# In-state: adiabatic switching vs wavepackets

1. Apr 17, 2014

### geoduck

Am I correct to say that adiabatic switching and wavepackets are two completely different ways to derive the same S-matrix? Because in the adiabatic switching method, the Hamiltonian is dependent on time, equal to the noninteracting Hamiltonian when |t|=∞ and the interacting Hamiltonian at all other times. In the wavepacket method, the Hamiltonian is always interacting.

The In-state is defined as the eigenstate of the fully-interacting Hamiltonian that looks like an eigenstate of non-interacting Hamiltonian in the far past. In particular in NR QM, scattering of a single particle off an elastic potential:

$$\psi_{k,in}(x)=e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r}$$

According to the adiabatic method, you prepare the eigenstate $e^{i\vec{k}\cdot \vec{x}}$ of H0 at t=-∞. At some large but finite time -T, the interaction gradually turns on so that the Hamiltonian gradually goes from H0 to H. From the adiabatic theorem, $e^{i\vec{k}\cdot \vec{x}}$ gradually turns into $\psi_{k,in}(x)=e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r}$.

Instead, for the wavepacket method, you create a wavepacket:

$$\int d^3k \, g(\vec{k})\psi_{k,in}(x)=\int d^3k \, g(\vec{k}) \left(e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r}\right)$$

Since $\psi_{k,in}$ is a true eigenstate of the fully-interacting Hamiltonian, it's easy to find the wavefunction at t=-∞:

$$\int d^3k \, e^{-i\hbar k^2t/2\mu}g(\vec{k})\psi_{k,in}(x)=\int d^3k \, e^{-i\hbar k^2t/2\mu} g(\vec{k}) \left(e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r}\right)$$

The last term vanishes due to the fact that the stationary point $k=(\mu r/t)\hbar$ is not in the integration region for negative time (t=-∞) and positive r, so that:

$$\int d^3k \, e^{-i\hbar k^2t/2\mu}g(\vec{k})\psi_{k,in}(x)=\int d^3k \, e^{-i\hbar k^2t/2\mu} g(\vec{k}) e^{i\vec{k}\cdot \vec{x}}$$

In other words, the Hamiltonian is always interacting, but somehow a wavepacket constructed of eigenstates of H0 is the same as a wavepacket of eigenstates of H, when t→-∞

Is my interpretation of the two formalisms correct? One involves eigenstates evolving adiabatically, the other involves eigenstates that don't evolve (other than a phase) but instead the wavepacket construction is somehow not sensitive to whether you use the eigenstates of H0 or H for your basis, when t→-∞?

2. Apr 17, 2014

### Einj

Yes, it seems correct to me. I think that the adiabatic technique to construct the S-matrix is actually a little bit old-fashioned. In particular there is no reason to assume that the interaction turns off at very early times. The in-state formalism is the most rigorous one.

If you are looking for a good reference on that you should take a look to Weinberg's book "Lectures on Quantum Mechanics". In my opinion he does an amazing treatment of the in-states and S-matrix formalism.

3. Apr 17, 2014

### geoduck

I think the adiabatic technique might be more intuitive however. Maybe I'm wrong about this, but if you have electrons, and a lattice, and combine them to interact to form a metal, you'd like to initially describe the electrons by free electrons and the lattice by a free lattice, with the interaction between the lattice and the electrons turned off. With the wavepacket technique, you would be describing the electrons with eigenstates of electrons+lattice, which is kind of weird (but the wavepacket will ensure that this mixture looks like a wavepacket of just free electrons at early times).

In high-energy physics, as you say the adiabatic technique makes no sense and in fact we know that even for a single free electron, you can't shut off the E&M interaction, as the electron always interacts with its photon cloud and this in fact renormalizes the charge and mass. The Wikipedia article on the LSZ formalism ( http://en.wikipedia.org/wiki/LSZ_reduction#In_and_Out_fields ) even mentions that adiabatic technique is not correct:

"These interactions do not fade away as particles drift apart, so much care must be used in establishing asymptotic relations between the interacting field and the in field."

$$\varphi(x)\sim\sqrt Z\varphi_{\mathrm{in}}(x)\quad \mathrm{as}\quad x^0\rightarrow-\infty$$

and the prescription is sort of a wavepacket method:

"The correct prescription, as developed by Lehmann, Symanzik and Zimmermann, requires two normalizable states"

$$\lim_{x^0\rightarrow-\infty} \int \mathrm{d}^3x \langle\alpha|f(x)\overleftrightarrow\partial_0\varphi(x)|\beta\rangle= \sqrt Z \int \mathrm{d}^3x \langle\alpha|f(x)\overleftrightarrow\partial_0\varphi_{\mathrm{in}}(x)|\beta\rangle$$

Although I vaguely recall skimming through QFT textbooks that don't mention the LSZ-reduction, so maybe the adiabatic technique can even work for field theory? The adiabatic technique does seem to give the same results of wavepackets for NR QM.

