change of basis

[Series intro] An introduction to geometric algebra.

July 25, 2020 Geometric Algebra for Electrical Engineers , , , , , , , , , , , , , , , , ,

What’s in the pipe.

It’s been a while since I did any math or physics writing. This is the first post in a series where I plan to work my way systematically from an introduction of vectors, to the axioms of geometric algebra.  I plan to start with an introduction of vectors as directed “arrows”, building on that to discuss coordinates, tuples, and column matrix representations, and representation independent ideas. With those basics established, I’ll remind the reader about how generalized vector and dot product spaces are defined and give some examples. Finally, with the foundation of vectors and vector spaces in place, I’ll introduce the concept of a multivector space, and the geometric product, and start unpacking the implications of the axioms that follow naturally from this train of thought.

The applications that I plan to include in this series will be restricted to Euclidean spaces (i.e. where length is given by the Pythagorean law), primarily those of 2 and 3 dimensions.  However, it will be good to also lay the foundations for the non-Euclidean spaces that we encounter in relativistic electromagnetism (there is actually no other kind), and in computer graphics applications of geometric algebra, especially since we can do so nearly for free.  I plan to try to introduce the requisite ideas (i.e. the metric, which allows for a generalized dot product) by discussing Euclidean non-orthonormal bases.  Such bases have applications in condensed matter physics where there are useful for modelling crystal and lattice structure, and provide a hands conceptual bridge to a set of ideas that might otherwise seem abstract and without “real world” application.

Motivation.

Many introductions to geometric algebra start by first introducing the dot product, then bivectors and the wedge product, and eventually define the product of two vectors as the synthetic sum of the dot and wedge
\begin{equation}\label{eqn:multivector:20}
\Bx \By = \Bx \cdot \By + \Bx \wedge \By.
\end{equation}
It takes a fair amount of work to do this well. In the seminal work [4] a few pages are taken for each of the dot and wedge products, showing the similarities and building up ideas, before introducing the geometric product in this fashion. In [2] the authors take a phenomenal five chapters to build up the context required to introduce the geometric product.  I am not disparaging the authors for taking that long to build up the ideas, as their introduction of the subject is exceedingly clear and thorough, and they do a lot more than the minumum required to define the geometric product.

The strategy to introduce the geometric product as a sum of dot and wedge can result in considerable confusion, especially since the wedge product is often defined in terms of the geometric product
\begin{equation}\label{eqn:multivector:40}
\Bx \wedge \By =
\inv{2} \lr{
\Bx \By – \By \Bx
}.
\end{equation}
The whole subject can appear like a chicken and egg problem. I personally found the subject very confusing initially, and had considerable difficulty understanding which of the many identities of geometric algebra were the most fundamental. For this reason, I found the axiomatic approach of [1] very refreshing. The cavaet with that work is that is is exceptionally terse, as they jammed a reformulation of most of physics using geometric algebra into that single book, and it would have been thousands of pages had they tried to make it readable by mere mortals.

When I wrote my own book on geometric algebra, I had the intuition that the way to introduce the subject ought to be like the vector space in abstract linear algebra. The construct of a vector space is a curious and indirect way to define a vector. Vectors are not defined as entities, but simply as members of a vector space, a space that is required to have a set of properties. I thought that the same approach would probably work with multivectors, which could be defined as members of a multivector space, a mathematical construction with a set of properties.

I did try this approach, but was not fully satisfied with what I wrote. I think that dissatisfaction was because I tried to define the multivector first. To define the multivector, I first introduced a whole set of prerequisite ideas (bivector, trivector, blade, k-vector, vector product, …), but that was also problematic, since the vector multiplication idea required for those concepts wasn’t fully defined until the multivector space itself was defined.

My approach shows some mathematical cowardness. Had I taken the approach of the vector space fully to heart, the multivector could have been defined as a member of a multivector space, and all the other ideas follow from that. In this multi-part series, I’m going to play with this approach anew, and see how it works out.  If it does work, I’ll see if I can incorporate this approach into a new version of my book.

