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I Isospin conservation - an application

  1. May 6, 2016 #1
    Hi guys!

    I'm trying to understand how to use the fact that the total isospin is conserved in all strong processes in the particular case (vide Griffiths, pages 118 and 119):
    $$ p+p \rightarrow d+\pi^+ $$
    Griffiths first argues that the deuteron d has isospin I=0, because of experimental reasons, basically. He uses this to conclude that "the isospin states on the right are |11>, |10>,
    and |11>, respectively, whereas those on the left are I1 1>, $$\frac{1}{\sqrt{2}}(|10>+|00>)$$
    and |1 - 1>"
    I do not understand how he concludes this. Could you please enlighten me?
  2. jcsd
  3. May 6, 2016 #2
    Deuterium is an “iso-singlet”, I=0 state is |0,0>

    In terms of isospin states we have:

    p p=|1/2,1/2> and |1/2,+1/2> ; d pion+ =|0,0> and |1,1>

    If we use the same techniques as is used to combine angular momentum in QM then

    we can go from 1/2 basis to the 1 basis.

    For pp, and d ,pion+ there is only one way

    to combine the spin states:

    p and p=|1/2,1/2> and |1/2,+1/2>=|1,1>

    d and pion+=|0,0> and |1,1> = |1,1>
  4. May 7, 2016 #3
    I'm confused.

    I agree with p = |1/2, 1/2> . With d = |0,0> I also agree. I do not know why you say that ##\pi^+=|1,1>##. What does this have to do with "the isospin states on the right are |11>, |10>,
    and |11>, respectively" that Griffiths said?


    I do not see why. Let me see if I got this right:
    When we combine the angular momentums of the two protons (##Î = Î_{p1}+Î_{p2}##), we get 1 singlet with I=0 and 1 triplet with I=1. These are the kets of the eigenbasis common to ##Î^2## and ##Î_3## of our state space. We have another basis, which is the eigenbasis common to ##Î_3(p1)## and ##Î_3(p2)##. We can write the kets of the first basis in terms of the kets of the second:
    $$|0,0> = \frac{1}{\sqrt{2}} (|1/2,-1/2>-|-1/2,1/2>)$$
    $$|1,1> = |1/2,1/2>$$
    $$|1,-1> = |-1/2,-1/2>$$
    $$|1,0> = \frac{1}{\sqrt{2}} (|1/2,-1/2>+|-1/2,1/2>) $$
    I'm not understanding how to relate this with what you said:
    Honestly, I didn't even understand what you mean...

    Thank you!
    Last edited: May 7, 2016
  5. May 7, 2016 #4
    i think you are combining isospin in the same manner as you combine spin in
    a two particle bound state say two electrons in He.

    but here the reaction is a scattering event - the p and p is not forming a combined state of the system
    that is if you take [ 1/2, -1/2 ] as possible state for a particle then it will represent neutron and the
    reaction will be different a p n scattering will give rise to d and pi(0)
    for example

    p +n = d + pi(0)

    p and n=|1/2,1/2> and |1/2,-1/2> and on the rt hand side we have d and pi(0) =|0,0> and |1,0>

    well i think the singlet and triplet states of isospin of a pp or dpi+ can be described if the system is a bound state on nuclear time scale.
  6. May 7, 2016 #5


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    I could not make sense of that statement by Griffiths until I looked in his book. You did not include the whole statement! In my copy, he gives THREE different reactions and then says that the left hand side correspond, respectively, to |11>, |10>, and |11>. The ##p+p \rightarrow d+\pi^+## is the first reaction so he is saying that the lhs is |11> (and ONLY |11>) and the rhs is also |11> for that reaction! The other values he lists for the lhs and the rhs correspond to the other two reactions listed below.
  7. May 7, 2016 #6
    You're absolutely right!
    Even so, I do not understand how he concludes that the lhs is |11> and that the rhs is |11> too! In fact, I also don't understand why he says that in the other two reactions
    $$p+n\rightarrow d+\pi^0$$
    $$n+n\rightarrow d+\pi^-$$
    he says that the isospin states are |10> and |1-1> on the lhs and ##\frac{1}{\sqrt{2}} (|10> + |00>)## and |1-1> on the rhs, respectively.
    What does he mean by saying that "the isospin state on the lhs is..."? You have two particles (p and n for example) which will react. The only way I see to have an isospin state on the lhs is by adding the isospins just as I did above...
  8. May 7, 2016 #7


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    Just by combining the isospin like we combine spin 1/2. On the lhs we are combining ##|1/2,1/2 \rangle ## with ##|1/2, 1/2 \rangle##. This automatically results in the state ##|1,1 \rangle##.
    On the rhs, the deuteron is an isosinglet so combining it with the ##\pi^+## automatically gives the isospin of the ##\pi^+##, which is ##|1 ,1 \rangle##.

