How to calculate density matrix for the GHZ state

[Agrippa]

The GHZ state is:

$|\psi> = \frac{|000> + |111>}{\sqrt2}$

To calculate density matrix we go from:

$GHZ = \frac{1}{2}(|000> + |111>)(<000| + <111|)$
$GHZ = \frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|)$

To:

$GHZ<br /> = 1/2[<br /> \left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> \end{array} \right)<br /> +<br /> \left( \begin{array}{cc}<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> \end{array} \right)<br /> +<br /> \left( \begin{array}{cc}<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> \end{array} \right)<br /> +<br /> \left( \begin{array}{cc}<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> \end{array} \right)<br /> <br /> ]$

And finally to:

$GHZ = 1/2\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> \end{array} \right)$

But I see another author (p2) separates the Hilbert space into two subsystems GHZ_A⊗GHZ_BC and gets a "reduced" density matrix:

$GHZ_A⊗GHZ_{BC} = 1/4\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> \end{array} \right)$

Can anyone explain what this final matrix represents, and how one calculates it?

 

[jfizzix]

That density matrix represents the mixed state:
$\frac{1}{4}\big(|0,0,0\rangle\langle 0,0,0|+|0,1,1\rangle\langle 0,1,1|+|1,0,0\rangle\langle 1,0,0|+|1,1,1\rangle\langle 1,1,1|\big)$
or more simply:
$\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)$
To figure that out, system $A$ separates the density matrix into four equal square blocks (of size 4x4)
Of each block, system $B$ separates the density matrix into four equal square sub-blocks (of size 2x2)
Of each sublock, the state of system $C$ associated to those particular amplitudes of A and B are given.

 

[Agrippa]

That density matrix represents the mixed state:
$\frac{1}{4}\big(|0,0,0\rangle\langle 0,0,0|+|0,1,1\rangle\langle 0,1,1|+|1,0,0\rangle\langle 1,0,0|+|1,1,1\rangle\langle 1,1,1|\big)$

Thanks. I can (roughly, without yet drawing it up) see how you've constructed this expression from the final matrix I drew up.

or more simply:
$\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)$

Hmmm, I'm trying to work through this equation but am getting the wrong result:
$1/4\left( \begin{array}{cc}<br /> 1 & 0\\<br /> 0 & 1\\<br /> \end{array} \right)<br /> \otimes<br /> 1/4\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 1\\<br /> \end{array} \right)<br /> = 1/4\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> \end{array} \right)$
But perhaps I'm just making a simple mistake somewhere. At least, I can kind of see what you've done: you've tried to find a density matrix that corresponds to system A, and a density matrix corresponding to system BC, such that putting them together ($\otimes$) gives you the final matrix from my first post.

To figure that out, system $A$ separates the density matrix into four equal square blocks (of size 4x4)
Of each block, system $B$ separates the density matrix into four equal square sub-blocks (of size 2x2)
Of each sublock, the state of system $C$ associated to those particular amplitudes of A and B are given.

Unfortunately I don't quite follow you here. In particular, the talk of systems A and B separating the matrix into blocks, and the talk of the state of C being given (but not A or B (or A versus BC)?). Does it relate in some way to the calculation I just made above?
My main confusion is this: take the original density matrix for the pure state. I understand that the diagonal terms represent probabilities for measurement outcomes: 0.5 and 0.5. I also understand (to a lesser degree) that the off-diagonal terms represent the system's ability to exhibit interference. So the density matrix for a pure state like GHZ has a clear interpretation. But how does one interpret the "reduced" density matrix for GHZ_A⊗GHZ_BC?

 

[jfizzix]

Hmmm, I'm trying to work through this equation but am getting the wrong result:
$1/4\left( \begin{array}{cc}<br /> 1 & 0\\<br /> 0 & 1\\<br /> \end{array} \right)<br /> \otimes<br /> 1/4\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 1\\<br /> \end{array} \right)<br /> = 1/4\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> \end{array} \right)$
But perhaps I'm just making a simple mistake somewhere.

It is a simple mistake, easily rectified.
What's here the kronecker product in reverse order. If $X$ is the first matrix you have here, and $Y$ is the second matrix, then to get $X\otimes Y$, you should make a big matrix out of (in this case 4) copies of $Y$, and then multiply each copy of $Y$ by the corresponding entry in $X$.
It looks like that you've done instead is make a big matrix out of... 16 copies of $X$, and then multiplied them by the corresponding entries of $Y$.
In short, you've found $Y\otimes X$ instead of $X\otimes Y$.

