Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Unitary exponential sandwich

Posted by peeterjoot on September 27, 2010

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


One of the chapter II exercises in [1] involves a commutator exponential sandwich of the form

\begin{aligned}e^{i F} B e^{-iF}\end{aligned} \hspace{\stretch{1}}(1.1)

where F is Hermitian. Asking about commutators on physicsforums I was told that such sandwiches (my term) preserve expectation values, and also have a Taylor series like expansion involving the repeated commutators. Let’s derive the commutator relationship.


Let’s expand a sandwich of this form in series, and shuffle the summation order so that we sum over all the index plane diagonals k + m = \text{constant}. That is

\begin{aligned}e^{A} B e^{-A}&=\sum_{k,m=0}^\infty \frac{1}{{k!m!}} A^k B (-A)^m \\ &=\sum_{r=0}^\infty \sum_{m=0}^r \frac{1}{{(r-m)!m!}} A^{r-m} B (-A)^m \\ &=\sum_{r=0}^\infty \frac{1}{{r!}} \sum_{m=0}^r \frac{r!}{(r-m)!m!} A^{r-m} B (-A)^m \\ &=\sum_{r=0}^\infty \frac{1}{{r!}} \sum_{m=0}^r \binom{r}{m} A^{r-m} B (-A)^m.\end{aligned}

Assuming that these interior sums can be written as commutators, we’ll shortly have an induction exercise. Let’s write these out for a couple values of r to get a feel for things.


\begin{aligned}\binom{1}{0} A B + \binom{1}{1} B (-A) &= \left[{A},{B}\right]\end{aligned}


\begin{aligned}\binom{2}{0} A^2 B + \binom{2}{1} A B (-A) + \binom{2}{2} B (-A)^2 &= A^2 B - 2 A B A + B A\end{aligned}

This compares exactly to the double commutator:

\begin{aligned}\left[{A},{\left[{A},{B}\right]}\right]&= A(A B - B A) -(A B - B A)A \\ &= A^2 B - A B A - A B A + B A^2 \\ &= A^2 B - 2 A B A + B A^2 \\ \end{aligned}


\begin{aligned}\binom{3}{0} A^3 B + \binom{3}{1} A^2 B (-A) + \binom{3}{2} A B (-A)^2 + \binom{3}{3} B (-A)^3 &= A^3 B - 3 A^2 B A + 3 A B A^2 - B A^3.\end{aligned}

And this compares exactly to the triple commutator

\begin{aligned} [A, [A, [A, B] ] ] &= A^3 B - 2 A^2 B A + A B A^2 -(A^2 B A - 2 A B A^2 + B A^3) \\ &= A^3 B - 3 A^2 B A + 3 A B A^2 -B A^3 \\ \end{aligned}

The induction pattern is clear. Let’s write the r fold commutator as

\begin{aligned}C_r(A,B) &\equiv \underbrace{[A, [A, \cdots, [A,}_{r \text{times}} B]] \cdots ] = \sum_{m=0}^r \binom{r}{m} A^{r-m} B (-A)^m,\end{aligned} \hspace{\stretch{1}}(2.2)

and calculate this for the r+1 case to verify the induction hypothesis. We have

\begin{aligned}C_{r+1}(A,B) &= \sum_{m=0}^r \binom{r}{m} \left( A^{r-m+1} B (-A)^m-A^{r-m} B (-A)^{m} A \right) \\ &= \sum_{m=0}^r \binom{r}{m} \left( A^{r-m+1} B (-A)^m+A^{r-m} B (-A)^{m+1} \right) \\ &= A^{r+1} B+ \sum_{m=1}^r \binom{r}{m} A^{r-m+1} B (-A)^m+ \sum_{m=0}^{r-1} \binom{r}{m} A^{r-m} B (-A)^{m+1} + B (-A)^{r+1} \\ &= A^{r+1} B+ \sum_{k=0}^{r-1} \binom{r}{k+1} A^{r-k} B (-A)^{k+1}+ \sum_{m=0}^{r-1} \binom{r}{m} A^{r-m} B (-A)^{m+1} + B (-A)^{r+1} \\ &= A^{r+1} B+ \sum_{k=0}^{r-1} \left( \binom{r}{k+1} + \binom{r}{k} \right) A^{r-k} B (-A)^{k+1}+ B (-A)^{r+1} \\ \end{aligned}

We now have to sum those binomial coefficients. I like the search and replace technique for this, picking two visibly distinct numbers for r, and k that are easy to manipulate without abstract confusion. How about r=7, and k=3. Using those we have

\begin{aligned}\binom{7}{3+1} + \binom{7}{3} &=\frac{7!}{(3+1)!(7-3-1)!}+\frac{7!}{3!(7-3)!} \\ &=\frac{7!(7-3)}{(3+1)!(7-3)!}+\frac{7!(3+1)}{(3+1)!(7-3)!} \\ &=\frac{7! \left( 7-3 + 3 + 1 \right) }{(3+1)!(7-3)!} \\ &=\frac{(7+1)! }{(3+1)!((7+1)-(3+1))!}.\end{aligned}

Straight text replacement of 7 and 3 with r and k respectively now gives the harder to follow, but more general identity

\begin{aligned}\binom{r}{k+1} + \binom{r}{k} &=\frac{r!}{(k+1)!(r-k-1)!}+\frac{r!}{k!(r-k)!} \\ &=\frac{r!(r-k)}{(k+1)!(r-k)!}+\frac{r!(k+1)}{(k+1)!(r-k)!} \\ &=\frac{r! \left( r-k + k + 1 \right) }{(k+1)!(r-k)!} \\ &=\frac{(r+1)! }{(k+1)!((r+1)-(k+1))!} \\ &=\binom{r+1}{k+1}\end{aligned}

For our commutator we now have

\begin{aligned}C_{r+1}(A,B) &= A^{r+1} B+ \sum_{k=0}^{r-1} \binom{r+1}{k+1} A^{r-k} B (-A)^{k+1} + B (-A)^{r+1} \\ &= A^{r+1} B+ \sum_{s=1}^{r} \binom{r+1}{s} A^{r+1-s} B (-A)^{s} + B (-A)^{r+1} \\ &= \sum_{s=0}^{r+1} \binom{r+1}{s} A^{r+1-s} B (-A)^{s} \qquad\square\end{aligned}

That completes the inductive proof and allows us to write

\begin{aligned}e^A B e^{-A}&=\sum_{r=0}^\infty \frac{1}{{r!}} C_{r}(A,B),\end{aligned} \hspace{\stretch{1}}(2.3)

Or, in explicit form

\begin{aligned}e^A B e^{-A}&=B + \frac{1}{{1!}} \left[{A},{B}\right]+ \frac{1}{{2!}} \left[{A},{\left[{A},{B}\right]}\right]+ \cdots\end{aligned} \hspace{\stretch{1}}(2.4)


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


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