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# Posts Tagged ‘harmonic oscillator’

## A final pre-exam update of my notes compilation for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 22, 2013

Here’s my third update of my notes compilation for this course, including all of the following:

April 21, 2013 Fermi function expansion for thermodynamic quantities

April 20, 2013 Relativistic Fermi Gas

April 10, 2013 Non integral binomial coefficient

April 10, 2013 energy distribution around mean energy

April 09, 2013 Velocity volume element to momentum volume element

April 04, 2013 Phonon modes

April 03, 2013 BEC and phonons

April 03, 2013 Max entropy, fugacity, and Fermi gas

April 02, 2013 Bosons

April 02, 2013 Relativisitic density of states

March 28, 2013 Bosons

plus everything detailed in the description of my previous update and before.

## PHY452H1S Basic Statistical Mechanics. Lecture 21: Phonon modes. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on April 4, 2013

[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.

These are notes for the last class, which included a lot of discussion not captured by this short set of notes (as well as slides which were not transcribed).

# Phonon modes

If we model a solid as a set of interconnected springs, as in fig. 1.1, then the potentials are of the form

Fig 1.1: Solid oscillator model

\begin{aligned}V = \frac{1}{{2}} C \sum_n \left( {u_n - u_{n+1}} \right)^2,\end{aligned} \hspace{\stretch{1}}(1.2.1)

with kinetic energies

\begin{aligned}K = \sum_n \frac{p_n^2}{2m}.\end{aligned} \hspace{\stretch{1}}(1.2.2)

It’s possible to introduce generalized forces

\begin{aligned}F = -\frac{\partial {V}}{\partial {u_n}}\end{aligned} \hspace{\stretch{1}}(1.2.3)

Can differentiate

\begin{aligned}m \frac{d^2 u_n}{dt^2} = - C \left( { u_n - u_{n+1}} \right)- C \left( { u_n - u_{n-1}} \right)\end{aligned} \hspace{\stretch{1}}(1.2.4)

Assuming a Fourier representation

\begin{aligned}u_n = \sum_k \tilde{u}(k) e^{i k n a},\end{aligned} \hspace{\stretch{1}}(1.2.5)

we find

\begin{aligned}m \frac{d^2 \tilde{u}(k)}{dt^2} = - 2 C \left( { 1 - \cos k a} \right)\tilde{u}(k)\end{aligned} \hspace{\stretch{1}}(1.2.6)

This looks like a harmonic oscillator with

\begin{aligned}\omega(k) = \sqrt{ \frac{2 C}{m} ( 1 - \cos k a)}.\end{aligned} \hspace{\stretch{1}}(1.2.7)

This is plotted in fig. 1.2. In particular note that for for $k a \ll 1$ we can use a linear approximation

Fig 1.2: Angular frequency of solid oscillator model

\begin{aligned}\omega(k) \approx \sqrt{ \frac{C}{m} a^2 } \left\lvert {k} \right\rvert.\end{aligned} \hspace{\stretch{1}}(1.2.8)

Experimentally, looking at specific for a complex atomic structure like Gold, we find for example good fit for a model such as

\begin{aligned}C \sim \underbrace{A T}_{\text{Contribution due to electrons.}}+ \underbrace{B T^3}_{\text{Contribution due to phonon like modes where there are linear energy momenum relations.}}.\end{aligned} \hspace{\stretch{1}}(1.2.9)

## An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

## Some problems from Kittel chapter 3

Posted by peeterjoot on February 10, 2013

[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.)]

## Question: Classical gas partition function

[1] expresses the classical gas partition function (3.77) as

\begin{aligned}Z_1 \propto \int \exp\left( - \frac{p_x^2 + p_y^2 + p_z^2 }{2 M \tau}\right) dp_x dp_y dp_z\end{aligned} \hspace{\stretch{1}}(1.0.1)

Show that this leads to the expected $3 \tau/2$ result for the thermal average energy.

Let’s use the adjustment technique from the text for the $N$ partition case and write

\begin{aligned}Z_N = \frac{1}{{N!}} Z_1^N,\end{aligned} \hspace{\stretch{1}}(1.0.2)

with $Z_1$ as above. This gives us

\begin{aligned}U &= \tau^2 \frac{\partial {}}{\partial {\tau}} \ln Z_N \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}} \left(N \ln Z_1 - \ln N!\right) \\ &= N \tau^2 \frac{\partial {\ln Z_1 }}{\partial {\tau}} \\ &= N \tau^2 \frac{\partial {}}{\partial {\tau}} \sum_{k = 1}^{3} \ln\int \exp\left( - \frac{p_k^2 }{2 M \tau}\right) dp_k \\ &= N \tau^2\sum_{k = 1}^{3}\frac{\frac{\partial {}}{\partial {\tau}} \int \exp\left( - \frac{p_k^2 }{2 M \tau} \right) dp_k}{\int \exp\left( - \frac{p_k^2 }{2 M \tau} \right) dp_k} \\ &= N \tau^2\sum_{k = 1}^{3}\frac{\frac{\partial {}}{\partial {\tau}} \sqrt{ 2 \pi M \tau }}{\sqrt{ 2 \pi M \tau}} \\ &= 3 N \tau^2\frac{\frac{1}{{2}} \tau^{-1/2}}{\sqrt{ \tau}} \\ &= \frac{3}{2} N \tau \\ &= \frac{3}{2} N k_{\mathrm{B}} T\end{aligned} \hspace{\stretch{1}}(1.0.3)

## Question: Two state system

[1] problem 3.1.

Find an expression for the free energy as a function of $\tau$ of a system with two states, one at energy $0$ and one at energy $\epsilon$. From the free energy, find expressions for the energy and entropy of the system.

Our partition function is

\begin{aligned}Z = 1 + e^{-\epsilon /\tau}\end{aligned} \hspace{\stretch{1}}(1.0.4)

The free energy is just

\begin{aligned}F = -\tau \ln Z = -\tau \ln (1 + e^{-\epsilon/\tau})\end{aligned} \hspace{\stretch{1}}(1.0.5)

The entropy follows immediately

\begin{aligned}\sigma \\ &= -\frac{\partial {F}}{\partial {\tau}} \\ &= \frac{\partial {}}{\partial {\tau}}\left( \tau \ln \left( 1 + e^{-\epsilon/\tau} \right) \right) \\ &= \ln \left( 1 + e^{-\epsilon/\tau} \right)-\tau \epsilon \frac{-1}{\tau^2} \frac{1}{{1 + e^{-\epsilon/\tau}}} \\ &= \ln \left( 1 + e^{-\epsilon/\tau} \right)+\frac{\epsilon}{\tau} \frac{e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}}\end{aligned} \hspace{\stretch{1}}(1.0.6)

The energy is

\begin{aligned}U \\ &= F + \tau \sigma \\ &= -\tau \ln (1 + e^{-\epsilon/\tau}) + \tau \sigma \\ &= \tau\left( \not{{\ln \left( 1 + e^{-\epsilon/\tau} \right)}} + \frac{\epsilon}{\tau} \frac{e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}} -\not{{\ln (1 + e^{-\epsilon/\tau}) }} \right)\end{aligned} \hspace{\stretch{1}}(1.0.7)

This is

\begin{aligned}U=\frac{\epsilon e^{-\epsilon/\tau}}{1 + e^{-\epsilon/\tau}}=\frac{\epsilon}{1 + e^{\epsilon/\tau}}.\end{aligned} \hspace{\stretch{1}}(1.0.8)

These are all plotted in (Fig 1).

Fig1: Plots for two state system

\imageFigure{kittelCh3Problem1PlotsFig1}{Plots for two state system}{fig:kittelCh3Problem1Plots:kittelCh3Problem1PlotsFig1}{0.2}

## Question: Magnetic susceptibility

[1] problem 3.2.

Use the partition function to find an exact expression for the magnetization $M$ and the susceptibility $\chi = dM/dB$ as a function of temperature and magnetic field for the model system of magnetic moments in a magnetic field. The result for the magnetization, found by other means, was $M = n m \tanh( m B/\tau)$, where $n$ is the particle concentration. Find the free energy and express the result as a function only of $\tau$ and the parameter $x = M/nm$. Show that the susceptibility is $\chi = n m^2/\tau$ in the limit $m B \ll \tau$.

