# How to calculate density matrix for the GHZ state

• Agrippa
In summary, the GHZ state is a maximally entangled state expressed as |\psi> = \frac{|000> + |111>}{\sqrt2}. To calculate the density matrix, one can go from GHZ = \frac{1}{2}(|000> + |111>)(<000| + <111|) to the final matrix form GHZ = 1/2\left( \begin{array}{cc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 &
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:

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

And finally to:

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

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

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

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

Last edited:
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
jfizzix said:
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.
jfizzix said:
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:
$1/4\left( \begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array} \right) \otimes 1/4\left( \begin{array}{cc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right) = 1/4\left( \begin{array}{cc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{array} \right)$
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.
jfizzix said:
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 GHZA⊗GHZBC?

Agrippa said:
Hmmm, I'm trying to work through this equation but am getting the wrong result:
$1/4\left( \begin{array}{cc} 1 & 0\\ 0 & 1\\ \end{array} \right) \otimes 1/4\left( \begin{array}{cc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right) = 1/4\left( \begin{array}{cc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{array} \right)$
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 $X\otimes Y$.

jfizzix said:
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?

Agrippa said:
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:
$\frac{1}{2}\big(|0\rangle\langle 0|\otimes|0,0\rangle\langle 0,0|+|1\rangle\langle 1|\otimes |1,1\rangle\langle 1,1|\big)$

jfizzix said:
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 GHZA⊗GHZBC 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 'GHZA⊗GHZBC' 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.)

jfizzix said:
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:
$1/2\left( \begin{array}{cc} 1 & 0\\ 0 & 0\\ \end{array} \right) \otimes 1/2\left( \begin{array}{cc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{array} \right) + 1/2\left( \begin{array}{cc} 0 & 0\\ 0 & 1\\ \end{array} \right) \otimes 1/2\left( \begin{array}{cc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ \end{array} \right) = 1/2\left( \begin{array}{cc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{array} \right)$
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 GHZA⊗GHZBC represent; and why the diagonal contains all ones in the case of GHZA⊗GHZB⊗GHZC = I/8.

## 1. How do I define the GHZ state for calculating density matrix?

The GHZ state is a maximally entangled state of three qubits, with the state vector given by |GHZ⟩ = (|000⟩ + |111⟩)/√2. This state is also known as the Greenberger-Horne-Zeilinger state.

## 2. What is the density matrix and why is it used for the GHZ state?

The density matrix is a mathematical representation of the state of a quantum system. It is used for the GHZ state because it allows for the calculation of probabilities for different measurement outcomes, as well as the study of entanglement and other quantum properties.

## 3. How do I calculate the density matrix for the GHZ state?

The density matrix for the GHZ state can be calculated by taking the outer product of the state vector |GHZ⟩ with its conjugate transpose, resulting in the matrix ρ = |GHZ⟩⟨GHZ|. This matrix will have dimensions of 8x8, as it represents the state of three qubits.

## 4. What are the main properties of the density matrix for the GHZ state?

The density matrix for the GHZ state has several important properties, including symmetry, purity, and trace. It is also a Hermitian matrix, meaning it is equal to its own conjugate transpose. These properties can be used to analyze the entanglement and other quantum properties of the GHZ state.

## 5. Can the density matrix for the GHZ state be used to calculate entanglement measures?

Yes, the density matrix for the GHZ state can be used to calculate entanglement measures, such as concurrence and entanglement entropy. These measures can provide insight into the amount of entanglement present in the GHZ state and how it changes under different operations or measurements.

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