Differentiation of an operator equation in paper by Chen, Lee, Pereira 1979

Question

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.

Answer 1

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.

Answer 2

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 $\text{Tr}(BL^{n-1}A) = \text{Tr}(ABL^{n-1})$.