Our partition function for a unit volume containing $n$ spins is

\begin{aligned}Z=\frac{\left( e^{-m B/\tau} +e^{m B/\tau} \right)^n}{n!}=2 \frac{\left( \cosh\left( m B/\tau \right) \right)^n}{n!},\end{aligned} \hspace{\stretch{1}}(1.0.9)

so that the Free energy is

\begin{aligned}F = -\tau\left( \ln 2 - \ln n! + n \ln \cosh\left( m B/\tau \right) \right).\end{aligned} \hspace{\stretch{1}}(1.0.15)

The energy, magnetization and magnetic field were interrelated by

\begin{aligned}- M B &= U \\ &= \tau^2 \frac{\partial {}}{\partial {\tau}}\left( -\frac{F}{\tau} \right) \\ &= \tau^2 n\frac{\partial {}}{\partial {\tau}}\ln \cosh\left( m B/\tau \right) \\ &= \tau^2 n \frac{ -m B/\tau^2\sinh\left( m B/\tau \right)}{\cosh\left( m B/\tau \right)} \\ &= - m B n \tanh \left( m B/\tau \right).\end{aligned} \hspace{\stretch{1}}(1.0.11)

This gives us

\begin{aligned}M = m n \tanh \left( m B/\tau \right),\end{aligned} \hspace{\stretch{1}}(1.0.12)

so that

\begin{aligned}\chi = \frac{dM}{dB}= \frac{m^2 n}{\tau \cosh^2 \left( m B/\tau \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.13)

For $m B/\tau \ll 1$, the cosh term goes to unity, so we have

\begin{aligned}\chi \approx= \frac{m^2 n}{\tau},\end{aligned} \hspace{\stretch{1}}(1.0.14)

as desired.

With $x = M/nm$, or $m = M/nx$, the free energy is

\begin{aligned}F =-\tau\left( \ln 2/n! + n \ln \cosh\left( \frac{M B}{n x \tau} \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.15)

That last expression isn’t particularly illuminating. What was the point of that substitution?

## Question: Free energy of a harmonic oscillator

[1] problem 3.3.

A one dimensional harmonic oscillator has an infinite series of equally spaced energy states, with $\epsilon_s = s \hbar \omega$, where $s$ is a positive integer or zero, and $\omega$ is the classical frequency of the oscillator. We have chosen the zero of energy at the state $s = 0$. Show that for a harmonic oscillator the free energy is

\begin{aligned}F = \tau \ln\left( 1 - e^{-\hbar \omega/\tau} \right).\end{aligned} \hspace{\stretch{1}}(1.0.16)

Note that at high temperatures such that $\tau \gg \hbar \omega$ we may expand the argument of the logarithm to obtain $F \approx \tau \ln (\hbar \omega/\tau)$. From 1.0.16 show that the entropy is

\begin{aligned}\sigma = \frac{\hbar\omega/\tau}{e^{\hbar \omega/\tau} - 1} -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)\end{aligned} \hspace{\stretch{1}}(1.0.17)

I found it curious that this problem dropped the factor of $\hbar\omega/2$ from the energy. Including it we have

\begin{aligned}\epsilon_s = \left( s + \frac{1}{{2}} \right) \hbar \omega,\end{aligned} \hspace{\stretch{1}}(1.0.18)

So that the partition function is

\begin{aligned}Z= \sum_{s = 0}^\infty e^{-\left( s + \frac{1}{{2}} \right) \hbar \omega/\tau }=e^{-\hbar \omega/2\tau}\sum_{s = 0}^\infty e^{-s \hbar \omega/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.19)

The free energy is

\begin{aligned}F &= -\tau \ln Z \\ &= -\tau\left( -\frac{\hbar \omega}{2\tau} + \ln \left( \sum_{s = 0}^\infty e^{-s \hbar \omega/\tau} \right) \right) \\ &= \frac{\hbar \omega}{2} +\ln\left( \sum_{s = 0}^\infty e^{-s \hbar \omega/\tau} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

We see that the contribution of the $\hbar \omega/2$ in the energy of each state just adds a constant factor to the free energy. This will drop out when we compute the entropy. Dropping that factor now that we know why it doesn’t contribute, we can complete the summation, so have, by inspection

\begin{aligned}F = -\tau \ln Z=\tau \ln\left( 1 - e^{-\hbar \omega/\tau} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

Taking derivatives for the entropy we have

\begin{aligned}\sigma &= -\frac{\partial {F}}{\partial {\tau}} \\ &= -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)+\tau\frac{\hbar \omega}{\tau^2} \frac{e^{-\hbar \omega/\tau}}{1 - e^{-\hbar \omega/\tau}} \\ &= -\ln\left( 1 - e^{-\hbar \omega/\tau} \right)+\frac{\frac{\hbar \omega}{\tau}}{e^{\hbar \omega/\tau} - 1}\end{aligned} \hspace{\stretch{1}}(1.0.22)

## Question: Energy fluctuation

[1] problem 3.4.

Consider a system of fixed volume in thermal contact with a reservoir. Show that the mean square fluctuation in the energy of the system is

\begin{aligned}\left\langle{{ (\epsilon - \left\langle{\epsilon}\right\rangle)^2 }}\right\rangle = \tau^2\left( \frac{\partial {U}}{\partial {\tau}} \right)_V\end{aligned} \hspace{\stretch{1}}(1.0.23)

Here $U$ is the conventional symbol for $\left\langle{{\epsilon}}\right\rangle$. Hint: Use the partition function $Z$ to relate ${\partial {U}}/{\partial {t}}$ to the mean square fluctuation. Also, multiply out the term $(\cdots)^2$.

With a probability of finding the system in state $s$ of

\begin{aligned}P_s = \frac{e^{-\epsilon_s/\tau}}{Z}\end{aligned} \hspace{\stretch{1}}(1.0.24)

the average energy is

\begin{aligned}U &= \left\langle{{\epsilon}}\right\rangle \\ &= \sum_s P_s \epsilon_s \\ &= \sum_s \epsilon_s \frac{e^{-\epsilon_s/\tau}}{Z} \\ &= \frac{1}{{Z}} \sum_s \epsilon_s e^{-\epsilon_s/\tau}\end{aligned} \hspace{\stretch{1}}(1.0.25)

So we have

\begin{aligned}\tau^2 \frac{\partial {U}}{\partial {\tau}} \\ &= -\frac{\tau^2}{Z^2} \frac{dZ}{d\tau}\sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{\tau^2}{Z}\sum_s \frac{\epsilon_s^2}{\tau^2} e^{-\epsilon_s/\tau} \\ &= -\frac{\tau^2}{Z^2} \frac{dZ}{d\tau}\sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{1}{{Z}}\sum_s \epsilon_s^2 e^{-\epsilon_s/\tau}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

But

\begin{aligned}\frac{dZ}{d\tau}=\frac{d}{d\tau} \sum_s e^{-\epsilon_s/\tau}=\sum_s \frac{\epsilon_s}{\tau^2} e^{-\epsilon_s/\tau},\end{aligned} \hspace{\stretch{1}}(1.0.27)

giving

\begin{aligned}\tau^2 \frac{\partial {U}}{\partial {\tau}} &= \frac{1}{Z^2}\sum_s \epsilon_s e^{-\epsilon_s/\tau} \sum_s \epsilon_s e^{-\epsilon_s/\tau}+ \frac{1}{{Z}}\sum_s \epsilon_s^2 e^{-\epsilon_s/\tau} \\ &= -\left\langle{{ \epsilon }}\right\rangle^2 + \left\langle{{\epsilon^2}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(1.0.28)

which shows 1.0.23 as desired.

# References

[1] C. Kittel and H. Kroemer. Thermal physics. WH Freeman, 1980.

## 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.

## Curious problem using the variational method to find the ground state energy of the Harmonic oscillator (cont.)

Posted by peeterjoot on October 7, 2011

Picking up where this problem was last abandoned.

[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.)]

# Calculating the Fourier terms

Our wave function, with $\beta=1$ is plotted in figure (\ref{fig:expMinusBetsAbsX})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{expMinusBetsAbsX}
\caption{Exponential trial function with absolute exponential die off.}
\end{figure}

The zeroth order fitting using the gaussian exponential is found to be

\begin{aligned}\psi_0(x) = \sqrt{2 \beta} \text{erfc}\left(\frac{\beta }{\sqrt{2} \alpha }\right)e^{- \alpha^2 x^2/2 +\beta^2/(2 \alpha^2)} \end{aligned} \hspace{\stretch{1}}(6.16)

With $\alpha = \beta = 1$, this is plotted in figure (\ref{fig:expMinusBetsAbsXfirstOrderFitting}) and can be seen to match fairly well

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{expMinusBetsAbsXfirstOrderFitting}
\caption{First ten orders, fitting harmonic oscillator wavefunctions to this trial function.}
\end{figure}

The higher order terms get small fast, but we can see in figure (\ref{fig:expMinusBetsAbsXtenthOrderFitting}), where a tenth order fitting is depicted that it would take a number of them to get anything close to the sharp peak that we have in our exponential trial function.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{expMinusBetsAbsXtenthOrderFitting}
\caption{Tenth order harmonic oscillator wavefunction fitting.}
\end{figure}

Note that the all the brakets of even orders in $n$ with the trial function are zero, which is why the tenth order approximation is only a sum of six terms.

