16
$\begingroup$

I was following Schwarz's book on quantum field theory. There he defines the asymptotic momentum eigenstates $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ in the S-matrix element $\langle f|S|i\rangle$ as the eigenstates of the full Hamiltonian i.e., $H=H_0+H_{int}$. Therefore, the states $|i\rangle=|k_1 k_2\rangle$ is defined as

$$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|\Omega\rangle$$

where $|\Omega\rangle$ is the vacuum of the full interacting theory. Then the LSZ reduction formula connects the S-matrix element $\langle f|S|i\rangle$ to the Green' functions of the interaction theory defined as $$G^{(n)}(x_1,x_2,...x_n)=\langle \Omega|T[\phi(x_1)\phi(x_2)...\phi(x_n)]|\Omega\rangle.$$ Here are a few doubts.

Doubt 1 When the particles are far away, the interaction can be considered to be adiabatically switched off. Therefore, at $t=\pm\infty$ the states are really free particle states and should have been written as

$$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|0\rangle$$

and $$|k_3 k_4\rangle=a_{k_3}^{\dagger}(+\infty) a_{k_4}^{\dagger}(+\infty)|0\rangle$$ where $|0\rangle$ is the vacuum of the free thoery. I do not understand why these the states $|i\rangle$ and $|f\rangle$ are derived from $|\Omega\rangle$ instead of $|0\rangle$.

Doubt 2 The initial and final states were derived from the vacuum of the interacting theory $|\Omega\rangle$. According to my understanding, this suggests that the states $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ are eigenstates of the full Hamiltonian $H$. Since then there is no perturbation, there should not be any scattering or transition at all.


More references Even Peskin and Schroeder, Bjorken and Drell, Srednicki take the same approach as Schwartz; they too define the external momentum eigenstates to the eigenstate of the full Hamiltonian $H$. However, if the system was initially in a stationary state why should it undergo a transition in absence of any perturbation?

$\endgroup$
  • 1
    $\begingroup$ I remember this is an issue that puzzled me for a long while as well but I don't remember how I got around it. Hmm, wouldn't $\Omega$ reduce to $|0\rangle$ anyways, if $t\to\pm\infty$? $\endgroup$ – gented Oct 4 '15 at 18:32
  • 2
    $\begingroup$ You should work with wave-packets because states that are too sharply localized in energy (i.e. exact energy eigenstate) or in momenta (i.e. exact momentum eigenstates) are fully delocalized in time or space respectively, and one can't therefore switch-off the interactions at large times or distances (given that the states would still overlap). There is a nice explanation in Weinberg volume I of QFT, chapter 3. $\endgroup$ – TwoBs Oct 6 '15 at 21:28
  • $\begingroup$ @GennaroTedesco- After going through several books, I have the impression that in the limit $t\rightarrow \pm\infty$, the vacuum $|\Omega\rangle$ will not reduce to free theory vacuum $|0\rangle$ because there are self-interactions which can never be switched off even when the particles are infinitely far apart. However, I'm not sure whether this is the correct answer to question (i). $\endgroup$ – SRS Oct 3 '16 at 14:16
  • $\begingroup$ I also think that my confusion stems from the fact that many authors use the same notation i.e., $|0\rangle$ for interacting and free vacuum. But I'm kind of convinced that in LSZ formalism, the vacuum state from which asymptotic in state and out states are built, is the interacting vacuum. $\endgroup$ – SRS Oct 3 '16 at 14:26
  • $\begingroup$ But then again, Schwartz's book (and Itzykson-Zuber's book too) pretends that this adiabatic switching is possible and yet uses the interacting vacuum $|\Omega\rangle$ to build up asymptotic states. It does not mention that self-interactions can never be turned off and yet uses yet uses the $|\Omega\rangle$ to build up asymptotic states. $\endgroup$ – SRS Oct 3 '16 at 14:45
11
+50
$\begingroup$

The first question we have to ask is: what is a one particle state in an interacting theory? It is reasonable to require that they are states that are both momentum eigenstates and energy eigenstates. (In fact, as the Hamiltonian and the momentum operator commute, these are not two different conditions.) Weinberg, in his famous textbook, says that particle states are those which transform under an irreducible representation of the Poincare group, but we need not fuss around with the Poincare group here.