 

[Agrippa]

It is a simple mistake, easily rectified.
What's here the kronecker product in reverse order. If $X$ is the first matrix you have here, and $Y$ is the second matrix, then to get $X\otimes Y$, you should make a big matrix out of (in this case 4) copies of $Y$, and then multiply each copy of $Y$ by the corresponding entry in $X$.
It looks like that you've done instead is make a big matrix out of... 16 copies of $X$, and then multiplied them by the corresponding entries of $Y$.
In short, you've found $Y\otimes X$ instead of $X\otimes Y$.

Yes I see the mistake now, thanks. What I still don't understand is the relationship between original density matrix and the reduced density matrix. In particular, we know that the density matrix of the GHZ state is:
$\frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|)$
We now want to break the system down into two systems (A and BC). But how do we know that this:
$\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)$
is a valid way to decompose the system? Why, for example, do we associate $\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)$ with subsystem A? How do we know that this expression in any way represents subsystem A?

 

[jfizzix]

Yes I see the mistake now, thanks. What I still don't understand is the relationship between original density matrix and the reduced density matrix. In particular, we know that the density matrix of the GHZ state is:
$\frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|)$
We now want to break the system down into two systems (A and BC). But how do we know that this:
$\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)$
is a valid way to decompose the system? Why, for example, do we associate $\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)$ with subsystem A? How do we know that this expression in any way represents subsystem A?

The product
$\frac{1}{4}\big(|0\rangle\langle 0|+|1\rangle\langle 1|\big)\otimes\big(|0,0\rangle\langle 0,0|+|1,1\rangle\langle 1,1|\big)$

is not a valid decomposition of the GHZ state:
$\frac{1}{2}( |000><000| + |111><111| + |111><000| + |000><111|)$

The GHZ state is not separable into such a product or any other product. It's about as entangled as three-particle spin states get.

If one were to measure particle A of the GHZ state (in the 0/1 basis), the state one would get as a result would be:
$\frac{1}{2}\big(|0\rangle\langle 0|\otimes|0,0\rangle\langle 0,0|+|1\rangle\langle 1|\otimes |1,1\rangle\langle 1,1|\big)$

 

[Agrippa]

The GHZ state is not separable into such a product or any other product. It's about as entangled as three-particle spin states get.

But in decoherence theory, isn't it standard to separate an entangled state into system state and environment state? I thought that something like that must be going on here. For example, I thought that GHZ_A⊗GHZ_BC amounted to something like treating systems B&C as the environment. Is that not right? I'm trying to figure out how to calculate the (reduced) density matrix called 'GHZ_A⊗GHZ_BC' from just the density matrix of the GHZ state.

(In case it helps, I'm ultimately trying to understand a particular definition of (integrated) information of a quantum state (see e.g. page 2), which is calculated via a function of the density matrix of the state and the reduced density matrices of its decompositions.)

If one were to measure particle A of the GHZ state (in the 0/1 basis), the state one would get as a result would be:
$\frac{1}{2}\big(|0\rangle\langle 0|\otimes|0,0\rangle\langle 0,0|+|1\rangle\langle 1|\otimes |1,1\rangle\langle 1,1|\big)$

That's useful, thanks. I've calculated it as follows:
$1/2\left( \begin{array}{cc}<br /> 1 & 0\\<br /> 0 & 0\\<br /> \end{array} \right)<br /> \otimes<br /> 1/2\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> \end{array} \right)<br /> +<br /> 1/2\left( \begin{array}{cc}<br /> 0 & 0\\<br /> 0 & 1\\<br /> \end{array} \right)<br /> \otimes<br /> 1/2\left( \begin{array}{cc}<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 1\\<br /> \end{array} \right)<br /> = 1/2\left( \begin{array}{cc}<br /> 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\<br /> 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\<br /> \end{array} \right)$
I interpret the diagonal values as representing probabilities (0.5 chance for |000> and 0.5 chance for |111>) and the off-diagonal values as capacity for interference, which is zero given that we now have an eigenstate (either |000> or |111>).
But then I wonder what the two additional 1's on the diagonal of GHZ_A⊗GHZ_BC represent; and why the diagonal contains all ones in the case of GHZ_A⊗GHZ_B⊗GHZ_C = I/8.