Details for this \href{https://github.com/peeterjoot/physicsplay/blob/master/notes/phy456/gaussian\

The question of interest is why if we can approximate the trial function so nicely (except at the origin) even with just a first order approximation (polynomial times gaussian functions where the polynomials are Hankel functions), and we can get an exact value for the lowest energy state using the first order approximation of our trial function, why do we get garbage if we include enough terms that the peak is sharp? It must therefore be important to consider the origin, but how do we give some meaning to the derivative of the absolute value function? The key, supplied when asking in office hours for the course, is to express the absolute value function in terms of Heavyside step functions, for which the derivative can be identified as the delta function.

# Correcting, treating the origin this way.

Here’s how we can express the absolute value function using the Heavyside step

\begin{aligned}{\left\lvert{x}\right\rvert} = x \theta(x) - x \theta(-x),\end{aligned} \hspace{\stretch{1}}(7.17)

where the step function is zero for $x 0$ as plotted in figure (\ref{fig:stepFunction}).

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

Expressed this way, with the identification $\theta'(x) = \delta(x)$, we have for the derivative of the absolute value function

\begin{aligned}{\left\lvert{x}\right\rvert}' &= x' \theta(x) - x' \theta(-x) + x \theta'(x) - x \theta'(-x) \\ &= \theta(x) - \theta(-x) + x \delta(x) + x \delta(-x) \\ &= \theta(x) - \theta(-x) + x \delta(x) + x \delta(x) \\ \end{aligned}

Observe that we have our expected unit derivative for $x > 0$, and $-1$ derivative for $x < 0$. At the origin our $\theta$ contributions vanish, and we are left with

\begin{aligned}{\left.{{{\left\lvert{x}\right\rvert}' }}\right\vert}_{{x=0}}= 2 {\left.{{x \delta(x)}}\right\vert}_{{x=0}} \\ \end{aligned}

We’ve got zero times infinity here, so how do we give meaning to this? As with any delta functional, we’ve got to apply it to a well behaved (square integrable) test function $f(x)$ and integrate. Doing so we have

\begin{aligned}\int_{-\infty}^\infty dx {\left\lvert{x}\right\rvert}' f(x) &= 2 \int_{-\infty}^\infty dx x \delta(x) f(x) \\ &= 2 (0) f(0)\end{aligned}

This equals zero for any well behaved test function $f(x)$. Since the delta function only picks up the contribution at the origin, we can therefore identify ${\left\lvert{x}\right\rvert}'$ as zero at the origin.

Using the same technique, we can express our trial function in terms of steps

\begin{aligned}\psi = e^{-\beta {\left\lvert{x}\right\rvert}} = \theta(x) e^{-\beta x} + \theta(-x) e^{\beta x}.\end{aligned} \hspace{\stretch{1}}(7.18)

This we can now take derivatives of, even at the origin, and find

\begin{aligned}\psi' &= \theta'(x) e^{-\beta x} + \theta'(-x) e^{\beta x} -\beta \theta(x) e^{-\beta x} + \beta \theta(-x) e^{\beta x} \\ &= \delta(x) e^{-\beta x} - \delta(-x) e^{\beta x} -\beta \theta(x) e^{-\beta x} + \beta \theta(-x) e^{\beta x} \\ &= \not{{\delta(x) e^{-\beta x} - \delta(x) e^{\beta x}}} -\beta \theta(x) e^{-\beta x} + \beta \theta(-x) e^{\beta x} \\ &= \beta \left( -\theta(x) e^{-\beta x} + \theta(-x) e^{\beta x} \right)\end{aligned}

Taking second derivatives we find

\begin{aligned}\psi''&= \beta \left( -\theta'(x) e^{-\beta x} + \theta'(-x) e^{\beta x} +\beta \theta(x) e^{-\beta x} + \beta \theta(-x) e^{\beta x} \right) \\ &= \beta \left( -\delta(x) e^{-\beta x} - \delta(-x) e^{\beta x} +\beta \theta(x) e^{-\beta x} + \beta \theta(-x) e^{\beta x} \right) \\ &= \beta^2 \psi - 2 \beta \delta(x)\end{aligned}

Now application of the Hamiltonian operator on our trial function gives us

\begin{aligned}H \psi = -\frac{\hbar^2}{2m} \left( \beta^2 \psi - 2 \delta(x) \right) + \frac{1}{{2}} m \omega^2 x^2 \psi,\end{aligned} \hspace{\stretch{1}}(7.19)

so

\begin{aligned}{\langle {\psi} \rvert} H {\lvert {\psi} \rangle} &= \int_{-\infty}^\infty \left( -\frac{\hbar^2 \beta^2}{2m} + \frac{1}{{2}} m \omega^2 x^2 \right) e^{-2 \beta {\left\lvert{x}\right\rvert} }+ \frac{\hbar^2 \beta}{m}\int_{-\infty}^\infty \delta(x)e^{- \beta {\left\lvert{x}\right\rvert}} \\ &=-\frac{\beta \hbar^2}{2m} + \frac{m \omega^2}{4 \beta^3} + \frac{\hbar^2 \beta}{m} \\ &=\frac{\beta \hbar^2}{2m} + \frac{m \omega^2}{4 \beta^3}.\end{aligned}

Normalized we have

\begin{aligned}E[\beta] = \frac{{\langle {\psi} \rvert} H {\lvert {\psi} \rangle}}{\left\langle{{\psi}} \vert {{\psi}}\right\rangle} = \frac{\beta^2 \hbar^2}{2m} + \frac{m \omega^2}{4 \beta^2}.\end{aligned} \hspace{\stretch{1}}(7.20)

This is looking much more promising. We’ll have the sign alternation that we require to find a positive, non-complex, value for $\beta$ when $E[\beta]$ is minimized. That is

\begin{aligned}0 = \frac{\partial {E}}{\partial {\beta}} = \frac{\beta \hbar^2}{m} - \frac{m \omega^2}{2 \beta^3},\end{aligned} \hspace{\stretch{1}}(7.21)

so the extremum is found at

\begin{aligned}\beta^4 = \frac{m^2 \omega^2}{2 \hbar^2}.\end{aligned} \hspace{\stretch{1}}(7.22)

Plugging this back in we find that our trial function associated with the minimum energy (unnormalized still) is

\begin{aligned}\psi = e^{-\sqrt{\frac{m \omega x^2}{\sqrt{2} \hbar}}},\end{aligned} \hspace{\stretch{1}}(7.23)

and that energy, after substitution, is

\begin{aligned}E[\beta_{\text{min}}] = \frac{\hbar \omega}{2} \sqrt{2}\end{aligned} \hspace{\stretch{1}}(7.24)

We have something that’s $1.4 \times$ the true ground state energy, but is at least a ball park value. However, to get this result, we have to be very careful to treat our point of singularity. A derivative that we’d call undefined in first year calculus, is not only defined, but required, for this treatment to work!

## My submission for PHY356 (Quantum Mechanics I) Problem Set 5.

Posted by peeterjoot on December 7, 2010

# Problem.

## Statement

A particle of mass m moves along the x-direction such that $V(X)=\frac{1}{{2}}KX^2$. Is the state

\begin{aligned}u(\xi) = B \xi e^{+\xi^2/2},\end{aligned} \hspace{\stretch{1}}(2.1)

where $\xi$ is given by Eq. (9.60), $B$ is a constant, and time $t=0$, an energy eigenstate of the system? What is probability per unit length for measuring the particle at position $x=0$ at $t=t_0>0$? Explain the physical meaning of the above results.

## Solution

### Is this state an energy eigenstate?

Recall that $\xi = \alpha x$, $\alpha = \sqrt{m\omega/\hbar}$, and $K = m \omega^2$. With this variable substitution Schr\”{o}dinger’s equation for this harmonic oscillator potential takes the form

\begin{aligned}\frac{d^2 u}{d\xi^2} - \xi^2 u = \frac{2 E }{\hbar\omega} u\end{aligned} \hspace{\stretch{1}}(2.2)

While we can blindly substitute a function of the form $\xi e^{\xi^2/2}$ into this to get

\begin{aligned}\frac{1}{{B}} \left(\frac{d^2 u}{d\xi^2} - \xi^2 u\right)&=\frac{d}{d\xi} \left( 1 + \xi^2 \right) e^{\xi^2/2} - \xi^3 e^{\xi^2/2} \\ &=\left( 2 \xi + \xi + \xi^3 \right) e^{\xi^2/2} - \xi^3 e^{\xi^2/2} \\ &=3 \xi e^{\xi^2/2}\end{aligned}

and formally make the identification $E = 3 \omega \hbar/2 = (1 + 1/2) \omega \hbar$, this isn’t a normalizable wavefunction, and has no physical relevance, unless we set $B = 0$.

