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SU(N) symmetry breaking by non-trivial parity.

  1. May 17, 2015 #1
    I would like to prove the following:
    Suppose we have the diagonal matrix ##P = diag(1,\ldots,1, -1,\ldots, 1)##, with ##N_+## elements of ##1## and ##N_-## elements of ##-1## such as ##N_+ + N_- = N## and ##N_+, N_- \geq 1##.
    This matrix is a non trivial parity matrix since it is not proportional to ##I##.
    If ##T^\alpha## are the generators of ##SU(N)## and ##[T^\alpha, P] = 0##, I would like to prove that ##SU(N)## breaks as:
    ##SU(N) \rightarrow SU(N_+) \otimes SU(N_-) \otimes U(1).##
    I am particularly interested in the case of ##SU(3) \rightarrow SU(2) \otimes U(1).##
    Thanks in advance!
     
  2. jcsd
  3. May 17, 2015 #2
    Let's decompose your parity operator into two projectors ##P=P_{+} - P_{-}## with ##P_{+}+P_{-1}=1##, ##P_{+/-}^2 = P_{+/-}## and ##P_{+}P_{-}=0##. Then an element ##g## from the Lie algebra of ##SU(N)## commutes with ##P## like ##[g,P]=[g,P_{+}-P_{-}]=[g,P_{+}]-[g,P_{-}]=gP_{+}-P_{+}g - gP_{-}+P_{-}g##. Multiply with ##1=P_{+}+P_{-}## from left and right to get ##[g,P]=2(P_{-}gP_{+}-P_{+}gP_{-})##. Because [A,B]=0 implies that P_{+/-}[A,B]=0, your condition of [g,P]=0 implies ##P_{-}gP_{+}=0## and ##P_{+}gP_{-}=0## independently. This means the generators commuting with your parity operator live entirely in the complement of the kernel of either projector.

    The lie algebra ##u(n)## is spanned by all Hermitian operators on ##\mathbb{C}^n##. So we're left to show that ##P_{+/-}HP_{+/-}## for all trace-free Hermitian ##H## spans exactly the space of all Hermitian operators on the subspace generated by the respective projector. Clearly, ##QHQ## is Hermitian for any projector Q and any Hermitian H. It's also simple to see that we cover all: Let ##S## be a Hermitian operator on the subspace, then we can write it as ##S=QSQ=Q(S+K)Q## where ##K## is Hermitian and lives in the complement. That means ##S+K## is a Hermitian operator in the superspace and we can be sure to reach ##S## from there. We are also sure that ##K## can be chosen so that ##S+K## is trace free. That means your subspace generators span ##u(N_{+/-})##.

    So your conditions breaks ##SU(N)## into ##U(N_+)## and ##U(N_-)##. The ##U(1)## factors of both unitary groups are coupled as fixing one determines the other. (If you can't see this directly, look at the direct sums of the corresponding lie algebras and the global trace that splits into two subtraces that sum to 0.) That means in fact they both share the same ##U(1)##, and you have

    ##SU(N) \mapsto SU(N_+)\otimes SU(N_-) \otimes U(1)##

    as you expected.

    Cheers,

    Jazz
     
    Last edited: May 17, 2015
  4. May 19, 2015 #3
    Could you please explain:
    1. What is meant by the direct product of the sub-groups?
    2. What is the role of U(1), in this direct product?
     
  5. May 19, 2015 #4
    These are very basic questions, and I would expect that you know the concepts before asking about symmetry breaking. I can't give you a math lecture here, so I suggest you pick up a good textbook introducing group theory, Lie groups and representation theory. There are many such books with a specific focus on physics and for all levels of abstraction. If you have questions about my proof above, I'll be happy to answer them.

