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## Second form of adiabatic approximation.

Posted by peeterjoot on December 11, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

In class we were shown an adiabatic approximation where we started with (or worked our way towards) a representation of the form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k c_k(t) e^{-i \int_0^t (\omega_k(t') - \Gamma_k(t')) dt' } {\left\lvert {\psi_k(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.1)

where ${\left\lvert {\psi_k(t)} \right\rangle}$ were normalized energy eigenkets for the (slowly) evolving Hamiltonian

\begin{aligned}H(t) {\left\lvert {\psi_k(t)} \right\rangle} = E_k(t) {\left\lvert {\psi_k(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.2)

In the problem sets we were shown a different adiabatic approximation, where are starting point is

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_k c_k(t) {\left\lvert {\psi_k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(1.3)

For completeness, here’s a walk through of the general amplitude derivation that’s been used.

# Guts

We operate with our energy identity once again

\begin{aligned}0 &=\left(H - i \hbar \frac{d{{}}}{dt} \right) \sum_k c_k {\left\lvert {k} \right\rangle} \\ &=\sum_k c_k E_k {\left\lvert {k} \right\rangle} - i \hbar c_k' {\left\lvert {k} \right\rangle} - i \hbar c_k {\left\lvert {k'} \right\rangle} ,\end{aligned}

where

\begin{aligned}{\left\lvert {k'} \right\rangle} = \frac{d{{}}}{dt} {\left\lvert {k} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.4)

Bra’ing with ${\left\langle {m} \right\rvert}$, and split the sum into $k = m$ and $k \ne m$ parts

\begin{aligned}0 =c_m E_m - i \hbar c_m' - i \hbar c_m \left\langle{{m}} \vert {{m'}}\right\rangle - i \hbar \sum_{k \ne m} c_k \left\langle{{m}} \vert {{k'}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.5)

Again writing

\begin{aligned}\Gamma_m = i \left\langle{{m}} \vert {{m'}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.6)

We have

\begin{aligned}c_m' = \frac{1}{{i \hbar}} c_m (E_m - \hbar \Gamma_m) - \sum_{k \ne m} c_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.7)

In this form we can make an “Adiabatic” approximation, dropping the $k \ne m$ terms, and integrate

\begin{aligned}\int \frac{d c_m'}{c_m} = \frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \end{aligned} \hspace{\stretch{1}}(2.8)

or

\begin{aligned}c_m(t) = A \exp\left(\frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \right).\end{aligned} \hspace{\stretch{1}}(2.9)

Evaluating at $t = 0$, fixes the integration constant for

\begin{aligned}c_m(t) = c_m(0) \exp\left(\frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \right).\end{aligned} \hspace{\stretch{1}}(2.10)

Observe that this is very close to the starting point of the adiabatic approximation we performed in class since we end up with

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k c_k(0) e^{-i \int_0^t (\omega_k(t') - \Gamma_k(t')) dt' } {\left\lvert {k(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.11)

So, to perform the more detailed approximation, that started with 1.1, where we ended up with all the cross terms that had both $\omega_k$ and Berry phase $\Gamma_k$ dependence, we have only to generalize by replacing $c_k(0)$ with $c_k(t)$.

## One more adiabatic pertubation derivation.

Posted by peeterjoot on December 8, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

I liked one of the adiabatic pertubation derivations that I did to review the material, and am recording it for reference.

# Build up.

In time dependent pertubation we started after noting that our ket in the interaction picture, for a Hamiltonian $H = H_0 + H'(t)$, took the form

\begin{aligned}{\left\lvert {\alpha_S(t)} \right\rangle} = e^{-i H_0 t/\hbar} {\left\lvert {\alpha_I(t)} \right\rangle} = e^{-i H_0 t/\hbar} U_I(t) {\left\lvert {\alpha_I(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.1)

Here we have basically assumed that the time evolution can be factored into a portion dependent on only the static portion of the Hamiltonian, with some other operator $U_I(t)$, providing the remainder of the time evolution. From 2.1 that operator $U_I(t)$ is found to behave according to

\begin{aligned}i \hbar \frac{d{{U_I}}}{dt} = e^{i H_0 t/\hbar} H'(t) e^{-i H_0 t/\hbar} U_I,\end{aligned} \hspace{\stretch{1}}(2.2)

but for our purposes we just assumed it existed, and used this for motivation. With the assumption that the interaction picture kets can be written in terms of the basis kets for the system at $t=0$ we write our Schr\”{o}dinger ket as

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i H_0 t/\hbar} a_k(t) {\left\lvert {k} \right\rangle}= \sum_k e^{-i \omega_k t/\hbar} a_k(t) {\left\lvert {k} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.3)

where ${\left\lvert {k} \right\rangle}$ are the energy eigenkets for the initial time equation problem

\begin{aligned}H_0 {\left\lvert {k} \right\rangle} = E_k^0 {\left\lvert {k} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.4)

For the adiabatic problem, we assume the system is changing very slowly, as described by the instantanious energy eigenkets

\begin{aligned}H(t) {\left\lvert {k(t)} \right\rangle} = E_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.5)

Can we assume a similar representation to 2.3 above, but allow ${\left\lvert {k} \right\rangle}$ to vary in time? This doesn’t quite work since ${\left\lvert {k(t)} \right\rangle}$ are no longer eigenkets of $H_0$

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i H_0 t/\hbar} a_k(t) {\left\lvert {k(t)} \right\rangle}\ne \sum_k e^{-i \omega_k t} a_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.6)

Operating with $e^{i H_0 t/\hbar}$ does not give the proper time evolution of ${\left\lvert {k(t)} \right\rangle}$, and we will in general have a more complex functional dependence in our evolution operator for each ${\left\lvert {k(t)} \right\rangle}$. Instead of an $\omega_k t$ dependence in this time evolution operator let’s assume we have some function $\alpha_k(t)$ to be determined, and can write our ket as

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i \alpha_k(t)} a_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.7)

Operating on this with our energy operator equation we have

\begin{aligned}0 &=\left(H - i \hbar \frac{d}{dt} \right) {\left\lvert {\psi} \right\rangle} \\ &=\left(H - i \hbar \frac{d}{dt} \right) \sum_k e^{-i \alpha_k} a_k {\left\lvert {k} \right\rangle} \\ &=\sum_k e^{-i \alpha_k(t)} \left( \left( E_k a_k-i \hbar (-i \alpha_k' a_k + a_k')\right) {\left\lvert {k} \right\rangle}-i \hbar a_k {\left\lvert {k'} \right\rangle}\right) \\ \end{aligned}

Here I’ve written ${\left\lvert {k'} \right\rangle} = d{\left\lvert {k} \right\rangle}/dt$. In our original time dependent pertubaton the $-i \alpha_k'$ term was $-i \omega_k$, so this killed off the $E_k$. If we assume this still kills off the $E_k$, we must have

\begin{aligned}\alpha_k = \frac{1}{{\hbar}} \int_0^t E_k(t') dt',\end{aligned} \hspace{\stretch{1}}(3.8)

and are left with

\begin{aligned}0=\sum_k e^{-i \alpha_k(t)} \left( a_k' {\left\lvert {k} \right\rangle}+a_k {\left\lvert {k'} \right\rangle}\right).\end{aligned} \hspace{\stretch{1}}(3.9)

Bra’ing with ${\left\langle {m} \right\rvert}$ we have

\begin{aligned}0=e^{-i \alpha_m(t)} a_m' +e^{-i \alpha_m(t)} a_m \left\langle{{m}} \vert {{m'}}\right\rangle+\sum_{k \ne m} e^{-i \alpha_k(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.10)

or

\begin{aligned}a_m' +a_m \left\langle{{m}} \vert {{m'}}\right\rangle=-\sum_{k \ne m} e^{-i \alpha_k(t)} e^{i \alpha_m(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.11)

The LHS is a perfect differential if we introduce an integration factor $e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle}$, so we can write

\begin{aligned}e^{-\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle} ( a_m e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } )'=-\sum_{k \ne m} e^{-i \alpha_k(t)} e^{i \alpha_m(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.12)

This suggests that we want to form a new function

\begin{aligned}b_m = a_m e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } \end{aligned} \hspace{\stretch{1}}(3.13)

or

\begin{aligned}a_m = b_m e^{-\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } \end{aligned} \hspace{\stretch{1}}(3.14)

Plugging this into our assumed representation we have a more concrete form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- \int_0^t dt' ( i \omega_k + \left\langle{{k}} \vert {{k'}}\right\rangle ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.15)

Writing

\begin{aligned}\Gamma_k = i \left\langle{{k}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.16)

this becomes

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.17)

## A final pass.

Now that we have what appears to be a good representation for any given state if we wish to examine the time evolution, let’s start over, reapplying our instantaneous energy operator equality

\begin{aligned}0 &=\left(H - i \hbar \frac{d}{dt} \right){\left\lvert {\psi} \right\rangle} \\ &=\left(H - i \hbar \frac{d}{dt} \right)\sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k {\left\lvert {k} \right\rangle} \\ &=- i \hbar \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } \left(i \Gamma_kb_k {\left\lvert {k} \right\rangle} +b_k' {\left\lvert {k} \right\rangle} +b_k {\left\lvert {k'} \right\rangle} \right).\end{aligned}

Bra’ing with ${\left\langle {m} \right\rvert}$ we find

\begin{aligned}0&=e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } i \Gamma_mb_m +e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } b_m' \\ &+e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } b_m \left\langle{{m}} \vert {{m'}}\right\rangle +\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle \end{aligned}

Since $i \Gamma_m = \left\langle{{m}} \vert {{m'}}\right\rangle$ the first and third terms cancel leaving us just

\begin{aligned}b_m'=-\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_{km} - \Gamma_{km} ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.18)

where $\omega_{km} = \omega_k - \omega_m$ and $\Gamma_{km} = \Gamma_k - \Gamma_m$.

