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I Spontaneous Symmetry breaking of multiplet of scalar fields

  1. Apr 18, 2016 #1

    CAF123

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    Consider a theory with two multiplets of real scalar fields ##\phi_i## and ##\epsilon_i##, where ##i### runs
    from 1 to N. The Lagrangian is given by: $$\mathcal L = \frac{1}{2} (\partial_{\mu} \phi_i) (\partial^{\mu} \phi_i) + \frac{1}{2} (\partial_{\mu} \epsilon_i) (\partial^{\mu} \epsilon_i) − \frac{m^2}{2}[\phi_i \phi_i+ \epsilon_i \epsilon_i] − \frac{g}{8}[(\phi_i \phi_i)(\phi_j \phi_j ) + (\epsilon_i \epsilon_i)(\epsilon_j \epsilon_j)] − \frac{λ}{2}(\phi_i \epsilon_i)(\phi_j \epsilon_j ),$$ where ##m^2 < 0, g > 0 ## and ##\lambda > −g/2.## Summation over repeated indices is implied.

    Is the following accurate? The lagrangian can be written in vector notation and we can see it is then invariant under a simultaneous transformation of ##\phi## and ##\epsilon## such that ##\epsilon \rightarrow R_{ij}\epsilon_j## and ##\phi_i \rightarrow R_{ij} \phi_j## if ##R_{ik} R_{ij} = \delta_{kj}## The symmetry group is then ##O(N) \otimes O(N)## with generators ##T_a^{O(N) \otimes O(N)} = T_a^{O(N)} \otimes \text{Id}_{N \times N} + \text{Id}_{N \times N} \otimes T_a^{O(N)}## so there are ##2 \cdot \text{dim}O(N)## number of generators.

    The vacua of the theory can be found as the minimum of the potential $$V(\phi, \epsilon) = \frac{m^2}{2} ( \vec \phi^T \vec \phi + \vec \epsilon^T \vec \epsilon) + \frac{g}{8} ((\vec \phi^T \vec \phi)^2 + (\vec \epsilon^T \epsilon)^2) + \frac{\lambda}{2} (\vec \phi^T \vec \epsilon)^2$$ I am a bit confused here - to find the vacua I think I could write $$\frac{\partial V}{\partial \phi^T \phi} = \frac{\partial V}{\partial \epsilon^T \epsilon} \overset{!}{=} 0$$ but what happens to the term proportional to ##\lambda##?

    Thanks!
     
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  3. Apr 18, 2016 #2

    ChrisVer

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    1. The Lagrangian can be written in vector notation.
    2. I am not sure about the symmetry group... it looks fine [but I didn't really think about it]. Why not SO(N) and O(N)xO(N) instead? Are you applying a different O(N) on each field ? then how is your lambda term (that mixes the phi and epsilon) invariant?
    In order for that term to be invariant I'd say you have to apply the same rotation on phi and epsilon, so that their product [itex] \phi \cdot \epsilon[/itex] will remain invariant. Am I wrong?
    3. For the minimum of the potential (or better said the "extreme"), you have:
    [itex] \frac{\partial V}{\partial \phi} = \frac{\partial V}{\partial \epsilon}=0[/itex]
    In general it's sometimes useful to proceed up to the second derivatives as well..
    Why did you take the derivatives wrt the fields' product?
    Sometimes it's always better to return to the known-grounds of the old-loved "Higgs-like-but-real-fiield"; and write:
    [itex]V = \mu^2 \Phi^T \Phi + \frac{1}{2}\lambda (\Phi^T \Phi)^2[/itex]
    the minimum of which you obtained by [itex]\frac{\partial V}{\partial \Phi} =0[/itex]
    no?
    The reason behind this in general (and also a good reminder to help you out with those stuff): you have a function with two variables [itex]f(x,y)[/itex], how do you procceed in finding the extreme points? if not by taking :[itex]|\nabla f(x,y)|=0[/itex]?
     
    Last edited: Apr 18, 2016
  4. Apr 19, 2016 #3

    CAF123

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    Hi ChrisVer
    I should have said that ##R## is also a real matrix but there was no condition on the determinant of R given so I could only be dealing with O(N) safely I suppose. It's the same O(N) matrix ##R## applied to each multiplet of fields so does this guarantee that the symmetry group is indeed O(N) and not a tensor product of two copies?
    And just to check something: If the symmetry group had been O(N) x O(N), then the number of generators associated to this group would have been twice the number of generators associated to O(N)? If correct, this would follow the same logic that so(3,1) (lorentz group) is isomorphic to two copies of su(2) and thus has 2 dim(su(2)) = 2 x 3 = 6 generators, which is indeed comprises the three rotations and three boosts. However, I don't see how it follows from the generic formula for generators in a product group: $$T_a^{R_1 \otimes R_2} = T_a^{R_1} \otimes \text{Id} + \text{Id} \otimes T_a^{R_2}$$ - because a is common to both sides and, as is clear on the rhs, a runs from 1 up to the dimension of the representation. I understand that the rhs produces a matrix which is bigger than that of the ##T_a^{R_i}## but I don't see why this means an increased number of generators for the lhs.

