# Reduced Density Matrix

1. Jul 10, 2005

### shomey

http://citebase.eprints.org/cgi-bin...pdf&identifier=oai:arXiv.org:quant-ph/9708045
(which is by the way very interesting)
i've encountered two unknown terms:
"reduced denstiy matrix" --- i have never heard this term before...
projectors --- i think i know this one, but i'm not quite sure so if you can clear this up for me it would be great...

Thanks allot,
Ron

2. Jul 10, 2005

### heimdall0

I think I've seen reduced density matrix used to refer to the density matrix with off diagonal terms set to 0.

My guess is that "projector" refers to a function that projects a quantum state onto a basis vector of the Hilbert space.

3. Jul 11, 2005

### vanesch

Staff Emeritus
Nonono !

The reduced density matrix appears in the frame of composed systems, say, A and B, each with their hilbert space Ha and Hb ; so the total hilbert space of the composed system is Ha x Hb. In a tensor product basis |psiA_i> x |psiB_j>, shortly, |ij>, the (true) density matrix rho is written as:

rho = sum_{ij} sum_{kl} rho_{ij}{kl} |ij><kl|

If we are only concerned with system A, and we don't care about system B, we can define the REDUCED DENSITY MATRIX for A:

rho_A = sum_{i} sum{k} sigma_ik |i><k|

with sigma_ik = sum{j} rho_{ij}{kj}

This last sum is called: taking the partial trace over B.

All observables which are only related to A can be calculated with only rho_A, the reduced density matrix of A. Of course, if you want to calculate observables of B, or correlations between A and B, you'll need the original density matrix rho ; but if you limit your attention to things measured or done on A, all you need is rho_A.

cheers,
Patrick.

4. Jul 11, 2005

### shomey

Patrik,
I've tried to understand what you've written but i've got lost...
if it's not too much trouble, could you give me an example of a system with two states? (say - spin for example)

thanks allot,
Ron

5. Jul 11, 2005

### dextercioby

Hmmm, funny thing, you had a reference to Landau's book right for that word.

Daniel.

6. Jul 11, 2005

### shomey

That help allot!
thank you very much

Ron

7. Jul 11, 2005

### dextercioby

You might compare this treatment with the one occuring naturally in classical statistics and the discussion of the BBGKY chain.

Daniel.

8. Jul 11, 2005

### shomey

I have another question on that matter
(I hope I'm not annoying you all...)

What is the meaning of this reduced density matrix?
i mean, it can't really be a representation of a state, because it is a matrix and not a vector. however, it does give, in a way, a representation of the system's state...
so, what exactly is it?

thanks again,
Ron

9. Jul 11, 2005

### dextercioby

The "reduced density matrix" is a very deceiving name. It stands for a trace class,densly defined, bounded,sefadjoint,positively defined linear operator acting on separable Hilbert space associated to a quantum physical system.

That's what it really is.

Daniel.

10. Jul 11, 2005

### mikeu

No worries Ron, glad to be able to help... Now, from a physical interpretation (rather than purely mathematical since Daniel already provided that) the reduced density matrix is really just the density matrix for the subsystem you are looking at, given that you don't know anything specific about the other part of the system (besides what to density matrix of the whole system tells you). For example, in the system I described above, the reduced density matrix for the second qubit was

$$\hat{\rho}_2 = \frac{1}{2}\left(\left|0\right\rangle\left\langle0\right| + \left|1\right\rangle\left\langle1\right| \right) = \sum_i p_i\left|\psi_i\right\rangle\left\langle\psi_i\right|$$

so you can just read off that there is a probability $p_0=0.5$ of the particle being in state $\left|\psi_0\right\rangle=\left|0\right\rangle$ and a probability $p_1=0.5$ of the particle being in state $\left|\psi_1\right\rangle=\left|1\right\rangle$. However, since this is a mixed state (i.e. not 100% probability of a single state, at least in the chosen basis) you can't write down the single state vector that the qubit is in - because it's not in one. This is the nature of entanglement, that neither particle is on its own in a single definable state.

You'll find that if you were to make a measurement on the first qubit and obtain the result $\left|0\right\rangle$, the total system would have collapsed into $\left|00\right\rangle$. If you then recalculated the reduced density matrix for the second qubit, it would simply be $\left|0\right\rangle\left\langle0\right|$, telling you that the particle now has probability $p_0=1$ of being in state $\left|0\right\rangle$ and probability $p_1=0$ of being in state $\left|1\right\rangle$. This is exactly the result expected from the Bell state - measuring one of the qubits determines the state of the other.

Mike

11. Jul 12, 2005

### QMrocks

sigma_ik = sum{j=l} rho_{ij}{kj}

assuming that |psiB_j> are othogonal basis set.

12. Jul 12, 2005

### shomey

Well... thank you all for the answers, you have helped me allot.
one thing is still not clear to me: (the article is about quantum reference systems):
in the end of the article, he claimbs to explain the vilation of Bell's inequallity.
But he actually shows that the correlation will always violate Bell's inequallity.

I'm sure I missed something, but I'm trying for hours to find what it is with no success.

if someone is familiar with this method of Reference Systems, could you clear this up for me?

thanks,
Ron

13. Jul 13, 2005

### vanesch

Staff Emeritus
That's why I already replaced the l by a j in the expression (where's yours ?)