All we will say is that, in the interacting theory, there are some single particle states, labelled by

$$|\lambda k \rangle$$

where $k$ is the four-momentum, and $\lambda$ is whatever other labels we need for our particles. (In this answer I will be working with just a real scalar field, but even in the spin-0 case there can still be extra data that distinguishes our particles in an interacting theory.)

Now, we know know we have a set of momentum and energy eigenstates $|\lambda k\rangle$ that represent the stable particles of our theory. We can now "smudge out" these definite momentum states into wave packets, using a Gaussian window function $f_W$ that has some momentum uncertainty $\kappa$. We will denote these smudged out, approximate energy and momentum eigenstates with a subscript $W$ for "window."

$$|\lambda k\rangle_W \equiv \int d^3{\mathbf{k}'} f_W(\mathbf{k} - \mathbf{k}') |\lambda k\rangle$$

We will come back to these.

Now, the free vacuum $|0\rangle$ of $\hat H_0$ and the true vacuum $|\Omega\rangle$ of $\hat H = \hat H_0 + \hat H_{\rm int}$ are very different states. Particles in the interacting theory must indeed be defined to be formed from the action of the "creation operator" on the true vacuum, as long as we properly define what we mean by the "creation operator" in the interacting theory.

To create an annihilate particles, we will use the Klein Gordon inner product. (We suppress $\hbar$ and $c$.)

$$(\psi_1, \psi_2)_{KG} \equiv i \int d^3 x (\psi^*_1 \partial_t \psi_2 - \partial_t \psi^*_1 \psi_2)$$

The motivation for defining this is that in the FREE theory, the Klein Gordon inner product gives us an inner product between single particle states. If we have two single particle states (in the free theory) $|\Psi_1\rangle$ and $|\Psi_2 \rangle$, we have

$$\langle \Psi_1 | \Psi_2 \rangle = ( \psi_1, \psi_2 )_{KG}$$

where we used the "single particle wave functions" of the states defined by

$$\psi_i(x) \equiv \langle 0| \hat \phi(x) |\Psi_i\rangle$$

The niceties of free field theory come from the simple algebra of the creation and annihilation operators, combined with the fact that the annihilation operator annihilates the vacuum. We will try to recreate those relationships using the Klein Gordon inner product. However, to do this, we will need to use widely separated wave packets.

From here on out, everything will be in the interacting theory.

For a given function $\psi$, we define the creation and annihilation operators that "create" the state corresponding to that wave function as follows. $$ \hat a^\dagger_i (t) \equiv -\big( \psi^*_i(t, \cdot), \hat \phi(t, \cdot) \big)_{KG} $$ $$\hat a_i(t) = \big( \psi_i(t, \cdot), \hat \phi(t, \cdot) \big)_{KG}$$

(In the free theory, this creation operator literally would create the single particle state with the single particle wave function $\psi_1$.)

(Something I must mention about these operators is their time evolution. It is a point of notational confusion that $\hat a^\dagger_{1}(t)$ depends explicitly on a time $t$, given that we usually have defined time dependence such that $e^{i \hat H t} \hat{O}(t') e^{-i \hat H t} = \hat {\mathcal{O}}(t'+t)$. This is not the case here.)

Now, sadly, in the interacting theory, the annihilation operator defined above will not annihilate the vacuum. However, we can recover something close:

$$\langle \Omega| \hat a_1(t) |\Omega\rangle = i \int d^3{x}\langle \Omega| \big( \psi_1^*(t, \vec x) \partial_t \hat \phi (t, \vec x) - \partial_t \psi_1^*(t, \vec x) \hat \phi(t, \vec x) \big) |\Omega\rangle$$ $$= i \int d^3{x} \big( \psi_1^*(t, \vec x) \partial_t \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle - \partial_t \psi^*_1(t, \vec x) \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle \big)$$ $$= i \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle \int d^3{x} (- \partial_t \psi_1^*(t, \vec x))$$

The fact that $\partial_t \langle \Omega| \hat \phi(t, \vec x) |\Omega\rangle = 0$ follows directly from the fact that the vacuum state has zero energy, so $e^{- i \hat H t} |\Omega\rangle = |\Omega\rangle$. Now as we want $\langle \Omega| \hat a_1(t) |\Omega\rangle = 0$ for any $\psi_1$, we can see that this is achieved if and only if $\langle \Omega| \hat \phi(x) |\Omega\rangle = \langle \Omega| \hat \phi(0) |\Omega\rangle = 0$. We will assume this is the case.