    I see. In the Higgs doublet case, because of the functional form, I just redefined ##\chi = \phi^T \phi## say and then took derivatives wrt ##\chi##. This would miss the trivial solution but that is irrelevant anyway for SSB. Better yet, I could have called ##\chi^2 = \phi^T \phi##. But yup I see that this redefinition is not of use here because of the coupled term in the two multiplets proportional to lambda. But proceeding with ##\partial V/\partial \phi_k = 0## I obtain $$m^2 \phi_k + \frac{g}{2} \phi_k (\phi_j \phi_j)+ \lambda \epsilon_k (\phi_j \epsilon_j) = 0$$ which in matrix notation is $$m^2 \vec \phi + \frac{g}{2} \vec \phi \vec \phi^2 + \lambda \vec \epsilon (\vec \phi^T \cdot \vec \epsilon) = 0$$ and similar equation for ##\epsilon##. I get the trivial solution ##(\vec \phi, \vec \epsilon) = (0,0)## and if I multiply the equation I get by ##\phi^T## and the one for epsilon by ##\epsilon^T## I get $$m^2 \phi^2 + \frac{g}{2} (\phi^2)^2 + \lambda( \phi^T \cdot \epsilon)^2 = 0$$ and $$m^2 \epsilon^2 + \frac{g}{2} (\epsilon^2)^2 + \lambda( \phi^T \cdot \epsilon)^2 = 0$$ Assuming that's fine, I then get $$m^2 \phi^2 + \frac{g}{2} (\phi^2)^2 = m^2 \epsilon^2 + \frac{g}{2} (\epsilon^2)^2 $$ Since the fields are independent this must be equal to a constant. Perhaps there is a flaw here somewhere because my notes tells me to distinguish relative parallel and orthogonal directions of the fields to find the vacua.

    Thanks!
     
  5. Apr 19, 2016 #4

    ChrisVer

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    I guess the flaw is that you have kept them as fields after you took the vacuum condition... they should have been vevs, and then you have 2 equations with 2 unknowns I think it's solvable.

    Speculative comment:
    I don't know what exactly you have to do, but taking parallel and orthogonal is always possible; you can rotate some of them out (I think 1 with each rotation?)...
    I mean the dot products would lead you to a [itex]\epsilon_{//} \phi_{//} + \epsilon_{T} \phi_{T} [/itex] (Vectors) and then you would have to find each vacuum...
    [itex]<\epsilon>= \begin{pmatrix} <\epsilon_{//}> // <\epsilon_{T}> \end{pmatrix}[/itex]
    ?
     
  6. Apr 19, 2016 #5

    CAF123

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    Ok, so I should write e.g $$m^2 \langle \phi \rangle + \frac{g}{2} \langle \phi \rangle \langle \phi^2 \rangle + \lambda \langle \epsilon \rangle \langle \phi^T \cdot \epsilon \rangle = 0 $$ with the ##\langle, \rangle## denoting the vev. This is just a scalar equation for ##\langle \phi \rangle## and ## \langle \epsilon \rangle##. So together with the other equation for epsilon, I have two equations with these two unknowns. Is that what you meant?

    Ok I see so decompose the dot product ## \vec \phi^T \cdot \vec \epsilon = \vec \phi^T_{//} \cdot \vec \epsilon_{//} + \vec \phi^T_{T} \cdot \vec \epsilon_T## but what to do about ##\langle \vec \phi \rangle ## etc for example? This seems to have introduced four variables and now only have two equations.
    Is there a typo in that equation? I didn't understand what it meant.

    Thanks! :)
     
  7. Apr 19, 2016 #6

    ChrisVer

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    the thing is that you have [itex]N[/itex] [itex]\phi[/itex] fields and [itex]N[/itex] [itex]\epsilon[/itex] fields...
    I guess writing them in the following way may help in seeing it?
    [itex] \phi = \begin{pmatrix} \phi_1 \\ \phi_2 \\ ... \\ \phi_N \end{pmatrix}[/itex], [itex] \epsilon= \begin{pmatrix} \epsilon_1 \\ \epsilon_2 \\ ... \\ \epsilon_N \end{pmatrix}[/itex]

    Now the [itex]\phi \cdot \epsilon[/itex] can be written as a parallel + transverse product by the following:
    [itex] \phi = \begin{pmatrix} \vec{\phi}_{//} \\ \vec{\phi}_T \end{pmatrix}[/itex], [itex] \epsilon= \begin{pmatrix} \vec{\epsilon}_{//} \\ \vec{\epsilon}_T \end{pmatrix}[/itex]

    And you can see the parallel and transverse parts as seperate fields... when you get the vev of your fields it will be the vev of your parallel and transverse fields, like:
    [itex] <\phi> = \begin{pmatrix} <\phi_{//}> \\ <\phi_T> \end{pmatrix}[/itex]
    Which is what I wrote but I mistyped parenthesis instead of \begin or end{pmatrix}

    Recall that for example the Higgs field vev is:
    [itex]<H> = \begin{pmatrix} <\phi^+> \\ <\phi^0> \end{pmatrix}[/itex]
    the only difference of which is that you can gauge away the several phi fields in the above giving you just the vev [itex]v_F[/itex] in the lower component/
     
  8. Apr 19, 2016 #7

    ChrisVer

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    yup... only that you don't take the vev of each field but substitute to each field its vev value... so things like [itex]<\phi^2>[/itex] would better be written as [itex]<\phi>^2[/itex] I guess...
     
  9. Apr 19, 2016 #8

    ChrisVer

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    The number of generators for the O(N) x O(N) will indeed be the number of genetors of O(N)*2
    But before going any further here, I don't see why your symmetry group is as given, so I am not sure how to explain it well... again I would only see an O(N) group (or SO(N) as I mentioned before), and not a product of the two....So there is something I am missing myself....
    However if your group is O(N)xO(N), it means that your field representation is given as (o1, o2)... on the first acts your first O(N) group generators and on the second act your second O(N)...
    I guess the main difference between this and the Lorentz group is that the Lorentz Group is not SU(2)xSU(2), as you also pointed out they only have equivalent Lie algebras. Note that the SU(2) xSU(2) is a compact group something that doesn't work well with boosts.
    So let me rewrite your group as [itex]O_1(N) \times O_2(N)[/itex] and each has the "same" generators [itex]T^a_1, T^a_2[/itex].
    Now a field that exists in the [itex](o_1, o_2)[/itex] representation of your group, under its action it will get its first components (those in [itex]o_1[/itex]) transformed by the action of [itex]T^a_1[/itex] and leaving those in [itex]o_2[/itex] unchanged (thus the Identity) and the same for the second components transformed by the action of [itex]T^a_2[/itex] and leaving the [itex]o_1[/itex] unchanged... Does this help?
    How many generators do you have??? well... at a first thought I would say 2 times the gnerators of O(N)... but they come in pairs...
    .
     
  10. Apr 20, 2016 #9

    ChrisVer

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    if you read my last post, you will also understand why I am not very convinced about you group choice (btw do you have a reference?).... I don't see which field of yours transforms under the 1 or the 2 group... Obviously you can allow for phi to be in the repr of 1, while the epsilon to be in the repr of 2... But then if you then make a transformation you will have:
    [itex](\epsilon_i \phi_i)(\epsilon_r \phi_r) \rightarrow (\epsilon^\prime_i \phi^\prime_i )(\epsilon_r^\prime \phi_r^\prime) = \big(R_2 (\theta)\big)_{ji} \big(R_1(\Theta)\big)_{ik} \big(R_2 (\theta)\big)_{mr} \big(R_1(\Theta)\big)_{rn} (\epsilon_j \phi_k)(\epsilon_m \phi_n)[/itex]
    to my understanding in order for this to be invariant you must have that you rotate them with the same angle and also that the two transformations have to be identical...with the last I mean that you can't rotate the 1st and 2nd component of phi, and the 5th and 6th component of epsilon...
    Otherwise I would say that the mixing term [itex]\lambda[/itex] is explicitly breaking your symmetry. Something like [itex]O_1(N) \times O_2(N) \rightarrow O(N)[/itex].
    If you are right, I would like someone to point out my "mistake".

    Another possibility would be for phi,epsilon to be in mixtures of 1 and 2 ...I haven't put much thought on this but I find it unlikely to happen...because they wouldn't have N-components I guess...

    @samalkhaiat maybe can help (if he has the time :rolleyes:)?
     
