- #1
geoduck
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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:
[tex]\psi_{k,in}(x)=e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r} [/tex]
According to the adiabatic method, you prepare the eigenstate [itex]e^{i\vec{k}\cdot \vec{x}}[/itex] 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, [itex]e^{i\vec{k}\cdot \vec{x}}[/itex] gradually turns into [itex]\psi_{k,in}(x)=e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r} [/itex].
Instead, for the wavepacket method, you create a wavepacket:
[tex]\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) [/tex]
Since [itex]\psi_{k,in}[/itex] is a true eigenstate of the fully-interacting Hamiltonian, it's easy to find the wavefunction at t=-∞:
[tex]\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) [/tex]
The last term vanishes due to the fact that the stationary point [itex]k=(\mu r/t)\hbar[/itex] is not in the integration region for negative time (t=-∞) and positive r, so that:
[tex]\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}} [/tex]
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→-∞?
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:
[tex]\psi_{k,in}(x)=e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r} [/tex]
According to the adiabatic method, you prepare the eigenstate [itex]e^{i\vec{k}\cdot \vec{x}}[/itex] 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, [itex]e^{i\vec{k}\cdot \vec{x}}[/itex] gradually turns into [itex]\psi_{k,in}(x)=e^{i\vec{k}\cdot \vec{x}}+\frac{f(\theta, \phi)e^{ikr}}{r} [/itex].
Instead, for the wavepacket method, you create a wavepacket:
[tex]\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) [/tex]
Since [itex]\psi_{k,in}[/itex] is a true eigenstate of the fully-interacting Hamiltonian, it's easy to find the wavefunction at t=-∞:
[tex]\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) [/tex]
The last term vanishes due to the fact that the stationary point [itex]k=(\mu r/t)\hbar[/itex] is not in the integration region for negative time (t=-∞) and positive r, so that:
[tex]\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}} [/tex]
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→-∞?