By changing the problem, this state could be physically relevant. We’d require a potential of the form

\begin{aligned}V(x) =\left\{\begin{array}{l l}f(x) & \quad \mbox{iflatex x < a} \\ \frac{1}{{2}} K x^2 & \quad \mbox{if $a < x b$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.3)

For example, $f(x) = V_1, g(x) = V_2$, for constant $V_1, V_2$. For such a potential, within the harmonic well, a general solution of the form

\begin{aligned}u(x,t) = \sum_n H_n(\xi) \Bigl(A_n e^{-\xi^2/2} + B_n e^{\xi^2/2} \Bigr) e^{-i E_n t/\hbar},\end{aligned} \hspace{\stretch{1}}(2.4)

is possible since normalization would not prohibit non-zero $B_n$ values in that situation. For the wave function to be a physically relevant, we require it to be (absolute) square integrable, and must also integrate to unity over the entire interval.

### Probability per unit length at $x=0$.

We cannot answer the question for the probability that the particle is found at the specific $x=0$ position at $t=t_0$ (that probability is zero in a continuous space), but we can answer the question for the probability that a particle is found in an interval surrounding a specific point at this time. By calculating the average of the probability to find the particle in an interval, and dividing by that interval’s length, we arrive at plausible definition of probability per unit length for an interval surrounding $x = x_0$

\begin{aligned}P = \text{Probability per unit length nearlatex x = x_0} =\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 – \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.5)

By this definition, the probability per unit length is just the probability density itself, evaluated at the point of interest.

Physically, for an interval small enough that the probability density is constant in magnitude over that interval, this probability per unit length times the length of this small interval, represents the probability that we will find the particle in that interval.

### Probability per unit length for the non-normalizable state given.

It seems possible, albeit odd, that this question is asking for the probability per unit length for the non-normalizable $E_1$ wavefunction 2.1. Since normalization requires $B=0$, that probability density is simply zero (or undefined, depending on one’s point of view).

### Probability per unit length for some more interesting harmonic oscillator states.

Suppose we form the wavefunction for a superposition of all the normalizable states

\begin{aligned}u(x,t) = \sum_n A_n H_n(\xi) e^{-\xi^2/2} e^{-i E_n t/\hbar}\end{aligned} \hspace{\stretch{1}}(2.6)

Here it is assumed that the $A_n$ coefficients yield unit probability

\begin{aligned}\int {\left\lvert{u(x,0)}\right\rvert}^2 dx = \sum_n {\left\lvert{A_n}\right\rvert}^2 = 1\end{aligned} \hspace{\stretch{1}}(2.7)

For the impure state of 2.6 we have for the probability density

\begin{aligned}{\left\lvert{u}\right\rvert}^2&=\sum_{m,n}A_n A_m^{*} H_n(\xi) H_m(\xi) e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar} \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+\sum_{m \ne n}A_n A_m^{*} H_n(\xi) H_m(\xi) e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar} \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+\sum_{m \ne n}A_n A_m^{*} H_n(\xi) H_m(\xi) e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar} \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+\sum_{m < n}H_n(\xi) H_m(\xi)\left(A_n A_m^{*}e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar}+A_m A_n^{*}e^{-\xi^2} e^{-i (E_m - E_n)t_0/\hbar}\right) \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+2 \sum_{m < n}H_n(\xi) H_m(\xi)e^{-\xi^2}\Re \left(A_n A_m^{*}e^{-i (E_n - E_m)t_0/\hbar}\right) \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2} \\ &\quad+2 \sum_{m < n}H_n(\xi) H_m(\xi)e^{-\xi^2}\left(\Re ( A_n A_m^{*} ) \cos( (n - m)\omega t_0)+\Im ( A_n A_m^{*} ) \sin( (n - m)\omega t_0)\right) \\ \end{aligned}

Evaluation at the point $x = 0$, we have

\begin{aligned}{\left\lvert{u(0,t_0)}\right\rvert}^2=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(0))^2 +2 \sum_{m < n} H_n(0) H_m(0) \left( \Re ( A_n A_m^{*} ) \cos( (n - m)\omega t_0) +\Im ( A_n A_m^{*} ) \sin( (n - m)\omega t_0)\right)\end{aligned} \hspace{\stretch{1}}(2.8)

It is interesting that the probability per unit length only has time dependence for a mixed state.

For a pure state and its wavefunction $u(x,t) = N_n H_n(\xi) e^{-\xi^2/2} e^{-i E_n t/\hbar}$ we have just

\begin{aligned}{\left\lvert{u(0,t_0)}\right\rvert}^2=N_n^2 (H_n(0))^2 = \frac{\alpha}{\sqrt{\pi} 2^n n!} H_n(0)^2\end{aligned} \hspace{\stretch{1}}(2.9)

This is zero for odd $n$. For even $n$ is appears that $(H_n(0))^2$ may equal $2^n$ (this is true at least up to n=4). If that’s the case, we have for non-mixed states, with even numbered energy quantum numbers, at $x=0$ a probability per unit length value of ${\left\lvert{u(0,t_0)}\right\rvert}^2 = \frac{\alpha}{\sqrt{\pi} n!}$.

## 1D forced harmonic oscillator. Quick solution of non-homogeneous problem.

Posted by peeterjoot on February 19, 2010

[Click here for a PDF of this post with nicer formatting]Note that this PDF file is formatted in a wide-for-screen layout that is probably not good for printing.

# Motivation.

In [1] equation (25) we have a forced harmonic oscillator equation

\begin{aligned}m \ddot{x} + m \omega^2 x = \gamma(t).\end{aligned} \hspace{\stretch{1}}(1.1)

The solution of this equation is provided, but for fun lets derive it.

# Guts

Writing

\begin{aligned}\omega u = \dot{x},\end{aligned} \hspace{\stretch{1}}(2.2)

we can rewrite the second order equation as a first order linear system

\begin{aligned}\dot{u} + \omega x &= \gamma(t)/m \omega \\ \dot{x} - \omega u &= 0,\end{aligned} \hspace{\stretch{1}}(2.3)

Or, with $X = (u, x)$, in matrix form

\begin{aligned}\dot{X} + \omega \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}X&=\begin{bmatrix}\gamma(t)/m \omega \\ 0\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.5)

The two by two matrix has the same properties as the complex imaginary, squaring to the identity matrix, so the equation to solve is now of the form

\begin{aligned}\dot{X} + \omega i X &= \Gamma.\end{aligned} \hspace{\stretch{1}}(2.6)

The homogeneous part of the solution is just the matrix

\begin{aligned}X &= e^{-i \omega t} A \\ &= \left( \cos(\omega t) \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}- \sin(\omega t)\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}\right) A,\end{aligned}

where $A$ is a two by one column matrix of constants. Assuming for the specific solution $X = e^{-i \omega t} A(t)$, and substuiting we have

\begin{aligned}e^{-i \omega t} \dot{A} = \Gamma(t).\end{aligned} \hspace{\stretch{1}}(2.7)

This integrates directly, fixing the unknown column vector function $A(t)$

\begin{aligned}A(t) = A(0) + \int_0^t e^{i \omega \tau} \Gamma(\tau).\end{aligned} \hspace{\stretch{1}}(2.8)

Thus the non-homogeneous solution takes the form

\begin{aligned}X = e^{-i \omega t} A(0) + \int_0^t e^{i \omega (\tau - t)} \Gamma(\tau).\end{aligned} \hspace{\stretch{1}}(2.9)

Note that $A(0) = (\dot{x}_0/\omega, x_0)$. Multiplying this out, and discarding all but the second row of the matrix product gives $x(t)$, and Feynman’s equation (26) follows directly.

# References

[1] L.M. Brown, G.D. Carson, L.F. Locke, W.W. Spirduso, S.J. Silverman, D. Holtom, E. Fisher, J.E. Mauch, J.W. Birch, K.L. Turabian, et al. Feynman’s thesis: A New approach to quantum theory. Houghton Mifflin, 1954.

## Electrodynamic field energy for vacuum (reworked)

Posted by peeterjoot on December 21, 2009

# Previous version.

Reducing the products in the Dirac basis makes life more complicated then it needs to be (became obvious when attempting to derive an expression for the Poynting integral).

# Motivation.

From Energy and momentum for Complex electric and magnetic field phasors [PDF] how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism is now understood. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$, where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation

\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)

This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$, as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$, is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor $T(a)$ was found to be

\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)

and in particular for $a = \gamma^0$, this four vector divergence takes the form

\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)

relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [2]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

# Setup

Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$. That is

\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)

where summation is over all angular wave number triplets $\mathbf{k} = 2 \pi (k_1/\lambda_1, k_2/\lambda_2, k_3/\lambda_3)$. The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$, and the associated bivector field $F$.