# Summary

We assumed that a ket for the system has a representation in the form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i \alpha_k(t) } a_k(t) {\left\lvert {k(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.20)

where $a_k(t)$ and $\alpha_k(t)$ are given or to be determined. Application of our energy operator identity provides us with an alternate representation that simplifes the results

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(4.20)

With

\begin{aligned}{\left\lvert {m'} \right\rangle} &= \frac{d}{dt} {\left\lvert {m} \right\rangle} \\ \Gamma_k &= i \left\langle{{m}} \vert {{m'}}\right\rangle \\ \omega_{km} &= \omega_k - \omega_m \\ \Gamma_{km} &= \Gamma_k - \Gamma_m\end{aligned} \hspace{\stretch{1}}(4.21)

we find that our dynamics of the coefficients are related by

\begin{aligned}b_m'=-\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_{km} - \Gamma_{km} ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.25)

## Review of Quantum mechanics approximation results.

Posted by peeterjoot on November 10, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

Here I’ll summarize what I’d put on a cheat sheet for the tests or exam, if one would be allowed. While I can derive these results, memorization unfortunately appears required for good test performance in this class, and this will give me a good reference of what to memorize.

This set of review notes covers all the approximation methods we covered except for Fermi’s golden rule.

# Variational method

We can find an estimate of our ground state energy using

\begin{aligned}\boxed{\frac{{\left\langle {\Psi} \right\rvert} H {\left\lvert {\Psi} \right\rangle}}{\left\langle{{\Psi}} \vert {{\Psi}}\right\rangle}\ge E_0}\end{aligned} \hspace{\stretch{1}}(2.1)

# Time independent perturbation

Given a perturbed Hamiltonian and an associated solution for the unperturbed state

\begin{aligned}\boxed{\begin{aligned}H &= H_0 + \lambda H', \qquad \lambda \in [0,1] \\ H_0 {\left\lvert {{\psi_{m\alpha}}^{(0)}} \right\rangle} &= {E_m}^{(0)} {\left\lvert {{\psi_{m\alpha}}^{(0)}} \right\rangle},\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.2)

we assume a power series solution for the energy

\begin{aligned}E_m = {E_m}^{(0)} + \lambda {E_m}^{(1)} + \lambda^2 {E_m}^{(2)} + \cdots\end{aligned} \hspace{\stretch{1}}(3.3)

For a non-degenerate state ${\left\lvert {\psi_m} \right\rangle} = {\left\lvert {\psi_{m1}} \right\rangle}$, with an unperturbed value of ${\left\lvert {\psi_{m}^{(0)}} \right\rangle} = {\left\lvert {\psi_{m1}^{(0)}} \right\rangle}$, we seek a power series expansion of this ket in the perturbed system

\begin{aligned}\begin{aligned}{\left\lvert {\psi_m} \right\rangle} &= \sum_{n,\alpha} {c_{n\alpha;m}}^{(0)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} +\lambda\sum_{n,\alpha} {c_{n\alpha;m}}^{(1)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} + \lambda^2\sum_{n,\alpha} {c_{n\alpha;m}}^{(2)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} + \cdots \\ &\propto{\left\lvert {{\psi_m}^{(0)}} \right\rangle} + \lambda\sum_{n \ne m, \alpha} {\bar{c}_{n\alpha;m}}^{(1)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} +\lambda^2\sum_{n \ne m, \alpha} {\bar{c}_{n\alpha;m}}^{(2)} {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle} + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(3.4)

Any states $n \ne m$ are allowed to have degeneracy. For this case, we found to second order in energy and first order in the kets

\begin{aligned}\boxed{\begin{aligned}E_m &= E_m^{(0)} + \lambda {H_{m1;m1}}' + \lambda^2 \sum_{n \ne m, \alpha} \frac{{\left\lvert{{H_{n\alpha;m1}}'}\right\rvert}^2 }{ E_m^{(0)} - E_n^{(0)} } + \cdots\\ {\left\lvert {\psi_m} \right\rangle} &\propto {\left\lvert {{\psi_m}^{(0)}} \right\rangle} + \lambda\sum_{n \ne m, \alpha} \frac{{H_{n\alpha;m1}}'}{ E_m^{(0)} - E_n^{(0)} } {\left\lvert {{\psi_{n\alpha}}^{(0)}} \right\rangle}+ \cdots \\ H_{n\alpha;s\beta}' &={\left\langle {{\psi_{n\alpha}}^{(0)}} \right\rvert}H'{\left\lvert {{\psi_{s\beta}}^{(0)}} \right\rangle}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(3.5)

# Degeneracy.

When the initial energy eigenvalue $E_m$ has a degeneracy $\gamma_m > 1$ we use a different approach to compute the perturbed energy eigenkets and perturbed energy eigenvalues. Writing the kets as ${\left\lvert {m\alpha} \right\rangle}$, then we assume that the perturbed ket is a superposition of the kets in the degenerate energy level

\begin{aligned}{\left\lvert {m \alpha} \right\rangle}' = \sum_i c_i {\left\lvert {m i} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(4.6)

We find that we must have

\begin{aligned}\left( (E^0 - E)I + \lambda \begin{bmatrix} H_{mi;mj}' \end{bmatrix} \right)\begin{bmatrix}c_1 \\ c_2 \\ \dot{v}s \\ c_{\gamma_m}\end{bmatrix}= 0.\end{aligned} \hspace{\stretch{1}}(4.7)

Diagonalizing this matrix $\begin{bmatrix} H_{mi;mj}' \end{bmatrix}$ (a subset of the complete $H'$ matrix element)

\begin{aligned}\begin{bmatrix}{\left\langle {m i} \right\rvert} H' {\left\lvert {m j} \right\rangle}\end{bmatrix}= U_m \begin{bmatrix}\delta_{ij} \mathcal{H}_{m,i}'\end{bmatrix} U_m^\dagger,\end{aligned} \hspace{\stretch{1}}(4.8)

we find, by taking the determinant, that the perturbed energy eigenvalues are in the set

\begin{aligned}\boxed{E = E_m^0 + \lambda \mathcal{H}_{m,i}', \quad i \in [1, \gamma_m]}\end{aligned} \hspace{\stretch{1}}(4.9)

To compute the perturbed kets we must work in a basis for which the block diagonal matrix elements are diagonal for all $m$, as in

\begin{aligned}\begin{bmatrix}{\left\langle {m i} \right\rvert} H' {\left\lvert {m j} \right\rangle}\end{bmatrix}= \begin{bmatrix}\delta_{ij} \mathcal{H}_{m,i}'\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(4.10)

If that is not the case, then the unitary matrices of 4.8 can be computed, and the matrix

\begin{aligned}U = \begin{bmatrix}U_1 & & & \\ & U_2 & & \\ & & \ddots & \\ & & & U_N \\ \end{bmatrix},\end{aligned} \hspace{\stretch{1}}(4.11)

can be formed. The kets

\begin{aligned}{\left\lvert {\overline{m \alpha}} \right\rangle} = U^\dagger {\left\lvert {m \alpha} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.12)

will still be energy eigenkets of the unperturbed Hamiltonian

\begin{aligned}H_0 {\left\lvert {\overline{m \alpha}} \right\rangle} = E_m^0 {\left\lvert {\overline{m \alpha}} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.13)

but also ensure that the partial diagonalization condition of 4.8 is satisfied. In this basis, dropping overbars, the first order perturbation results found previously for perturbation about a non-degenerate state also hold, allowing us to write

\begin{aligned}\boxed{{\left\lvert {s \alpha} \right\rangle}' = {\left\lvert {s \alpha} \right\rangle} + \lambda \sum_{m \ne s, \beta} \frac{{H'}_{m \beta ; s \alpha}}{ E_s^{(0)} - E_m^{(0)} } {\left\lvert {m \beta} \right\rangle}+ \cdots}\end{aligned} \hspace{\stretch{1}}(4.14)

