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Glashow model left/right hypercharge ratio

  • #1

Homework Statement



a)

Consider Glashow's model with general hypercharge assignments [tex]Y_L[/tex], [tex]Y_R[/tex], for the left-handed and right-handed fields, respectively. By demanding the correct electromagnetic couplings of the electron and neutrino to the photon, determine the ratio [tex]Y_L/Y_R[/tex].
[8 marks]

b)

Why are [tex]Y_L[/tex] and [tex]Y_R[/tex] not fixed by this requirement?
[2 marks]


Homework Equations



None given, but I suspect these will be relevant

[tex] Y=2 \left( Q+T^3 \right) [/tex]

where Y is hypercharge, Q is electric charge and T^3 is the 3rd component of weak isospin.

and

[tex]\frac{1}{g^2}+\frac{1}{g'^2}=\frac{1}{e^2}.[/tex]

where g is the SU(2) coupling of the W fields, g' is the U(1) coupling of the B field and e is the electron charge?

The Attempt at a Solution



Well I tried taking the ratio of the above formula for left and right, but came across problems. Firstly The neutrino doesn't couple with the photon. Secondly there are no right handed neutrinos (in this model). Lastly the neutrino doesn't have a charge. So I'm unsure if I'm even taking the right route or how the second equation comes into play.
 

Answers and Replies

  • #2
fzero
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Well I tried taking the ratio of the above formula for left and right, but came across problems. Firstly The neutrino doesn't couple with the photon. Secondly there are no right handed neutrinos (in this model). Lastly the neutrino doesn't have a charge. So I'm unsure if I'm even taking the right route or how the second equation comes into play.
I think you're misunderstanding the Glashow model completely. There is an electroweak gauge group [tex]SU(2)_L \times U(1)_Y[/tex]. This symmetry is broken to a subgroup [tex]U(1)_{\text{em}}[/tex] which is a linear combination of [tex]U(1)_Y[/tex] and the [tex] U(1) \subset SU(2)_L[/tex] that is generated by the generator [tex]T^3[/tex]. The charge of an object under [tex]U(1)_{\text{em}}[/tex] is

[tex]Q = T^3 + \frac{1}{2} Y,[/tex]

where [tex]T^3[/tex] is the eigenvalue of the [tex]T^3[/tex] generator and [tex]Y[/tex] is the charge under [tex]U(1)_Y[/tex].

The neutrino and left-handed electron are assembled into an [tex]SU(2)_L[/tex] doublet

[tex]\begin{pmatrix} \nu_e \\ e_L \end{pmatrix},[/tex]

while the right-handed electron is an [tex]SU(2)_L[/tex] singlet. The [tex]T^3[/tex] charges are 1/2 for [tex]\nu_e[/tex], [tex]-1/2[/tex] for [tex]e_L[/tex] and 0 for [tex]e_R[/tex]. For your problem you are to assign [tex]U(1)_Y[/tex] charges [tex]Y_{L,R}[/tex] separately to the left and right-handed particles.

I don't seem to agree with the conclusions stated in the problem. Demanding that the neutrino is neutral will fix [tex]Y_R[/tex] and therefore the electron charge, which will in turn fix [tex]Y_R[/tex]. Perhaps you should check your notes on the Glashow model. I could be introducing additional electroweak physics that is not included in that model.
 
  • #3
I could be introducing additional electroweak physics that is not included in that model.
You could well be, as we are yet to cover electroweak symmetry breaking.

I could be wrong, but are the terms in the Gell-Mann-Nishijima relation matrices for the left-handed case and numbers for the right handed? If this is true, with the values you gave, [tex]Y_L= \left( \stackrel{1}{-2} \right) [/tex] and [tex]Y_R = -1[/tex]. But as you said these are now fixed.

I don't see how I can have one left handed hypercharge if there are two particles it is related to.
 
  • #4
fzero
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You could well be, as we are yet to cover electroweak symmetry breaking.

