Time independent perturbation theory with degeneracy

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Time independent perturbation with degeneracy.

In class it was claimed that if we repeated the derivation of the first order pertubation with degenerate states, then we’d get into (divide by zero) trouble if the state we were perturbing had degeneracy. Here I alter the previous derivation to show this explicitly.

The setup

Like the non-degenerate case, we are covering the time independent perturbation methods from section 16.1 of the text [1].

We start with a known Hamiltonian , and alter it with the addition of a “small” perturbation

For the original operator, we assume that a complete set of eigenvectors and eigenkets is known

We seek the perturbed eigensolution

and assumed a perturbative series representation for the energy eigenvalues in the new system

Note that we do not assume that the perturbed energy states, if degenerate in the original system, are still degenerate after pertubation.

Given an assumed representation for the new eigenkets in terms of the known basis

and a pertubative series representation for the probability coefficients

so that

Setting requires

for

We rescale our kets

where

The normalization of the rescaled kets is then

One can then construct a renormalized ket if desired

so that

The meat.

We continue by renaming terms in 1.10

where

Now we act on this with the Hamiltonian

or

Expanding this, we have

We want to write this as

This is

So we form

and so forth.

Zeroth order in

Since , this first condition on is not much more than a statement that .

First order in

How about ? For this to be zero we require that both of the following are simultaneously zero

This first condition is

With

or

From the second condition we have

Utilizing the Hermitian nature of we can act backwards on

We note that . We can also expand the , which is

I found that reducing this sum wasn’t obvious until some actual integers were plugged in. Suppose that , and , then this is

Observe that we can also replace the superscript with in the above manipulation without impacting anything else. That and putting back in the abstract indexes, we have the general result

Utilizing this gives us, for

Here we see our first sign of the trouble hinted at in lecture 5. Just because does not mean that . For example, with and we would have

We’ve got a unless additional restrictions are imposed!

If we return to 1.33, we see that, for the result to be valid, when , and there exists degeneracy for the state, we require

(then 1.33 becomes a equality, and all is still okay)

And summarizing what we learn from our conditions we have

Second order in

Doing the same thing for we form (or assume)

We need to know what the is, and find that it is zero

Utilizing that we have

From 1.37, treating the case carefully, we have

Again, only if for do we have a result we can use. If that is the case, the first sum is killed without a divide by zero, leaving

We can now summarize by forming the first order terms of the perturbed energy and the corresponding kets

Notational discrepency: OOPS. It looks like I used different notation than in class for our matrix elements for the placement of the indexes.

References

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

vi (vim) trick of the day: replace commas on current line with newlines

:, ! tr , '\n'

PHY456H1F: Quantum Mechanics II. Lecture 6 (Taught by Prof J.E. Sipe). Interaction picture.

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Disclaimer.

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

Interaction picture.

Recap.

Recall our table comparing our two interaction pictures

A motivating example.

While fundamental Hamiltonians are independent of time, in a number of common cases, we can form approximate Hamiltonians that are time dependent. One such example is that of Coulomb excitations of an atom, as covered in section 18.3 of the text [1], and shown in figure (\ref{fig:qmTwoL6fig1}).

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL6fig1}
\caption{Coulomb interaction of a nucleus and heavy atom.}
\end{figure}

We consider the interaction of a nucleus with a neutral atom, heavy enough that it can be considered classically. From the atoms point of view, the effects of the heavy nucleus barreling by can be described using a time dependent Hamiltonian. For the atom, that interaction Hamiltonian is

Here and is the position vector for the heavy nucleus, and is the position to each charge within the atom, where ranges over all the internal charges, positive and negative, within the atom.

Placing the origin close to the atom, we can write this interaction Hamiltonian as

The first term vanishes because the total charge in our neutral atom is zero. This leaves us with

where is the electric field at the origin due to the nucleus.

Introducing a dipole moment operator for the atom

the interaction takes the form

Here we have a quantum mechanical operator, and a classical field taken together. This sort of dipole interaction also occurs when we treat a atom placed into an electromagnetic field, treated classically as depicted in figure (\ref{fig:qmTwoL6fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL6fig2}
\caption{atom in a field}
\end{figure}

In the figure, we can use the dipole interaction, provided , where is the “width” of the atom.

