spherical polar coordinates

Derivatives of spherical polar vector representation.

December 6, 2023 math and physics play No comments , , , , ,

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On discord, on the bivector server, ‘stationaryactionprinciple’ asked a question that I really liked.
It’s a question that nagged me before too, but I hadn’t taken the time to puzzle through it properly.

The main character in this question is the spherical polar form of a radial vector, which has the form
\begin{equation}\label{eqn:dexpquestion:20}
\begin{aligned}
i &= \Be_{12} \\
j &= \Be_{31} e^{i\phi} \\
\Bx(r,\theta,\phi) &= r \Be_3 e^{j \theta},
\end{aligned}
\end{equation}
as illustrated in Fig. 1

Fig. 1. Spherical polar conventions.

Notice that all the \( \phi \) dependence comes from the bivector \( j = j(\phi) \), which makes life a bit tricky. We can take \( r, \theta \) or \( \phi \) partials of \( \Bx \), but need to be particularly careful how we do this for the \( \phi \) partials of the exponential factor.

One correct way to compute such a partial is to first expand the exponential in its trig constituents, as
\begin{equation}\label{eqn:dexpquestion:120}
e^{j \theta} = \cos\theta + j \sin\theta,
\end{equation}
and then take the derivative with respect to \(\phi\). If we do so, we get
\begin{equation}\label{eqn:dexpquestion:140}
\PD{\phi}{} e^{j\theta} = \PD{\phi}{j} \sin\theta.
\end{equation}
On the other hand, should we just directly take derivatives of the exponential, one might think that the result is
\begin{equation}\label{eqn:dexpquestion:160}
\PD{\phi}{} e^{j\theta} = \PD{\phi}{(j\theta)} e^{j\theta} = \theta \PD{\phi}{j} e^{j\theta}.
\end{equation}
but this is not correct, for a subtle reason. To understand why, we can step back to the power series representation of the exponential, and compute
\begin{equation}\label{eqn:dexpquestion:60}
\begin{aligned}
\PD{\phi}{e^{j\theta}}
&= \sum_{k = 0}^\infty \PD{\phi}{} \frac{ (j \theta)^k }{k!} \\
&= \sum_{k = 1}^\infty \PD{\phi}{j^k} \frac{ \theta^k }{k!}.
\end{aligned}
\end{equation}
If you treat \( j \) as a complex number, this then reduces to
\begin{equation}\label{eqn:dexpquestion:80}
\begin{aligned}
\PD{\phi}{e^{j\theta}}
&= \sum_{k = 1}^\infty k \PD{\phi}{j} j^{k-1} \frac{ \theta^k }{k!} \\
&=
\theta \PD{\phi}{j} \sum_{k = 1}^\infty \frac{ (j\theta)^{k-1} }{(k-1)!} \\
&=
\theta \PD{\phi}{j} e^{j\theta}.
\end{aligned}
\end{equation}
But, as we have said, this is wrong. The reason that this is wrong is because \( \PDi{\phi}{j} \) does not commute with \( j \), so
\begin{equation}\label{eqn:dexpquestion:100}
\PD{\phi}{j^k} = \PD{\phi}{j} j^{k-1} + j \PD{\phi}{j} j^{k-2} + \cdots,
\end{equation}
not \( k (\PDi{\phi}{j}) j^{k-1} \).

This non-commutativity, sneakily hiding in the power series for the exponential, messes us up. If we are careful, though, we should still be able to compute the correct result using the power series representation of the exponential. To do so, we need to understand the commutation relations for \( j \) and \( j’ \). Writing \( j’ = \PDi{\phi}{j} \), those two bivectors are
\begin{equation}\label{eqn:dexpquestion:180}
\begin{aligned}
j &= \Be_{31} e^{i\phi} \\
j’ &= \Be_{32} e^{i\phi},
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:dexpquestion:200}
\begin{aligned}
j j’
&= \Be_{31} e^{i\phi} \Be_{32} e^{i\phi} \\
&= \Be_{3132} e^{-i\phi} e^{i\phi} \\
&= -\Be_{12},
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:dexpquestion:220}
\begin{aligned}
j’ j
&= \Be_{32} e^{i\phi} \Be_{31} e^{i\phi} \\
&= \Be_{3231} e^{-i\phi} e^{i\phi} \\
&= \Be_{12}.
\end{aligned}
\end{equation}
We find that \( j \) and \( j’ \), in this case, anticommute
\begin{equation}\label{eqn:dexpquestion:240}
j j’ = -j’ j.
\end{equation}
We can now compute
\begin{equation}\label{eqn:dexpquestion:260}
\begin{aligned}
\PD{\phi}{j^k}
&= j’ j^{k-1} + j j’ j^{k-2} + j^2 j’ j^{k-3} \cdots \\
&= j’ j^{k-1} – j’ j^{k-1} + (-1)^2 j’ j^{k-1} \cdots
\end{aligned}
\end{equation}
This is zero for any even \( k \) and \( j’ j^{k-1} \) for odd \( k \).

Plugging this back into our Taylor series for the derivative (before we messed it up), we find
\begin{equation}\label{eqn:dexpquestion:280}
\begin{aligned}
\PD{\phi}{e^{j\theta}}
&= \sum_{k = 1, k \in \mathrm{odd}}^\infty j’ j^{k-1} \frac{ \theta^k }{k!} \\
&= j’ \inv{j}
\sum_{k = 1,\, k \in \mathrm{odd}}^\infty \frac{ (j\theta)^k }{k!} \\
&= j’ \inv{j} \sinh( j \theta ) \\
&= j’ \inv{j} j \sin( \theta ) \\
&= j’ \sin( \theta ).
\end{aligned}
\end{equation}
This is exactly the result that we had when we expanded \( e^{j\theta} \) in it’s cis form, and then took derivatives, so we have now reconciled the two different approaches.

Observe that, as a side effect of this exploration, we know also know how to compute the derivative of \( e^{j\theta} \) for the special case where \( j j’ = -j’ j \), which will be the case for any \( j \) where \( j^2 = \mathrm{constant} \).

Update to old phy356 (Quantum Mechanics I) notes.

February 12, 2015 math and physics play , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.

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