On page 108 of Peskin Shroeder.
If the formula $$ |\mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle \propto \lim_{T\to \infty(1-i\epsilon)} e^{-iHT}\, | \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle_0 \tag{4.88} $$ (where $|\mathbf{P} \rangle $ is a one-particle state of momentum $\mathbf{P}$ in the interacting theory, $|\mathbf{P} \rangle_0 =\sqrt{2E_{\mathbf{p}}}a_{\mathbf{p}}^{\dagger} |0\rangle$, and $|0\rangle$ is the vacum of the free theory.)
could somehow be justified, then we could rewrite an $S$-matrix element as \begin{alignat}{2} \langle \mathbf{P_1}\cdots\mathbf{P_n} |S| \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle &=&& \phantom{}_{\text{out}}\langle \mathbf{P_1}\cdots\mathbf{P_n} | \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle_{\text{in}} \\ &=&& \lim_{T\to \infty} \langle \mathbf{P_1}\cdots\mathbf{P_n} |e^{-iH(2T)}\,| \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle . \tag{4.87} \\ & \propto && \lim_{T\to \infty(1-i\epsilon)} \phantom{}_0 \langle\mathbf{P_1}\cdots\mathbf{P_n} |e^{-iH(2T)}\,| \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle_0 \\ & \propto && \lim_{T\to \infty(1-i\epsilon)} \phantom{}_0 \langle\mathbf{P_1}\cdots\mathbf{P_n} | T\left(\exp\left[ -i\int_{-T}^T dt\,H_I(t) \right]\right) | \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle_0 , \tag{4.89} \end{alignat} where $H_I(t)=e^{iH_0 (t-t_0)} \,(H_{\text{int}})\, e^{-iH_0 (t-t_0)},\, H=H_0+H_{\text{int}},\, H_0$ is the Hamiltonian of the free theory.
Why can we get the last line from the third line?
Recalling the time-evolution operator (the interaction picture propagator) $U(t,t')$ is \begin{alignat}{2} U(t,t') &=&& T\left(\exp\left[ -i\int_{t'}^t d\tilde{t}\,H_I(\tilde{t}) \right]\right) \tag{4.23} , \\ U(t,t') &=&& e^{iH_0 (t-t_0)}\,e^{-iH(t-t^{\prime})}\,e^{-iH_0 (t^{\prime}-t_0)} \tag{4.25} , \end{alignat} where $t_0$ is our reference time when we expand $\phi$ in terms of ladder operators at page 83: $$ \phi(t_0,\mathbf{x})=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\mathbf{p}}}} \left( a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}} + a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot\mathbf{x}} \right). $$ Was the last line of (4.89) derived because $$ e^{-iH_0 (-T-t_0)} \, a_{\mathbf{P_\cal{A}}}^{\dagger}a_{\mathbf{P_\cal{B}}}^{\dagger}|0\rangle =a_{\mathbf{P_\cal{A}}}^{\dagger}a_{\mathbf{P_\cal{B}}}^{\dagger}|0\rangle \quad ?$$ Thanks.


1 Answer 1


Review of the picture transform

So the "picture transform" is a quantum coordinate transform which basically starts from the usual facts of quantum mechanics. These are roughly: systems are described by vectors in a Hilbert space; everything you can observe is modeled by a Hermitian operator on that Hilbert space, with eigenvalues being the actual values you observe; the only prediction of QM in any particular case is an average value of some observable $A$ in some state $\psi$ given by $\langle A \rangle_\psi = \langle \psi|\hat A |\psi\rangle$; some observable corresponding to total energy $\hat H$ also governs the time evolution of the state vectors in the Hilbert space as $i\hbar \partial_t|\psi\rangle = \hat H |\psi\rangle$.

The picture transform says: hey, that above equation is solved by $|\psi\rangle = U(t) |\psi_0\rangle$ for some unitary $U$ given by $\hat H$ as $$i \hbar \partial_t U = \hat H ~U.$$But suppose the Hamiltonian were instead $\hat h$ for some arbitrary Hermitian $\hat h.$ Then we would find $|\psi\rangle = u(t) |\psi_0\rangle$ for some other unitary $u$. Now insert $u u^\dagger = 1$ into the above to find, $$\langle A\rangle_\psi = \langle \psi_0| U^\dagger~u u^\dagger ~\hat A ~u u^\dagger ~U |\psi_0\rangle.$$Defining $A' = u^\dagger \hat A u$ and $|\psi'\rangle = u^\dagger U|\psi_0\rangle$ must therefore preserve all predictions of quantum mechanics, with the only cost being that now we must consider $$i\hbar \partial_t A' = u^\dagger (\hat A \hat h - \hat h \hat A) u=A'h' - h' A',$$ while for our Hamiltonian we find,$$ i\hbar \partial_t|\psi'\rangle = u^\dagger (- \hat h + \hat H) U |\psi_0\rangle = (H' - h')|\psi'\rangle.$$(Both formulas straightforwardly map to the right-hand-sides after strategically inserting $uu^\dagger = 1$ terms.)