(and yes, I'm assuming orthogonal bases in the two hilbert spaces)

cheers,
Patrick.

14. Jul 13, 2005

### QMrocks

15. Jul 13, 2005

### vanesch

Staff Emeritus
16. Jul 13, 2005

### vanesch

Staff Emeritus
the reduced density matrix and invariance under "collapse"

I thought I posted something like this already but can't find it. So I'll try to write it down here. The point I want to make is that, if you have a composite system A + B, then the reduced density matrix of A doesn't change when we apply a la Copenhagen, a measurement on B.

First of all, what does it mean to have a "collapse" from the density matrix point of view ?
Let us first consider a simple 2-state system: |psi> = a |1> + b|2>. The associated density matrix of this (pure state) is: rho = |psi><psi|, or:

( a a* ; a b* )
( b a* ; b b* )

Performing a measurement in the basis {|1>, |2>} comes down to PUTTING THE NON-DIAGONAL ELEMENTS TO 0 IN THE DENSITY MATRIX IN THIS BASIS.

Indeed, after measurement, we have:

( a a* ; 0 )
( 0 ; b b* )

which comes down to a statistical mixture with probability |a|^2 to be in state |1> and a probability |b|^2 to be in state |2>, exactly what the Born rule prescribes for the state |psi>.
Important remark: this operation is of course sensibly dependent on IN WHAT BASIS we apply this zeroing ! We know indeed that the Born rule is dependent on in what basis we apply it: we're supposed to do so in the eigenbasis of the observable we're measuring.

Now, let us switch to an entangled system A + B, and an overall state |psi>.

|psi> can be written as Sum_{ij} a_{ij} |i> |j>, with |i> an orthogonal basis in the hilbert space H_A of system A, and |j> an orthogonal basis in the hilbert space H_B of system B.

The density matrix corresponding to it is then:

rho = |psi><psi| = Sum_{ij} Sum{kl} a_{ij} a*_{kl} |i>|j><k|<l|

Let us assume that we observe a property of system A, of which |i> is an eigenbasis. We have then that the probability for each |i> to be realised is given by Sum_j a_{ij}a*_{ij}. This is in fact what can be obtained if we define a LOCAL density matrix rho_A:

rho_A_{i,k} = Sum_j a_{ij} a*_{kj}, which is nothing else but the partial trace of rho: each block with given i and k is a submatrix (with indices l and j), and we take the trace of that submatrix.

In the case we perform a measurement in the basis |i>, we have, as said before, that we only keep the diagonal elements of rho_A ; if we change basis {|i>} in H_A (but keeping the basis |j> in H_B for the moment), it should be clear that this comes down to transforming the matrix rho_A{i,k} in the new basis. In THAT basis, if we perform a measurement (another observable of A), we take again the diagonal elements of that transformed matrix rho_A.
So it should be clear that rho_A is the correct density matrix resulting in the right diagonal elements (the probabilities of outcome), when written in the measurement basis of A.
One can even show that, for any observable O_A, acting only on H_A, the expectation value is given <O_A> = Tr( O_A rho_A).

But let us now see what happens to rho_A when we do a measurement on system *B*. This comes down to setting each non-diagonal element in the |j> basis to 0 in the density matrix rho. However, this DOESN'T AFFECT THE PARTIAL TRACE because that only uses the DIAGONAL elements, which are not touched upon!
Let us consider now what happens if we have another observable for B: we have to change the basis |j> (while keeping our basis |i> for A). Again, the trace is INVARIANT UNDER CHANGE OF BASIS.
So we see that the partial traces over the basis in H_B do not change, whether we perform a measurement or not, and in that case, no matter in what basis we perform a measurement.
The matrix rho_A being made up only of partial traces over B, this means that every individual element of the matrix rho_A remains intact, whether we perform a measurement or not, and in that case, no matter in what basis we perform a measurement.

cheers,
Patrick.

17. Jul 13, 2005

### mikeu

I agree with this if someone makes a measurement on B but whoever is examining A is unaware of the result (or fact that measurement was made). That way Alice can't obtain any information from Bob when he makes a measurement on B far away from A.

However, if you know the result then you know the state of A, so A has a different density matrix, no? If A+B starts in the Bell state |00>+|11> and you measure B and get |0>, you have to know that A is also in |0>. So as far as you are concerned its density matrix should just be |0X0|, shouldn't it? Similarly, the reduced density matrix for A, with A+B now in |00>, gives |0X0| as well...

18. Jul 13, 2005

### vanesch

Staff Emeritus
Yes, of course, because for each single event, this knowledge changes the statistical distribution and hence the density matrix. But in fact, in the long run, as taken over the entire sample, if you don't use this knowledge to calculate CORRELATIONS between A and B, you will end up again with a mixture that is described by the original reduced density matrix of A.
The only thing that you now have is extra knowledge from which you can calculate correlations, and those are of course NOT described by the reduced density matrix.

cheers,
Patrick.

19. Oct 8, 2011

### Muneer QAU

can anyone give me the info how can i get information about my system from density matrix????????

thanks

20. Oct 10, 2011

### kith

Expectation values for observables A are calculated by <A>=tr{ρA}.

The probability of a measurement outcome can be expressed as the expectation value of the projection operator |a><a|, projecting on the corresponding eigenvalue a, so P(a)=tr{ρ|a><a|}=<a|ρ|a>.