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I Maximally entangled state expression and generalization

  1. May 5, 2017 #1
    Hey everybody,,

    I have question related to the expression of the maximally entangled state..

    I know that that for W-state its given by ( For three qubits) :


    and for Disentangled state ( for three qubits) its given as :


    Where one and 0 represent the up down state.

    ** What about Maximally entangled state ?


    and How can I generalize it for 5 qubits for example ?
    and why they call it maximally entangled ?
  2. jcsd
  3. May 5, 2017 #2
    How you've presented your question is a little confusing for me. I would argue that the first equation your wrote is itself a maximally entangled Bell state, and you can generalise it by replacing your three numbers in the ket vectors (a,b,c) with three (a,b,c) or five numbers (a,b,c,d,e) and so on. You would then need to include all permutations of these states with your 0's and 1's, i.e. with 3 qubits you have three terms, and so with 5 qubits there will be five terms. Finally, to ensure normalization, the 'n' parameter in your first equation must be set to the number of qubits you are considering. A state is maximally entangled when it is in this format, and has a weaker degree of entanglement when the system is close to, but not exactly, that form for W you expressed.

    Does this help?
  4. May 5, 2017 #3

    First thanks for your reply,,

    Basically and depending on my knowledge, we consider the first equation present in my question as a representation of Partially entangled state, and the third one as a Maximally entangled state because of the product present in it which will hold different states together inseparably (in the isolated perfect system).

    Actually I am facing problem with generalizing MAX ent. state because I am trying to perform a program to study a system starting with Maximally entangled state under the influence of the environment using Heisenberg spin Chains.

    Thanks again for you help!
  5. May 5, 2017 #4
    I'm sorry but as i said, that first equation IS maximally entangled. Maybe I'm just confused by your notation. I think perhaps you are using two different representations for your equations from eq 1. to eq 3?
  6. May 5, 2017 #5
    I am sure that You are right !! Thank You!
    And the first one is the representation of the Max. entg. :)

    The question now, How can I distinguish between the W state and Max state by looking to the expression?
    Are they have different representations ?

    Thanks in advance !
  7. May 5, 2017 #6

    Simon Phoenix

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    Hi Noora,

    I'm going to disagree with Ben here and say that the Werner state of 3 qubits $$ | \psi \rangle = \frac {1} { \sqrt {3} } ( | 001 \rangle + | 010 \rangle + |100 \rangle ) $$ is not a maximally entangled state of 3 qubits. One way to see this is to note that the reduced density operator for each qubit is not maximally mixed. Compare this to the GHZ state for 3 qubits : $$ | \psi \rangle = \frac {1} { \sqrt {2} } ( | 000 \rangle + | 111 \rangle ) $$where now the reduced density operator for each qubit is maximally mixed.

    The GHZ state is the unique state of ##n## qubits that is maximally entangled in that it's only a state of the GHZ form that will maximise the information content of the multipartite correlation. The general form for a maximally entangled state of ##n## qubits (a GHZ state) is of course $$ | \psi \rangle = \frac {1} { \sqrt {2} } ( | 000...0 \rangle + | 111...1 \rangle ) $$ Hope that helps.
  8. May 5, 2017 #7
    Thank You Sir for your great help.
    You clarify the whole notion.

    Thanks again,,
  9. May 5, 2017 #8
    One question pops in my mind and I searched about it and didn't get it!

    How can I get the reduced density matrix of each qubit ?

    Can anyone give me an idea using simplest explanation ?
  10. May 6, 2017 #9

    Simon Phoenix

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    The reduced density matrix is obtained by what is called 'tracing out' over the system, or systems, you want to ignore.

    To give a kind of classical analogy think of Brownian motion - where a large pollen grain is suspended in some fluid, say water. Now the pollen grain is being hit by the zillions and zillions of water molecules. Sometimes there's more water molecules hitting from one side than another and this gives the pollen grain a little 'kick'. Watch the grain for a time and you'll see it jiggle about from all of these random kicks. The grain does what's called a random walk. Now in principle in classical mechanics we can write down an equation of motion for the pollen grain and all of the zillions of water molecules and solve it. But that's a little bit difficult, to say the least. First off we'll have zillions of coupled differential equations to solve, and secondly how are we going to figure out precise initial conditions for every single object involved in the dynamics?

    So we write down ##x(t)## for the pollen grain and develop a stochastic model for its evolution by considering a kind of 'averaged' effect of all the water molecules. In effect we've 'traced out' all the detailed dynamics of the individual water molecules and just considered our system of interest (the pollen grain) to get an equation of motion for it alone.

    That's the spirit behind reduced density operators (or matrices); we focus on the object we're interested in and 'average' out the behaviour of those objects we're not interested in. What we're left with is a description, just for our object of interest, that contains the 'effect' of all the detailed interactions.

    OK - there's a lot I've glossed over there and sort of fudged a bit. I'm just trying to get across an intuitive feel for what it's about.

    So what is the procedure technically? Well let's suppose we have two quantum systems ##A## and ##B##. We can express the state of system ##A## using some orthonormal basis states ## | a \rangle ## with orthonormal basis states ## | b \rangle ## for the system ##B##. The total density matrix is ## \rho ##. The reduced density matrix (operator) for system ##A## is then formed by taking ## \langle b | \rho | b \rangle ## and summing over all ##b##. Any orthonormal basis for ##B## will do here - so pick one that's most convenient. In symbols we have $$ {\rho}_A = \sum_{b} \langle b | \rho | b \rangle $$ It's just summing the diagonal elements of the matrix with elements ## \langle b | \rho | b' \rangle ## - hence the name 'trace'.

    Have a go at working out the reduced density matrix for one qubit of the entangled state of 2 qubits given by $$ | \psi \rangle = a |00 \rangle + b |11 \rangle $$You should get $$ { \rho }_A = |a|^2 |0 \rangle \langle 0| +|b|^2 |1 \rangle \langle 1| $$Hope that helps
  11. May 6, 2017 #10

    Yes this helps a lot !!!
    Thank You so much Sir.

    sincerely grateful,,
    Last edited: May 6, 2017
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