etc to evaluate correlation functions. When I use the above, I end up with factors of $$e^{i H_S(t_0) T} e^{-i T \int_{0}^{T} H_S(t^\prime) dt^\prime}$$ all over the place (for example, when I try to introduce the unitary operators in (4.28)), since all of my other expressions have the integrated Hamiltonian in the exponential.

Therefore, the expectation values (4.31) don't simplify to expressions with just the unitary operator as it does when the Schrodinger Hamiltonian does not have explicit time dependence. It seems to me that the (4.31) and following, including Wick's theorem, need to be modified for when the Schrodinger Hamiltonian is explicitly time-dependent (although this is entirely glossed over in problem 4.1).

Does anyone know to work out the perturbation theory for an explicit time-dependent Schrodinger interaction Hamiltonian?

Answer 1

We consider a Hamilton operator (in the Schrödinger picture) of the form $$ H(t)= H_0 +H_1(t), \tag{1} \label{1} $$ where $H_0$ is time independent and the time dependent part $H_1(t)$ is supposed to vanish for $t\to \pm \infty$. In the Schrödinger picture, the time evolution of a state vector $|\psi_S(t)\rangle$ is determined by the Schrödinger equation $$ i \frac{d}{dt} | \psi_S(t) \rangle = (H_0+H_1(t)) |\psi_S(t)\rangle. \tag{2} \label{2} $$ The state vector $| \psi_S(0)\rangle$ at $t=0$ and the state vector $|\psi_S(t) \rangle$ at time $t$ are related by a unitary transformation. It is convenient to disentangle the time evolution due to $H_0$ and the contribution due to $H_1(t)$ by writing this unitary transformation in the form $$ |\psi_S(t)\rangle = e^{-iH_0t} U(t) |\psi_S(0)\rangle, \tag{3} \label{3} $$ where \eqref{2} implies $$ i \frac{d}{dt} U(t) =\underbrace{e^{iH_0t} H_1(t) e^{-iH_0t} }_{H_{1, IP}(t)} \,U(t), \qquad U(0)=\mathbb{1}, \tag{4} \label{4} $$ for the unitary operator $U(t)$. The subscript $IP$ used in \eqref{4} refers to the interaction picture. The (formal) solution of \eqref{4} is given by the time ordered exponential $$ U(t)= T e^{-i\int\limits_0^t \! d\tau \, H_{1, IP}(\tau)}, \tag{5} \label{5} $$ where $T$ is the time-ordering symbol.

The time evolution in the Schrödinger picture and the one in the interaction picture are related by $$ |\psi_S(t)\rangle = e^{-iH_0t} \underbrace{U(t) |\psi_S(0)\rangle}_{|\psi_{IP}(t)\rangle} \quad \Leftrightarrow \quad |\psi_{IP}(t) \rangle = e^{i H_0 t} |\psi_S(t) \rangle. \tag{6} \label{6} $$ Heisenberg picture and interaction picture are related by $$ | \psi_{IP}(t) \rangle = U(t) |\psi_H\rangle \quad \Leftrightarrow \quad |\psi_H\rangle = U(t)^\dagger |\psi_{IP}(t) \rangle. \tag{7} \label{7} $$

Suppose that the motion starts in the remote past in the ground state $|0\rangle$ of the unperturbed system with $$H_0 |0\rangle=0, \tag{8} \label{8}$$ i.e. $$ |\psi_S(-\infty) \rangle =\lim\limits_{t\to -\infty} e^{-iH_0t} |\psi_{IP}(t) \rangle = |0 \rangle, \tag{9} \label{9} $$ being equivalent to $$ \lim\limits_{t\to -\infty} | \psi_{IP}(t) \rangle = |0\rangle \tag{10}\label{10} $$ or $$ U(-\infty) \! \! \!\underbrace{|0, {\rm in}\rangle}_{\rm Heisenberg \; state} \! \! \! = |0\rangle \quad \Leftrightarrow \quad |0, {\rm in}\rangle= Te^{-i \int\limits_{-\infty}^0 \! d\tau \, H_{1, IP}(\tau)} |0\rangle. \tag{11} \label{11} $$ In the course of time, the state vector moves away from the ground state, $$ | \psi_{IP}(t) \rangle = T e^{-i \int\limits_{-\infty}^t \! d\tau \, H_{1, IP}(\tau)} |0 \rangle, \tag{12} \label{12} $$ until, for sufficiently large times, it again remaines put because $H_1(t)$ tends to zero. For $H_1 \equiv 0$, the vector $|\psi_{IP}(+\infty)\rangle$ would coincide with the ground state of $H_0$, but in the presence of $H_1(t)$ it does in general not present an eigenstate of $H_0$. The perturbation violates time-translation invariance and the energy fails to be conserved. The vector $|\psi_{IP}(+\infty)\rangle$ does not coincide with the ground state but is a superposition of energy eigenstates of $H_0$. The probability amplitude that the system is found in the ground state at $t\to \infty$ after the perturbation described by $H_1(t)$ is given by $$ \langle 0 | T e^{-i \int\limits_{-\infty}^{+\infty} \! d\tau \, H_{1, IP}(\tau)} |0\rangle. \tag{13} \label{13} $$ Using $$ |0, {\rm out} \rangle = T e^{i \int\limits_0^\infty \! d\tau \, H_{1,IP}(\tau)} |0\rangle, \tag{14} \label{14} $$ we may also write $$ \langle 0 |T e^{-i \int\limits_{-\infty}^{+\infty} \! d\tau \, H_{1, IP}(\tau)} |0\rangle = \langle 0, {\rm out} | 0, {\rm in} \rangle_{H_1(t)}, \tag{15} \label{15} $$ being referred to as the vacuum-to-vacuum transition amplitude in the presence of the time dependent perturbation $H_1(t)$.

In problem 4.1 of Peskin & Schroeder, the time dependent perturbation caused by the (classical) external source $j(x)$ corresponds to the Hamiltonian $$ H_1(t)= -\int d^3 x \, j(t, \mathbf{x}) \phi(\mathbf{x}). \tag{16} \label{16} $$ In this case, \eqref{15} assumes the form $$ \langle 0, {\rm out} | 0, {\rm in} \rangle_j = \langle 0 | T e^{i \int d^4x \, j(x)\phi_{IP}(x)} |0 \rangle, \tag{17} $$ in accordance with 4.1 (a) in your text book.