Fourier inversion, with $V = \lambda_1 \lambda_2 \lambda_3$, follows from

\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{ i \mathbf{k}' \cdot \mathbf{x}} e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)

but only this orthogonality relationship and not the Fourier coefficients themselves

\begin{aligned}A_\mathbf{k} = \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{- i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)

will be of interest here. Evaluating the curl for this potential yields

\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)

Since the four vector potential has been expressed using an explicit split into time and space components it will be natural to re express the bivector field in terms of scalar and (spatial) vector potentials, with the Fourier coefficients. Writing $\sigma_m = \gamma_m \gamma_0$ for the spatial basis vectors, ${A_\mathbf{k}}^0 = \phi_\mathbf{k}$, and $\mathbf{A} = A^k \sigma_k$, this is

\begin{aligned}A_\mathbf{k} = (\phi_\mathbf{k} + \mathbf{A}_\mathbf{k}) \gamma_0.\end{aligned} \quad\quad\quad(9)

The Faraday bivector field $F$ is then

\begin{aligned}F = \sum_\mathbf{k} \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(10)

This is now enough to express the energy momentum tensor $T(\gamma^\mu)$

\begin{aligned}T(\gamma^\mu) &= -\frac{\epsilon_0}{2} \sum_{\mathbf{k},\mathbf{k}'}\text{Real} \left(\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}'})}}^{*} + i \mathbf{k}' {{\phi_{\mathbf{k}'}}}^{*} - i \mathbf{k}' \wedge {{\mathbf{A}_{\mathbf{k}'}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) e^{i (\mathbf{k} -\mathbf{k}') \cdot \mathbf{x}}\right).\end{aligned} \quad\quad\quad(11)

It will be more convenient to work with a scalar plus bivector (spatial vector) form of this tensor, and right multiplication by $\gamma_0$ produces such a split

\begin{aligned}T(\gamma^\mu) \gamma_0 = \left\langle{{T(\gamma^\mu) \gamma_0}}\right\rangle + \sigma_a \left\langle{{ \sigma_a T(\gamma^\mu) \gamma_0 }}\right\rangle\end{aligned} \quad\quad\quad(12)

The primary object of this treatment will be consideration of the $\mu = 0$ components of the tensor, which provide a split into energy density $T(\gamma^0) \cdot \gamma_0$, and Poynting vector (momentum density) $T(\gamma^0) \wedge \gamma_0$.

Our first step is to integrate (12) over the volume $V$. This integration and the orthogonality relationship (6), removes the exponentials, leaving

\begin{aligned}\int T(\gamma^\mu) \cdot \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0 }}\right\rangle \\ \int T(\gamma^\mu) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \gamma^\mu \left( -\frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} - i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) \gamma_0}}\right\rangle \end{aligned} \quad\quad\quad(13)

Because $\gamma_0$ commutes with the spatial bivectors, and anticommutes with the spatial vectors, the remainder of the Dirac basis vectors in these expressions can be eliminated

\begin{aligned}\int T(\gamma^0) \cdot \gamma_0&= -\frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(15)

\begin{aligned}\int T(\gamma^0) \wedge \gamma_0&= -\frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(16)

\begin{aligned}\int T(\gamma^m) \cdot \gamma_0&= \frac{\epsilon_0 V }{2} \sum_{\mathbf{k}}\text{Real} \left\langle{{\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \end{aligned} \quad\quad\quad(17)

\begin{aligned}\int T(\gamma^m) \wedge \gamma_0&= \frac{\epsilon_0 V}{2} \sum_{\mathbf{k}}\text{Real} \sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \sigma_m\left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle.\end{aligned} \quad\quad\quad(18)

# Expanding the energy momentum tensor components.

## Energy

In (15) only the bivector-bivector and vector-vector products produce any scalar grades. Except for the bivector product this can be done by inspection. For that part we utilize the identity

\begin{aligned}\left\langle{{ (\mathbf{k} \wedge \mathbf{a}) (\mathbf{k} \wedge \mathbf{b}) }}\right\rangle= (\mathbf{a} \cdot \mathbf{k}) (\mathbf{b} \cdot \mathbf{k}) - \mathbf{k}^2 (\mathbf{a} \cdot \mathbf{b}).\end{aligned} \quad\quad\quad(19)

This leaves for the energy $H = \int T(\gamma^0) \cdot \gamma_0$ in the volume

\begin{aligned}H = \frac{\epsilon_0 V}{2} \sum_\mathbf{k} \left(\frac{1}{{c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2 +\mathbf{k}^2 \left( {\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right) - {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ \frac{2}{c} \text{Real} \left( i {{\phi_\mathbf{k}}}^{*} \cdot \dot{\mathbf{A}}_\mathbf{k} \right)\right)\end{aligned} \quad\quad\quad(20)

We are left with a completely real expression, and one without any explicit Geometric Algebra. This does not look like the Harmonic oscillator Hamiltonian that was expected. A gauge transformation to eliminate $\phi_\mathbf{k}$ and an observation about when $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ equals zero will give us that, but first lets get the mechanical jobs done, and reduce the products for the field momentum.

## Momentum

Now move on to (16). For the factors other than $\sigma_a$ only the vector-bivector products can contribute to the scalar product. We have two such products, one of the form

\begin{aligned}\sigma_a \left\langle{{ \sigma_a \mathbf{a} (\mathbf{k} \wedge \mathbf{c}) }}\right\rangle&=\sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) - \sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) \\ &=\mathbf{c} (\mathbf{a} \cdot \mathbf{k}) - \mathbf{k} (\mathbf{a} \cdot \mathbf{c}),\end{aligned}

and the other

\begin{aligned}\sigma_a \left\langle{{ \sigma_a (\mathbf{k} \wedge \mathbf{c}) \mathbf{a} }}\right\rangle&=\sigma_a (\mathbf{k} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{c}) - \sigma_a (\mathbf{c} \cdot \sigma_a) (\mathbf{a} \cdot \mathbf{k}) \\ &=\mathbf{k} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{k}).\end{aligned}

The momentum $\mathbf{P} = \int T(\gamma^0) \wedge \gamma_0$ in this volume follows by computation of

\begin{aligned}&\sigma_a \left\langle{{ \sigma_a\left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} - i \mathbf{k} \wedge {{\mathbf{A}_{\mathbf{k}}}}^{*} \right) \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} + i \mathbf{k} \wedge \mathbf{A}_\mathbf{k} \right) }}\right\rangle \\ &= i \mathbf{A}_\mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{k} \right) - i \mathbf{k} \left( \left( -\frac{1}{{c}} {{(\dot{\mathbf{A}}_{\mathbf{k}})}}^{*} + i \mathbf{k} {{\phi_{\mathbf{k}}}}^{*} \right) \cdot \mathbf{A}_\mathbf{k} \right) \\ &- i \mathbf{k} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right) + i {{\mathbf{A}_{\mathbf{k}}}}^{*} \left( \left( \frac{1}{{c}} \dot{\mathbf{A}}_\mathbf{k} + i \mathbf{k} \phi_\mathbf{k} \right) \cdot \mathbf{k} \right)\end{aligned}

All the products are paired in nice conjugates, taking real parts, and premultiplication with $-\epsilon_0 V/2$ gives the desired result. Observe that two of these terms cancel, and another two have no real part. Those last are

\begin{aligned}-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i {{(\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k}+\dot{\mathbf{A}}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)&=-\frac{\epsilon_0 V \mathbf{k}}{2 c} \text{Real} \left( i \frac{d}{dt} \mathbf{A}_\mathbf{k} \cdot {{\mathbf{A}_\mathbf{k}}}^{*} \right)\end{aligned}

Taking the real part of this pure imaginary $i {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$ is zero, leaving just

\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)+ \mathbf{k}^2 \phi_\mathbf{k} {{ \mathbf{A}_\mathbf{k} }}^{*}- \mathbf{k} {{\phi_\mathbf{k}}}^{*} (\mathbf{k} \cdot \mathbf{A}_\mathbf{k})\right)\end{aligned} \quad\quad\quad(21)

I am not sure why exactly, but I actually expected a term with ${\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2$, quadratic in the vector potential. Is there a mistake above?