    For the other two reactions, we just have to combine the isospin to get his results.
  9. May 7, 2016 #8


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    By the way, he does not say that the isospin state of the lhs of the second reaction is |1,0>. He says that it is ## |p,n> = \frac{1}{\sqrt{2}} (|1,0\rangle + |0,0 \rangle ) ##.
  10. May 7, 2016 #9
    I thought it automatically resulted in 1 singlet ##|0,0\rangle## and a triplet ##|1,-1\rangle,\,|1,0\rangle,\, |1,-1\rangle##. Why do we select the state ##|1,1\rangle##?

    I have the same question about the pion+: why is it ##|1,1\rangle##? The addition of the up-quark and anti-down-quark isospins give us one singlet and one triplet... Why do we select ##|1,1\rangle##?

    Note: In my previous post, I wrote lhs when I should have written rhs and vice-versa.
  11. May 7, 2016 #10


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    Let's say we combine the states ##|s_1=\frac{1}{2}, m_{s_1}=\frac{1}{2} \rangle ## with ##|s_1=\frac{1}{2}, m_{s_1}=\frac{1}{2} \rangle ##, what we get is ##|s=1,m_s=1 \rangle ## and nothing else. We can see this by using Clebsch-Gordan tables but we can also see this directly: the two z components must add up so that ##m_S = \frac{1}{2} + \frac{1}{2} = 1 ## and since the z component is 1, the total spin must be 1 (it cannot be zero). So we get |1,1>.
  12. May 7, 2016 #11
    But what you're doing is writing the state ##|m_1=1/2,m_2=1/2\rangle## in terms of the basis kets {##|S,M>##}. i.e. you're changing the basis. I have no problems with that.
    However, i think your answer did not justify why one should choose ##|1,1>##. You simply explained why ##|1,1> = |1/2,1/2>## (where the rhs is in the eigenbasis of (Sz1, Sz2) and the lhs is in the eigenbasis of (S, Sz), with S=S1+S2. I repeat: this only explains how the change of basis was carried out. It does not explain why one should choose ##|1,1>## (or ##|m_1=1/2,m_2=1/2\rangle## - it's the same thing, only in a different basis) as the isospin state of the system in the first place.

    I think I'm beginning to understand what my problem is:
    When you combine two particles, each described by a state which "lives" in a two dimensional state space ##\epsilon_i## spanned by ##{|+\rangle,|-\rangle}## (like in the spin(or isospin) 1/2 case), then your combined state "lives" in the state space given by the tensor product of those two state spaces, giving a state space ##\epsilon = \epsilon_1 \otimes \epsilon_2## spanned by ##{|+,+\rangle, |+,-\rangle, |-,+\rangle, |-,-\rangle}##. Until now I think everything is correct. Now, my mistake: I was thinking of the proton (and the neutron, for that matter) as a particle "living" in a 2-dimensional isospin state space, like the ones I described above. That is, I was thinking as if the state of the proton could be written as an arbitrary linear combination of the basis kets:
    $$|p\rangle = \alpha |+\rangle + \beta |-\rangle$$
    But this is incorrect! In fact: ##|p\rangle = |+\rangle##. i.e. the state of the proton isn't described by some arbitrary ket in the isospin state space. It is a concrete ket belonging to that space. That said, I need not find all the basis kets which span ##\epsilon##. I apply the tensor product between two kets which are fixed. And that's easy:
    In the case with a proton and a neutron on the lhs, for example:
    $$|p\rangle \otimes |n\rangle = |1/2, -1/2\rangle$$
    And now I'm ready do apply the change of basis. In the case of the proton+neutron:
    $$|1/2,-1/2\rangle = \frac{1}{\sqrt{2}}(|1,0\rangle+|0,0\rangle)$$

    I'm sorry if this got a little confusing, but I think it's correct...
    Last edited: May 7, 2016
  13. May 7, 2016 #12


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    That's it.