Just like you can mentally understand $U$ as being some sort of time-ordered continuous product and write down a helpful mnemonic,$$U(t) = \mathcal T \exp\left(\frac1{i\hbar} \int_{t_0}^t d\tau ~ \hat H(\tau)\right),$$ if you define $\eta' = H' - h'$ you will find a helpful mnemonic that the new unitary evolution operator looks like,$$U'(t) = u^\dagger U = \mathcal T \exp\left(\frac1{i\hbar} \int_{t_0}^t d\tau ~ \eta'(\tau)\right).$$I claim that this last expression is exactly what was intended in the expression you have above, with $H_I(t)$ taking the place of $\eta'.$

Why the S-matrix tells us everything

The S-matrix is just a matrix expansion of $U'.$ (Equivalently, $U$, when taking the picture transform as the identity transform, the so-called "Schrödinger picture.") But suppose we have a basis $|n\rangle$ for the Hilbert space, $1 = \sum_n |n\rangle\langle n|.$ Then our above expression for the expectation value can be calculated as:$$\langle A\rangle_\psi = \sum_{mnpq} \langle \psi_0|m\rangle\langle m|{U'}^\dagger|n\rangle\langle n|A'|p\rangle\langle p|U'|q\rangle\langle q|\psi_0\rangle. = (\psi_0)_m^*~U^\dagger_{mn}~A_{np} ~U_{pq}~(\psi_0)_q.$$Therefore $U_{pq} = \langle p | U' | q\rangle$ really does in some sense contain all of the mystery of the time-evolution. And in your case it seems like you're probably dealing with, say, plane waves coming in from infinity getting scattered out to other plane waves travelling out to infinity: this is a great set of states to get a very simple picture, "here is what my Hamiltonian does."

Mixing these together in the interaction picture.

Usually when we're doing this we have some Hamiltonian which we can solve analytically, in your case it seems to be something like $\hat h = \hbar \omega_p a^\dagger_p a_p,$ where $p$ indexes a particle in a definite momentum-state.

When you do this, the presciption above basically just puts phase factors $e^{\pm~i\omega_p t}$ on all of your creation/annihilation operators, and on your definite-momentum states.

If that's right then your question becomes a non-issue: if you paid very careful attention to the $\propto$ sign, and tried to convert it into an $=$ sign, you would have to insert phase factors all over the place in order to keep this thing sane. I think Peskin and Schroeder are probably implicitly saying "hey, you've got this big complicated phase factor out front which does technically matter, but we are explicitly not considering it as important for the physics that we are about to discuss."

I don't have the book but that's where my gut would point me. They are proportional except for some sort of awkward phase factor which would make the equation too long to fit cleanly on one line, so it got absorbed into a $\propto$ symbol.

  • $\begingroup$ $\hat{A}$ is an operator. Do you assume $\partial_{t}\hat{A}=\hat{A}\partial_{t}$ in calculating $i\hbar \partial_t A' = u^\dagger (\hat A \hat h - \hat h \hat A) u$? $\endgroup$
    – GotchaP
    Commented Jan 22, 2017 at 3:44
  • $\begingroup$ I am indeed assuming that and if you didn't want to you'd get an extra $u^\dagger (\hat A \hat h - \hat h \hat A + i \hbar \dot A) u$ so just an $\dot A'$ term appears there. $\endgroup$
    – CR Drost
    Commented Jan 23, 2017 at 5:21
  • $\begingroup$ What is $\hat{A}$ in my case? And in (4.89), $\,|\psi'\rangle = T\left(\exp\left[ -i\int_{-T}^T dt\,H_I(t) \right]\right) | \mathbf{P_\cal{A}}\mathbf{P_\cal{B}} \rangle_0 $ , $ \langle \psi_0 |=\phantom{}_0 \langle\mathbf{P_1}\cdots\mathbf{P_n} |$ ? $\endgroup$
    – GotchaP
    Commented Jan 24, 2017 at 0:53
  • 1
    $\begingroup$ @GotchaP I don't know precisely what your book is doing but it is probably just calculating a matrix element $U_{pq}$, which as I noted above gives you the essentials of every expectation value in principle. That's how I would interpret your expression. Indeed you're calling this thing the S-matrix and the usual definition is to have a set of outbound states $\langle o_m|$ and some inbound states $|i_n\rangle$ with the definition being $S_{mn} = \lim_{t\to\infty} \langle o_m| U(t) U^\dagger(-t) |i_n\rangle.$ $\endgroup$
    – CR Drost
    Commented Jan 24, 2017 at 5:16
  • $\begingroup$ Since $ [H,a_{\mathbf{p}}^{\dagger}]=w_{\mathbf{P}}a_{\mathbf{p}}^{\dagger} \,(2.32)$, $ \,[e^{cH_0},a_{\mathbf{p}}^{\dagger}]=e^{cw_{\mathbf{P}}} $. Then indeed we can get straightforwardly the last line in (4.89). I learned (2.32) years ago, and I had forgotten it. $\endgroup$
    – GotchaP
    Commented Jan 24, 2017 at 21:26

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