Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Understanding the difference between |01>+|10> and |01>-|01>

  1. Nov 6, 2016 #1
    Hi,

    I am a student from the Netherlands in 6V, the year before university, and I'm doing research on Quantum Computing.

    However, I have difficulty understanding the four states of two entangled qubits,
    $$\left|00\right> \\
    \left|01\right> + \left|10\right> \\
    \left|01\right> - \left|10\right> \\
    \left|11\right>$$

    This Veritasium video states that two qubits have these four states, and calls the ##\left|01\right> + \left|10\right>## the ##\left|T_0\right>## state, and ##\left|01\right> - \left|10\right>## the singlet state, ##\left|S\right>##.
    However, what does the difference between the minus and plus signify? I understand that in both states the qubits have the property of being opposite to one another, but that's it.

    It would be greatly appreciated if someone could help me understand this difference!

    Kind regards,

    Isaiah van Hunen
     
  2. jcsd
  3. Nov 6, 2016 #2

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    Their inner product is zero, so if you prepare one of these states and test for the other you'll never get a positive response.
     
  4. Nov 6, 2016 #3

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    For one thing, a difference in statistics between the two states. Try exchanging the two qubits (i.e., reversing the order of the digits in each component of the state vector). What do you get with the ##T_0## state? What do you get with the ##S## state?
     
  5. Nov 6, 2016 #4
    As regards the minus and plus states ...that can be understood in terms of the composition of two spin states and the result is a symmetric and another anti symmetric combination.
    here you are handling 'entangled' qubits and i not being an expert can easily direct you to consult a paper which deals with the entangled qubits
    reference
    https://inst.eecs.berkeley.edu/~cs191/fa08/lectures/lecture2.pdf
    i hope it can guide you how to handle their superposition.
     
  6. Nov 6, 2016 #5
    First of all, thank you all for your responses.

    I don't quite understand what you mean, to be honest. ##0\times1## and ##1\times0## both equal ##0##, if that is what you mean.
    However, I fail to see how that explains the difference between ##\left|T_0\right>## state and ##\left|S\right>## state.

    Do you mean this?
    $$\left|00\right> \\

    \left|10\right> + \left|01\right> \\
    \left|10\right> - \left|01\right> \\
    \left|11\right>$$
    I understand that exchanging the qubit only has an effect on the ##\left|T_0\right>## state, but I still fail to see what the difference between ##\left|T_0\right>## and ##\left|S\right>## is.

    ##\left|00\right> + \left|11\right>##, as I understand it, refers to the state where both qubits are either ##0##, or both are ##1##.
    Am I correct in assuming that ##\left|01\right> + \left|10\right>## refers to the state where both qubits are always of opposite direction, the first being ##0## and the second ##1##, or the other way around?
    At least then I understand the ##\left|T_0\right>## state.

    Thank you for your reference. I read it, but didn't really find an answer to my question in there.

    Maybe I understand these states better with the associated quantum circuit?
    Is this how one would create the ##\left|T_0\right>## state?
    download.png
     
  7. Nov 6, 2016 #6

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The key to understanding this is the concept of "probability amplitude", which can be negative as well as positive. You might find this a useful insight into QM:

    http://www.scottaaronson.com/democritus/lec9.html
     
  8. Nov 6, 2016 #7
    Thank you for your reply.
    I try to imagine qubits as arrows, and I understand that a qubit can have a negative amplitude (but once squared, a positive probability).
    However, how would I then picture the ##\left|S\right>## state?

    Is this image relevant to the four states?
    hVZ2zzg.gif

    It comes from this explanation of Grover's Search Algorithm, and there are four states shown.
    Is ##\left|01\right> + \left|10\right>##, ##\left|T_0\right>## the mixed state of the initially 'on' wire shown, and ##\left|01\right> - \left|10\right>##, ##\left|S\right>## the mixed state of the initially 'off' wire shown?
     
  9. Nov 6, 2016 #8

    PeroK

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't know anything about quantum computing. In terms of spin, the singlet state has a total spin of ##0##, whereas the others (collectively the triplet) have a total spin of 1. But, I don't know how this is reflected in your computer processing.

