This 1979 paper by Chen, Lee, and Pereira gives an operator $L$ satisfying

$$\dot L = [A, L],\tag{1}$$

where $A$ is another operator, and the dot denotes time differentiation. They then define $I_n = \text{Tr}(L^n)$, and with the cyclic property of the trace one easily finds that $\dot I_n = 0$.

It is the next part I have trouble with. They define

$$J_n = \text{Tr}(BL^{n-1}),\tag{2}$$

where $B$ is an operator that fulfills

$$\dot B = [A, B] + L,\tag{3}$$

and they claim that therefore

$$\dot J_n = I_n.\tag{4}$$

When I try to verify (4) it starts promising with

$$ \begin{split} \dot J_n &= \text{Tr} \left( \dot BL^{n-1} + B(n-1)L^{n-2} \dot L \right) \\ &= \text{Tr} \left( [A,B]L^{n-1} + (n-1)BL^{n-2}[A,L] \right) + I_n, \end{split}\tag{5} $$

so I have been trying to show that the trace in the final expression is zero, but without success. One idea I had was to use an inductive argument, because (4) is certainly true for $n = 1,2$, but I haven't even been able to show it for $n=3$.

What am I missing? The article states (4) as if it is an obvious consequence of (2) and (3), but I don't see it.

  • $\begingroup$ The last line of your (5) is nonsense! [A,L] does not commute with L. $\endgroup$ Commented Jan 29, 2022 at 15:12
  • $\begingroup$ But the authors are right: (4) is obvious by inspection. You may find sophisticated "proofs", but you'll get lost in abstraction. Expand the commutators of the Trace to vanish in two lines, + and -, and observe/intuit the systematics of cancelations: each term on the upper line cancels with the term on the lower one to its left, and the first one on the upper line with the last one of the lower one. It's cyclic. Once you "see it" there are fancy/elegant ways to compact it, but you must first appreciate the obviousness.... $\endgroup$ Commented Jan 29, 2022 at 15:20

2 Answers 2


The authors are absolutely right: (4) is obvious by inspection. You are seeking to prove that $$ \text{Tr}([A,B]L^{n-1} + B [A,L]L^{n-2}+ BL[A,L]L^{n-3} +...+BL^{n-2}[A,L] )=0.$$

Write the commutators as differences, and use cyclicity of the trace to lead B in the front, $$ \text{Tr}\left (B(L^{n-1}A +AL^{n-1}+ LAL^{n-2}+...+L^{n-2}AL )\\ -B(AL^{n-1}+ LAL^{n-2}+...+ L^{n-1}A)\right )=0 ~~. $$ Note the cyclic cancelation pattern between the upper and lower lines.

Your formula (5) is wrong, since $\dot L$ does not commute with L, so you misapplied the chain rule for operators in your shortcut bogus evaluation of the derivative of a power. Check that explicitly for a small n.

  • $\begingroup$ Thank you. Yes, I agree it is obvious after correcting my chain rule mistake. I guess I was fooled because when differentiating $I_n$ naively applying the bogus derivative rule happens to work. $\endgroup$
    – ummg
    Commented Jan 29, 2022 at 18:54

After Cosmas Zachos corrected my chain rule mistake and showed his solution I came up with a nice alternate proof. One notes that if

$$\frac{d}{dt} L^n = [A, L^n]\tag{A}$$

for $n = 1, 2, \dotsc, m$ then

$$ \begin{split} \frac{d}{dt} L^{m+1} &= \dot L L^m + L \frac{d}{dt} L^m \\ &= [A,L] L^m + L [A, L^m] \\ &= ALL^m - LAL^m + LAL^m - LL^mA \\ &= [A, L^{m+1}]. \end{split} $$

We know that (A) holds for $n=1$, so it holds for all $n$ by (strong) induction.

Armed with (A) it is very straightforward to verify (4):

$$ \begin{split} \dot J_n &= \text{Tr}(\dot BL^{n-1} + B \frac{d}{dt} L^{n-1}) \\ &= \text{Tr}([A,B]L^{n-1} + LL^{n-1} + B[A, L^{n-1}]) \\ &= \text{Tr}(ABL^{n-1} - BAL^{n-1} + L^n + BAL^{n-1} - BL^{n-1}A) \\ &= \text{Tr}(L^n) = I_n, \end{split} $$

where the cyclic property of the trace was used to write $\text{Tr}(BL^{n-1}A) = \text{Tr}(ABL^{n-1})$.


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