    I don't understand that line, what one should write is

    $$ |p\rangle \otimes |n\rangle = |\frac{1}{2}, \frac{1}{2} \rangle \otimes | \frac{1}{2}, - \frac{1}{2} \rangle = \frac{1}{\sqrt{2}} \bigl( |1,0\rangle + | 0,0 \rangle
  14. May 7, 2016 #13
    I wrote the proton+proton in the end, instead of the proton+neutron... I'll correct that.

    Let me try to be clearer, anyway:

    $$|p\rangle \otimes |n\rangle = |I_1=1/2, i_1=1/2\rangle \otimes |I_2=1/2, i_2=-1/2\rangle = |I_1=1/2, i_1=1/2; |I_2=1/2, i_2=-1/2\rangle = |i_1=1/2, i_2=-1/2\rangle = |1/2,-1/2\rangle$$
    i.e. I've omitted the total isospin since it's always 1/2. I'm working in the eigenbasis common to ##î_1## and ##î_2##.
    I can now write it in the eigenbasis common to ##Î## and ##î=î_3##:
    $$|1/2,-1/2\rangle = \frac{1}{\sqrt{2}}(|1,0\rangle+|0,0\rangle)$$
  15. May 7, 2016 #14


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    Ah ok, I did not understand the notation you were using. Yes, you got it.
  16. May 7, 2016 #15
    A little off topic but...

    I never understood isospin at all until I had a proper understanding of representations of Lie Groups - after that it was very easy!
    I would strongly recommend first learning the basics of representations for SU(2) and SU(3), all will become clear to you.
    If you learn how to construct tensor product representations and do tensor product decompositions, then calculations like these can be accomplished using a diagrammatic technique (Young Tableaux). It is also nice to see the quark model states as points on the weight lattice.
  17. May 7, 2016 #16