## Gauge transformation to simplify the Hamiltonian.

In (20) something that looked like the Harmonic oscillator was expected. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential $A \rightarrow A + \nabla \psi$, then Maxwell’s equation takes the form

\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(22)

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(23)

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$. That is

\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(24)

Or,

\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(25)

With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (20) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just

\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2+ {\left\lvert{ c \mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(26)

Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just

\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ V }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(27)

The desired harmonic oscillator form would be had in (26) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$, the lack of an $A^0$ component means that this can be inverted as

\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(29)

The gradient can also be factored scalar and spatial vector components

\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(30)

So, with this $A^0 = 0$ gauge choice the bivector field $F$ is

\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(31)

From the left the gradient action on $A$ is

\begin{aligned}\stackrel{ \rightarrow }{\nabla} A &= ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0 (-\gamma_0 \mathbf{A}) \\ &= ( -\partial_0 + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} ) \mathbf{A},\end{aligned}

and from the right

\begin{aligned}A \stackrel{ \leftarrow }{\nabla}&= \mathbf{A} \gamma_0 \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \mathbf{A} ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \partial_0 \mathbf{A} + \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \end{aligned}

Taking the difference we have

\begin{aligned}F &= \frac{1}{{2}} \Bigl( -\partial_0 \mathbf{A} + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \partial_0 \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \Bigr).\end{aligned}

Which is just

\begin{aligned}F = -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}.\end{aligned} \quad\quad\quad(32)

For this vacuum case, premultiplication of Maxwell’s equation by $\gamma_0$ gives

\begin{aligned}0 &= \gamma_0 \nabla ( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= (\partial_0 + \boldsymbol{\nabla})( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \partial_0 \boldsymbol{\nabla} \cdot \mathbf{A} - \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A} + \partial_0 ( \boldsymbol{\nabla} \wedge \mathbf{A} ) + \underbrace{\boldsymbol{\nabla} \cdot ( \boldsymbol{\nabla} \wedge \mathbf{A} ) }_{\boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A})}+ \underbrace{\boldsymbol{\nabla} \wedge ( \boldsymbol{\nabla} \wedge \mathbf{A} )}_{=0} \\ \end{aligned}

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in $\mathbf{A}$

\begin{aligned}0 &= \partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) \\ 0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) \end{aligned} \quad\quad\quad(33)

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} &=\sum_{\mathbf{k} \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_{\mathbf{k} \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \\ &=i \sum_\mathbf{k} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{i \mathbf{k} \cdot \mathbf{x}} \end{aligned}

Since $\mathbf{A}_\mathbf{k} = \mathbf{A}_\mathbf{k}(t)$, there are two ways for $\partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) = 0$. For each $\mathbf{k}$ there must be a requirement for either $\mathbf{A}_\mathbf{k} \cdot \mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k} = \text{constant}$. The constant $\mathbf{A}_\mathbf{k}$ solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$, the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

\begin{aligned}0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} \end{aligned} \quad\quad\quad(35)

Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes

\begin{aligned}H = \frac{\epsilon_0}{c^2} V \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} (c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(36)

How does the gauge choice alter the Poynting vector? From (21), all the $\phi_\mathbf{k}$ dependence in that integrated momentum density is lost

\begin{aligned}\mathbf{P} &= \epsilon_0 V \sum_{\mathbf{k}}\text{Real} \left(i \mathbf{A}_\mathbf{k} \left( \frac{1}{{c}} {{\dot{\mathbf{A}}_\mathbf{k}}}^{*} \cdot \mathbf{k} \right)\right).\end{aligned} \quad\quad\quad(37)

The $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ solutions to Maxwell’s equation are seen to result in zero momentum for this infinite periodic field. My expectation was something of the form $c \mathbf{P} = H \hat{\mathbf{k}}$, so intuition is either failing me, or my math is failing me, or this contrived periodic field solution leads to trouble.

# Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.

This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential $\phi$ upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.

As a bit of followup, I’d like to consider the constant $\mathbf{A}_\mathbf{k}$ case in more detail, and any implications of the freedom to pick $\mathbf{A}_0$.

The general calculation of $T^{\mu\nu}$ for the assumed Fourier solution should be possible too, but was not attempted. Doing that general calculation with a four dimensional Fourier series is likely tidier than working with scalar and spatial variables as done here.

Now that the math is out of the way (except possibly for the momentum which doesn’t seem right), some discussion of implications and applications is also in order. My preference is to let the math sink-in a bit first and mull over the momentum issues at leisure.

# References

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

## Electrodynamic field energy for vacuum.

Posted by peeterjoot on December 19, 2009

# Motivation.

We now know how to formulate the energy momentum tensor for complex vector fields (ie. phasors) in the Geometric Algebra formalism. To recap, for the field $F = \mathbf{E} + I c \mathbf{B}$, where $\mathbf{E}$ and $\mathbf{B}$ may be complex vectors we have for Maxwell’s equation

\begin{aligned}\nabla F = J/\epsilon_0 c.\end{aligned} \quad\quad\quad(1)

This is a doubly complex representation, with the four vector pseudoscalar $I = \gamma_0 \gamma_1 \gamma_2 \gamma_3$ acting as a non-commutatitive imaginary, as well as real and imaginary parts for the electric and magnetic field vectors. We take the real part (not the scalar part) of any bivector solution $F$ of Maxwell’s equation as the actual solution, but allow ourself the freedom to work with the complex phasor representation when convenient. In these phasor vectors, the imaginary $i$, as in $\mathbf{E} = \text{Real}(\mathbf{E}) + i \text{Imag}(\mathbf{E})$, is a commuting imaginary, commuting with all the multivector elements in the algebra.

The real valued, four vector, energy momentum tensor $T(a)$ was found to be

\begin{aligned}T(a) = \frac{\epsilon_0}{4} \Bigl( {{F}}^{*} a \tilde{F} + \tilde{F} a {{F}}^{*} \Bigr) = -\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} a F \Bigr).\end{aligned} \quad\quad\quad(2)

To supply some context that gives meaning to this tensor the associated conservation relationship was found to be

\begin{aligned}\nabla \cdot T(a) &= a \cdot \frac{1}{{ c }} \text{Real} \left( J \cdot {{F}}^{*} \right).\end{aligned} \quad\quad\quad(3)

and in particular for $a = \gamma^0$, this four vector divergence takes the form

\begin{aligned}\frac{\partial {}}{\partial {t}}\frac{\epsilon_0}{2}(\mathbf{E} \cdot {\mathbf{E}}^{*} + c^2 \mathbf{B} \cdot {\mathbf{B}}^{*})+ \boldsymbol{\nabla} \cdot \frac{1}{{\mu_0}} \text{Real} (\mathbf{E} \times {\mathbf{B}}^{*} )+ \text{Real}( \mathbf{J} \cdot {\mathbf{E}}^{*} ) = 0,\end{aligned} \quad\quad\quad(4)

relating the energy term $T^{00} = T(\gamma^0) \cdot \gamma^0$ and the Poynting spatial vector $T(\gamma^0) \wedge \gamma^0$ with the current density and electric field product that constitutes the energy portion of the Lorentz force density.

Let’s apply this to calculating the energy associated with the field that is periodic within a rectangular prism as done by Bohm in [1]. We do not necessarily need the Geometric Algebra formalism for this calculation, but this will be a fun way to attempt it.

# Setup

Let’s assume a Fourier representation for the four vector potential $A$ for the field $F = \nabla \wedge A$. That is

\begin{aligned}A = \sum_{\mathbf{k}} A_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}},\end{aligned} \quad\quad\quad(5)

where summation is over all wave number triplets $\mathbf{k} = (p/\lambda_1,q/\lambda_2,r/\lambda_3)$. The Fourier coefficients $A_\mathbf{k} = {A_\mathbf{k}}^\mu \gamma_\mu$ are allowed to be complex valued, as is the resulting four vector $A$, and the associated bivector field $F$.

Fourier inversion follows from

\begin{aligned}\delta_{\mathbf{k}', \mathbf{k}} =\frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} e^{2 \pi i \mathbf{k}' \cdot \mathbf{x}} e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(6)

but only this orthogonality relationship and not the Fourier coefficients themselves

\begin{aligned}A_\mathbf{k} = \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} A(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3,\end{aligned} \quad\quad\quad(7)

will be of interest here. Evaluating the curl for this potential yields

\begin{aligned}F = \nabla \wedge A= \sum_{\mathbf{k}} \left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \sum_{m=1}^3 \gamma^m \wedge A_\mathbf{k} \frac{2 \pi i k_m}{\lambda_m} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}.\end{aligned} \quad\quad\quad(8)

We can now form the energy density

\begin{aligned}U = T(\gamma^0) \cdot \gamma^0=-\frac{\epsilon_0}{2} \text{Real} \Bigl( {{F}}^{*} \gamma^0 F \gamma^0 \Bigr).\end{aligned} \quad\quad\quad(9)

With implied summation over all repeated integer indexes (even without matching uppers and lowers), this is

\begin{aligned}U =-\frac{\epsilon_0}{2} \sum_{\mathbf{k}', \mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}'}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}'}}}^{*} \frac{2 \pi i k_m'}{\lambda_m} \right) e^{-2 \pi i \mathbf{k}' \cdot \mathbf{x}}\gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}}\gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(10)

