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For a bipartite quantum system evolving under some master equation, is the time derivative of the reduced density matrix equal to the partial trace of the time derivative of the matrix?

In other words, is the following true:

$\dot{\rho}_{A} = Tr_B(\dot{\rho}_{AB})$

(Where $\rho_A = Tr_B(\rho_{AB})$)

If not, is there some other simple method to find $\dot{\rho}_{A}$ from $\dot{\rho}_{AB}$?

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up vote 3 down vote accepted

Yes, as long as $B$ doesn't depend on time.

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Does not the cyclic invariance of the trace cancel the possible temporal variation of the basis B? –  juanrga Dec 4 '12 at 21:15
    
@juanrga: I don't think so. –  Arnold Neumaier Dec 5 '12 at 9:07
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Could you write the total equation when B depends on time? –  juanrga Dec 5 '12 at 9:49
    
There will be an extra additive term $d/dB~ Tr B(\rho_{AB})$ times $dB/dt$, but what this precisely should mean depends on the way $B$ changes. –  Arnold Neumaier Dec 5 '12 at 10:07
    
Just to check, by B you're referring to the basis being time-dependent, rather than the subsystem evolving in time? –  Ben Aaronson Dec 5 '12 at 15:21
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By definition,

$$\rho_A = \mathrm{Tr}_B (\rho_{AB}) = \sum_B \langle B| \rho_{AB} |B \rangle$$

Differentiating both sides

$$\dot{\rho}_A = \sum_B \langle B| \dot{\rho}_{AB} |B \rangle = \mathrm{Tr}_B (\dot{\rho}_{AB})$$

because the basis are time-independent. I have checked the literature I do not know any practical case where time-dependent basis are used to compute averages; but, you ask above, in a comment, about using a time-dependent basis. Let us see what happen if you chose a time-dependent basis $| B(t) \rangle$. In this case the above equation is augmented, a priori, by a term

$$\sum_B \langle \dot{B} | \rho_{AB} |B(t) \rangle + \sum_B \langle B(t)| \rho_{AB} | \dot{B} \rangle$$

but using $|\dot{B} \rangle = (H/i\hbar) | B(t) \rangle$ and the conjugate

$$-\sum_B \langle B(t) | (H/i\hbar) \rho_{AB} |B(t) \rangle + \sum_B \langle B(t)| \rho_{AB} (H/i\hbar) | B(t) \rangle = 0 \>\>\>\>\>\>\>\> (1)$$

where, in the last step, I have used the cyclic invariance of the trace.

Therefore the equation $\dot{\rho}_A = \mathrm{Tr}_B (\dot{\rho}_{AB})$ is valid for both time-dependent and time-independent basis.

EDIT: In response to mistaken comments I am adding some extra details. The left-hand side of equation (1) can be written as

$$-\mathrm{Tr}_B \{(H/i\hbar)\rho_{AB}\} + \mathrm{Tr}_B \{\rho_{AB}(H/i\hbar)\}$$

Using now the cyclic invariance of the trace $\mathrm{Tr}_B \{XY\} = \mathrm{Tr}_B \{YX\}$ for $X = (H/i\hbar)$ and $Y = \rho_{AB}$, we obtain

$$-\mathrm{Tr}_B \{\rho_{AB} (H/i\hbar)\} + \mathrm{Tr}_B \{\rho_{AB}(H/i\hbar)\} = 0$$

explaining the zero in equation (1).

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Unfortunately, you assume in your last displayed equation that $<a|AB|b>=<a|BA|b>$, which is generally false. Note that you eliminated the trace already in your top equation, so later you don't have a trace at your disposal. –  Arnold Neumaier Dec 6 '12 at 14:37
    
@ArnoldNeumaier: Although I received a "-1", you are completely wrong. First, I am not assuming what you believe and, second, I have not eliminated the trace. The last displayed equation is just $-\mathrm{Tr}_B ((H/i\hbar)\rho_{AB}) + \mathrm{Tr}_B (\rho_{AB}(H/i\hbar)) = 0$. and then the cyclic invariance of the trace $\mathrm{Tr}_B (XY) = \mathrm{Tr}_B (YX)$ was used. –  juanrga Dec 6 '12 at 18:27
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Thanks for the update; now I see what you meant. I think your extra comment is important to make the argument intelligible. - But calling me ''completely wrong'' has no more factual content than saying simply ''no'', whereas it creates unnecessary emotional strain. –  Arnold Neumaier Dec 6 '12 at 18:53
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Nice conclusion. The time-dependent basis may be relevant when the environment traced over has a time-dependent state space, as in this case no time-independent basis exists. –  Arnold Neumaier Dec 6 '12 at 18:57
    
@ArnoldNeumaier: Thank you for reverting the -1 and for the nice comments. Maybe my wording in the above comment can be considered a bit rude by some people, but as counterpart I do not give negative points. –  juanrga Dec 6 '12 at 19:52
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