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    Since you are starting to study particle physics, it is good time to learn about group theory and its representations. After all particle multiplets are nothing but irreducible representations of the group in question. The group action (i.e., transformation) causes the states (or particles) of an irreducible representation to mix with each other. But particles belonging to different irreducible representations transform differently. That is to say that irreducible representations form invariant vector spaces (i.e., the group acts differently in each representation space). For example for [itex]SU(2)[/itex] the nucleon [itex]N[/itex] or quark [itex]q[/itex] (doublets)
    [tex]N(q) = \begin{pmatrix} p(u) \\ n(d) \end{pmatrix} \equiv \begin{pmatrix} | \frac{1}{2} , \frac{1}{2} \rangle \\ | \frac{1}{2} , - \frac{1}{2} \rangle \end{pmatrix} ,[/tex] belong to 2-dimensional representation space ([itex]I = 1/2[/itex]) called the fundamental (or defining) representation, hence the name “doublet”. And the Pions (triplet) belong to 3-dimentional irreducible representation space ([itex]I = 1[/itex]) called the adjoint representation
    [tex]\Pi = \begin{pmatrix} \pi^{+} \\ \pi^{0} \\ \pi^{-} \end{pmatrix} \equiv \begin{pmatrix} |1,1\rangle \\ |1,0\rangle \\ |1,-1\rangle \end{pmatrix} .[/tex]
    The convention here is as follow, the first entry is the particle of the highest electric charge or the state with the greater [itex]I_{3}[/itex]. The two are related by the Gell-Mann-Nishijima relation [tex]Q = I_{3} + \frac{B}{2} ,[/tex] where [itex]B[/itex] is the baryon number. Notice that the dimension of a representation space is given by [itex]D = 2I + 1[/itex] which is number of different values [itex]I_{3}[/itex] can take in the representation.
    By taking the tensor product of irreducible representation we obtain higher-dimensional representation. However, in general the tensor product representations are not irreducible. So, to obtain irreducible representations we decompose the tensor product into irreducible ones by (for example) decomposing the tensor into symmetric and antisymmetric parts, and/or subtracting an invariant trace.
    Let me show you how this is done for of the nucleon and the quark doubles.
    Let us consider [itex]N \otimes N^{T}[/itex] (which we regard as direct product of matrices)
    [tex]\begin{pmatrix} p \\ n \end{pmatrix} \otimes \begin{pmatrix} p & n \end{pmatrix} = \begin{pmatrix} pp & pn \\ np & nn \end{pmatrix} .[/tex]
    Now, any matrix has a unique decomposition into symmetric and antisymmetric matrices
    [tex]\sqrt{2}M_{ij} = \frac{1}{\sqrt{2}} (M_{ij} + M_{ji}) + \frac{1}{\sqrt{2}} (M_{ij} - M_{ji}) .[/tex]
    Using this we can write
    [tex]\sqrt{2}\begin{pmatrix} pp & pn \\ np & nn \end{pmatrix} = \begin{pmatrix} \sqrt{2} pp & \frac{1}{\sqrt{2}}(pn + np) \\ \frac{1}{\sqrt{2}}(pn + np) & \sqrt{2} nn \end{pmatrix} + \begin{pmatrix} 0 & \frac{1}{\sqrt{2}}(pn - np) \\ \frac{-1}{\sqrt{2}}(pn - np) & 0 \end{pmatrix} . \ (1)[/tex]
    The symmetric matrix on the RHS has three independent and symmetric components (or states). Thus, these 3 states span a 3-dimentional vector space [itex][3_{S}][/itex]. This (from [itex]D = 2I +1[/itex]) implies [itex]I = 1[/itex]. The second (antisymmetric) matrix on the RHS of (1) has one antisymmetric component hence a one dimensional vector space [itex][1_{A}][/itex] corresponding to [itex]I = 0[/itex].
    In the representation theory, Eq(1) is written as equation between (the dimension of) vector spaces [tex][2] \otimes [2] = [3] \oplus [1] .[/tex] Or, In the language of angular momentum, it corresponds to the vector addition of two [itex]I = 1/2[/itex] nucleons
    [tex]\vec{\frac{1}{2}} + \vec{\frac{1}{2}} = \vec{1} + \vec{0} .[/tex]
    Now, using the above mentioned convention regarding the charges of the particles and the [itex]I_{3}[/itex] of the states (which we can obtain from [itex]Q = I_{3} + B/2[/itex]), we are led to the identification
    [tex]\begin{pmatrix} pp \\ \frac{1}{\sqrt{2}} ( pn + np ) \\ nn \end{pmatrix} = \begin{pmatrix} |1,1\rangle \\ |1,0\rangle \\ |1,-1\rangle \end{pmatrix} , \ \ \ \ \ (2)[/tex]
    [tex]\frac{1}{\sqrt{2}} ( pn - np ) = | 0 , 0 \rangle . \ \ \ (3)[/tex]
    Solving (2) and (3) gives us
    [tex]| pn \rangle \equiv | \frac{1}{2} , \frac{1}{2} \rangle | \frac{1}{2} , - \frac{1}{2} \rangle = \frac{1}{\sqrt{2}} \left( | 1 , 0 \rangle + | 0 , 0 \rangle \right) ,[/tex] [tex]| np \rangle \equiv | \frac{1}{2} , - \frac{1}{2} \rangle | \frac{1}{2} , \frac{1}{2} \rangle = \frac{1}{\sqrt{2}} \left( | 1 , 0 \rangle - | 0 , 0 \rangle \right) . [/tex]
    Okay, I leave you to apply the same method to the quark and antiquark doublets to obtain the quarks content and the [itex]|I , I_{3} \rangle[/itex] states of the Pions. You need to observe just one mathematical issue regarding the equivalence of the two fundamental representations of [itex]SU(2)[/itex]. That is the two doublets [itex]q[/itex] and [itex]\bar{q}[/itex] transform in the same way. So, it is necessary to take the conjugate representation to be
    [tex]\bar{q} = \left( - \bar{d} , \bar{u} \right) = \left( | \frac{1}{2} , \frac{1}{2} \rangle \ , \ | \frac{1}{2} , - \frac{1}{2} \rangle \right) .[/tex]
    Now do the same thing to the matrix
    [tex]q \otimes \bar{q} = \begin{pmatrix} - u \bar{d} & u \bar{u} \\ - d \bar{d} & d \bar{u} \end{pmatrix} .[/tex]
    Good luck.
  18. May 9, 2016 #17


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    It's quiet the same as with spins...
    If you have 2 particles with spin [itex]S^{(i)}= 1/2[/itex] (their [itex]S^{(i)}_z = \pm 1/2[/itex]) with [itex]i=1,2[/itex] for particle 1 and 2 respectively...
    What's the total spin of the system you get? Well you get [itex]S^{(1)}+ S^{(2)} = 1 + 0[/itex]
    or the triplet and the singlet...
    In the general case where there is ambiguity in the spin orientation of the particles your triplet would have the |1 1>, |1 0> and |1 -1> reps in {S,Sz}basis.
    The protons are though Iz=+1/2 particles, and so (like adding two spin-ups) you get the |1 1> out of them.
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