    In terms of visualising things, spin is more complex than just an arrow. All four states are mutually orthogonal.

    You probably need someone who knows quantum computing to help you further.
     
  10. Nov 6, 2016 #9

    Strilanc

    User Avatar
    Science Advisor

    Those aren't the right states. The Bell basis is:

    $$\begin{align}
    B_{00} &= |00\rangle + |11\rangle \\
    B_{10} &= |00\rangle - |11\rangle \\
    B_{01} &= |01\rangle + |10\rangle \\
    B_{11} &= |01\rangle - |10\rangle
    \end{align}$$

    (Edit: I don't mean that the video is wrong. They're talking about something slightly different. I'm saying you should care about the Bell basis for now, not the thing they are talking about.)

    I'll call the Bell basis states "agree add", "agree subtract", "disagree add", and "disagree subtract" respectively. You can switch between these states by hitting either one of the involved entangled qubits with a Z gate (which switches the add/subtract bit) or with an X gate (which switches the agree/disagree bit).

    These are all entangled states, but when you measure the involved qubits you will find that the specific state you're in affects whether the measurements along each axis will always agree or always disagree.

    In the "disagree subtract" state, i.e. the singlet state, measurements give opposite answers along every axis.

    SingletParities.png

    In the "agree add" state, measurements along the X and Z axes will agree but measurements along the Y axis will disagree.

    AgreeAddParity.png

    The other two states cause XY-agree/Z-disagree and YZ-agree/X-disagree respectively. Also you can create states where an X-axis measurement of one qubit will agree with a *Y*-axis measurement of the other (simply rotate one of the qubits 90 degrees). There's a whole spectrum of entangled states!

    These kinds of details are important in quantum algorithms. For example, superdense coding puts information into the X and Z parities. If you don't know what the starting parities were, you won't be able to tell if the sender toggled them or not.

    Oh hey, I wrote that post and made those images.

    That animation is a representation of operations being applied to a single qubit. The bell basis states involve two qubits. So no, it's not directly relevant.
     
    Last edited: Nov 6, 2016
  11. Nov 6, 2016 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    No, it does not change the ##T_0## state. It reverses the sign of the ##S## state.

    The ##T_0## state is ##\vert01\rangle + \vert10\rangle##. If you swap the qubits in each term you get ##\vert10\rangle + \vert01\rangle##, which is the same state ##T_0## (since adding the terms is commutative).

    The ##S## state is ##\vert01\rangle + \vert10\rangle##. If you swap the qubits in each term you get ##\vert10\rangle - \vert01\rangle##, which is not the same state ##S##; it's ##- S##.

    This describes both the ##T_0## and the ##S## state. But you have to be careful about exactly what this description describes. It describes what happens if you measure the spin of both qubits along the same axis--the axis along which the single qubit basis states ##0## and ##1## are defined. Call this the ##z## axis. In both the ##T_0## and the ##S## states, the two spins will always be opposite (whereas in the states ##\vert00\rangle## and ##\vert11\rangle## both spins will always be the same with this measurement).

    In other words, this measurement alone is not enough to distinguish the ##T_0## and ##S## states. You need to make measurements along other spin directions. (Strilanc's post describes how you might do this.)
     
  12. Nov 6, 2016 #11

    zonde

    User Avatar
    Gold Member

    One entangled state describes particles that give the same measurement result when both are rotated clockwise or both are rotated counter clockwise by the same angle. The other one gives the same measurement result when one is rotated clockwise but other is rotated counter clockwise by the same angle.
    In other words in complementary basis measurements are correlated for one state and anticorrelated for the other one.
     
  13. Nov 7, 2016 #12

    A. Neumaier

    User Avatar
    Science Advisor
    2016 Award

    ##||01\rangle## and ##||10\rangle## are orthogonal unit vectors. If you calculate the inner product of their sum and their difference you get zero. The square of the amplitude is the probability of the event I was referring to, hence it is zero. Thus the two states have nothing to do with each other - they are maximally dissimilar.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted