Partial trace and the reduced density matrix

[yucheng]

TL;DR

Trace paradox?

From Rand Lectures on Light, we have, in the interaction picture, the equation of motion of the reduced density matrix:
$$i \hbar \rho \dot_A (t) = Tr_B[V(t), \rho_{AB}(t)] = \Sigma_b \langle \phi_b | V \rho_{AB} -\rho_{AB} V | \phi_b \rangle = \Sigma_b \phi_b | \langle V \rho_{AB} | \phi_b \rangle - \langle \phi_b| \rho_{AB} V | \phi_b \rangle = Tr_B(V \rho_AB) - Tr_B(\rho_AB V) = 0???$$

 

[Demystifier]

TL;DR Summary: Trace paradox?

I have corrected your LaTeX formula to make it readable and meaningful:
$$i \hbar \dot{\rho}_A (t) = Tr_B[V(t), \rho_{AB}(t)] = \Sigma_b \langle \phi_b | V \rho_{AB} -\rho_{AB} V | \phi_b \rangle $$
$$= \Sigma_b \langle\phi_b | V \rho_{AB} | \phi_b \rangle - \langle \phi_b| \rho_{AB} V | \phi_b \rangle = Tr_B(V \rho_{AB}) - Tr_B(\rho_{AB} V) = 0???$$

 

[Demystifier]

The solution of the paradox is that partial trace is not cyclic.
https://physics.stackexchange.com/questions/489202/is-a-partial-trace-cyclic

 

[Demystifier]

By the way, there is also another instructive trace paradox. Since $[x,p]=i\hbar 1$, we have
$${\rm Tr} [x,p] ={\rm Tr}(i\hbar 1)=i\hbar {\rm Tr}1=i\hbar\infty$$
but also
$${\rm Tr} [x,p] ={\rm Tr} (xp) - {\rm Tr} (px) =0$$
so
$$0=i\hbar\infty$$
Can you resolve this one?

Hint: The solution of this paradox is entirely unrelated to the solution of the previous one. The key is to understand the meaning of ${\rm Tr}1=\infty$, can we pretend that it is actually a big but finite number?

 

[yucheng]

@Demystifier
OMG I did not even realize it was published! I thought I just left it as a draft, but thanks for replying!

After doing some other problems, I realized that a partial trace is defined for a composite Hilbert space, which means that taking the trace with respect to the $| b \rangle$ basis breaks the common argument for commutation under the trace i,e, using the resolution of the identity because we have $\Sigma \langle b' |\rho_{AB}| a,b \rangle \langle a,b| V |b' \rangle$ instead.