Review and background.

In this series, I’m going to assume a reader interested in geometric algebra, is probably also familiar with a wide variety of concepts, including but not limited to

  • vectors,
  • coordinates,
  • matrices,
  • basis,
  • change of basis,
  • dot product,
  • real and complex numbers,
  • rotations and translations,
  • vector spaces, and
  • linear transformations.

Despite those assumptions, as mentioned above, I’m going to attempt to build up the basics of vector representation and vector spaces in a systematic fashion, starting from a very elementary level.

My reasons for doing so are mainly to explore the logical sequencing of the ideas required.  I’ve always found well crafted pedagogical sequences rewarding, and will hopefully construct one here that is appreciated by anybody who chooses to follow along.

Next time.

As preparation for the next article in this series, the reader is asked to watch a short lesson from Vector, not so supervillain extraordinaire (Despicable Me).

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] L. Dorst, D. Fontijne, and S. Mann. Geometric Algebra for Computer Science. Morgan Kaufmann, San Francisco, 2007.

[4] D. Hestenes. New Foundations for Classical Mechanics. Kluwer Academic Publishers, 1999.

Quantum SHO ladder operators as a diagonal change of basis for the Heisenberg EOMs

August 19, 2015 phy1520 , , , , , , , , , , , , , , , , , , ,

[Click here for a PDF of this post with nicer formatting]

Many authors pull the definitions of the raising and lowering (or ladder) operators out of their butt with no attempt at motivation. This is pointed out nicely in [1] by Eli along with one justification based on factoring the Hamiltonian.

In [2] is a small exception to the usual presentation. In that text, these operators are defined as usual with no motivation. However, after the utility of these operators has been shown, the raising and lowering operators show up in a context that does provide that missing motivation as a side effect.
It doesn’t look like the author was trying to provide a motivation, but it can be interpreted that way.

When seeking the time evolution of Heisenberg-picture position and momentum operators, we will see that those solutions can be trivially expressed using the raising and lowering operators. No special tools nor black magic is required to find the structure of these operators. Unfortunately, we must first switch to both the Heisenberg picture representation of the position and momentum operators, and also employ the Heisenberg equations of motion. Neither of these last two fit into standard narrative of most introductory quantum mechanics treatments. We will also see that these raising and lowering “operators” could also be introduced in classical mechanics, provided we were attempting to solve the SHO system using the Hamiltonian equations of motion.

I’ll outline this route to finding the structure of the ladder operators below. Because these are encountered trying to solve the time evolution problem, I’ll first show a simpler way to solve that problem. Because that simpler method depends a bit on lucky observation and is somewhat unstructured, I’ll then outline a more structured procedure that leads to the ladder operators directly, also providing the solution to the time evolution problem as a side effect.

The starting point is the Heisenberg equations of motion. For a time independent Hamiltonian \( H \), and a Heisenberg operator \( A^{(H)} \), those equations are

\begin{equation}\label{eqn:harmonicOscDiagonalize:20}
\ddt{A^{(H)}} = \inv{i \Hbar} \antisymmetric{A^{(H)}}{H}.
\end{equation}

Here the Heisenberg operator \( A^{(H)} \) is related to the Schrodinger operator \( A^{(S)} \) by

\begin{equation}\label{eqn:harmonicOscDiagonalize:60}
A^{(H)} = U^\dagger A^{(S)} U,
\end{equation}

where \( U \) is the time evolution operator. For this discussion, we need only know that \( U \) commutes with \( H \), and do not need to know the specific structure of that operator. In particular, the Heisenberg equations of motion take the form

\begin{equation}\label{eqn:harmonicOscDiagonalize:80}
\begin{aligned}
\ddt{A^{(H)}}
&= \inv{i \Hbar}
\antisymmetric{A^{(H)}}{H} \\
&= \inv{i \Hbar}
\antisymmetric{U^\dagger A^{(S)} U}{H} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} U H
– H U^\dagger A^{(S)} U
} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} H U
– U^\dagger H A^{(S)} U
} \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{A^{(S)}}{H} U.
\end{aligned}
\end{equation}

