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Questions tagged [ghz-state]

For questions involving GHZ state

14 questions
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Difference between a star graph state and GHZ graph state

I saw in a class that star state and GHZ state are local-Clifford equivalent (Hadamard on n-1 qubits for a n star state). But then, when I wanted to draw a GHZ state and check on the litterature whether or not I got it right, it seemed as a lot of…
MohamedSU
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What is the general form for GHZ and W class states?

I am reading the following paper on mixed three-qubit states. It states that any three–qubit pure state can be written as (equation 1 of the paper) $|\psi_{GHZ}\rangle=\lambda_0 |000\rangle+\lambda_1 e^{i\theta} |100\rangle+\lambda_2…
Anindita Sarkar
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Cirq: n-qubit GHZ state

I am trying to write $n$ party GHZ state but at the end of the day, it seems like bi partite state. I am missing a detail. Here is my code number =6 qubits = cirq.LineQubit.range(number) def n_party_GHZ_circuit(qubits) GHZ_circuit =…
quest
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Output of my quantum circuit to create 3-qubit GHZ-like state does not make sense mathematically

I want to create the GHZ-like state, $|\Psi\rangle = \frac{1}{\sqrt{2}} \left(|011\rangle - |100 \rangle \right)$. I build my circuit in the following way. apply the x gate to the first and third qubits to make the input as $|101\rangle$ apply the…
molan li
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What states are valid GHZ-game states?

The GHZ game involves three non-communicating players, Alice, Bob, and Carol. They each receive a bit, $r,s,t$ with the guarantee that $r\oplus s\oplus t=0$ (i.e. with the guarantee that there are an even number of 1s). Their goal is to output…
quanticbolt
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Are GHZ states absolutely maximally entangled?

The definition of an absolutely maximally entangled state is as follows: "A multipartite state $|\psi\rangle$ of a system S is called absolutely maximally entangled if for any bipartition A|B of S, the reduced density operator is maximally mixed …
reza
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Transferring GHZ state onto some qubits

This question is just my effort that I made by understanding the previous answers to my questions . I have a GHZ state $|000\rangle+|111\rangle$ (please ignore the normalizing constant). For teleportaion of these three qubits (GHZ state) I make use…
Upstart
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GHZ depth proving by density matrices

How to prove, without using light cone and entanglement, that if we have GHZ circuit with any 3 qubit gates, so if the circuit depth is less than $\log_3(n)/2$, so there are two qubits that their measurement will give independent events? (By using…
Kirill Adamov
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Clarifications on stabilisers/generators for any state

An n-qubit state will have n generators. If I have a spin entangled state like $|0\rangle|11\rangle + |1\rangle|00\rangle$, I would say the generators are $XXX, IZZ, ZZI$. First: Is $ZZI$ correct for the above state? I.e., can I have stabilisers…
user23402
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Producing $|ghz\rangle\langle ghz|$ State in Cirq

I could manage to produce 3 qubit ghz state in Cirq. But I don't know how I can produce $|GHZ\rangle\langle GHZ|$ in Cirq Here is my code for 3 qubit ghz state Can you help me please for improving my code from $|GHZ\rangle$ to $|GHZ\rangle\langle…
quest
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Correlation decay in GHZ states, IBM quantum video

I watched an IBM quantum tutorial "Hello World" video. It assumed and then demonstrated that the correlation decays with the distance between two Pauli Z observables. I'm not able to derive the decay, even though intuitively it makes sense. (It's a…
Grigor Peradze
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GHZ Definition - Does it all have to be $|000\ldots\rangle$ or $|111\ldots\rangle$ or can it be $|010\ldots\rangle$ or $|101\ldots\rangle$

Trying to dig a bit into GHZ States and the proper definition. Trying to determine if GHZ state is strictly $|000\ldots\rangle$ or $|111\ldots\rangle$ or can it be $|010\ldots\rangle$ or $|101\ldots\rangle$. This is with the consideration that the…
Lunarlife
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How can I find the probabilities corresponding to measurement results of an observable of the GHZ states?

I'm working on a problem involving the calculation of probabilities for outcomes of a measurement on a quantum state perturbed by an error. The state in question is a GHZ state $|\text{GHZ}\rangle = \alpha|000\rangle + \beta|111\rangle$, with the…
Alex1111
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Why does measuring a GHZ state in the computational basis destroy the entanglement?

Measuring in a basis other than the computational (Z) basis will generally destroy any entanglement present in the GHZ state, as the entanglement is defined with respect to specific basis states. Why? (FYI, if I measure GHZ in z state, it will give…
Rayhan Kabir
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