# Interaction picture.

We split of the Hamiltonian into time independent and time dependent parts, and also factorize the time evolution operator

\begin{aligned}\boxed{\begin{aligned}H &= H_0 + H_I(t) \\ {\left\lvert {\alpha_S} \right\rangle} &= e^{-i H_0 t/\hbar } {\left\lvert {\alpha_I(t)} \right\rangle} = e^{-i H_0 t/\hbar } U_I(t) {\left\lvert {\alpha_I(0)} \right\rangle} .\end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.15)

Plugging into Schr\”{o}dinger’s equation we find

\begin{aligned}\boxed{\begin{aligned}i \hbar \frac{d{{}}}{dt} {\left\lvert {\alpha_I(t)} \right\rangle} &= H_I(t) {\left\lvert {\alpha_I(t)} \right\rangle} \\ i \hbar \frac{d{{U_I}}}{dt} &= H_I' U_I \\ H_I'(t) &= e^{i H_0 t/\hbar } H_I(t) e^{-i H_0 t/\hbar } \end{aligned}}\end{aligned} \hspace{\stretch{1}}(5.16)

# Time dependent perturbation.

We moved on to time dependent perturbations of the form

\begin{aligned}\boxed{\begin{aligned}H(t) &= H_0 + H'(t) \\ H_0 {\left\lvert {\psi_n^{(0)} } \right\rangle} &= \hbar \omega_n {\left\lvert {\psi_n^{(0)} } \right\rangle}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(6.17)

where $\hbar \omega_n$ are the energy eigenvalues, and ${\left\lvert {\psi_n^{(0)} } \right\rangle}$ the energy eigenstates of the unperturbed Hamiltonian.

Use of the interaction picture led quickly to the problem of seeking the coefficients describing the perturbed state

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_n c_n(t) e^{-i \omega_n t} {\left\lvert {\psi_n^{(0)} } \right\rangle},\end{aligned} \hspace{\stretch{1}}(6.18)

and plugging in we found

\begin{aligned}\boxed{\begin{aligned}i \hbar \cdot_s &= \sum_n H_{sn}'(t) e^{i \omega_{sn} t} c_n(t) \\ \omega_{sn} &= \omega_s - \omega_n \\ H_{sn}'(t) &= {\left\langle {\psi_s^{(0)}} \right\rvert} H'(t) {\left\lvert {\psi_n^{(0)} } \right\rangle},\end{aligned}}\end{aligned} \hspace{\stretch{1}}(6.19)

## Perturbation expansion in series.

Introducing a $\lambda$ parametrized dependence in the perturbation above, and assuming a power series expansion of our coefficients

\begin{aligned}\boxed{\begin{aligned}H'(t) &\rightarrow \lambda H'(t) \\ c_s(t) &= c_s^{(0)}(t) + \lambda c_s^{(1)}(t) + \lambda^2 c_s^{(2)}(t) + \cdots\end{aligned}}\end{aligned} \hspace{\stretch{1}}(6.20)

we found, after equating powers of $\lambda$ a set of coupled differential equations

\begin{aligned}\begin{aligned}i \hbar \cdot_s^{(0)}(t) &= 0 \\ i \hbar \cdot_s^{(1)}(t) &= \sum_{n} H_{sn}'(t) e^{i \omega_{sn} t} c_n^{(0)}(t) \\ i \hbar \cdot_s^{(2)}(t) &= \sum_{n} H_{sn}'(t) e^{i \omega_{sn} t} c_n^{(1)}(t) \\ &\dot{v}s\end{aligned}\end{aligned} \hspace{\stretch{1}}(6.21)

Of particular value was the expansion, assuming that we started with an initial state in energy level $m$ before the perturbation was “turned on” (ie: $\lambda = 0$).

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = e^{-i \omega_m t} {\left\lvert {\psi_m^{(0)} } \right\rangle}\end{aligned} \hspace{\stretch{1}}(6.22)

So that $c_n^{(0)}(t) = \delta_{nm}$. We then found a first order approximation for the transition probability coefficient of

\begin{aligned}\boxed{i \hbar \cdot_m^{(1)} = H_{ms}'(t) e^{i \omega_{ms} t}}\end{aligned} \hspace{\stretch{1}}(6.23)

# Sudden perturbations.

The idea here is that we integrate Schr\”{o}dinger’s equation over the small interval containing the changing Hamiltonian

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = {\left\lvert {\psi(t_0)} \right\rangle} + \frac{1}{{i\hbar}} \int_{t_0}^t H(t') {\left\lvert {\psi(t')} \right\rangle} dt'\end{aligned} \hspace{\stretch{1}}(7.24)

and find

\begin{aligned}\boxed{{\left\lvert {\psi_\text{after}} \right\rangle} = {\left\lvert {\psi_\text{before}} \right\rangle}.}\end{aligned} \hspace{\stretch{1}}(7.25)

An implication is that, say, we start with a system measured in a given energy, that same system after the change to the Hamiltonian will then be in a state that is now a superposition of eigenkets from the new Hamiltonian.

Given a Hamiltonian that turns on slowly at $t=0$, a set of instantaneous eigenkets for the duration of the time dependent interval, and a representation in terms of the instantaneous eigenkets

\begin{aligned}\boxed{\begin{aligned}H(t) &= H_0, \qquad t \le 0 \\ H(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} &= E_n(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ {\left\lvert {\psi} \right\rangle} &= \sum_n \bar{b}_n(t) e^{-i\alpha_n + i \beta_n} {\left\lvert {\hat{\psi}_n} \right\rangle} \\ \alpha_n(t) &= \frac{1}{{\hbar}} \int_0^t dt' E_n(t'),\end{aligned}}\end{aligned} \hspace{\stretch{1}}(8.26)

plugging into Schr\”{o}dinger’s equation we find

\begin{aligned}\boxed{\begin{aligned}\frac{d{{\bar{b}_m}}}{dt} &= - \sum_{n \ne m} \bar{b}_n e^{-i \gamma_{nm} } {\left\langle {\hat{\psi}_m(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ \gamma_{nm}(t) &= \alpha_n(t) - \alpha_m(t) - (\beta_n(t) - \beta_m(t)) \\ \beta_n(t) &= \int_0^t dt' \Gamma_n(t') \\ \Gamma_n(t) &= i {\left\langle {\hat{\psi}_n(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ \end{aligned}}\end{aligned} \hspace{\stretch{1}}(8.27)

## Evolution of a given state.

Given a system initially measured with energy $E_m(0)$ before the time dependence is “turned on”

\begin{aligned}\boxed{{\left\lvert {\psi(0)} \right\rangle} = {\left\lvert {\hat{\psi}_m(0)} \right\rangle},}\end{aligned} \hspace{\stretch{1}}(8.28)

we find that the first order Taylor series expansion for the transition probability coefficients are

\begin{aligned}\boxed{\bar{b}_s(t) = \delta_{sm} - t (1 - \delta_{sm}) {\left\langle {\hat{\psi}_s(0)} \right\rvert} {\left.{{\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}}}\right\vert}_{{t=0}}.}\end{aligned} \hspace{\stretch{1}}(8.29)

If we introduce a $\lambda$ perturbation, separating all the (slowly changing) time dependent part of the Hamiltonian $H'$ from the non time dependent parts $H_0$ as in

\begin{aligned}H(t) = H_0 + \lambda H'(t)\end{aligned} \hspace{\stretch{1}}(8.30)

then we find our perturbed coefficients are

\begin{aligned}\boxed{\bar{b}_s(t) =\delta_{ms}(1 + \lambda \text{constant})- (1-\delta_{ms}) \lambda\int_0^t dt'e^{i \gamma_{sm}(t') } {\left\langle {\hat{\psi}_s(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {\hat{\psi}_m(t')} \right\rangle} }\end{aligned} \hspace{\stretch{1}}(8.31)

# WKB.

We write Schr\”{o}dinger’s equation as

\begin{aligned}\boxed{\begin{aligned}0 &= \frac{d^2 U}{dx^2} + k^2 U \\ k^2 &= -\kappa^2 = \frac{2m (E - V)}{\hbar}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(9.32)

and seek solutions of the form $U \propto e^{i\phi}$. Schr\”{o}dinger’s equation takes the form

\begin{aligned}- (\phi'(x))^2 + i \phi''(x) + k^2(x) = 0.\end{aligned} \hspace{\stretch{1}}(9.33)

Initially setting $\phi'' = 0$ we refine our approximation to find

\begin{aligned}\phi'(x) = k(x) \sqrt{ 1 + i \frac{k'(x)}{k^2(x)} } .\end{aligned} \hspace{\stretch{1}}(9.34)

To first order, this gives us

\begin{aligned}\boxed{U(x) \propto \frac{1}{{\sqrt{k(x)}}} e^{\pm i \int dx k(x)} }\end{aligned} \hspace{\stretch{1}}(9.35)

What we didn’t cover in class, but required in the problems was the Bohr-Sommerfeld condition described in section 24.1.2 of the text [1].

\begin{aligned}\boxed{\int_{x_1}^{x_2} dx \sqrt{ 2m (E - V(x))} = \left( n + \frac{1}{{2}} \right) \pi.}\end{aligned} \hspace{\stretch{1}}(9.36)

This was found from the WKB connection formulas, themselves found my some Bessel function arguments that I have to admit that I didn’t understand.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## Revisiting adiabatic approximation for expansion around an initial pure state.