    Last edited: Apr 20, 2016
  11. Apr 20, 2016 #10

    CAF123

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    Apologies, I should have made myself more clear - I am in agreement now with you that the symmetry group is not O(N) x O(N) but just O(N). From what has been said, this is because the fields phi and epsilon transform according to the same R in O(N). I don't say SO(N) because there was no condition given in the paragraph about whether det R=1 so I just say O(N). The discussion about O(N) x O(N) (carried on below) was simply just to clarify some questions I have about generators of product group.

    Yup this is the formula I posted in one of my replies - ##T_a^{R_1 \otimes R_2} = T_a^{R_1} \otimes \text{Id} + \text{Id} \otimes T_a^{R_2}##. ##R_1 = R_2 ## stand for a representation of O(N) and the ##T_a^{R_i}## are the corresponding representation of the generators. a runs from 1 to dim O(N). This equation gives a representation of the generators in the tensor product group given a representation of the O(N) generators. My question is, why does this equation imply I can find 2 * O(N) number of generators for the product group?

    Thanks!
     
  12. Apr 20, 2016 #11

    ChrisVer

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    Because you have dim[O(N)] [itex]T_1^a[/itex] and dim[O(N)] [itex]T_2^a[/itex].
     
  13. Apr 20, 2016 #12

    CAF123

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    Ok so, given $$V(\phi, \epsilon) = \frac{m^2}{2} (\phi_i \phi_i + \epsilon_i \epsilon_i) + \frac{g}{8} ((\phi_i \phi_i) (\phi_j \phi_j) + (\epsilon_i \epsilon_i) (\epsilon_j \epsilon_j)) + \frac{\lambda}{2} (\phi_i \epsilon_i) (\phi_j \epsilon_j)$$ which in vector notation is $$V(\phi, \epsilon) = \frac{m^2}{2} (\phi^T \phi + \epsilon^T \epsilon) + \frac{g}{8}((\phi^T \phi)^2 + (\epsilon^T \epsilon)^2) + \frac{\lambda}{2} (\phi^T \epsilon)^2$$ I can then decompose it into its parallel and orthogonal components: $$V = \frac{m^2}{2} (\phi_{//}^2 + \phi_T^2 + \epsilon_{//}^2 + \epsilon_T^2) + \frac{g}{8} ((\phi_{//}^2 + \phi_T^2 )^2 + (\epsilon_{//}^2 + \epsilon_T^2)^2) + \frac{\lambda}{2} (\phi_{//} \cdot \epsilon_{//} + \phi_T \cdot \epsilon_T)^2$$ Now should I then proceed by finding $$\frac{\partial V}{\partial \phi_{//}} = 0 = \frac{\partial V}{\partial \phi_T}?$$ Upon differentiating wrt ##\phi_{//}## I get $$0 = m^2 \langle \phi_{//} \rangle + \frac{g}{2}( \langle \phi_{//}^2 + \phi_T^2) \rangle \langle \phi_{//} \rangle + \lambda (\langle \phi_{//} \cdot \epsilon_{//} + \phi_T \cdot \epsilon_T \rangle) \langle \epsilon_{//} \rangle?$$ I guess I am just getting confused with the notation here. Thanks :)
     
  14. Apr 20, 2016 #13

    ChrisVer

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    yes... I would write the vevs with a letter [itex]x_{p},x_{t}[/itex] for [itex]\phi[/itex] and [itex]y_p,y_t[/itex] for [itex]\epsilon[/itex] and p/t denoting the parallel/transverse.
    That way your equations become:
    [itex] m^2 x_p + \frac{g}{2} ( x_p^2 + x_t^2) x_p + \lambda ( x_p y_p + x_t y_t) y_p =0[/itex]
    Similarily other 3 equations.
     
  15. Apr 20, 2016 #14

    CAF123

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    I see, this is going to be a coupled set of four (non linear) equations in four variables. Looks like it might be a pain to solve
     
  16. Apr 20, 2016 #15

    ChrisVer

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    I guess you could rotate your fields so that you can choose [itex]\phi_{//}=0[/itex] ??? what do you think?
     
  17. Apr 20, 2016 #16

    CAF123

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    So this would correspond to setting some of the N fields to be vanishing on the vacuum? I suppose the parallel and orthogonal are terms meaning parallel and orthogonal relative to the vacuum state? But the field ##\phi## is scalar so how can it have parallel and orthogonal components in the first place?
     
  18. Apr 20, 2016 #17

    ChrisVer

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    The general term " scalar" means that it's invariant under Lorentz Transformations.
    You rotate it away before looking for the vacuum.
    It's a little difficult to explain the parallel and orthogonal terms... you have a N-dimensional space (or equivalently N scalar fields)... you can choose some phi's component to represent the "parallel" direction in that space...
    What you do by rotating then your N-vector [itex]\phi[/itex] is bringing it to such a configuration where the parallel component vanishes?.
     