The Hamiltonian for the harmonic oscillator, with Schrodinger-picture position and momentum operators \( x, p \) is

\begin{equation}\label{eqn:harmonicOscDiagonalize:40}
H = \frac{p^2}{2m} + \inv{2} m \omega^2 x^2,
\end{equation}

so the equations of motions are

\begin{equation}\label{eqn:harmonicOscDiagonalize:100}
\begin{aligned}
\ddt{x^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{\frac{p^2}{2m}} U \\
&= \inv{2 m i \Hbar} U^\dagger \lr{ i \Hbar \PD{p}{p^2} } U \\
&= \inv{m } U^\dagger p U \\
&= \inv{m } p^{(H)},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:harmonicOscDiagonalize:120}
\begin{aligned}
\ddt{p^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{\inv{2} m \omega^2 x^2 } U \\
&= \frac{m \omega^2}{2 i \Hbar} U^\dagger \lr{ -i \Hbar \PD{x}{x^2} } U \\
&= -m \omega^2 U^\dagger x U \\
&= -m \omega^2 x^{(H)}.
\end{aligned}
\end{equation}

In the Heisenberg picture the equations of motion are precisely those of classical Hamiltonian mechanics, except that we are dealing with operators instead of scalars

\begin{equation}\label{eqn:harmonicOscDiagonalize:140}
\begin{aligned}
\ddt{p^{(H)}} &= -m \omega^2 x^{(H)} \\
\ddt{x^{(H)}} &= \inv{m } p^{(H)}.
\end{aligned}
\end{equation}

In the text the ladder operators are used to simplify the solution of these coupled equations, since they can decouple them. That’s not really required since we can solve them directly in matrix form with little work

\begin{equation}\label{eqn:harmonicOscDiagonalize:160}
\ddt{}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix}
=
\begin{bmatrix}
0 & -m \omega^2 \\
\inv{m} & 0
\end{bmatrix}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix},
\end{equation}

or, with length scaled variables

\begin{equation}\label{eqn:harmonicOscDiagonalize:180}
\begin{aligned}
\ddt{}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}
&=
\begin{bmatrix}
0 & -\omega \\
\omega & 0
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\sigma_y
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}.
\end{aligned}
\end{equation}

Writing \( y = \begin{bmatrix} \frac{p^{(H)}}{m \omega} \\ x^{(H)} \end{bmatrix} \), the solution can then be written immediately as

\begin{equation}\label{eqn:harmonicOscDiagonalize:200}
\begin{aligned}
y(t)
&=
\exp\lr{ -i \omega \sigma_y t } y(0) \\
&=
\lr{ \cos \lr{ \omega t } I – i \sigma_y \sin\lr{ \omega t } } y(0) \\
&=
\begin{bmatrix}
\cos\lr{ \omega t } & \sin\lr{ \omega t } \\
-\sin\lr{ \omega t } & \cos\lr{ \omega t }
\end{bmatrix}
y(0),
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:harmonicOscDiagonalize:220}
\begin{aligned}
\frac{p^{(H)}(t)}{m \omega} &= \cos\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \sin\lr{ \omega t } x^{(H)}(0) \\
x^{(H)}(t) &= -\sin\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \cos\lr{ \omega t } x^{(H)}(0).
\end{aligned}
\end{equation}

This solution depends on being lucky enough to recognize that the matrix has a Pauli matrix as a factor (which squares to unity, and allows the exponential to be evaluated easily.)