The grade selection used here doesn’t change the result since we already have a scalar, but will just make it convenient to filter out any higher order products that will cancel anyways. Integrating over the volume element and taking advantage of the orthogonality relationship (6), the exponentials are removed, leaving the energy contained in the volume

\begin{aligned}H = -\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2}\sum_{\mathbf{k}} \text{Real} \left\langle{{\left( \frac{1}{{c}} \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} - \gamma^m \wedge {{A_{\mathbf{k}}}}^{*} \frac{2 \pi i k_m}{\lambda_m} \right) \gamma^0\left( \frac{1}{{c}} \gamma^0 \wedge \dot{A}_\mathbf{k} + \gamma^n \wedge A_\mathbf{k} \frac{2 \pi i k_n}{\lambda_n} \right) \gamma^0}}\right\rangle.\end{aligned} \quad\quad\quad(11)

# First reduction of the Hamiltonian.

Let’s take the products involved in sequence one at a time, and evaluate, later adding and taking real parts if required all of

\begin{aligned}\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 (\gamma^0 \wedge \dot{A}_\mathbf{k}) \gamma^0 }}\right\rangle &=-\frac{1}{{c^2}}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k}) }}\right\rangle \end{aligned} \quad\quad\quad(12)

\begin{aligned}- \frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) \gamma^0}}\right\rangle &=\frac{2 \pi i k_m}{c \lambda_m} \left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) ( \gamma^0 \wedge \dot{A}_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(13)

\begin{aligned}\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) \gamma^0 ( \gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle &=-\frac{2 \pi i k_n}{c \lambda_n} \left\langle{{ ( \gamma^0 \wedge {{\dot{A}_{\mathbf{k}}}}^{*} ) ( \gamma^n \wedge A_\mathbf{k} ) }}\right\rangle \end{aligned} \quad\quad\quad(14)

\begin{aligned}-\frac{4 \pi^2 k_m k_n}{\lambda_m \lambda_n}\left\langle{{ (\gamma^m \wedge {{A_{\mathbf{k}}}}^{*} ) \gamma^0(\gamma^n \wedge A_\mathbf{k} ) \gamma^0}}\right\rangle. &\end{aligned} \quad\quad\quad(15)

The expectation is to obtain a Hamiltonian for the field that has the structure of harmonic oscillators, where the middle two products would have to be zero or sum to zero or have real parts that sum to zero. The first is expected to contain only products of ${\left\lvert{{\dot{A}_\mathbf{k}}^m}\right\rvert}^2$, and the last only products of ${\left\lvert{{A_\mathbf{k}}^m}\right\rvert}^2$.

While initially guessing that (13) and (14) may cancel, this isn’t so obviously the case. The use of cyclic permutation of multivectors within the scalar grade selection operator $\left\langle{{A B}}\right\rangle = \left\langle{{B A}}\right\rangle$ plus a change of dummy summation indexes in one of the two shows that this sum is of the form $Z + {{Z}}^{*}$. This sum is intrinsically real, so we can neglect one of the two doubling the other, but we will still be required to show that the real part of either is zero.

Lets reduce these one at a time starting with (12), and write $\dot{A}_\mathbf{k} = \kappa$ temporarily

\begin{aligned}\left\langle{{ (\gamma^0 \wedge {{\kappa}}^{*} ) (\gamma^0 \wedge \kappa }}\right\rangle &={\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma^0 \gamma_m \gamma^0 \gamma_{m'} }}\right\rangle \\ &=-{\kappa^m}^{{*}} \kappa^{m'}\left\langle{{ \gamma_m \gamma_{m'} }}\right\rangle \\ &={\kappa^m}^{{*}} \kappa^{m'}\delta_{m m'}.\end{aligned}

So the first of our Hamiltonian terms is

\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}\left\langle{{ (\gamma^0 \wedge {{\dot{A}_\mathbf{k}}}^{*} ) (\gamma^0 \wedge \dot{A}_\mathbf{k} }}\right\rangle &=\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2}{\left\lvert{{{\dot{A}}_{\mathbf{k}}}^m}\right\rvert}^2.\end{aligned} \quad\quad\quad(16)

Note that summation over $m$ is still implied here, so we’d be better off with a spatial vector representation of the Fourier coefficients $\mathbf{A}_\mathbf{k} = A_\mathbf{k} \wedge \gamma_0$. With such a notation, this contribution to the Hamiltonian is

\begin{aligned}\frac{\epsilon_0 \lambda_1 \lambda_2 \lambda_3}{2 c^2} \dot{\mathbf{A}}_\mathbf{k} \cdot {{\dot{\mathbf{A}}_\mathbf{k}}}^{*}.\end{aligned} \quad\quad\quad(17)

To reduce (13) and (13), this time writing $\kappa = A_\mathbf{k}$, we can start with just the scalar selection

\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) ( \gamma^0 \wedge \dot{\kappa} ) }}\right\rangle &=\Bigl( \gamma^m {{(\kappa^0)}}^{*} - {{\kappa}}^{*} \underbrace{(\gamma^m \cdot \gamma^0)}_{=0} \Bigr) \cdot \dot{\kappa} \\ &={{(\kappa^0)}}^{*} \dot{\kappa}^m\end{aligned}

Thus the contribution to the Hamiltonian from (13) and (13) is

\begin{aligned}\frac{2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \pi k_m}{c \lambda_m} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \dot{A_\mathbf{k}}^m \Bigl)=\frac{2 \pi \epsilon_0 \lambda_1 \lambda_2 \lambda_3}{c} \text{Real} \Bigl( i {{(A_\mathbf{k}^0)}}^{*} \mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k} \Bigl).\end{aligned} \quad\quad\quad(18)

Most definitively not zero in general. Our final expansion (15) is the messiest. Again with $A_\mathbf{k} = \kappa$ for short, the grade selection of this term in coordinates is

\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- {{\kappa_\mu}}^{*} \kappa^\nu \left\langle{{ (\gamma^m \wedge \gamma^\mu) \gamma^0 (\gamma_n \wedge \gamma_\nu) \gamma^0 }}\right\rangle\end{aligned} \quad\quad\quad(19)

Expanding this out yields

\begin{aligned}\left\langle{{ (\gamma^m \wedge {{\kappa}}^{*} ) \gamma^0 (\gamma^n \wedge \kappa ) \gamma^0 }}\right\rangle&=- ( {\left\lvert{\kappa_0}\right\rvert}^2 - {\left\lvert{A^a}\right\rvert}^2 ) \delta_{m n} + {{A^n}}^{*} A^m.\end{aligned} \quad\quad\quad(20)

The contribution to the Hamiltonian from this, with $\phi_\mathbf{k} = A^0_\mathbf{k}$, is then

\begin{aligned}2 \pi^2 \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \Bigl(-\mathbf{k}^2 {{\phi_\mathbf{k}}}^{*} \phi_\mathbf{k} + \mathbf{k}^2 ({{\mathbf{A}_\mathbf{k}}}^{*} \cdot \mathbf{A}_\mathbf{k})+ (\mathbf{k} \cdot {{\mathbf{A}_k}}^{*}) (\mathbf{k} \cdot \mathbf{A}_k)\Bigr).\end{aligned} \quad\quad\quad(21)

A final reassembly of the Hamiltonian from the parts (17) and (18) and (21) is then

\begin{aligned}H = \epsilon_0 \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2 c^2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{2 \pi}{c} \text{Real} \Bigl( i {{ \phi_\mathbf{k} }}^{*} (\mathbf{k} \cdot \dot{\mathbf{A}}_\mathbf{k}) \Bigl)+2 \pi^2 \Bigl(\mathbf{k}^2 ( -{\left\lvert{\phi_\mathbf{k}}\right\rvert}^2 + {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 ) + {\left\lvert{\mathbf{k} \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(22)

This is finally reduced to a completely real expression, and one without any explicit Geometric Algebra. All the four vector Fourier vector potentials written out explicitly in terms of the spacetime split $A_\mathbf{k} = (\phi_\mathbf{k}, \mathbf{A}_\mathbf{k})$, which is natural since an explicit time and space split was the starting point.

# Gauge transformation to simplify the Hamiltonian.

While (22) has considerably simpler form than (11), what was expected, was something that looked like the Harmonic oscillator. On the surface this does not appear to be such a beast. Exploitation of gauge freedom is required to make the simplification that puts things into the Harmonic oscillator form.