Posted by peeterjoot on November 10, 2011

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# New info. How to do the $\lambda$ expansion.

Asking about this, Federico nicely explained. “The reason why you are going in circles when trying the lambda expansion is because you are not assuming the term ${\left\langle {\psi(t)} \right\rvert} (d/dt) {\left\lvert {\psi(t)} \right\rangle}$ to be of order lambda. This has to be assumed, otherwise it doesn’t make sense at all trying a perturbative approach. This assumption means that the coupling between the level $s$ and the other levels is assumed to be small because the time dependent part of the Hamiltonian is small or changes slowly with time. Making a Taylor expansion in time would be sensible only if you are interested in a short interval of time. The lambda-expansion approach would work for any time as long as the time dependent piece of the Hamiltonian doesn’t change wildly or is too big.”

In the tutorial he outlined another way to justify this. We’ve written so far

\begin{aligned}H = \left\{\begin{array}{l l}H(t) & \quad \mbox{latex t > 0} \\ H_0 & \quad \mbox{$t < 0$} \end{array}\right.\end{aligned} \hspace{\stretch{1}}(5.23)

where $H(0) = H_0$. We can make this explicit, and introduce a $\lambda$ factor into the picture if we write

\begin{aligned}H(t) = H_0 + \lambda H'(t),\end{aligned} \hspace{\stretch{1}}(5.24)

where $H_0$ has no time dependence, so that our Hamiltonian is then just the “steady-state” system for $\lambda = 0$.

Now recall the method from [1] that we can use to relate our bra-derivative-ket to the Hamiltonian. Taking derivatives of the energy identity, braketed between two independent kets ($m \ne n$) we have

\begin{aligned}0 &= {\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} \left(H(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} - \hbar \omega_n {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \right) \\ &= {\left\langle {\hat{\psi}_m(t)} \right\rvert}\left(\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} +H(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} -\hbar \frac{d{{\omega_n}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} -\hbar \omega_n \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \right) \\ &= \hbar (\omega_m - \omega_n) {\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} -\not{{\hbar \frac{d{{\omega_n}}}{dt} \delta_{mn}}}+{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \end{aligned}

So for $m \ne n$ we find a dependence between the bra-derivative-ket and the time derivative of the Hamiltonian

\begin{aligned}{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} =\frac{{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} }{\hbar (\omega_n - \omega_m) }\end{aligned} \hspace{\stretch{1}}(5.25)

Referring back to 5.24 we see the $\lambda$ dependence in this quantity, coming directly from the $\lambda$ dependence imposed on the time dependent part of the Hamiltonian

\begin{aligned}{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} =\lambda\frac{{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{H'(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} }{\hbar (\omega_n - \omega_m) }\end{aligned} \hspace{\stretch{1}}(5.26)

Given this $\lambda$ dependence, let’s revisit the perturbation attempt of 3.11. Our first order factors of $\lambda$ are now

\begin{aligned}\frac{d{{}}}{dt} \bar{b}_s^{(1)}(t) &= - \sum_{n \ne s} \delta_{mn} e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ &= \left\{\begin{array}{l l}0 & \quad \mbox{iflatex m = s} \\ – e^{i \gamma_{sm}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle} &\quad \mbox{if $m \ne s$} \\ \end{array}\right.\end{aligned}

So we find to first order

\begin{aligned}\bar{b}_s(t) =\delta_{ms}(1 + \lambda \text{constant})+ - (1-\delta_{ms}) \lambda\int_0^t dt'e^{i \gamma_{sm}(t') } {\left\langle {\hat{\psi}_s(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {\hat{\psi}_m(t')} \right\rangle} \end{aligned} \hspace{\stretch{1}}(5.27)

A couple observations of this result. One is that the constant factor in the $m = s$ case makes sense. This would likely be a negative contribution since we have to decrease the probability coefficient for finding our wavefunction in the $m = s$ state after perturbation, since we are increasing the probability for finding it elsewhere by changing the Hamiltonian.

Also observe that since $e^{i\gamma_{sm}} \sim 0$ for small $t$ this is consistent with the first order Taylor series expansion where we found our first order contribution was

\begin{aligned}- (1 - \delta_{ms}) t {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(5.28)

Also note that this $-e^{i \gamma_{sm}(t') } {\left\langle {\hat{\psi}_s(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {\hat{\psi}_m(t')} \right\rangle}$ is exactly the difference from $0$ that was mentioned in class when the trial solution of $\bar{b}_s = \delta_{sm}$ was tested by plugging it into 2.5, so it’s not too surprising that we should have a factor of exactly this form when we refine our approximation.

A question to consider should we wish to refine the $\lambda$ perturbation to higher than first order in $\lambda$: is there any sort of $\lambda$ dependence in the $e^{i \gamma_{sm}}$ coming from the $\Gamma_{sm}$ term in that exponential?

# References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

## Second order time evolution for the coefficients of an initially pure ket with an adiabatically changing Hamiltonian.

Posted by peeterjoot on November 6, 2011

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# Motivation.

In lecture 9, Prof Sipe developed the equations governing the evolution of the coefficients of a given state for an adiabatically changing Hamiltonian. He also indicated that we could do an approximation, finding the evolution of an initially pure state in powers of $\lambda$ (like we did for the solutions of a non-time dependent perturbed Hamiltonian $H = H_0 + \lambda H'$). I tried doing that a couple of times and always ended up going in circles. I’ll show that here and also develop an expansion in time up to second order as an alternative, which appears to work out nicely.

# Review.

We assumed that an adiabatically changing Hamiltonian was known with instantaneous eigenkets governed by

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} = \hbar \omega_n {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.1)

The problem was to determine the time evolutions of the coefficients $\bar{b}_n(t)$ of some state ${\left\lvert {\psi(t)} \right\rangle}$, and this was found to be

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} &= \sum_n \bar{b}_n(t) e^{-i \gamma_n(t)} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ \gamma_s(t) &= \int_0^t dt' (\omega_s(t') - \Gamma_s(t')) \\ \Gamma_s(t) &= i {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_s(t)} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.2)

where the $\bar{b}_s(t)$ coefficient must satisfy the set of LDEs

\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} = - \sum_{n \ne s} \bar{b}_n(t) e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.5)

where

\begin{aligned}\gamma_{sn}(t) = \gamma_{s}(t) - \gamma_{n}(t).\end{aligned} \hspace{\stretch{1}}(2.6)

Solving these in general doesn’t look terribly fun, but perhaps we can find an explicit solution for all the $\bar{b}_s$‘s, if we simplify the problem somewhat. Suppose that our initial state is found to be in the $m$th energy level at the time before we start switching on the changing Hamiltonian.

\begin{aligned}{\left\lvert {\psi(0)} \right\rangle} = \bar{b}_m(0) {\left\lvert {\hat{\psi}_m(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.7)

We therefore require (up to a phase factor)

\begin{aligned}\begin{array}{l l}\bar{b}_m(0) = 1 & \\ \bar{b}_s(0) = 0 & \quad \mbox{iflatex s \ne m}.\end{array}\end{aligned} \hspace{\stretch{1}}(2.8)

Equivalently we can write

\begin{aligned}\bar{b}_s(0) = \delta_{ms}\end{aligned} \hspace{\stretch{1}}(2.9)

# Going in circles with a $\lambda$ expansion.

In class it was hinted that we could try a $\lambda$ expansion of the following form to determine a solution for the $\bar{b}_s$ coefficients at later times

\begin{aligned}\bar{b}_s(t) = \delta_{ms} + \lambda \bar{b}^{(1)}_s(t) + \cdots\end{aligned} \hspace{\stretch{1}}(3.10)

I wasn’t able to figure out how to make that work. Trying this first to first order, and plugging in, we find

\begin{aligned}\lambda \frac{d{{}}}{dt} \bar{b}^{(1)}_s(t) = - \sum_{n \ne s} ( \delta_{mn} + \lambda \bar{b}^{(1)}_n(t) ) e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(3.11)

equating powers of $\lambda$ yields two equations

\begin{aligned}\frac{d{{}}}{dt} \bar{b}_s^{(1)}(t) &= - \sum_{n \ne s} \bar{b}^{(1)}_n(t) e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ 0 &= - \sum_{n \ne s} \delta_{mn} e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.12)

Observe that the first identity is exactly what we started with in 2.5, but has just replaced the $\bar{b}_n$‘s with $\bar{b}^{(1)}_n$‘s. Worse is that the second equation is only satisfied for $s = m$, and for $s \ne m$ we have

\begin{aligned}0 = - e^{i \gamma_{sm}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.14)

So this $\lambda$ power series only appears to work if we somehow had ${\left\lvert {\hat{\psi}_s(t)} \right\rangle}$ always orthonormal to the derivative of ${\left\lvert {\hat{\psi}_m(t)} \right\rangle}$. Perhaps this could be done if the Hamiltonian was also expanded in powers of $\lambda$, but such a beastie seems foreign to the problem. Note that we don’t even have any explicit dependence on the Hamiltonian in the final $\bar{b}_n$ differential equations, as we’d probably need for such an expansion to work out.