Because it is great for examples, we will see this dipole interaction a lot.

The interaction picture.

Having talked about both the Schr\”{o}dinger and Heisenberg pictures, we can now move on to describe a hybrid, one where our Hamiltonian has been split into static and time dependent parts

We will formulate an approach for dealing with problems of this sort called the interaction picture.

This is also covered in section 3.3 of the text, albeit in a much harder to understand fashion (the text appears to try to not pull the result from a magic hat, but the steps to get to the end result are messy). It would probably have been nicer to see it this way instead.

In the Schr\”{o}dinger picture our dynamics have the form

How about the Heisenberg picture? We look for a solution

We want to find this operator that evolves the state from the state as some initial time , to the arbitrary later state found at time . Plugging in we have

This has to hold for all , and we can equivalently seek a solution of the operator equation

where

the identity for the Hilbert space.

Suppose that was independent of time. We could find that

If depends on time could you guess that

holds? No. This may be true when is a number, but when it is an operator, the Hamiltonian does not necessarily commute with itself at different times

So this is wrong in general. As an aside, for numbers, 2.13 can be verified easily. We have

Expectations

Suppose that we do find . Then our expectation takes the form

Put

and form

so that our expectation has the familiar representations

New strategy. Interaction picture.

Let’s define

or

Let’s see how this works. We have

Define

so that our operator equation takes the form

Note that we also have the required identity at the initial time

Without requiring us to actually find all of the dynamics of the time dependent interaction are now embedded in our operator equation for , with all of the simple interaction related to the non time dependent portions of the Hamiltonian left separate.

Connection with the Schr\”{o}dinger picture.

In the Schr\”{o}dinger picture we have

With a definition of the interaction picture ket as

the Schr\”{o}dinger picture is then related to the interaction picture by

Also, by multiplying 2.22 by our Schr\”{o}dinger ket, we remove the last vestiges of and from the dynamical equation for our time dependent interaction

Interaction picture expectation.

Inverting 2.25, we can form an operator expectation, and relate it the interaction and Schr\”{o}dinger pictures

With a definition

we have

As before, the time evolution of our interaction picture operator, can be found by taking derivatives of 2.28, for which we find

Summarizing the interaction picture.

Given

and initial time states

we have

where

and

or

Our interaction picture Hamiltonian is

and for Schr\”{o}dinger operators, independent of time, we have the dynamical equation

Justifying the Taylor expansion above (not class notes).

Multivariable Taylor series

As outlined in section 2.8 () of [2], we want to derive the multi-variable Taylor expansion for a scalar valued function of some number of variables

consider the displacement operation applied to the vector argument

We can Taylor expand a single variable function without any trouble, so introduce

where

We have

so that

The multivariable Taylor series now becomes a plain old application of the chain rule, where we have to evaluate

so that

Assuming an Euclidean space we can write this in the notationally more pleasant fashion using a gradient operator for the space

To handle the higher order terms, we repeat the chain rule application, yielding for example

Thus the Taylor series associated with a vector displacement takes the tidy form

Even more fancy, we can form the operator equation

Here a dummy variable has been retained as an instruction not to differentiate the part of the directional derivative in any repeated applications of the operator.

That notational cludge can be removed by swapping and

where .

Having derived this (or for those with lesser degrees of amnesia, recall it), we can see that 2.2 was a direct application of this, retaining no second order or higher terms.

Our expression used in the interaction Hamiltonian discussion was

which we can see has the same structure as above with some variable substitutions. Evaluating it we have

and at we have

We see in this direction derivative produces the classical electric Coulomb field expression for an electrostatic distribution, once we take the and multiply it with the factor.

With algebra.

A different way to justify the expansion of 2.2 is to consider a Clifford algebra factorization (following notation from [3]) of the absolute vector difference, where is considered small.

Neglecting the term, we can then Taylor series expand this scalar expression

Observe this is what was found with the multivariable Taylor series expansion too.

References

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

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

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

PHY456H1F: Quantum Mechanics II. Lecture 5 (Taught by Prof J.E. Sipe). Pertubation theory and degeneracy. Review of dynamics

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Disclaimer.