In the free theory, $\langle 0| \hat a_1(t) \hat a_2^\dagger(t) |0\rangle = \langle \Psi_1 | \Psi_2\rangle = (\psi_1, \psi_2)_{KG}$. In an interacting theory, for any $\hat a_1$ and state $|\Psi_2\rangle$ (not just a single particle state) we have

$$\langle \Omega| \hat a_1(t) |\Psi_2\rangle = \langle \Omega| \big( \psi_1(t, \cdot) , \hat \phi(t, \cdot) \big)_{KG}|\Psi_2\rangle$$ $$= \big( \psi_1(t, \cdot), \langle \Omega| \hat \phi(t, \cdot) |\Psi_2\rangle\big)_{KG} $$ $$= \big( \psi_1(t, \cdot), \psi_2(t, \cdot) \big)_{KG}$$

$$\langle\Psi_2| \hat a_1(t) |\Omega\rangle = \big( \psi_1(t, \cdot) , \psi_2^*(t, \cdot) \big)_{KG}$$

Remember our single particle states? We're now going consider the "single particle wave function" of those states. Namely, they have to be plane waves.

$$\langle \Omega| \hat \phi(x) |\lambda k\rangle = C_\lambda e^{-ikx}$$

where $C_\lambda$ is a constant that depends on $\lambda$.

We now want to see what our states $\hat a^\dagger_1|\Omega\rangle$ have to do with these true particle wave packets $| \lambda k \rangle_W$. To do this, we will see what the inner product of these two states are. Just from our simple algebra above, for an annihilation operator $\hat a_{\lambda_1 k_1} = (\psi_{k_1}, \hat \phi)_{KG}$ where $k_1^2 = m_{\lambda_1}^2$, we have

\begin{equation*} \begin{split} \langle \Omega | \hat a_{\lambda_1 k_1} (t) |\lambda_2 k_2\rangle_W = \big( \psi_{ k_1}(t, \cdot), \langle \Omega |\hat \phi(t, \cdot)|\lambda_2 k_2\rangle_W \big)_{KG} = C_{\lambda_2}\big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }(t, \cdot) \big)_{KG}\\ {}_W \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1}(t) |\Omega\rangle = \big( \psi_{ k_1}(t, \cdot), {}_W \langle \lambda_2 k_2 |\hat \phi(t, \cdot) |\Omega\rangle \big)_{KG} = C_{\lambda_2}\big(\psi_{ k_1}(t, \cdot), \psi^*_{ k_2 }(t, \cdot) \big)_{KG}. \end{split} \end{equation*} We desire for the top expression to be $\propto \delta_{\lambda_1 \lambda_2} \delta^3(\mathbf{k}_1 - \mathbf{k}_2)$ and for the bottom expression to be 0. If this were the case, then the only single particle state $\hat a^\dagger_{ k_1}(t) |\Omega\rangle$ would overlap with would be $|\lambda_1 k_1\rangle$, and $\hat a_{k_1}(t)$ could still functionally "annihilate" the vacuum, even though we need to keep ${}_W \langle \lambda k |$ on the left. Defining $\omega_{\lambda k} \equiv (m^2_{\lambda} + \mathbf{k}^2)^\frac{1}{2}$, we have

\begin{equation*} \begin{split} \big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }(t, \cdot) \big)_{KG} = (2 \pi)^3 \int d^3{\mathbf{k}} f_W(\mathbf{k}_1 - \mathbf{k}) f_W(\mathbf{k}_2 - \mathbf{k}) (\omega_{\lambda_1 k} + \omega_{\lambda_2 k})e^{it(\omega_{\lambda_1 k} - \omega_{\lambda_2 k})} \\ \big(\psi_{ k_1}(t, \cdot), \psi_{ k_2 }^*(t, \cdot) \big)_{KG} = (2 \pi)^3 \int d^3{\mathbf{k}} f_W(\mathbf{k}_1 - \mathbf{k}) f_W(\mathbf{k}_2 + \mathbf{k}) (\omega_{\lambda_1 k} - \omega_{\lambda_2 k})e^{it(\omega_{\lambda_1 k} + \omega_{\lambda_2 k})}. \\ \end{split} \end{equation*}