# A Taylor series expansion in time.

What we can do is to expand the $\bar{b}_n$‘s in a power series parametrized by time. That is, again, assuming we started with energy equal to $\hbar \omega_m$, form

\begin{aligned}\bar{b}_s(t) = \delta_{sm} + \frac{t}{1!} \left( {\left.{{ \frac{d{{}}}{dt}\bar{b}_s(t) }}\right\vert}_{{t=0}} \right)+ \frac{t^2}{2!} \left( {\left.{{ \frac{d^2}{dt^2} \bar{b}_s(t) }}\right\vert}_{{t=0}} \right)+ \cdots\end{aligned} \hspace{\stretch{1}}(4.15)

The first order term we can grab right from 2.5 and find

\begin{aligned}{\left.{{\frac{d{{\bar{b}_s(t)}}}{dt}}}\right\vert}_{{t=0}} &= - \sum_{n \ne s} \bar{b}_n(0) {\left.{{{\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle}}}\right\vert}_{{t=0}} \\ &= - \sum_{n \ne s} \delta_{nm}{\left.{{{\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle}}}\right\vert}_{{t=0}} \\ &=\left\{\begin{array}{l l}0 & \quad \mbox{latex s = m} \\ – {\left.{{{\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}}}\right\vert}_{{t=0}} \\ & \quad \mbox{$s \ne m$} \\ \end{array}\right.\end{aligned}

Let’s write

\begin{aligned}{\left\lvert {n} \right\rangle} &= {\left\lvert {\hat{\psi}_n(0)} \right\rangle} \\ {\left\lvert {n'} \right\rangle} &= {\left.{{ \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_n(t)} \right\rangle} }}\right\vert}_{{t=0}}\end{aligned} \hspace{\stretch{1}}(4.16)

So we can write

\begin{aligned}{\left.{{\frac{d{{\bar{b}_s(t)}}}{dt}}}\right\vert}_{{t=0}} =- (1 - \delta_{sm}) \left\langle{{s}} \vert {{m'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.18)

and form, to first order in time our approximation for the coefficient is

\begin{aligned}\bar{b}_s(t) =\delta_{sm} - t (1 - \delta_{sm}) \left\langle{{s}} \vert {{m'}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(4.19)

Let’s do the second order term too. For that we have

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} &= - \sum_{n \ne s} {\left.{{\left(\left(\frac{d{{}}}{dt} \bar{b}_n(t) +\delta_{nm} i \frac{d{{\gamma_{sn}(t)}}}{dt}\right)\left\langle{{s}} \vert {{n'}}\right\rangle+\delta_{nm} \frac{d{{}}}{dt} \left( {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \right) \right)}}\right\vert}_{{t=0}}\end{aligned}

For the $\gamma_{sn}$ derivative we note that

\begin{aligned}{\left.{{\frac{d{{}}}{dt} \gamma_s(t)}}\right\vert}_{{t=0}} = \omega_s(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.20)

So we have

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} &= - \sum_{n \ne s} \Bigl(- (1 - \delta_{nm}) \left\langle{{n}} \vert {{m'}}\right\rangle+\delta_{nm} i (\omega_{sn}(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle + i\left\langle{{n}} \vert {{n'}}\right\rangle)\Bigr)\left\langle{{s}} \vert {{n'}}\right\rangle+\delta_{nm} \Bigl( \left\langle{{s'}} \vert {{n'}}\right\rangle+\left\langle{{s}} \vert {{n''}}\right\rangle\Bigr)\end{aligned}

Again for $s = m$, all terms are killed. That’s somewhat surprising, but suggests that we will need to normalize the coefficients after the perturbation calculation, since we have unity for one of them.

For $s \ne m$ we have

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} &= \sum_{n \ne s} \Bigl(\left\langle{{n}} \vert {{m'}}\right\rangle-\delta_{nm} i (\omega_{sn}(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle + i\left\langle{{n}} \vert {{n'}}\right\rangle)\Bigr)\left\langle{{s}} \vert {{n'}}\right\rangle-\delta_{nm} \Bigl( \left\langle{{s'}} \vert {{n'}}\right\rangle+\left\langle{{s}} \vert {{n''}}\right\rangle\Bigr) \\ &= -i (\omega_{sm}(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle + i\left\langle{{m}} \vert {{m'}}\right\rangle)\Bigr)\left\langle{{s}} \vert {{m'}}\right\rangle-\Bigl( \left\langle{{s'}} \vert {{m'}}\right\rangle+\left\langle{{s}} \vert {{m''}}\right\rangle\Bigr) +\sum_{n \ne s} \left\langle{{n}} \vert {{m'}}\right\rangle \left\langle{{s}} \vert {{n'}}\right\rangle.\end{aligned}

So we have, for $s \ne m$

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} = (\left\langle{{m}} \vert {{m'}}\right\rangle - \left\langle{{s}} \vert {{s'}}\right\rangle ) \left\langle{{s}} \vert {{m'}}\right\rangle-i \omega_{sm}(0) \left\langle{{s}} \vert {{m'}}\right\rangle-\left\langle{{s'}} \vert {{m'}}\right\rangle-\left\langle{{s}} \vert {{m''}}\right\rangle+\sum_{n \ne s} \left\langle{{n}} \vert {{m'}}\right\rangle \left\langle{{s}} \vert {{n'}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(4.21)

It’s not particularly illuminating looking, but possible to compute, and we can use it to form a second order approximate solution for our perturbed state.

\begin{aligned}\begin{aligned}\bar{b}_s(t) &=\delta_{sm} - t (1 - \delta_{sm}) \left\langle{{s}} \vert {{m'}}\right\rangle \\ &+(1 - \delta_{sm})\left((\left\langle{{m}} \vert {{m'}}\right\rangle - \left\langle{{s}} \vert {{s'}}\right\rangle ) \left\langle{{s}} \vert {{m'}}\right\rangle-i \omega_{sm}(0) \left\langle{{s}} \vert {{m'}}\right\rangle-\left\langle{{s'}} \vert {{m'}}\right\rangle-\left\langle{{s}} \vert {{m''}}\right\rangle+\sum_{n \ne s} \left\langle{{n}} \vert {{m'}}\right\rangle \left\langle{{s}} \vert {{n'}}\right\rangle\right) \frac{t^2}{2}\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.22)

## PHY456H1F: Quantum Mechanics II. Recitation 3 (Taught by Mr. Federico Duque Gomez). WKB method and Stark shift.

Posted by peeterjoot on October 28, 2011

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# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# WKB method.

Consider the potential

\begin{aligned}V(x) = \left\{\begin{array}{l l}v(x) & \quad \mbox{iflatex x \in [0,a]} \\ \infty & \quad \mbox{otherwise} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.1)

as illustrated in figure (\ref{fig:qmTwoR3:qmTwoR3fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoR3fig1}
\caption{Arbitrary potential in an infinite well.}
\end{figure}

Inside the well, we have

\begin{aligned}\psi(x) = \frac{1}{{\sqrt{k(x)}}} \left( C_{+} e^{i \int_0^x k(x') dx'}+C_{-} e^{-i \int_0^x k(x') dx'}\right)\end{aligned} \hspace{\stretch{1}}(2.2)

where

\begin{aligned}k(x) = \frac{1}{{\hbar}} \sqrt{ 2m( E - v(x) }\end{aligned} \hspace{\stretch{1}}(2.3)

With

\begin{aligned}\phi(x) = e^{\int_0^x k(x') dx'}\end{aligned} \hspace{\stretch{1}}(2.4)

We have

\begin{aligned}\psi(x) &= \frac{1}{{\sqrt{k(x)}}} \left( C_{+}(\cos \phi + i\sin\phi) + C_{-}(\cos\phi - i \sin\phi)\right) \\ &= \frac{1}{{\sqrt{k(x)}}} \left( (C_{+} + C_{-})\cos \phi + i(C_{+} - C_{-}) \sin\phi\right) \\ &= \frac{1}{{\sqrt{k(x)}}} \left( (C_{+} + C_{-})\cos \phi + i(C_{+} - C_{-}) \sin\phi\right) \\ &\equiv \frac{1}{{\sqrt{k(x)}}} \left( C_2 \cos \phi + C_1 \sin\phi\right),\end{aligned}

Where

\begin{aligned}C_2 &= C_{+} + C_{-} \\ C_1 &= i( C_{+} - C_{-})\end{aligned} \hspace{\stretch{1}}(2.5)

Setting boundary conditions we have

\begin{aligned}\phi(0) = 0\end{aligned} \hspace{\stretch{1}}(2.7)

Noting that we have $\phi(0) = 0$, we have

\begin{aligned}\frac{1}{{\sqrt{k(0)}}} C_2 = 0\end{aligned} \hspace{\stretch{1}}(2.8)

So

\begin{aligned}\psi(x) \sim\frac{1}{{\sqrt{k(x)}}} \sin\phi\end{aligned} \hspace{\stretch{1}}(2.9)

At the other boundary

\begin{aligned}\psi(a) = 0\end{aligned} \hspace{\stretch{1}}(2.10)

So we require

\begin{aligned}\sin \phi(a) = \sin(n \pi)\end{aligned} \hspace{\stretch{1}}(2.11)

or

\begin{aligned}\frac{1}{{\hbar}} \int_0^a \sqrt{2 m (E - v(x')} dx' = n \pi\end{aligned} \hspace{\stretch{1}}(2.12)

This is called the Bohr-Sommerfeld condition.

