logo

The adiabatic approximation

30 Nov, 2019
4 min read

The adiabatic approximation

Motivation

In time independent perturbation theory, we saw that a time dependent perturbation H′(t)H'(t) can cause transitions ∣ψa⟩→∣ψb⟩|\psi_a \rangle \rightarrow |\psi_b \rangle

For this to occur, frequency ω\omega of the perturbation had to match the spacing between those levels.

ω0=Eb−Eaℏ(transition frequency)\omega_0 = \frac{E_b - E_a}{\hbar} \quad \text{(transition frequency)}

But what if

ω≪Eb−Eaℏ\omega \ll \frac{E_b - E_a}{\hbar}

image-20191130154808887

This is a very slow perturbation, which is what we call the adiabatic limit. We expect system initially in eigen state ∣ψa⟩|\psi_a \rangle will stay in that state (no transitions), but eigenstate itself will slowly change in time.

∣Ψ(t)⟩∼∣Ψa(t)⟩|\Psi(t) \rangle \sim |\Psi_a (t) \rangle

Example. Harmonic oscillator with time dependence K(t)=K0cos⁡(ωt)K(t) = K_0 \cos(\omega t).

image-20191130154456050

Suppose system starts off in ground state at time t=0t=0.

Then,

K(0)=K0K(0) = K_0Ψ(x,0)=ψ0(x)=(mΩπℏ)14e−mΩ0x2/2ℏ\Psi(x, 0) = \psi_0(x) = \left(\frac{m \Omega}{\pi \hbar}\right)^{\frac{1}{4}} e^{-m \Omega_{0} x^{2} / 2 \hbar}Ω0=k0m\Omega_{0}=\sqrt{\frac{k_{0}}{m}}

image-20191130154746486

If spring constant changes very slowly, (ω≪Ω0)(\omega \ll \Omega_0) we expect the only effect is that Ω0\Omega_0 becomes time dependent Ω0→Ω(t)=K(t)/m\Omega_0 \rightarrow \Omega(t) = \sqrt{K(t)/m}

Ψ(x,t)=(mΩ(t)πℏ)14e−mΩ(t)x2/2ℏ\Psi(x, t)=\left(\frac{m \Omega(t)}{\pi \hbar}\right)^{\frac{1}{4}} e^{-m \Omega(t) x^{2} / 2 \hbar}

If the time dependence is slow, the solution of the time dependent Schrodinger equation

iℏd∣Ψ(t)⟩dt=H(t)∣Ψ(t)⟩i \hbar \frac{d|\Psi(t)\rangle}{d t}=H(t)|\Psi(t)\rangle

should be well approximated by a succession of eigenvalue problems.

H(t)∣Ψn(t)⟩=En(t)∣Ψn(t)⟩H(t)\left|\Psi_{n}(t)\right\rangle= E_{n}(t)\left|\Psi_{n}(t)\right\rangle

Solving the time independent Schrodinger equation for each tt,

∣Ψ(t)⟩∼∣Ψn(t)⟩|\Psi(t)\rangle \sim\left|\Psi_{n}(t)\right\rangle

if ∣Ψ(0)⟩=∣Ψn(0)⟩|\Psi(0) \rangle=\left|\Psi_{n}(0)\right\rangle, we have the following theorem:

The adiabatic theorem

If H(t)H(t) varies slowly in time with respect to level spacing, the system prepared in the nnth eigenstate ∣Ψn(0)⟩|\Psi_n(0)\rangle of H(0)H(0) will remain in the nnth instantaneous eigenstate ∣Ψ(t)|\Psi(t) of H(t)H(t), picking up only a phase factor.

∣Ψ(t)⟩=eiθn(t)eiγn(t)∣ψn(t)⟩\boxed{|\Psi(t) \rangle = e^{i\theta_n(t)}e^{i\gamma_n(t)}|\psi_n(t)\rangle}

We say that θn(t)\theta_n(t) is the dynamical phase, with En(t)E_n(t) the instantaneous eigenenergies.

θn(t)=−1ℏ∫0tdt′En(t′)\theta_n(t) = -\frac{1}{\hbar} \int_{0}^{t} d t^{\prime} E_{n}\left(t^{\prime}\right)

and γn(t)\gamma_n(t) is the geometrical phase.

γn(t)=i∫0tdt′⟨ψn(t′)∣∂∂t′∣ψn(t′)⟩\gamma_n(t) = i \int_0^t dt' \left\langle\psi_{n}\left(t^{\prime}\right)\left|\frac{\partial}{\partial t^{\prime}}\right|\psi_{n}\left(t^{\prime}\right)\right\rangle

Proof of the adiabatic theorem

Define ∣psin(θ)⟩|psi_n (\theta)\rangle and En(t)E_n(t) via the time independent schrodinger equation,

H(t)∣ψn⟩=En(t)∣ψn(t)⟩H(t) |\psi_n \rangle = E_n(t) |\psi_n(t)\rangle

and treat tt as a parameter and solve the time independent Schrodinger equation for each tt.