If we hadn’t been that observant, then the first tool we’d have used instead would have been to diagonalize the matrix. For such diagonalization, it’s natural to work in completely dimensionless variables. Such a non-dimensionalisation can be had by defining

\begin{equation}\label{eqn:harmonicOscDiagonalize:240}
x_0 = \sqrt{\frac{\Hbar}{m \omega}},
\end{equation}

and dividing the working (operator) variables through by those values. Let \( z = \inv{x_0} y \), and \( \tau = \omega t \) so that the equations of motion are

\begin{equation}\label{eqn:harmonicOscDiagonalize:260}
\frac{dz}{d\tau}
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
z.
\end{equation}

This matrix can be diagonalized as

\begin{equation}\label{eqn:harmonicOscDiagonalize:280}
A
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
=
V
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
V^{-1},
\end{equation}

where

\begin{equation}\label{eqn:harmonicOscDiagonalize:300}
V =
\inv{\sqrt{2}}
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}.
\end{equation}

The equations of motion can now be written

\begin{equation}\label{eqn:harmonicOscDiagonalize:320}
\frac{d}{d\tau} \lr{ V^{-1} z } =
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
\lr{ V^{-1} z }.
\end{equation}

This final change of variables \( V^{-1} z \) decouples the system as desired. Expanding that gives

\begin{equation}\label{eqn:harmonicOscDiagonalize:340}
\begin{aligned}
V^{-1} z
&=
\inv{\sqrt{2}}
\begin{bmatrix}
-i & 1 \\
i & 1
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i \frac{p^{(H)}}{m \omega} + x^{(H)} \\
i \frac{p^{(H)}}{m \omega} + x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
a^\dagger \\
a
\end{bmatrix},
\end{aligned}
\end{equation}

where
\begin{equation}\label{eqn:harmonicOscDiagonalize:n}
\begin{aligned}
a^\dagger &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ -i \frac{p^{(H)}}{m \omega} + x^{(H)} } \\
a &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ i \frac{p^{(H)}}{m \omega} + x^{(H)} }.
\end{aligned}
\end{equation}

Lo and behold, we have the standard form of the raising and lowering operators, and can write the system equations as

\begin{equation}\label{eqn:harmonicOscDiagonalize:360}
\begin{aligned}
\ddt{a^\dagger} &= i \omega a^\dagger \\
\ddt{a} &= -i \omega a.
\end{aligned}
\end{equation}

It is actually a bit fluky that this matched exactly, since we could have chosen eigenvectors that differ by constant phase factors, like

\begin{equation}\label{eqn:harmonicOscDiagonalize:380}
V = \inv{\sqrt{2}}
\begin{bmatrix}
i e^{i\phi} & -i e^{i \psi} \\
1 e^{i\phi} & e^{i \psi}
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:harmonicOscDiagonalize:341}
\begin{aligned}
V^{-1} z
&=
\frac{e^{-i(\phi + \psi)}}{\sqrt{2}}
\begin{bmatrix}
-i e^{i\psi} & e^{i \psi} \\
i e^{i\phi} & e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i e^{i\phi} \frac{p^{(H)}}{m \omega} + e^{i\phi} x^{(H)} \\
i e^{i\psi} \frac{p^{(H)}}{m \omega} + e^{i\psi} x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{i\phi} a^\dagger \\
e^{i\psi} a
\end{bmatrix}.
\end{aligned}
\end{equation}

To make the resulting pairs of operators Hermitian conjugates, we’d want to constrain those constant phase factors by setting \( \phi = -\psi \). If we were only interested in solving the time evolution problem no such additional constraints are required.

The raising and lowering operators are seen to naturally occur when seeking the solution of the Heisenberg equations of motion. This is found using the standard technique of non-dimensionalisation and then seeking a change of basis that diagonalizes the system matrix. Because the Heisenberg equations of motion are identical to the classical Hamiltonian equations of motion in this case, what we call the raising and lowering operators in quantum mechanics could also be utilized in the classical simple harmonic oscillator problem. However, in a classical context we wouldn’t have a justification to call this more than a change of basis.

References

[1] Eli Lansey. The Quantum Harmonic Oscillator Ladder Operators, 2009. URL http://behindtheguesses.blogspot.ca/2009/03/quantum-harmonic-oscillator-ladder.html. [Online; accessed 18-August-2015].

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics, chapter {Time Development of the Oscillator}. Pearson Higher Ed, 2014.