Check with $v(x) = 0$.

We have

\begin{aligned}\frac{1}{{\hbar}} \sqrt{2m E} a = n \pi\end{aligned} \hspace{\stretch{1}}(2.13)

or

\begin{aligned}E = \frac{1}{{2m}} \left(\frac{n \pi \hbar}{a}\right)^2\end{aligned} \hspace{\stretch{1}}(2.14)

# Stark Shift

Time independent perturbation theory

\begin{aligned}H = H_0 + \lambda H'\end{aligned} \hspace{\stretch{1}}(3.15)

\begin{aligned}H' = e \mathcal{E}_z \hat{Z}\end{aligned} \hspace{\stretch{1}}(3.16)

where $\mathcal{E}_z$ is the electric field.

To first order we have

\begin{aligned}{\left\lvert {\psi_\alpha^{(1)}} \right\rangle} = {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} + \sum_{\beta \ne \alpha} \frac{ {\left\lvert {\psi_\beta^{(0)}} \right\rangle} {\left\langle {\psi_\beta^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} }{E_\alpha^{(0)} -E_\beta^{(0)} }\end{aligned} \hspace{\stretch{1}}(3.17)

and

\begin{aligned}E_\alpha^{(1)} = {\left\langle {\psi_\alpha^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} \end{aligned} \hspace{\stretch{1}}(3.18)

With the default basis $\{{\left\lvert {\psi_\beta^{(0)}} \right\rangle}\}$, and $n=2$ we have a 4 fold degeneracy

\begin{aligned}l,m &= 0,0 \\ l,m &= 1,-1 \\ l,m &= 1,0 \\ l,m &= 1,+1\end{aligned}

but can diagonalize as follows

\begin{aligned}\begin{bmatrix}\text{nlm} & 200 & 210 & 211 & 21\,-1 \\ 200 & 0 & \Delta & 0 & 0 \\ 210 & \Delta & 0 & 0 & 0 \\ 211 & 0 & 0 & 0 & 0 \\ 21\,-1 & 0 & 0 & 0 & 0\end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.19)

FIXME: show.

where

\begin{aligned}\Delta = -3 e \mathcal{E}_z a_0\end{aligned} \hspace{\stretch{1}}(3.20)

We have a split of energy levels as illustrated in figure (\ref{fig:qmTwoR3:qmTwoR3fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoR3fig2}
\caption{Energy level splitting}
\end{figure}

Observe the embedded Pauli matrix (FIXME: missed the point of this?)

\begin{aligned}\sigma_x = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(3.21)

Proper basis for perturbation (FIXME:check) is then

\begin{aligned}\left\{\frac{1}{{\sqrt{2}}}( {\left\lvert {2,0,0} \right\rangle} \pm {\left\lvert {2,1,0} \right\rangle} ), {\left\lvert {2, 1, \pm 1} \right\rangle}\right\}\end{aligned} \hspace{\stretch{1}}(3.22)

and our result is

\begin{aligned}{\left\lvert {\psi_{\alpha, n=2}^{(1)}} \right\rangle} = {\left\lvert {\psi_{\alpha}^{(0)}} \right\rangle} +\sum_{\beta \notin \text{degenerate subspace}} \frac{ {\left\lvert {\psi_\beta^{(0)}} \right\rangle} {\left\langle {\psi_\beta^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} }{E_\alpha^{(0)} -E_\beta^{(0)} }\end{aligned} \hspace{\stretch{1}}(3.23)

Utilizing instantaneous eigenstates

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_{\alpha} b_\alpha(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.24)

where

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}= E_\alpha(t) {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.25)

We found

\begin{aligned}b_\alpha(t) = \bar{b}_\alpha(t) e^{-\frac{i}{\hbar} \int_0^t (E_\alpha(t') - \hbar \Gamma_\alpha(t')) dt'}\end{aligned} \hspace{\stretch{1}}(4.26)

where

\begin{aligned}\Gamma_\alpha = i{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.27)

and

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.36)

Suppose we start in a subspace

\begin{aligned}\text{span} \left\{\frac{1}{{\sqrt{2}}}( {\left\lvert {2,0,0} \right\rangle} \pm {\left\lvert {2,1,0} \right\rangle} ), {\left\lvert {2, 1, \pm 1} \right\rangle}\right\}\end{aligned} \hspace{\stretch{1}}(4.29)

Now expand the bra derivative kets

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}=\left({\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} +\sum_{\gamma} \frac{ {\left\langle {\psi_\gamma^{(0)}} \right\rvert} H' {\left\lvert {\psi_\alpha^{(0)}} \right\rangle} {\left\langle {\psi_\gamma^{(0)}} \right\rvert} }{E_\alpha^{(0)} -E_\gamma^{(0)} }\right)\frac{d{{}}}{dt}\left({\left\lvert {\psi_{\beta}^{(0)}} \right\rangle} +\sum_{\gamma'} \frac{ {\left\lvert {\psi_{\gamma'}^{(0)}} \right\rangle} {\left\langle {\psi_{\gamma'}^{(0)}} \right\rvert} H' {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\gamma'}^{(0)} }\right)\end{aligned} \hspace{\stretch{1}}(4.30)

To first order we can drop the quadratic terms in $\gamma,\gamma'$ leaving

\begin{aligned}\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}&\sim\sum_{\gamma'} \left\langle{{\psi_{\alpha}^{(0)}}} \vert {{\psi_{\gamma'}^{(0)}}}\right\rangle \frac{ {\left\langle {\psi_{\gamma'}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\gamma'}^{(0)} }&=\frac{ {\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\alpha}^{(0)} }\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.31)

so

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}\frac{ {\left\langle {\psi_{\alpha}^{(0)}} \right\rvert} \frac{d{{H'(t)}}}{dt} {\left\lvert {\psi_\beta^{(0)}} \right\rangle} }{E_\beta^{(0)} -E_{\alpha}^{(0)} }\end{aligned} \hspace{\stretch{1}}(4.32)

## A different way to this end result.

A result of this form is also derived in [1] section 20.1, but with a different approach. There he takes derivatives of

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} = E_\beta(t) {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.33)

\begin{aligned}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + H(t) \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} = \frac{d{{E_\beta(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}+ E_\beta(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(4.34)

Bra’ing ${\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}$ into this we have, for $\alpha \ne \beta$

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}H(t) \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} &= \not{{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{E_\beta(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle}}}+ {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}E_\beta(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} \\ {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} + E_\alpha(t) {\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} &=\end{aligned}

or

\begin{aligned}{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} =\frac{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} }{E_\beta(t) - E_\alpha(t)}\end{aligned} \hspace{\stretch{1}}(4.35)

so without the implied $\lambda$ perturbation of ${\left\lvert {\hat{\psi}_\alpha(t)} \right\rangle}$ we can from 4.36 write the exact generalization of 4.32 as

\begin{aligned}\frac{d{{}}}{dt}\bar{b}_\alpha(t)=-\sum_{\beta \ne \alpha} \bar{b}_\beta(t)e^{-\frac{i}{\hbar} \int_0^t (E_{\beta\alpha}(t') - \hbar \Gamma_{\beta\alpha}(t')) dt'}\frac{{\left\langle {\hat{\psi}_\alpha(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_\beta(t)} \right\rangle} }{E_\beta(t) - E_\alpha(t)}\end{aligned} \hspace{\stretch{1}}(4.36)

# References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.

## A different derivation of the adiabatic perturbation coefficient equation

Posted by peeterjoot on October 27, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

Professor Sipe’s adiabatic perturbation and that of the text [1] in section 17.5.1 and section 17.5.2 use different notation for $\gamma_m$ and take a slightly different approach. We can find Prof Sipe’s final result with a bit less work, if a hybrid of the two methods is used.

# Guts

Our starting point is the same, we have a time dependent slowly varying Hamiltonian

\begin{aligned}H = H(t),\end{aligned} \hspace{\stretch{1}}(2.1)

where our perturbation starts at some specific time from a given initial state

\begin{aligned}H(t) = H_0, \qquad t \le 0.\end{aligned} \hspace{\stretch{1}}(2.2)

We assume that instantaneous eigenkets can be found, satisfying

\begin{aligned}H(t) {\left\lvert {n(t)} \right\rangle} = E_n(t) {\left\lvert {n(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.3)

Here I’ll use ${\left\lvert {n} \right\rangle} \equiv {\left\lvert {n(t)} \right\rangle}$ instead of the ${\left\lvert {\hat{\psi}_n(t)} \right\rangle}$ that we used in class because its easier to write.