I've read most of that book, because I'm a grader and that's the book the class is using. Weinberg does everything with wave-packets so no mention of adiabatic technique is made (he does derive the adiabatic theorem, but he seems to only use it for the Berry phase and not scattering).

The terminology though is sometimes confusing. When Weinberg says that the In-state is an eigenstate of the interacting Hamiltonian that looks like the eigenstate of the free-particle Hamiltonian at very early times, I think he means the In-state integrated over a wavepacket looks like a wave-packet of the free-particle Hamiltonian, at early times. In the adiabatic-switching formalism, you could leave out the term wavepackets, since the eigenstate of the interacting Hamiltonian really does become the eigenstate of the free-Hamiltonian at early times, since they evolve adiabatically into one another, in 1-1 correspondence. Weinberg shows mathematically how a wavepacket of the free theory evolves into a wavepacket of the interacting theory, although it would have been nice if he mentioned that the adiabatic switching method directly has an eigenstate of the free theory evolving into an eigenstate of the interacting theory which is not physically sound as you mention, since you can't just turn off the interaction, but nevertheless is seen in the literature sometimes.

Also, even if a book uses the adiabatic technique , I think I've seen some still use the term "In-state" to describe the eigenstate of the Hamiltonian for t=0 (eigenstate of H(0)).

4. Apr 18, 2014

### Einj

Yes, the adiabatic technique is probably more intuitive, even though it is actually wrong. Take for example the case of a single electron in a static Coulomb potential. The potential itself is constant with time and so there is no physical reason to switch the potential off at t=-∞. What is actually changing is the behavior of the electron wave function far far away from the scattering center. Therefore is definitely more correct to say that it's the wave function that is basically a free particle wave function when the electron is very far away (i.e. t=-∞). However, I think that in ordinary QM the two treatment should lead to the same result.

Yes, you can use the adiabatic approximation in QFT as well (see for example Mandl-Shaw Ch. 6.2). However, as you said, this only work for free fields. Suppose for example that you have a $\lambda\phi^4$ theory. Even in absence of any other kind of interaction, your particle will always interact with themselves and so it's never correct to treat them as free fields. This is particularly wrong if you use the LSZ formulae which are essential when dealing with Feynman path integrals which are the only rigorous way to study interacting QFT.

5. Apr 19, 2014

### geoduck

You wouldn't happen to remember the reason that the expression for the scattering amplitude:

$$\int d^3y \, e^{-i\vec{k'}\cdot\vec{x}} V(y)\psi_{in, \,k}(y)$$

is not zero? If you replace the potential V(y) by H-H0, then it should vanish if k' and k have equal energies:

$$\int d^3y \, e^{-i\vec{k'}\cdot\vec{x}} \left(H-H_0\right)\psi_{in, \,k}(y)$$

since H acting on the in-state gives the energy, and H0 can act on the left on the free-state and give the energy, and so if k' and k have equal energies, the scattering amplitude is zero.

It's very similar the T-matrix given by:

$$T_{\beta \alpha} = \langle \phi_\beta |V | \psi_{in, \,\alpha}\rangle$$

where alpha is the in state, and β is a state of the free Hamiltonian. If you make the replacement V=H-H0, then if the free state and in state have the same energy, you should get zero.

6. Apr 19, 2014

### Einj

Why shouldn't that be the case? In fact:
$$\int d^3ye^{i\vec k'\cdot \vec y}V(\vec y)\psi_k^{in}(\vec y)=\int d^3y\langle \phi_{k'}|y\rangle V(y)\langle y|\psi_k^{in}\rangle=\langle\phi_{k'}|V|\psi_k^{in}\rangle.$$

7. Apr 19, 2014

### geoduck

Yes, but the potential sandwiched between a free state and an In-state is the T-matrix. So it would imply that the T-matrix is zero, which means that there can be no transitions between states of the same energy, which is clearly wrong.

8. Apr 19, 2014

### Einj

You know, I never thought about it. This is actually even weirder since the T-matrix always appear with a delta-function in front of it, so the energy must be the same. I'll think about it...

