# Partial Trace Question

1. Nov 16, 2011

### McLaren Rulez

Hi,

I am not able to understand something about partial tracing. We have a quantum state $\rho_{AB}$. The Hilbert Space is $H_{A}\otimes H_{B}$. For some observable $A$ in $H_{A}$, we have

$Tr_{A}(\rho_{A}A)=Tr_{AB}(\rho_{AB}(A\otimes 1)) =\sum\sum<a_{j}, b_{k}|\rho_{AB}(A\otimes 1_{B})|a_{j}, b_{k}>$

where the summation is over j and k. So here is the question: Why is the first equality true? What exactly is the information conveyed here? I have some idea but its a bit fuzzy so could you help me? Thank you!

And on a related, yet different note, what is the meaning of $Tr_{AB}(\rho_{AB}(A\otimes B))$. That is, we are measuring with some operator $A$ in $H_{A}$ and $B$ in $H_{B}$. The trace is some "expectation value" so what information does it have? I think that for non-entagled states, the answer is clear but when the state is an entangled one, then what? Thank you!

Last edited: Nov 16, 2011
2. Nov 16, 2011

### Fredrik

Staff Emeritus
The first equality follows from the definitions of the stuff on the right, in particular $|a_j,b_k\rangle=|a_j\rangle\otimes|b_k\rangle$.

I typed this up for my notes a couple of years ago, not with a lot of explanations I'm afraid. This is it, with a couple of minor edits:

Let $\{|\mu\rangle\}$be a basis for $\mathcal H_X$ and $\{|\alpha\rangle\}$ a basis for $\mathcal H_Y$. Then $\{|\mu\rangle\otimes|\alpha\rangle\}$ is a basis for a dense subset of $\mathcal H=\mathcal H_X\otimes\mathcal H_Y$. Define $|\mu\alpha\rangle=|\mu\rangle\otimes|\alpha\rangle$.

The reduced density operator for subsystem X is defined by $$\rho'=\sum_{\mu\nu}\sum_\alpha |\mu\rangle\langle\mu\alpha|\rho|\nu\alpha \rangle\langle\nu|.$$ This is equivalent to defining $\rho'$ by saying that its matrix elements in the $\{|\mu\rangle\}$ basis are $$\rho'_{\mu\nu}=\sum_\alpha \rho_{\mu\alpha,\nu\alpha} =\sum_\alpha\langle\mu\alpha|\rho|\nu\alpha\rangle.$$

\begin{align} \operatorname{Tr}(\rho(A\otimes I)) &=\sum_{\mu\alpha}\langle\mu\alpha| \rho(A\otimes I)|\mu\alpha\rangle =\sum_{\mu\alpha}\sum_{\nu\beta} \underbrace{\langle\mu\alpha| \rho|\nu\beta\rangle}_{\displaystyle =\rho_{\mu\alpha,\nu\beta}} \underbrace{\langle\nu\beta|A\otimes I|\mu\alpha\rangle}_{\displaystyle =\langle\nu|A|\mu\rangle \delta_{\beta\alpha}}\\ &=\sum_{\mu\nu}\bigg(\underbrace{\sum_\alpha\rho_{\mu\alpha,\nu\alpha}}_{\displaystyle =\rho'_{\mu\nu}}\bigg) \langle\nu|A|\mu\rangle=\sum_{\mu\nu} \langle\mu|\rho'|\nu\rangle\langle\nu|A|\mu\rangle =\operatorname{Tr}(\rho'A) \end{align}

3. Nov 16, 2011

### kith

<C> = tr{ρC} is the expectation value for an arbitrary operator C. If C can be decomposed into a direct product A⊗B, the corresponding physical process is a measurement where A acts only on one part of the system and B only on the rest. This is independent of entanglement. If your state is not entangled, you can additionally write <C>=<A><B>.

In general, neither ρ nor C can be decomposed into direct products, because of entanglement and well..."operator-entanglement"? I don't know if there's even a name for this. ;-)

4. Nov 16, 2011

### McLaren Rulez

Thank you for the replies, kith and Fredrik.

Fredrik, I think I get your mathematics. Basically, you are defining the reduced matrix $\rho'=\sum \langle{\alpha}|\rho|\alpha\rangle$ where the summation is over $\alpha$. And the rest of it is using identity relations to get your result. Am I correct? I think you've written it out so that it works if I express all the bras and kets as row and column vectors and the operators as matrices but essentially, we are using the definition $\rho'=\sum \langle{\alpha}|\rho|\alpha\rangle$

The notes I am reading appear to prove the definition from the equality while you've done the opposite. The justification given for the equality is rather brief. It says "Analogous to the idea of marginal statistics P(a) that can be extracted from a probability distribution P(a,b), for any observable A on Alice's particle (i.e the Hilbert space $H_{A}$), the partial state $\rho_{A}$ must allow Alice to compute her statistics, that is:

$Tr_{A}(\rho_{A}A)=Tr_{AB}(\rho_{AB}(A\otimes 1))$

So, this is my problem. How do I know beforehand, that there exists such a partial state which allows Alice to get her statistics? If someone claimed that there is no way to get talk about reduced states (i.e. you must consider the full system to get any information), how can that claim be refuted?

kith, thank you for your help as well. I'm very grateful to both of you.

5. Nov 16, 2011

### Fredrik

Staff Emeritus
Sort of, yes, but strictly speaking, the equality you just wrote down doesn't make sense. $\rho$ is a linear operator on a tensor product space $\mathcal H_1\otimes\mathcal H_2$, and that leaves $\rho|\alpha\rangle$ with $|\alpha\rangle\in\mathcal H_1$ undefined.

After the first line of the calculation I did, we can see that if we define the reduced state operator $\rho'$ the way I did it, we get the nice-looking equality that you started with. This provides some of the motivation for the definition.

Hm, I think the idea is that the left-hand side represents the average value of a series of measurements using the same measuring device A on identical objects, while the right-hand side represents...the same thing, but now the description also says that there's another object somewhere in the universe that just sits idly by without being involved in the measurements.

I don't think I can give you a better answer right now. I have a feeling that there are still several important things I don't understand about the concept of "subsystems".

In this specific case, the fact that the operator is $A\otimes 1$ means that we're talking about a measuring device (represented by A) that performs a measurement on the A system alone. So we seem to be talking about whether the existence of another system somewhere in the universe has any effect on measurements on the first system. I think this has to be considered an assumption that goes into the definition of a quantum theory. If the assumption is stupid, experiments should reveal that. For example, does your experiment have different results if you don't wear pants when you perform it.

6. Nov 17, 2011

### McLaren Rulez

Haha I see your point. Still not too comfortable with the assumption that the other subsystem has no effect on our measurements as long as we only have the identity acting on it but fair enough. Thank you :)