The top expression is not $\propto \delta_{\lambda_1 \lambda_2} \delta^3_W(\mathbf{k}_1 - \mathbf{k}_2)$ and the bottom expression is not $0$. However, if we take $\kappa \ll |\mathbf{k}_1 - \mathbf{k}_2|$ and also take $t \to \pm \infty$, they are! This hinges on our assumption that $m_{\lambda_1} \neq m_{\lambda_2}$ if $\lambda_1 \neq \lambda_2$. The $e^{it (\ldots)}$ term will oscillate wildly in both integrals if $\lambda_1 \neq \lambda_2$, causing them to be 0. In the top integral, this oscillation does not occur when $\lambda_1 = \lambda_2$. Furthermore, the top integral will be negligible unless $\mathbf{k}_1 = \mathbf{k}_2$. Taking the $f_W(\mathbf{k}) \to \delta^3(\mathbf{k})$ and $t \to \pm \infty$ limit, we can now write

\begin{equation*} \begin{split} \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1}^\dagger (\pm \infty) |\Omega\rangle = C_{\lambda_2} (2 \pi)^3 2 \omega_{\lambda_2 k_2} \delta_{\lambda_1 \lambda_2} \delta^3(\mathbf{k}_1 - \mathbf{k}_2) \\ \langle \lambda_2 k_2 | \hat a_{\lambda_1 k_1} (\pm \infty) |\Omega\rangle = 0. \end{split} \end{equation*} These properties are even more important than I let on. This is because the states $| \lambda k \rangle$ are so generally defined: they are just momentum eigenstates with all the extra necessary data stuffed into $\lambda$. As they diagonalize the momentum operator, they form a basis of our entire state space! Therefore, we can immediately see from the first equation that

$$\hat a^\dagger_{\lambda k}(\pm \infty) |\Omega\rangle = -C_\lambda \big( e^{ikx}, \hat \phi( x)\big)_{KG} |\Omega\rangle \vert_{t = \pm \infty} = | \lambda k \rangle$$ where we have chosen the normalization $\langle\lambda k | \lambda' k' \rangle = C_{\lambda}^* C_{\lambda'} (2 \pi)^3 (2 \omega_{\lambda k}) \delta_{\lambda \lambda'} \delta^3(\mathbf{k} - \mathbf{k}')$. From the second equation, we can immediately see that

$$\langle\Psi | \hat a_{\lambda k}(\pm \infty) |\Omega\rangle = 0 \hspace{0.15 cm} \text{ for all } \langle\Psi | \hspace{0.5 cm} \Longrightarrow \hspace{0.5 cm} \hat a_{\lambda k}(\pm \infty) |\Omega\rangle = 0.$$ Apparently our asymptotic creation and annihilation operators behave almost exactly like our good old creation and annihilation operators from the free theory!

There's another important property I must mention, which is that two creation/annihilation operators that have different $\lambda k$ data will commute. This is a direct consequence of the fact that our creation/annihilation operators are spacial integrals weighted by wave packets that are spatially separated at large times. (For operators with the same $k$ but different $\lambda$, as $m_\lambda$ is different the wave packets will propagate at different speeds and will still succeed to separate.) Note that spatial separation is a property of wave packets but not of plane waves. This is another place where it is necessary to view plane waves as a limit of wave packets in order to properly understand your theory. In fact, the operators will not commute unless they are defined with this limiting procedure.