I could be wrong, but are the terms in the Gell-Mann-Nishijima relation matrices for the left-handed case and numbers for the right handed? If this is true, with the values you gave, [tex]Y_L= \left( \stackrel{1}{-2} \right) [/tex] and [tex]Y_R = -1[/tex]. But as you said these are now fixed.

I don't see how I can have one left handed hypercharge if there are two particles it is related to.
In order that [tex]SU(2)_L[/tex] be a symmetry, the fields in the doublet must have the same value of [tex]Y[/tex].
 
  • #5
In the lecture today we've been given a hint. Apparently we need to write out the interaction Lagrangian (density), rewrite in terms of the [tex]A_\mu[/tex] and [tex]Z_\mu[/tex] fields and [tex]\theta_W[/tex]. Then isolate the electron terms, set the term coefficients equal to [tex]e[/tex] and lastly isolate the neutrino terms and set the coefficients equal to 0. We will then have expressions that will give the left/right hypercharge ratio. The value should be 1/2.

I can't find the general interaction Lagrangian though in my notes (in the same way the lecturer wrote it when giving the hint.)
 
  • #6
fzero
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I can't find the general interaction Lagrangian though in my notes (in the same way the lecturer wrote it when giving the hint.)
The gauge and matter sector is the same as in the Standard Model. The Glashow model really only differs from the Standard Model in that there's no Higgs sector, so the masses are added by hand.
 
  • #7
OK so I've done it, although I heavily relied on the hint given and I am dubious as to how well I understand this and how well I will do in the exam. Here's the answer nonetheless:

[tex]{\mathcal{L}}_I = \ldots - gJ_\mu ^3 W^{3\mu } - g'Y_L \bar \chi _L \gamma ^\mu \chi _L B_\mu - g'Y_R \bar e_R \gamma ^\mu e_R B_\mu [/tex]


[tex] = \ldots - g\left( {\frac{1}{2}\bar \nu _L \gamma ^\mu \nu _L + \bar e_L \gamma ^\mu e_L } \right)\left( {\cos \theta _W Z^\mu + \sin \theta _W A^\mu } \right)
- g'Y_L \bar \chi _L \gamma ^\mu \chi _L \left( { - \sin \theta _W Z_\mu + \cos \theta _W A_\mu } \right)
- g'Y_R \bar e_R \gamma ^\mu e_R \left( { - \sin \theta _W Z_\mu + \cos \theta _W A_\mu } \right) [/tex]

taking out the electron photon interaction term

[tex]A^\mu \left[ {\left( {\frac{1}{2}g\sin \theta _W - Y_L g'\cos \theta _W } \right)\bar e_L \gamma _\mu e_L - Y_R g'\cos \theta _W \bar e_R \gamma ^\mu e_R } \right] [/tex]

and demanding the coupling strength e

[tex]
\left( {\frac{1}{2}g\sin \theta _W - Y_L g'\cos \theta _W } \right) = - Y_R g'\cos \theta _W = e
[/tex]

then taking out the photon neutrino interaction and demanding no coupling

[tex]
A^\mu \left[ {\left( { - \frac{1}{2}g\sin \theta _W - Y_L g'\cos \theta _W } \right)\bar \nu _L \gamma ^\mu \nu _L } \right]
[/tex]

[tex]
- \frac{1}{2}g\sin \theta _W - Y_L g'\cos \theta _W = 0
[/tex]

subbing the above into the electron case we have

[tex]
- 2Y_L g'\cos \theta _W = - Y_R g'\cos \theta _W
[/tex]

or

[tex]Y_L/Y_R=\frac{1}{2}[/tex]
 
  • #8
Sorry the second line should read:

"[tex]=\ldots -g\frac{1}{2}\left(\bar{\nu}_L \right. \ldots[/tex]"

and the last gamma in the isolated electron-photon interaction term should have a lower mu index, as should that of the neutrino term
 
Last edited:

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