  • For each tt, {ψn(t)}\{\psi_n(t)\} forms a complete orthonormal basis.
  • Expand solution ∣Psi(t)⟩|Psi(t)\rangle of full time dependent Schrodinger equation into that basis, for each tt.
Ψn(t)=ψne−iEnt/h\Psi_{n}(t)=\psi_{n} e^{-i E_{n} t / h}

If the Hamiltonian changes with time, then the eigenfunctions and eigenvalues are time-dependant:

H(t)ψn(t)=En(t)ψn(t)H(t) \psi_{n}(t)=E_{n}(t) \psi_{n}(t)

But, they are at least orthonormal and complete:

⟨ψn(t)∣ψm(t)⟩=δmn\left\langle \psi_n(t) | \psi_{m}(t)\right\rangle=\delta_{m n}

Hence, we can take our most general state, the solution to the time dependant Schrodinger equation

iℏ∂∂tΨ(t)=H(t)Ψ(t)i \hbar \frac{\partial}{\partial t} \Psi(t)=H(t) \Psi(t)

and expand into the instantaneous basis:

Ψ(t)=∑ncn(t)ψn(t)eiθn(t)\Psi(t)=\sum_{n} c_{n}(t) \psi_{n}(t) e^{i \theta_{n}(t)}

where

θn(t)≡−1ℏ∫0tEn(t′)dt′\theta_{n}(t) \equiv-\frac{1}{\hbar} \int_{0}^{t} E_{n}\left(t^{\prime}\right) d t^{\prime}

Substitute the instantaneous basis into the time dependent Schrodinger equation:

iℏ∑n[c˙nψn+cnψ˙n+icnψnθ˙n]eiθn=∑ncn(Hψn)eiθni \hbar \sum_{n}\left[\dot{c}_{n} \psi_{n}+c_{n} \dot{\psi}_{n}+i c_{n} \psi_{n} \dot{\theta}_{n}\right] e^{i \theta_{n}}=\sum_{n} c_{n}\left(H \psi_{n}\right) e^{i \theta_{n}}∑nc˙nψneiθn=−∑ncnψ˙neiθn\sum_{n} \dot{c}_{n} \psi_{n} e^{i \theta_{n}}=-\sum_{n} c_{n} \dot{\psi}_{n} e^{i \theta_{n}}

The inner product and orthonormality gives:

∑nc˙nδmneiθn=−∑ncn⟨ψm∣ψ˙n⟩eiθn\sum_{n} \dot{c}_{n} \delta_{m n} e^{i \theta_{n}}=-\sum_{n} c_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle e^{i \theta_{n}}c˙m(t)=−∑ncn⟨ψm∣ψ˙n⟩ei(θn−θm)\dot{c}_{m}(t)=-\sum_{n} c_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}

Now, take the time derivative of Schrodinger equation

∂∂tH(t)∣ψn(t)⟩=∂∂l(En(t)∣ψn(t)⟩)\frac{\partial}{\partial t} H(t)\left|\psi_{n}(t)\right\rangle=\frac{\partial}{\partial l}\left(E_{n}(t)\left|\psi_{n}(t)\right\rangle\right)

gives

H˙ψn+Hψ˙n=E˙nψn+Enψ˙n\dot{H} \psi_{n}+H \dot{\psi}_{n}=\dot{E}_{n} \psi_{n}+E_{n} \dot{\psi}_{n}

Taking the inner product ⟨ψm(t)∣m≠n\langle \psi_m (t) | \quad m\neq n

⟨ψm∣H˙∣ψn⟩+⟨ψm∣H∣ψ˙n⟩=E˙nδmn+En⟨ψm∣ψ˙n⟩\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle+\left\langle\psi_{m}|H| \dot{\psi}_{n}\right\rangle=\dot{E}_{n} \delta_{m n}+E_{n}\left\langle\psi_{m} | \dot{\psi}_{n}\right\rangle

Therefore,

⟨ψm∣H˙∣ψn⟩=(En−Em)⟨ψm∣ψ˙n⟩\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle=\left(E_{n}-E_{m}\right)\left\langle\psi_{m} | \dot{\psi}_{n}\right\ranglec˙m(t)=−cm⟨ψm∣ψ˙m⟩−∑n≠mcn⟨ψm∣H˙∣ψn⟩En−Eme(−i/ℏ)∫0t[En(t′)−Eut(t′)]dt′\dot{c}_{m}(t)=-c_{m}\left\langle\psi_{m} | \dot{\psi}_{m}\right\rangle-\sum_{n \neq m} c_{n} \frac{\left\langle\psi_{m}|\dot{H}| \psi_{n}\right\rangle}{E_{n}-E_{m}} e^{(-i / \hbar)\int_{0}^{t}\left[E_{n}\left(t^{\prime}\right)-E_{u t}\left(t^{\prime}\right)\right] d t^{\prime}}

Now invoke the adiabatic approximation. Assume H˙\dot{H} is extremely small, and drop the second term. We now have:

c˙m(t)=−cm⟨ψn∣ψ˙m⟩\dot{c}_{m}(t)=-c_{m}\left\langle\psi_{n} | \dot{\psi}_{m}\right\rangle

the solution is:

cm(t)=cm(0)eiγm(t)c_{m}(t)=c_{m}(0) e^{i \gamma_{m}(t)}

where

γn(t)≡i∫0t⟨ψm(t′)∣∂∂t′∣ψm(t′)⟩dt′\gamma_{n}(t) \equiv i \int_{0}^{t}\left\langle\psi_{m}\left(t^{\prime}\right) | \frac{\partial}{\partial t^{\prime}} |\psi_{m}\left(t^{\prime}\right)\right\rangle d t^{\prime}

And so, the final answer is

Ψn(t)=eiθn(t)eiγn(t)ψn(t)\Psi_{n}(t)=e^{i \theta_{n}(t)} e^{i \gamma_{n}(t)} \psi_{n}(t)

Which is the statement of the adiabatic theorem.

#quantum mechanics#featured
Share This:

Related Posts

Maxwell's Stress Tensor

4 min read
#physics#electrodynamics#featured

The Origins of Quanta

8 min read
#physics#quantum mechanics#featured

Introduction to Second Quantization

12 min read
#physics#quantum mechanics#quantum field theory#animated

Get The Best Of All Hands Delivered To Your Inbox

Subscribe to our newsletter and stay updated.

Copyright © 2020

Zhi Han