Now suppose that we have some arbitrary state, expressed in terms of the instantaneous basis kets ${\left\lvert {n} \right\rangle}$

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_n \bar{b}_n(t) e^{-i\alpha_n + i \gamma_n} {\left\lvert {n} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}\alpha_n(t) = \frac{1}{{\hbar}} \int_0^t dt' E_n(t'),\end{aligned} \hspace{\stretch{1}}(2.5)

and $\gamma_n$ (using the notation in the text, not in class) is to be determined.

For this state, we have at the time just before the perturbation

\begin{aligned}{\left\lvert {\psi(0)} \right\rangle} = \sum_n \bar{b}_n(0) e^{-i\alpha_n(0) + i \gamma_n(0)} {\left\lvert {n(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

The question to answer is: How does this particular state evolve?

Another question, for those that don’t like sneaky bastard derivations, is where did that magic factor of $e^{-i\alpha_n}$ come from in our superposition state? We will see after we start taking derivatives that this is what we need to cancel the $H(t){\left\lvert {n} \right\rangle}$ in Schr\”{o}dinger’s equation.

Proceeding to plug into the evolution identity we have

\begin{aligned}0 &={\left\langle {m} \right\rvert} \left( i \hbar \frac{d{{}}}{dt} - H(t) \right) {\left\lvert {\psi} \right\rangle} \\ &={\left\langle {m} \right\rvert} \left(\sum_n e^{-i \alpha_n + i \gamma_n}(i \hbar) \left(\frac{d{{\bar{b}_n}}}{dt}+ \bar{b}_n \left(-i \not{{\frac{E_n}{\hbar}}} + i \dot{\gamma}_m \right)\right) {\left\lvert {n} \right\rangle}+ i \hbar \bar{b}_n \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}- \not{{E_n \bar{b}_n {\left\lvert {n} \right\rangle}}} \right)\\ &=e^{-i \alpha_m + i \gamma_m}(i \hbar) \frac{d{{\bar{b}_m}}}{dt}+e^{-i \alpha_m + i \gamma_m}(i \hbar) i \dot{\gamma}_m \bar{b}_m+ i \hbar \sum_n \bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}e^{-i \alpha_n + i \gamma_n} \\ &\sim\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \sum_n e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} \\ &=\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle}+\sum_{n \ne m} e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}\end{aligned}

We are free to pick $\gamma_m$ to kill the second and third terms

\begin{aligned}0 =i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.7)

or

\begin{aligned}\dot{\gamma}_m = i {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.8)

which after integration is

\begin{aligned}\gamma_m(t)= i \int_0^t dt' {\left\langle {m(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {m(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.9)

as in class we can observe that this is a purely real function. We are left with

\begin{aligned}\frac{d{{\bar{b}_m}}}{dt}=-\sum_{n \ne m} \bar{b}_n e^{-i \alpha_{nm} + i \gamma_{nm}}{\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} ,\end{aligned} \hspace{\stretch{1}}(2.10)

where

\begin{aligned}\alpha_{nm} &= \alpha_{n} -\alpha_m \\ \gamma_{nm} &= \gamma_{n} -\gamma_m \end{aligned} \hspace{\stretch{1}}(2.11)

The task is now to find solutions for these $\bar{b}_m$ coefficients, and we can refer to the class notes for that without change.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## PHY456H1F: Quantum Mechanics II. Lecture 8 (Taught by Prof J.E. Sipe). Time dependent pertubation (cont.)

Posted by peeterjoot on October 8, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Time dependent pertubation.

We’d gotten as far as calculating

\begin{aligned}c_m^{(1)}(\infty) = \frac{1}{{i \hbar}} \boldsymbol{\mu}_{ms} \cdot \mathbf{E}(\omega_{ms})\end{aligned} \hspace{\stretch{1}}(2.1)

where

\begin{aligned}\mathbf{E}(t) = \int \frac{d\omega}{2 \pi} \mathbf{E}(\omega) e^{-i \omega t},\end{aligned} \hspace{\stretch{1}}(2.2)

and

\begin{aligned}\omega_{ms} = \frac{E_m - E_s}{\hbar}.\end{aligned} \hspace{\stretch{1}}(2.3)

Graphically, these frequencies are illustrated in figure (\ref{fig:qmTwoL8fig0FrequenciesAbsorbtionAndEmission})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL8fig0FrequenciesAbsorbtionAndEmission}
\caption{Positive and negative frequencies.}
\end{figure}

The probability for a transition from $m$ to $s$ is therefore

\begin{aligned}\rho_{m \rightarrow s} = {\left\lvert{ c_m^{(1)}(\infty) }\right\rvert}^2= \frac{1}{{\hbar}}^2 {\left\lvert{\boldsymbol{\mu}_{ms} \cdot \mathbf{E}(\omega_{ms})}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.4)

Recall that because the electric field is real we had

\begin{aligned}{\left\lvert{\mathbf{E}(\omega)}\right\rvert}^2 = {\left\lvert{\mathbf{E}(-\omega)}\right\rvert}^2.\end{aligned} \hspace{\stretch{1}}(2.5)

Suppose that we have a wave pulse, where our field magnitude is perhaps of the form

\begin{aligned}E(t) = e^{-t^2/T^2} \cos(\omega_0 t),\end{aligned} \hspace{\stretch{1}}(2.6)

as illustated with $\omega = 10, T = 1$ in figure (\ref{fig:gaussianWavePacket}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{gaussianWavePacket}
\caption{Gaussian wave packet}
\end{figure}

We expect this to have a two lobe Fourier spectrum, with the lobes centered at $\omega = \pm 10$, and width proportional to $1/T$.

For reference, as calculated using Mathematica this Fourier transform is

\begin{aligned}E(\omega) = \frac{e^{-\frac{1}{4} T^2 (\omega_0+\omega )^2}}{2 \sqrt{\frac{2}{T^2}}}+\frac{e^{\omega_0 T^2 \omega -\frac{1}{4} T^2 (\omega_0+\omega )^2}}{2 \sqrt{\frac{2}{T^2}}}\end{aligned} \hspace{\stretch{1}}(2.7)

This is illustrated, again for $\omega_0 = 10, and T=1$, in figure (\ref{fig:FTgaussianWavePacket})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{FTgaussianWavePacket}
\caption{FTgaussianWavePacket}
\end{figure}

where we see the expected Gaussian result, since the Fourier transform of a Gaussian is a Gaussian.

FIXME: not sure what the point of this was?

# Sudden pertubations.

Given our wave equation

\begin{aligned}i \hbar \frac{d{{}}}{dt} {\lvert {\psi(t)} \rangle} = H(t) {\lvert {\psi(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.8)

and a sudden pertubation in the Hamiltonian, as illustrated in figure (\ref{fig:suddenStepHamiltonian})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{suddenStepHamiltonian}
\caption{Sudden step Hamiltonian.}
\end{figure}

Consider $H_0$ and $H_F$ fixed, and decrease $\Delta t \rightarrow 0$. We can formally integrate 3.8

\begin{aligned}\frac{d{{}}}{dt} {\lvert {\psi(t)} \rangle} = \frac{1}{{i \hbar}} H(t) {\lvert {\psi(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.9)

For

\begin{aligned}{\lvert {\psi(t)} \rangle} -{\lvert {\psi(t_0)} \rangle} = \frac{1}{{i \hbar}} \int_{t_0}^t H(t') {\lvert {\psi(t')} \rangle} dt'.\end{aligned} \hspace{\stretch{1}}(3.10)

While this is an exact solution, it is also not terribly useful since we don’t know ${\lvert {\psi(t)} \rangle}$. However, we can select the small interval $\Delta t$, and write

\begin{aligned}{\lvert {\psi(\Delta t/2)} \rangle} ={\lvert {\psi(-\Delta t/2)} \rangle}+ \frac{1}{{i \hbar}} \int_{t_0}^t H(t') {\lvert {\psi(t')} \rangle} dt'.\end{aligned} \hspace{\stretch{1}}(3.11)

Note that we could use the integral kernel iteration technique here and substitute ${\lvert {\psi(t')} \rangle} = {\lvert {\psi(-\Delta t/2)} \rangle}$ and then develop this, to generate a power series with $(\Delta t/2)^k$ dependence. However, we note that 3.11 is still an exact relation, and if $\Delta t \rightarrow 0$, with the integration limits narrowing (provided $H(t')$ is well behaved) we are left with just

\begin{aligned}{\lvert {\psi(\Delta t/2)} \rangle} = {\lvert {\psi(-\Delta t/2)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.12)

Or

\begin{aligned}{\lvert {\psi_{\text{after}}} \rangle} = {\lvert {\psi_{\text{before}}} \rangle},\end{aligned} \hspace{\stretch{1}}(3.13)

provided that we change the Hamiltonian fast enough. On the surface there appears to be no consequences, but there are some very serious ones!