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

Issues concerning degeneracy.

When the perturbed state is non-degenerate.

Suppose the state of interest is non-degenerate but others are

FIXME: diagram. states designated by dashes labeled , , degeneracy for energy .

This is no problem except for notation, and if the analysis is repeated we find

where

When the perturbed state is also degenerate.

FIXME: diagram. states designated by dashes labeled , , degeneracy for energy , and states designated by dashes labeled , , degeneracy for energy .

If we just blindly repeat the derivation for the non-degenerate case we would obtain

where

Note that the is NOT a typo, and why we run into trouble. There is one case where a perturbation approach is still possible. That case is if we happen to have

That may not be obvious, but if one returns to the original derivation, the right terms cancel so that one will not end up with the problem.

FIXME: do this derivation.

Diagonalizing the perturbation Hamiltonian.

Suppose that we do not have this special zero condition that allows the perturbation treatment to remain valid. What can we do. It turns out that we can make use of the fact that the perturbation Hamiltonian is Hermitian, and diagonalize the matrix

In the example of a two fold degeneracy, this amounts to us choosing not to work with the states

both some linear combinations of the two

In this new basis, once found, we have

Utilizing this to fix the previous, one would get if the analysis was repeated correctly

We see that a degenerate state can be split by applying perturbation.

FIXME: do this derivation.

FIXME: diagram. as one energy level without perturbation, and as two distinct levels with perturbation.

guess I’ll bet that this is the origin of the spectral line splitting, especially given that an atom like hydrogen has degenerate states.

Review of dynamics.

We want to move on to time dependent problems. In general for a time dependent problem, the answer follows provided one has solved for all the perturbed energy eigenvalues. This can be laborious (or not feasible due to infinite sums).

Before doing this, let’s review our dynamics as covered in section 3 of the text [1].

Schr\”{o}dinger and Heisenberg pictures

Our operator equation in the Schr\”{o}dinger picture is the familiar

and most of our operators are time independent.

where is the operator in the Schr\”{o}dinger picture, and is non time dependent.

Formally, the time evolution of any state is given by

so the expectation of an operator can be written

With the introduction of the Heisenberg ket

and Heisenberg operators

the expectation evolution takes the form

Note that because the Hamiltonian commutes with it’s exponential (it commutes with itself and any power series of itself), the Hamiltonian in the Heisenberg picture is the same as in the Schr\”{o}dinger picture

Time evolution and the Commutator

Taking the derivative of 3.20 provides us with the time evolution of any operator in the Heisenberg picture

We can write this as a commutator

Summarizing the two pictures.

References

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

PHY456H1F: Quantum Mechanics II. Lecture 4 (Taught by Prof J.E. Sipe). Time independent perturbation theory (continued)

[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 independent perturbation.

The setup

To recap, we were covering the time independent perturbation methods from section 16.1 of the text [1]. We start with a known Hamiltonian , and alter it with the addition of a “small” perturbation

For the original operator, we assume that a complete set of eigenvectors and eigenkets is known

We seek the perturbed eigensolution

and assumed a perturbative series representation for the energy eigenvalues in the new system

Given an assumed representation for the new eigenkets in terms of the known basis

and a pertubative series representation for the probability coefficients

so that

Setting requires

for

We rescaled our kets

where

The normalization of the rescaled kets is then

One can then construct a renormalized ket if desired

so that

The meat.

That’s as far as we got last time. We continue by renaming terms in 2.10

where

Now we act on this with the Hamiltonian

or

Expanding this, we have

We want to write this as

This is

So we form

and so forth.

Zeroth order in

Since , this first condition on is not much more than a statement that .