We are finally ready to define our incoming and outgoing multi-particle states. As our asymptotic creation operators only change the ground state in localized spatial regions and each spatial excitation is justifiably called a "particle state" we can say that acting with a few of them on the ground state will create a perfectly good multi-particle state. We will now define our incoming (created at $t = -\infty$) and outgoing (created at $t = + \infty$) multi-particle asymptotic states.

$$ |\lambda_1 k_1, \ldots, \lambda_n k_n\rangle_{\rm in} \equiv \hat a^\dagger_{\lambda_1 k_1}(-\infty) \ldots \hat a^\dagger_{\lambda_n k_n}(-\infty) |\Omega\rangle \\ |\lambda_1 k_1, \ldots, \lambda_n k_n\rangle_{\rm out} \equiv \hat a^\dagger_{\lambda_1 k_1}(+\infty) \ldots \hat a^\dagger_{\lambda_n k_n}(+\infty) |\Omega\rangle$$

The four-momenta $k_i$ will have masses $k_i^2 = m_{\lambda_i}^2$ and no $|\lambda_i k_i\rangle$ is allowed to equal another. Some people prefer to rescale $\hat \phi$ in order to hide those $C_\lambda$ prefactors but I will not. The nature of these prefactors will be explored much later. It is important to note that the total momenta of these states are approximately the sum of all $\mathbf{k}_i$, and the energy is approximately the sum of all $\omega_{\lambda_i k_i}$. This lends more credence to the notion that these are "multi-particle" states.

Now that we have successfully defined our incoming and outgoing asymptotic multi particle states and derived some important properties of our newly constructed asymptotic creation and annihilation operators, we have completed the framework necessary to derive the LSZ reduction formula. Using the properties defined here, you should be able to justifiably go through the steps as outlined in Srednicki.

To answer your doubt 2: In order to get out states to have the right properties, we needed these to be wave-packets that are widely deparated in the distant past and future. Therefore, these states are only approximately momentum and energy eigenstates (although you can get as close as you want). As they're not perfect energy eigenstates, some time evolution will occur. You particles will start far apart, come together, interact, then (different) particles will leave.

TLDR: If you define creation and annihilation operators properly, using the Klein Gordon inner product with widely separated wave packets in the far past/future, you will get your actual particle states when acting with these operators on the true vacuum $|\Omega\rangle$.

$\endgroup$
  • 2
    $\begingroup$ I know this question is very old, but I want to know which book you were reading? It sounds like a completely different theory from graduate textbooks. $\endgroup$ – Drake Marquis Jul 8 '18 at 3:46
  • $\begingroup$ When I took QFT my professor was frustrated by the poor quality of the discussion of the LSZ reduction formula in standard textbooks and so devised a very long problem set where we were forced to do it right. This answer is essentially a summarized form of that problem set with some of my own thoughts interjected. $\endgroup$ – user1379857 Sep 25 at 16:36
6
$\begingroup$

Doubt 1: You cannot simply put $t=\pm\infty$ since all formulas become meaningless unless the limit is carefully done. The proof of the LSZ theorem by Haag and Ruelle shows that one needs the interacting vacuum. You can understand the relativistic situation by first looking at the simpler nonrelativistic situation, where a paper by Sandhas [1] gives the nonrelativistic analogue of the treatment by Haag and Ruelle.

Doubt 2: The asymptotic single-particle states are eigenstates, but these don't scatter. You need more than one particle for nontrivial scattering, and the product states are no longer eigenstates.

In Thirring's Course in Mathematical Physics, Vol. 3, there is a clear discussion of asymptotic states, again in the nonrelativistic situation. They are not eigenstates of the Hamiltonian: The (generalized) eigenstates are not the asymptotic plane waves but the the solutions of the Lippmann-Schwinger equations. (Take 2 particles and view them in the center of mass frame, to see the connection.) This makes your Doubt 2 moot.

[1]: W. Sandhas, Definition and Existence of Multichannel Scattering States, Communications in Mathematical Physics 3.5 (1966): 358-374. https://projecteuclid.org/download/pdf_1/euclid.cmp/1103839514

$\endgroup$
  • $\begingroup$ But all the books I cited claim that they are eigenstates of the full Hamiltonian. So are they all wrong? @ArnoldNeumaier $\endgroup$ – SRS May 8 '18 at 9:48
  • 2
    $\begingroup$ @SRS: Please cite one of these books with page number and explicit claim (by editing your question). The asymptotic single-particle states are eigenstates, but these don't scatter. You need more than one particle for nontrivial scattering, and the product states are no longer eigenstates. $\endgroup$ – Arnold Neumaier May 8 '18 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.