## Example: Harmonic oscillator.

Consider our harmonic oscillator Hamiltonian, with

\begin{aligned}H_0 &= \frac{P^2}{2m} + \frac{1}{{2}} m \omega_0^2 X^2 \\ H_F &= \frac{P^2}{2m} + \frac{1}{{2}} m \omega_F^2 X^2\end{aligned} \hspace{\stretch{1}}(3.14)

Here $\omega_0 \rightarrow \omega_F$ continuously, but very quickly. In effect, we have tightened the spring constant. Note that there are cases in linear optics when you can actually do exactly that.

Imagine that ${\lvert {\psi_{\text{before}}} \rangle}$ is in the ground state of the harmonic oscillator as in figure (\ref{fig:suddenHamiltonianPertubationHO})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{suddenHamiltonianPertubationHO}
\caption{Harmonic oscillator sudden Hamiltonian pertubation.}
\end{figure}

and we suddenly change the Hamilontian with potential $V_0 \rightarrow V_F$ (weakening the “spring”). Professor Sipe gives us a graphical demo of this, by impersonating a constrained wavefunction with his arms, doing weak chicken-flapping of them. Now with the potential weakended, he wiggles and flaps his arms with more freedom and somewhat chaotically. His “wave function” arms are now bouncing around in the new limiting potential (initally over doing it and then bouncing back).

We had in this case the exact relation

\begin{aligned}H_0 {\lvert {\psi_0^{(0)}} \rangle} = \frac{1}{{2}} \hbar \omega_0 {\lvert {\psi_0^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(3.16)

but we also have

\begin{aligned}{\lvert {\psi_{\text{after}}} \rangle} = {\lvert {\psi_{\text{before}}} \rangle} = {\lvert {\psi_0^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(3.17)

and

\begin{aligned}H_F {\lvert {\psi_n^{(f)}} \rangle} = \frac{1}{{2}} \hbar \omega_F \left( n + \frac{1}{{2}} \right) {\lvert {\psi_n^{(f)}} \rangle}\end{aligned} \hspace{\stretch{1}}(3.18)

So

\begin{aligned}{\lvert {\psi_{\text{after}}} \rangle}&={\lvert {\psi_0^{(0)}} \rangle} \\ &=\sum_n {\lvert {\psi_n^{(f)}} \rangle}\underbrace{\left\langle{{\psi_n^{(f)}}} \vert {{\psi_0^{(0)}}}\right\rangle }_{c_n} \\ &=\sum_n c_n {\lvert {\psi_n^{(f)}} \rangle}\end{aligned}

and at later times

\begin{aligned}{\lvert {\psi(t)^{(f)}} \rangle}&={\lvert {\psi_0^{(0)}} \rangle} \\ &=\sum_n c_n e^{i \omega_n^{(f)} t} {\lvert {\psi_n^{(f)}} \rangle},\end{aligned}

whereas

\begin{aligned}{\lvert {\psi(t)^{(o)}} \rangle}&=e^{i \omega_0^{(0)} t} {\lvert {\psi_0^{(0)}} \rangle},\end{aligned}

So, while the wave functions may be exactly the same after such a sudden change in Hamiltonian, the dynamics of the situation change for all future times, since we now have a wavefunction that has a different set of components in the basis for the new Hamiltonian. In particular, the evolution of the wave function is now significantly more complex.

FIXME: plot an example of this.

FIXME: what does Adiabatic mean in this context. The usage in class sounds like it was just “really slow and gradual”, yet this has a definition “Of, relating to, or being a reversible thermodynamic process that occurs without gain or loss of heat and without a change in entropy”.

This is treated in section 17.5.2 of the text [1].

This is the reverse case, and we now vary the Hamiltonian $H(t)$ very slowly.

\begin{aligned}\frac{d{{}}}{dt} {\lvert {\psi(t)} \rangle} = \frac{1}{{i \hbar}} H(t) {\lvert {\psi(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(4.19)

We first consider only non-degenerate states, and at $t = 0$ write

\begin{aligned}H(0) = H_0,\end{aligned} \hspace{\stretch{1}}(4.20)

and

\begin{aligned}H_0 {\lvert {\psi_s^{(0)}} \rangle} = E_s^{(0)} {\lvert {\psi_s^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(4.21)

Imagine that at each time $t$ we can find the “instantaneous” energy eigenstates

\begin{aligned}H(t) {\lvert {\hat{\psi}_s(t)} \rangle} = E_s(t) {\lvert {\hat{\psi}_s(t)} \rangle} \end{aligned} \hspace{\stretch{1}}(4.22)

These states do not satisfy Schr\”{o}dinger’s equation, but are simply solutions to the eigen problem. Our standard strategy in pertubation is based on analysis of

\begin{aligned}{\lvert {\psi(t)} \rangle} = \sum_n c_n(t) e^{- i \omega_n^{(0)} t} {\lvert {\psi_n^{(0)} } \rangle},\end{aligned} \hspace{\stretch{1}}(4.23)

\begin{aligned}{\lvert {\psi(t)} \rangle} = \sum_n b_n(t) {\lvert {\hat{\psi}_n(t)} \rangle},\end{aligned} \hspace{\stretch{1}}(4.24)

we will expand, not using our initial basis, but instead using the instananeous kets. Plugging into Schr\”{o}dinger’s equation we have

\begin{aligned}H(t) {\lvert {\psi(t)} \rangle} &= H(t) \sum_n b_n(t) {\lvert {\hat{\psi}_n(t)} \rangle} \\ &= \sum_n b_n(t) E_n(t) {\lvert {\hat{\psi}_n(t)} \rangle} \end{aligned}

This was complicated before with matrix elements all over the place. Now it is easy, however, the time derivative becomes harder. Doing that we find

\begin{aligned}i \hbar \frac{d{{}}}{dt} {\lvert {\psi(t)} \rangle}&=i \hbar\frac{d{{}}}{dt} \sum_n b_n(t) {\lvert {\hat{\psi}_n(t)} \rangle} \\ &=i \hbar\sum_n \frac{d{{b_n(t)}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} + \sum_n b_n(t) \frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} \\ &= \sum_n b_n(t) E_n(t) {\lvert {\hat{\psi}_n(t)} \rangle} \end{aligned}

We bra ${\langle {\hat{\psi}_m(t)} \rvert}$ into this

\begin{aligned}i \hbar\sum_n \frac{d{{b_n(t)}}}{dt} \left\langle{{\hat{\psi}_m(t)}} \vert {{\hat{\psi}_n(t)}}\right\rangle+ \sum_n b_n(t) {\langle {\hat{\psi}_m(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} = \sum_n b_n(t) E_n(t) \left\langle{{\hat{\psi}_m(t)}} \vert {{\hat{\psi}_n(t)}}\right\rangle ,\end{aligned} \hspace{\stretch{1}}(4.25)

and find

\begin{aligned}i \hbar\frac{d{{b_m(t)}}}{dt} + \sum_n b_n(t) {\langle {\hat{\psi}_m(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} = b_m(t) E_m(t) \end{aligned} \hspace{\stretch{1}}(4.26)

If the Hamiltonian is changed very very slowly in time, we can imagine that ${\lvert {\hat{\psi}_n(t)} \rangle}'$ is also changing very very slowly, but we are not quite there yet. Let’s first split our sum of bra and ket products

\begin{aligned}\sum_n b_n(t) {\langle {\hat{\psi}_m(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} \end{aligned} \hspace{\stretch{1}}(4.27)

into $n \ne m$ and $n = m$ terms. Looking at just the $n = m$ term

\begin{aligned}{\langle {\hat{\psi}_m(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} \end{aligned} \hspace{\stretch{1}}(4.28)

we note

\begin{aligned}0 &=\frac{d{{}}}{dt} \left\langle{{\hat{\psi}_m(t)}} \vert {{\hat{\psi}_n(t)}}\right\rangle \\ &=\left( \frac{d{{}}}{dt} {\langle {\hat{\psi}_m(t)} \rvert} \right) {\lvert {\hat{\psi}_m(t)} \rangle} \\ + {\langle {\hat{\psi}_m(t)} \rvert} \frac{d{{}}}{dt} {\lvert {\hat{\psi}_m(t)} \rangle} \\ \end{aligned}

Something plus its complex conjugate equals 0

\begin{aligned}a + i b + (a + i b)^{*} = 2 a = 0 \implies a = 0,\end{aligned} \hspace{\stretch{1}}(4.29)

so ${\langle {\hat{\psi}_m(t)} \rvert} \frac{d{{}}}{dt} {\lvert {\hat{\psi}_m(t)} \rangle}$ must be purely imaginary. We write

\begin{aligned}{\langle {\hat{\psi}_m(t)} \rvert} \frac{d{{}}}{dt} {\lvert {\hat{\psi}_m(t)} \rangle} = -i \Gamma_s(t),\end{aligned} \hspace{\stretch{1}}(4.30)

where $\Gamma_s$ is real.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.