    Cheers,

    Jazz
     
  6. May 25, 2015 #5
    Thank you for your answers, however I cannot understand how U(1) comes along in this product!
     
  7. May 25, 2015 #6
    The Lie group U(n) factors into SU(n) and U(1). In terms of the corresponding Lie algebras this means, that u(n) is the direct sum of su(n) and u(1). The algebra u(n) is spanned by all Hermitian matrices in ##\mathbb{C}^{n\times n}## while su(n) is spanned by the trace free Hermitian matrices. You can decompose a Hermitian matrix ##H## into a traceless matrix ##H_0## and a scalar multiple of the identity matrix ##I##: $$H=H_0 + c\cdot I$$
    The space spanned by ##I## is an ##n## dimensional representation of the lie algebra ##u(1)## and therefore we have in terms of Lie algebras:
    $$u(n)=su(n)\oplus u(1)$$

    This is the situation we have for each of the sectors ##+## and ##-## as described in my earlier post. The only difference is that the decomposition happens per sector. Each sector contains a full ##U(N_{+/-})## representation, that means we have two algebras spanned by the Hermitian matrices in each sector. We can factor each sector in the same way as above, separating a multiple of the identity matrix from the Hermitian matrices leaving traceless matrices. However, and that's the important point here, the decomposition comes with a constraint: Since we are decomposing a group ##SU(N)##, the global trace must be ##0##. The sector traces however sum to the global trace, that means the individual sector traces must be chosen to sum to 0, one trace determining the other. In terms of the Lie algebras it means that the two sector-##u(1)## components cannot be picked independently, they live in the same 1-dimensional subspace and generate a single one dimensional Lie group. To be explicit:

    $$su(N) = su(N_+) \oplus su(N_-) \oplus u(1)$$

    I hope that makes it clearer,

    Jazz
     
  8. May 26, 2015 #7

    Strilanc

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    Science Advisor

    I have a concrete follow-up question, based on a misunderstanding.

    A controlled-Z operation on ##n## qubits is represented by the matrix ##I - 2 \left| T \right\rangle\left\langle T \right|## where ##T## is the all-qubits-on state. ##T## comes last in the state vector, so the ##n## bit operation's matrix is a ##2^n##-by-##2^n## matrix with 1s along the diagonal, except the bottom-right value is -1. For example, the 3-bit operation has a matrix like this:

    ##\begin{bmatrix}
    1 & 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 & 1 & 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 & 1 & 0 \\
    0 & 0 & 0 & 0 & 0 & 0 & 0 & -1
    \end{bmatrix}##

    The above matrix is an example of the matrices under question. Anyways, I was under the impression that breaking this operation into smaller operations required some of the operations to overlap. For example, you can break it into several singly-controlled nots and quarter-turn phase gates:

    PFOrWl8.png

    But it shouldn't be possible to split the operation into non-overlapping pieces because that would allow communication across wires without any interaction.

    Given that the controlled-Z is an example of the matrices being talked about, and apparently we can factor those matrices into ##SU(2) \otimes SU(1) \otimes U(1)## parts, why doesn't that allow us to factor the circuit operations into operation-affecting-first-2-wires, operation-affecting-last-wire, and ignored-global-phase-factor?
     
  9. May 26, 2015 #8

    Strilanc

    User Avatar
    Science Advisor

    Argh, I figured it out immediately after posting. It doesn't factor into ##SU(2) \otimes SU(1) \otimes U(1)##, it factors into ##SU(7) \otimes SU(1) \otimes U(1)##. I mixed up the number of wires being operated on with the number of +1s and -1s appearing on the diagonal and created some nonsense.
     
  10. Jun 6, 2015 #9
    This time, I would like to ask about the case of:
    ##SU(2)\otimes U(1) \rightarrow U(1)\otimes U(1),## spontaneous symmetry breaking.

    It is given that the Wilson Loop:
    ##W \equiv exp[ig\int_{-\pi R}^{\pi R}{B_y^1 T^1 dy}] = diag(-1, -1, 1).##

    And as we can check, the Wilson Loop does not commute with SU(3) generators ##T^6, T^7## but it still commutes with ##T^3, T^8.##
    How does this SSB occur?
    The parity in this case is ##diag(1, -1, -1).##
     
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