First order in

How about ? For this to be zero we require that both of the following are simultaneously zero

This first condition is

With

or

From the second condition we have

Utilizing the Hermitian nature of we can act backwards on

We note that . We can also expand the , which is

I found that reducing this sum wasn’t obvious until some actual integers were plugged in. Suppose that , and , then this is

More generally that is

Utilizing this gives us

And summarizing what we learn from our conditions we have

Second order in

Doing the same thing for we form (or assume)

We need to know what the is, and find that it is zero

Again, suppose that . Our sum ranges over all , so all the brakets are zero. Utilizing that we have

From 2.34 we have

We can now summarize by forming the first order terms of the perturbed energy and the corresponding kets

We can continue calculating, but are hopeful that we can stop the calculation without doing more work, even if . If one supposes that the

term is “small”, then we can hope that truncating the sum will be reasonable for . This would be the case if

however, to put some mathematical rigor into making a statement of such smallness takes a lot of work. We are referred to [2]. Incidentally, these are loosely referred to as the first and second testaments, because of the author’s name, and the fact that they came as two volumes historically.

References

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

[2] A. Messiah, G.M. Temmer, and J. Potter. Quantum mechanics: two volumes bound as one. Dover Publications New York, 1999.

On C operator precedence.

Saw the following code fragment today, illustrating a common precedence mistake:

         if ( ( ( A == type ) || ( B == type ) )
              && ( ( 0 == p->size1 ) && ( 0 == p->size2 ) ) ||
                 ( p->size1 < p->size2 ) )

This has the following structure

if ( typeMatches && isZero || isLess )

but should be:

if ( typeMatches && (isZero || isLess) )

This doesn’t behave in an expected fashion, which can be verified easily:

#include <stdio.h>

int main()
{
   for ( unsigned typeMatches = 0 ; typeMatches < 2 ; typeMatches++ )
   {
      for ( unsigned zero = 0 ; zero < 2 ; zero++ )
      {
         for ( unsigned less = 0 ; less < 2 ; less++ )
         {
            unsigned r = typeMatches && zero || less ;

            printf("typeMatches: %u; zero: %u; less: %u\t:%u\n", typeMatches, zero, less, r ) ;
         }
      }
   }

   return 0 ;
}

I’ve seen people argue vehemently that you need to memorize the whole C operator precedence table to avoid errors like this. I think my preference is to prefer simple code. My head hurt when I looked at this fragment. Wouldn’t it be easier to understand what the author was thinking if some temporary variables were used. Perhaps like so:

            if ( ( A == type ) || ( B == type ) )
            {
               bool foundZero = ( 0 == p->size1 ) && ( 0 == p->size2 ) ;
               bool foundLess = ( p->size1 < p->size2 ) ;

               if ( foundZero || foundLess )
               {
                  // ...
               }
            }

I like variable names (and these could be more specific) used to implicitly comment the code. I think I’d seen that suggestion, and the suggestion not to let a statement span multiple lines, years ago one of Code Complete, or The Pragmatic programmer. Both of those are great books that deserve a read by any programmer (I’m due for a more recent re-read myself).

search within search in vim

I was asked how to search within a function for something and report an error if not found (as opposed to continuing past that function for the pattern of interest).

I’d never done this. Something I have done a bazillion times is search and replace within a function. Suppose one has the following:

void foo( int c )
{
   for ( int i = 0 ; i < c ; i++ )
   {
      bar( i ) ;
   }
}

Then positioning yourself on the first line of the function, you can do something like:

:,/^}/ s/bar/Blah/

to replace bar’s with Blah’s on all lines in the range (current line), to curly brace anchored on the beginning of the line (although this requires a coding convention where that is done).

How would you search only within the current function? I tried:

:,/^}/ /bar/

but this appears to take me past my current function, to the next /bar/ match that is NOT in the specified range.

I admit this is something that would have been handy a few times. In particular, we’ve got one source file that’s now 45000 lines long. I’ve gotten messed up a few times by searching for something, and ending up in a completely different function.

Here’s what I came up with:

,/^}/ s/bar//c

Use a prompted search and replace using the /c modifier, and ‘q’uit the replace once a match is found. If no match is found in the range, one gets an error.

This seems like a somewhat abusive use of /c, but works. I’d be curious if there are any other ways?

PHY456H1F: Quantum Mechanics II. Lecture 3 (Taught by Prof J.E. Sipe). Perturbation methods

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

States and wave functions

Suppose we have the following non-degenerate energy eigenstates

and consider a state that is “very close” to .