If we are to change our four vector potential $A \rightarrow A + \nabla \psi$, then Maxwell’s equation takes the form

\begin{aligned}J/\epsilon_0 c = \nabla (\nabla \wedge (A + \nabla \psi) = \nabla (\nabla \wedge A) + \nabla (\underbrace{\nabla \wedge \nabla \psi}_{=0}),\end{aligned} \quad\quad\quad(23)

which is unchanged by the addition of the gradient to any original potential solution to the equation. In coordinates this is a transformation of the form

\begin{aligned}A^\mu \rightarrow A^\mu + \partial_\mu \psi,\end{aligned} \quad\quad\quad(24)

and we can use this to force any one of the potential coordinates to zero. For this problem, it appears that it is desirable to seek a $\psi$ such that $A^0 + \partial_0 \psi = 0$. That is

\begin{aligned}\sum_\mathbf{k} \phi_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} + \frac{1}{{c}} \partial_t \psi = 0.\end{aligned} \quad\quad\quad(25)

Or,

\begin{aligned}\psi(\mathbf{x},t) = \psi(\mathbf{x},0) -\frac{1}{{c}} \sum_\mathbf{k} e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \int_{\tau=0}^t \phi_\mathbf{k}(\tau).\end{aligned} \quad\quad\quad(26)

With such a transformation, the $\phi_\mathbf{k}$ and $\dot{\mathbf{A}}_\mathbf{k}$ cross term in the Hamiltonian (22) vanishes, as does the $\phi_\mathbf{k}$ term in the four vector square of the last term, leaving just

\begin{aligned}H = \frac{\epsilon_0}{c^2} \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} \Bigl((2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 + {\left\lvert{ ( 2 \pi c \mathbf{k}) \cdot \mathbf{A}_\mathbf{k}}\right\rvert}^2\Bigr)\right).\end{aligned} \quad\quad\quad(27)

Additionally, wedging (5) with $\gamma_0$ now does not loose any information so our potential Fourier series is reduced to just

\begin{aligned}\mathbf{A} &= \sum_{\mathbf{k}} \mathbf{A}_\mathbf{k}(t) e^{2 \pi i \mathbf{k} \cdot \mathbf{x}} \\ \mathbf{A}_\mathbf{k} &= \frac{1}{{ \lambda_1 \lambda_2 \lambda_3 }}\int_0^{\lambda_1}\int_0^{\lambda_2}\int_0^{\lambda_3} \mathbf{A}(\mathbf{x}, t) e^{-2 \pi i \mathbf{k} \cdot \mathbf{x}} dx^1 dx^2 dx^3.\end{aligned} \quad\quad\quad(28)

The desired harmonic oscillator form would be had in (27) if it were not for the $\mathbf{k} \cdot \mathbf{A}_\mathbf{k}$ term. Does that vanish? Returning to Maxwell’s equation should answer that question, but first it has to be expressed in terms of the vector potential. While $\mathbf{A} = A \wedge \gamma_0$, the lack of an $A^0$ component means that this can be inverted as

\begin{aligned}A = \mathbf{A} \gamma_0 = -\gamma_0 \mathbf{A}.\end{aligned} \quad\quad\quad(30)

The gradient can also be factored scalar and spatial vector components

\begin{aligned}\nabla = \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) = ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0.\end{aligned} \quad\quad\quad(31)

So, with this $A^0 = 0$ gauge choice the bivector field $F$ is

\begin{aligned}F = \nabla \wedge A = \frac{1}{{2}} \left( \stackrel{ \rightarrow }{\nabla} A - A \stackrel{ \leftarrow }{\nabla} \right) \end{aligned} \quad\quad\quad(32)

From the left the gradient action on $A$ is

\begin{aligned}\stackrel{ \rightarrow }{\nabla} A &= ( \partial_0 - \boldsymbol{\nabla} ) \gamma^0 (-\gamma_0 \mathbf{A}) \\ &= ( -\partial_0 + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} ) \mathbf{A},\end{aligned}

and from the right

\begin{aligned}A \stackrel{ \leftarrow }{\nabla}&= \mathbf{A} \gamma_0 \gamma^0 ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \mathbf{A} ( \partial_0 + \boldsymbol{\nabla} ) \\ &= \partial_0 \mathbf{A} + \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \end{aligned}

Taking the difference we have

\begin{aligned}F &= \frac{1}{{2}} \Bigl( -\partial_0 \mathbf{A} + \stackrel{ \rightarrow }{\boldsymbol{\nabla}} \mathbf{A} - \partial_0 \mathbf{A} - \mathbf{A} \stackrel{ \leftarrow }{\boldsymbol{\nabla}} \Bigr).\end{aligned}

Which is just

\begin{aligned}F = -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A}.\end{aligned} \quad\quad\quad(33)

For this vacuum case, premultiplication of Maxwell’s equation by $\gamma_0$ gives

\begin{aligned}0 &= \gamma_0 \nabla ( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= (\partial_0 + \boldsymbol{\nabla})( -\partial_0 \mathbf{A} + \boldsymbol{\nabla} \wedge \mathbf{A} ) \\ &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} - \partial_0 \boldsymbol{\nabla} \cdot \mathbf{A} - \partial_0 \boldsymbol{\nabla} \wedge \mathbf{A} + \partial_0 ( \boldsymbol{\nabla} \wedge \mathbf{A} ) + \underbrace{\boldsymbol{\nabla} \cdot ( \boldsymbol{\nabla} \wedge \mathbf{A} ) }_{\boldsymbol{\nabla}^2 \mathbf{A} - \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A})}+ \underbrace{\boldsymbol{\nabla} \wedge ( \boldsymbol{\nabla} \wedge \mathbf{A} )}_{=0} \\ \end{aligned}

The spatial bivector and trivector grades are all zero. Equating the remaining scalar and vector components to zero separately yields a pair of equations in $\mathbf{A}$

\begin{aligned}0 &= \partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) \\ 0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} + \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) \end{aligned} \quad\quad\quad(34)

If the divergence of the vector potential is constant we have just a wave equation. Let’s see what that divergence is with the assumed Fourier representation

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{A} &=\sum_{k \ne (0,0,0)} {\mathbf{A}_\mathbf{k}}^m 2 \pi i \frac{k_m}{\lambda_m} e^{2\pi i \mathbf{k} \cdot \mathbf{x}} \\ &=2 \pi i \sum_{k \ne (0,0,0)} (\mathbf{A}_\mathbf{k} \cdot \mathbf{k}) e^{2\pi i \mathbf{k} \cdot \mathbf{x}} \\ \end{aligned}

Since $\mathbf{A}_\mathbf{k} = \mathbf{A}_\mathbf{k}(t)$, there are two ways for $\partial_t (\boldsymbol{\nabla} \cdot \mathbf{A}) = 0$. For each $\mathbf{k} \ne 0$ there must be a requirement for either $\mathbf{A}_\mathbf{k} \cdot \mathbf{k} = 0$ or $\mathbf{A}_\mathbf{k} = \text{constant}$. The constant $\mathbf{A}_\mathbf{k}$ solution to the first equation appears to represent a standing spatial wave with no time dependence. Is that of any interest?

The more interesting seeming case is where we have some non-static time varying state. In this case, if $\mathbf{A}_\mathbf{k} \cdot \mathbf{k}$ for all $\mathbf{k} \ne 0$ the second of these Maxwell’s equations is just the vector potential wave equation, since the divergence is zero. That is

\begin{aligned}0 &= -\frac{1}{{c^2}} \partial_{tt} \mathbf{A} + \boldsymbol{\nabla}^2 \mathbf{A} \end{aligned} \quad\quad\quad(36)

Solving this isn’t really what is of interest, since the objective was just to determine if the divergence could be assumed to be zero. This shows then, that if the transverse solution to Maxwell’s equation is picked, the Hamiltonian for this field, with this gauge choice, becomes

\begin{aligned}H = \frac{\epsilon_0}{c^2} \lambda_1 \lambda_2 \lambda_3 \sum_\mathbf{k}\left(\frac{1}{{2}} {\left\lvert{\dot{\mathbf{A}}_\mathbf{k}}\right\rvert}^2+\frac{1}{{2}} (2 \pi c \mathbf{k})^2 {\left\lvert{\mathbf{A}_\mathbf{k}}\right\rvert}^2 \right).\end{aligned} \quad\quad\quad(37)

# Conclusions and followup.

The objective was met, a reproduction of Bohm’s Harmonic oscillator result using a complex exponential Fourier series instead of separate sine and cosines.

The reason for Bohm’s choice to fix zero divergence as the gauge choice upfront is now clear. That automatically cuts complexity from the results. Figuring out how to work this problem with complex valued potentials and also using the Geometric Algebra formulation probably also made the work a bit more difficult since blundering through both simultaneously was required instead of just one at a time.

This was an interesting exercise though, since doing it this way I am able to understand all the intermediate steps. Bohm employed some subtler argumentation to eliminate the scalar potential $\phi$ upfront, and I have to admit I did not follow his logic, whereas blindly following where the math leads me all makes sense.

As a bit of followup, I’d like to consider the constant $\mathbf{A}_\mathbf{k}$ case, and any implications of the freedom to pick $\mathbf{A}_0$. I’d also like to construct the Poynting vector $T(\gamma^0) \wedge \gamma_0$, and see what the structure of that is with this Fourier representation.

A general calculation of $T^{\mu\nu}$ for an assumed Fourier solution should be possible too, but working in spatial quantities for the general case is probably torture. A four dimensional Fourier series is likely a superior option for the general case.

# References

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