# Peeter Joot's (OLD) Blog.

• ## Archives

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 320,607

## 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.