  19. Apr 20, 2016 #18

    samalkhaiat

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    If nothing in the Lagrangian distinguishes between the [itex]\phi_{i}[/itex]'s and the [itex]\epsilon_{i}[/itex] fields, the model will be equivalent to an [itex]O(2n)[/itex] invariant theory with [tex]\Phi = \begin{pmatrix} \phi_{i} \\ \epsilon_{i} \end{pmatrix} \equiv \begin{pmatrix} \varphi_{1} \\ \varphi_{2} \\ \vdots \\ \varphi_{2n} \end{pmatrix} [/tex] In this case the symmetry breaking is just [itex]O(2n) \to O(2n-1)[/itex].
    If you keep the two fields in the Lagrangian, your model is still [itex]O(n)[/itex]-invariant. So, the [itex]O(n)[/itex]-invariant potential can depend only on the magnitudes of the vectors [itex]\vec{\phi} \cdot \vec{\phi}[/itex], [itex]\vec{\epsilon}\cdot \vec{\epsilon}[/itex], and the scalar product [itex]\vec{\phi} \cdot \vec{\epsilon} = \phi \epsilon \cos \theta[/itex] [tex]V = V(\phi , \epsilon , \cos \theta)[/tex]
    The minimum of [itex]V[/itex] determines the values of these variables [itex]\phi^{0} = v_{1}[/itex], [itex]\epsilon^{0} = v_{2}[/itex], [itex](\cos \theta)^{0} = \cos \alpha[/itex]. So, you have a plane defined by these minimas, which you can take it to be [itex](\phi_{n} , \phi_{n-1})[/itex] plane. The two vectors [itex]\vec{\phi}[/itex] and [itex]\vec{\epsilon}[/itex] can have non-vanishing components in the last two intries. As one of infinitely many choises, here is a simple one [tex]\vec{\phi} = (0,0, \cdots , v_{1}) , \ \ \ \vec{\epsilon} = v_{2}(0,0, \cdots , \sin \alpha , \cos \alpha )[/tex] These clearly are invariant under the rotations of the first [itex](n-2)[/itex] components, hence the symmetry beaking pattern [itex]O(n) \to O(n-2)[/itex]. Of course, it is possible to have [itex]\alpha = 0[/itex] as the minimum solution. This can happen when [itex]V[/itex] depends on even powers of [itex]\cos \theta[/itex] and the coefficient of the [itex]\cos^{2}\theta[/itex] is negative. In this case the two vectors are parallel and the plane simply become a line and we have [itex]O(n) \to O(n-1)[/itex].
    In general, having a model with [itex]m[/itex] [itex]O(n)[/itex]-vectors leads to [itex]O(n) \to O(n-m)[/itex] beaking pattern.
     
    Last edited: Apr 21, 2016
  20. Apr 21, 2016 #19

    CAF123

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    How do we know it is a plane of minima?
    I'll try to see if I understand what you wrote: In the general configuration, ##\vec \phi## and ##\vec \epsilon## are at some arbitrary angle to each other so that e.g, ##\vec \phi \cdot \vec \epsilon = \phi \epsilon \cos \theta##. At the vacuum this is then ##\langle \vec \phi \cdot \vec \epsilon \rangle = \langle \vec \phi \rangle \langle \vec \epsilon \rangle \langle \cos \theta \rangle = v_1 v_2 \cos \alpha##. Assuming the fields are parallel at the vacuum, ##\cos \alpha=1## and ##\vec \phi = (0,0,...,v_1)## and ##\vec \epsilon = (0,0,...., v_2)##. This means ##O(n) \rightarrow O(n-1)##. If othogonal at vacuum, then ##\cos \alpha = 0## and so ##\vec \phi = (0,0,..., v_1)## and ##\vec \epsilon = (0,0,..., v_2,0)## This would also imply a O(n-1) residue symmetry but in different planes for the ##\vec \phi ## and ##\vec \epsilon## fields?

    Could you explain also why ##\alpha=0## can only be possible if '[itex]V[/itex] depends on even powers of [itex]\cos \theta[/itex] and the coefficient of the [itex]\cos^{2}\theta[/itex] is negative'

    What I was trying to do was solve for ##v_1, v_2## explictly in terms of the input parameters into the lagrangian (g, lambda). Without any external assumptions about what these minima would look like I was getting non linear coupled equations. As ChrisVer intially suggested, would you suggest ( in the notation used) to set ##\phi_{//}=0=\epsilon_{//}## say, thereby simplifying the equations?

    Thanks!
     
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