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Qubits and angular momentum-like operators

  1. Sep 1, 2014 #1
    Hi guys, my quesion is quite simple but I think I need to give some background...
    Let's suppose I have 3 qubits, so the basis of the space is:

    [tex]\left\{{\left |{000}\right>,\left |{001}\right>,\left |{010}\right>,\left |{100}\right>,\left |{011}\right>,\left |{101}\right>,\left |{110}\right>,\left |{111}\right>}\right\}[/tex]

    i.e. the dimesion is [tex]2^{3}=8[/tex]

    I define the angular momentum-like operators:

    [tex]J_{z}=\left [{\frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right)\otimes I \otimes I}\right] + \left [{I \otimes \frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right) \otimes I}\right] + \left [{I \otimes I \otimes \frac{1}{2}\left ({\left |{1}\right>\left <{1}\right|-\left |{0}\right>\left <{0}\right|}\right)}\right][/tex]
    [tex]J_{+}=\left [{|1⟩⟨0|⊗I⊗I}\right]+\left [{I⊗|1⟩⟨0|⊗I}\right]+\left [{I⊗I⊗|1⟩⟨0|}\right][/tex]
    [tex]J_{-}=\left [{|0⟩⟨1|⊗I⊗I}\right]+\left [{I⊗|0⟩⟨1|⊗I}\right]+\left [{I⊗I⊗|0⟩⟨1|}\right][/tex]
    [tex]J^{2}=\frac{1}{2}\left ({J_{+}J_{-}+J_{-}J_{+}}\right)+J_{z}^{2}[/tex]

    And get the right commutation relations:


    So I deduce [tex]J^{2}[/tex] and [tex]J_{z}[/tex] share a common set of eigenvectors [tex]\left\{{\left |{j,m}\right>}\right\}[/tex] where [tex]\left |{m}\right |\leq{j}\leq{\displaystyle\frac{3}{2}}[/tex]

    But there are only 6 of this vectors:

    [tex]\left\{{\left |{j=\frac{3}{2},m=\frac{3}{2}}\right>,\left |{j=\frac{3}{2},m=\frac{1}{2}}\right>,\left |{j=\frac{3}{2},m=-\frac{1}{2}}\right>,\left |{j=\frac{3}{2},m=-\frac{3}{2}}\right>,\left |{j=\frac{1}{2},m=\frac{1}{2}}\right>,\left |{j=\frac{1}{2},m=-\frac{1}{2}}\right>}\right\}[/tex]

    So my question is, does the sharing set of eigenvectors between two commuting observables must be complete?

    Why isn't complete in this example?
    What are the other 2 missing common eigenvectors?

    Thanks in advance! :smile:
  2. jcsd
  3. Sep 1, 2014 #2


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    Each qbit is like one spin 1/2, i.e., your 3-qbit space is the direct product of three spins 1/2.

    Now you can use the angular-momentum-addition rules to reduce out the irreducible representations of the rotation group. Then you find
    [tex]\frac{1}{2} \otimes \frac{1}{2} \otimes \frac{1}{2}=(0 \oplus 1) \otimes \frac{1}{2} = \frac{1}{2} \oplus \frac{1}{2} \oplus \frac{3}{2},[/tex]
    i.e., except of the irreducible representation 3/2 you have two more spin-1/2 representations, which makes the correct dimension 4+2+2=8.
  4. Sep 1, 2014 #3
    I'm not really into representation theory, but as far as I understand, the only thing that I get from there is the multiplicity of the eigenvalues... Could you explain your answer using linear algebra?

    I'm looking for the other two common eigenvectors.

    Thank you
  5. Sep 1, 2014 #4

    king vitamin

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    In order to totally determine the state, you need a maximal set of commuting operators. Your analysis has basically told you that you haven't done this yet: j and m aren't sufficient to determine the state. Vanhees was trying to send you on the right track by suggesting you consider adding the three spins pairwise. His equation is written in group theory notation, but the idea is simple: adding two spin-1/2s result in a spin-0 (1 dimensional) and a spin-1 (3 dimensional). Then, you add the last spin-1/2 to both of these results. The spin-1/2 and spin-0 give you a spin-1/2 (2-dim). The spin-1/2 and spin-1 gives you both a spin-1/2 (2-dim) and spin-3/2 (4-dim). So 8-dimensional as required.

    Now think about the above procedure. Clearly, enumerating the number of final states involved required knowing what the total angular momentum of the first two spin-1/2s is. We seem to have a commuting observable!
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