We form projections onto “direction”. The difference from this projection will be written , as depicted in figure (\ref{fig:qmTwoL3fig1}). This illustration cannot not be interpreted literally, but illustrates the idea nicely.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL3fig1}
\caption{Pictorial illustration of ket projections}
\end{figure}

For the amount along the projection onto we write

so that the total deviation from the original state is

The varied ket is then

where

In terms of these projections our kets magnitude is

Because this is

Similarly for the energy expectation we have

Or

This gives

where

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL3fig2}
\caption{Illustration of variation of energy with variation of Hamiltonian}
\end{figure}

“small errors” in don’t lead to large errors in

It is reasonably easy to get a good estimate and , although it is reasonably hard to get a good estimate of . This is for the same reason, because is not terribly sensitive.

Excited states.

Suppose we wanted an estimate of . If we knew the ground state . For any trial form

We are taking out the projection of the ground state from an arbitrary trial function.

For a state written in terms of the basis states, allowing for an degeneracy

and

(note that there are some theorems that tell us that the ground state is generally non-degenerate).

Often don’t know the exact ground state, although we might have a guess .

for

but cannot prove that

Then

FIXME: missed something here.

Somewhat remarkably, this is often possible. We talked last time about the Hydrogen atom. In that case, you can guess that the excited state is in the orbital and and therefore orthogonal to the (?) orbital.

Time independent perturbation theory.

See section 16.1 of the text [1].

We can sometimes use this sort of physical insight to help construct a good approximation. This is provided that we have some of this physical insight, or that it is good insight in the first place.

This is the no-think (turn the crank) approach.

Here we split our Hamiltonian into two parts

where is a Hamiltonian for which we know the energy eigenstates and the eigenkets. The is the “perturbation” that is supposed to be small “in some sense”.

Prof Sipe will provide some references later that provide a more specific meaning to this “smallness”. From some ad-hoc discussion in the class it sounds like one has to consider sequences of operators, and look at the convergence of those sequences (is this L2 measure theory?)

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{qmTwoL3fig3}
\caption{Example of small perturbation from known Hamiltonian}
\end{figure}

We’d like to consider a range of problems of the form

where

So that when we have

the problem that we already know, but for we have

the problem that we’d like to solve.

We are assuming that we know the eigenstates and eigenvalues for . Assuming no degeneracy

We seek

(this is the case).

Once (if) found, when we will have

This we know we can do because we are assumed to have a complete set of states.

with

where

There’s a subtlety here that will be treated differently from the text. We write

Take

where

We have:

FIXME: I missed something here.

Note that this is no longer normalized.

References

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

PHY456H1F, Quantum Mechanics II. My solutions to problem set 1 (ungraded).

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Harmonic oscillator.

Consider

Since it’s been a while let’s compute the raising and lowering factorization that was used so extensively for this problem.

It was of the form

Why this factorization has an imaginary in it is a good question. It’s not one that is given any sort of rationale in the text ([1]).

It’s clear that we want and . The difference is then

That commutator is an value, but what was the sign? Let’s compute so we don’t get it wrong

So we have

Factoring out an produces the form of the Hamiltonian that we used before

The factors were labeled the uppering () and lowering () operators respectively, and written

Observe that we can find the inverse relations

Question
What is a good reason that we chose this particular factorization? For example, a quick computation shows that we could have also picked

I don’t know that answer. That said, this second factorization is useful in that it provides the commutator relation between the raising and lowering operators, since subtracting 1.11 and 1.6 yields

If we suppose that we have eigenstates for the operator of the form

then the problem of finding the eigensolution of reduces to solving this problem. Because commutes with , an eigenstate of is also an eigenstate of . Utilizing 1.12 we then have

so we see that is an eigenstate of with eigenvalue .

Similarly for the raising operator

and find that is also an eigenstate of with eigenvalue .

Supposing that there is a lowest energy level (because the potential has a lower bound of zero) then the state for which the energy is the lowest when operated on by we have

Thus

and

This seems like a small bit of slight of hand, since it sneakily supplies an integer value to where up to this point was just a label.