9. Apr 20, 2014

### vanhees71

This is a pretty delicate issue. The "adiabatic switching" is necessary to properly define asymptotic states and to calculate the S-Matrix as transition-probability amplitudes for scattering process with well-defined (asymptotically) free particles in the initial and final states. Energy-momentum conservation for the particles only holds in the sense of asymptotic states.

If you have a situation, where you can resolve the time evolution of quantum mechanical processes energy and momentum can become indeterminate at finite times, because the system is not in an eigenstate of the Hamiltonian and momentum operators.

Anyway, suppose we have a typical scattering process and you have evaluated the S-matrix. The S-matrix elements have the typical structure
$$S_{fi}=\delta_{fi} +(2 \pi)^4 \mathrm{i} \delta^{(4)}(P_f-P_i) T_{fi}.$$
The first term refers to the possibility that the particles are not scattered at all, while the 2nd term encodes a true scattering event. The $\delta$ distribution indicates the conservation of total energy and momentum for the asymptotically free particle states from the initial to the final states with well-defined momenta. It's written in terms of the energy-momentum Fock states for the appropriate asymptotically free particles in the initial and final states (symmetrized or antisymmetrized product states of energy-momentum-spin-component eigenstates or "plane waves" of free single particles).

Of course, plane waves are not true states. That's why it doesn't make sense to naively square the S-matrix element, leading to a squared Dirac-$\delta$ distribution which makes no sense whatsoever. The true situation is that you prepare your (usually 2) particles with energy and momentum (and perhaps polarization=spin state) which is accurate on a macroscopic scale. It's impossible to prepare a true energy-momentum eigenstate, because these simply don't exist. They are generalized states ("distributions"). So what you have to do to describe the true physical situation is to use wave packets that are "smeared" in energy in momentum to some extent. This momentum spread must be even large enougth to ensure that you have well-separated particles located at a large distance to ensure that the particles are really "asymptotically free", i.e., the interaction between them is negligible at the initial time, where you prepare them in order to have really well-defined asymptotically free particles. Then you place a detector suitable to measure the wanted final state far away from the region, where the particles interact so that again the single particles are again well separated and asymptotically free in the so defined final state.

You find a very detailed and careful discussion in the textbook by Messiah.

All that tacitly assumed that the interaction is short-ranged, in order to properly define asymptotic states. This is, e.g., not fulfilled for the Coulomb interaction (or more generally the electromagnetic interaction in relativistic QED), which is of long-range nature. Already in the non-relativistic Coulomb problem plane waves are never good asymptotic states, because the long-range nature of the interaction implies a non-trivial phase factor. You can solve this problem exactly. For perturbation theory you can use plane waves in the sense of some limiting procedure. You simply use a Yukawa-like potential which implements some effective screening of the Coulomb field and then, after evaluating "infrared save" quantities, which needs an appropriate "soft-photon resummation" take the limit of vanishing screening mass. In QED a similar procedure works, introducing a small photon mass (which in this case of an Abelian gauge theory doesn't even spoil gauge invariance and renormalizability and unitarity of the S-matrix elements thanks to the possibility to formulate a massive vector field in the Stueckelberg formalism) as an IR regulator, then doing the soft-photon resummations and taking the limit of vanishing photon mass (see Weinberg, QT of fields, Vol. 1 for details).

Alternatively you can work with the true asymptotic states of charged particles and photons (Kulish, Faddeev, Kibble,...) showing that, e.g., an electron is always surrounded by a non-trivial "soft-photon cloud" and a photon by an "electron-positron cloud" due to the long-ranged nature of the em. interaction.

10. Apr 20, 2014

### geoduck

I think I worked it out. You get using Weinberg's notation $(\phi_\beta V \psi_\alpha)= (\phi_\beta (H-H_0) \psi_\alpha)=(E_\alpha-E_\beta)(\phi_\beta, \psi_\alpha)$ where $\phi$ is the eigenstate of H0 and ψ the In state. The trick is that the scalar product $(\phi_\beta, \psi_\alpha)$ is proportional to 1/(Eα-Eβ):

$$(\phi_\beta, \psi_\alpha)=(\phi_\beta, \phi_\alpha+GV\phi_\alpha+GVGV \phi_\alpha+...)= \left[\delta(\beta-\alpha)+\frac{1}{E_\alpha-E_\beta+i\epsilon}\left(\phi_\beta, (V+VGV+VGVGV+...)\phi_\alpha\right)\right]$$

where G is the free-propagator. I'm confident that you can take the limit ε→0 first, and then the limit Eα →Eβ. For example you recover the Born approximation when the E's cancel and you are left with $(\phi_\beta, V \phi_\alpha)$ for your transition matrix.

11. Apr 20, 2014

Sounds good!