If the eigenvalue equation we are trying to solve for the Hamiltonian is

Then we must then have

Part (a)

We’ve now got enough context to attempt the first part of the question, calculation of

We’ve calculated things like this before, such as

To continue we need an exact relation between and . Recall that was an eigenstate of with eigenvalue . This implies that the eigenstates and are proportional

or

so that

Similarly let

or

so that

We can now return to 1.19, and find

Consider half of this braket

Squaring, utilizing the Hermitian nature of the operator

Part (b)

Find the ground state energy of the Hamiltonian for .

The new Hamiltonian has the form

where

The energy states of the Hamiltonian are thus

and the ground state of the modified Hamiltonian is thus

Part (c)

Find the ground state energy of the Hamiltonian .

With a bit of play, this new Hamiltonian can be factored into

where

From 1.29 we see that we have the same sort of commutator relationship as in the original Hamiltonian

and because of this, all the preceding arguments follow unchanged with the exception that the energy eigenstates of this Hamiltonian are shifted by a constant

where the states are simultaneous eigenstates of the operator

The ground state energy is then

This makes sense. A translation of the entire position of the system should not effect the energy level distribution of the system, but we have set our reference potential differently, and have this constant energy adjustment to the entire system.

Hydrogen atom and spherical harmonics.

We are asked to show that for any eigenkets of the hydrogen atom we have

The summary sheet provides us with the wavefunction

where is a real valued function defined in terms of Lagueere polynomials. Working with the expectation of the operator to start with we have

Recalling that the only dependence in is we can perform the integration directly, which is

We have the same story for the expectation which is

Our integral is then just

also zero. The expectation is a slightly different story. There we have

Within this last integral we can make the substitution

and the integral takes the form

Here we have the product of two even functions, times one odd function (), over a symmetric interval, so the end result is zero, completing the problem.

I wasn’t able to see how to exploit the parity result suggested in the problem, but it wasn’t so bad to show these directly.

Angular momentum operator.

Working with the appropriate expressions in Cartesian components, confirm that for each component of angular momentum , if is in fact only a function of .

In order to proceed, we will have to consider a matrix element, so that we can operate on in position space. For that matrix element, we can proceed to insert complete states, and reduce the problem to a question of wavefunctions. That is

With we have

We are left with an sum of a symmetric product with the antisymmetric tensor so this is zero for all .

References

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

scripted perl undo of a code modification to test using diff.

We’ve got some code that uses an external product that allows us to put some of our info in a generic structure with integer members name1, name2, name3, name4.

Unfortunately, when we started using this, we didn’t use macros or inlines to encode exactly what we’d chosen to put in these fields, so now we’ve got a lots of code that tries to compensate by using comments like so:

               sqmIncrObjMetricsCounter( agtCB,
                                         pCaWobEntry->n.name2, // poolID
                                         pCaWobEntry->n.name3, // objectID
                                         0,
                                         SQLB_OBJTYPE(pCaWobEntry->n.name4), // objectType
                                         1,
                                         &objMetric ) ;

I chose to retrofit in some inline helper methods to eliminate the requirement for adding (and relying on!) comments to describe what was going on. Although I did figure out some partially automated methods, I was unfortunately not clever enough to figure out how to do this code modification in an entirely scripted fashion. I came up with what I thought was a good test strategy, where I was able to write a script to undo my changes, and then run diff against the ancestor in our version control repository to verify that I’d not actually introduced any changes of substance. I thought the method was a neat technique, yet another example of using ‘perl -p -i’ ad-hoc scripting. Here’s a fragment of the undo script:

s/SAL_ExtractPageNumFromPageName *\( *&([^ \)]+) *\)/$1.name1/g;
s/SAL_ExtractPageNumFromPageName *\( *([^ \)]+) *\)/$1->name1/g;

s/SAL_AssignPageNumToPageName *\( *&([^ ,]+) *, *([^ \)]+) *\)/$1.name1 = $2/g;
s/SAL_AssignPageNumToPageName *\( *([^ ,]+) *, *([^ \)]+) *\)/$1->name1 = $2/g;

After running this on all the files that I modified I was able to make sure that all the name1′s I’d changed to access via a helper method were still name1′s and that I’d not made any copy and paste errors, using the wrong new inlines anywhere.

Of course this presumes that I didn’t make a corresponding bug in my undo script, but that’s a much smaller task to review than all the other code in question, and I’m now much more comfortable committing this change to our version control system.