5. The Massless Photon

Last time we looked at the potential for the Standard Model if it just had U(1) gauge symmetry or it just had SU(2) gauge symmetry.  [A gauge symmetry is a symmetry of the equations we use, but not a symmetry of nature.  See the first article in this series.] But to put them together to get the full SU(2)xU(1) gauge symmetry isn’t entirely simple.  Instead, we have to look deeper at how these symmetries work, and then, as a warmup to the Standard Model as a whole, we’ll look at U(1) x U(1) symmetry… and learn why the photon has no mass.

What Are Symmetries and Transformations?

What does it mean to say “an equation has a symmetry?” or better, “is invariant under a symmetry?” The equation relates various objects or concepts to one another.  There is a symmetry transformation under which many of the objects or concepts will change.  And yet the equation — the relation between these objects — will not.

Let’s take the simplest of examples: suppose I have the equation

  • x + y = 2

and I tell you that I happen to have a solution to this equation in which x = 1 and y = 1.  

“That’s great,” you may say, “I agree that 1+1=2, but I can see this equation has a symmetry, so there must be other solutions too.  The symmetry transformation is

  • x → x’ = x + a
  • y → y’ = y – a

where x’ and y’ are new values for x and y, shifted up and down by a constant a.  Under this transformation, both x and y have changed, but

  • x’ + y’ = (x+a) + (y-a) = x + y = 2

so the equation has not changed: x’ + y’ = 2.” 

So this is indeed a symmetry transformation under which x and y are not invariant, but the equation which relates them is invariant!  

And indeed, this means there are other solutions to the equation.  If we take a = 1, then x’ = 2 and y’ = 0, which solves the equation.  If we take a = 1/2, then x’ = 3/2 and y’ = 1/2, another solution.  And so on.

Notice that there’s not just one symmetry transformation.   There’s an infinite set, one for each value of a, which can be any real number.  Thus the set of all possible values for a is the real line, and so the equation x + y = 2 is symmetric under a real-line’s worth of symmetry transformations.

U(1) as a Set of Symmetry Transformations

What does it mean to say the Standard Model’s potential V (shown in red at the top of the previous page in this series) is invariant under an SU(2)xU(1) symmetry?  Let’s just focus on the U(1) for now.  To say there is a U(1) symmetry is to say that there is a whole class of symmetry transformations of the fields that appear in the potential, labeled by a single number α, such that although some or all of the fields will change when the symmetry transformation is applied, the potential V will not change.  

Let’s take a field F — remember that means it is a function of space and time, so F is shorthand for F(x,y,z,t) — which at each point in space and time is a complex number, with a real and imaginary part.

What happens if we do a U(1) symmetry transformation labeled by α? Then the field F will change, at each point in space and time, by multiplying F by the complex number e.  This exponential can also be written exp[iα], which is easier to read, so I’ll use that notation instead.  

  • F → F exp[iα] 

(Compare this with our example above, where the number x was shifted x → x+a; here the symmetry transformation is different in detail, but it too has just a single label α.)  

Notice that the complex conjugate of exp[iα] is exp[-iα], and so exp[iα] times its complex conjugate is 1.

  • exp[-iα] exp[iα] = exp[-iα+iα] = exp[0] = 1

The fact that we are multiplying F by just 1 complex number with absolute value 1 (“Unit magnitude”) is why this set of transformations is called U(1).

There’s a geometric way to understand what this transformation is doing, if we think of F as a number in the “complex plane.” In this plane, what we are doing is rotating F by an angle α , as shown below.

In the complex plane, the number F (at right) can be represented as a point whose value on the horizontal axis is Re(F) [the real part of F] and on the vertical axis is Im(F) [the imaginary part of F]. Then multiplying F by exp[iα] is the same as rotating the point F around the origin by the angle α. For later reference, formulas relating the new point to the old point are shown in the box. 

By the way, I mentioned last time that a U(1) symmetry is like a rotation on a circle, described by one angle.  We can see this now, because “α” acts like an angle on the complex plane.  You can take it anywhere from 0 to 2π, but if you go further you won’t get something new, because exp[2πi] = 1, and

  • exp[ i(α + 2π)] = exp[iα] exp[2πi] = exp[iα]   

So the symmetry transformations used here form a set that’s just like the set of points on a circle, or (equivalently, in this case) the set of rotations of a circle.

How to Obtain a Symmetric Potential for a Field Theory

If U(1) is to be a symmetry, then the potential must be unchanged when we do a symmetry transformation with label α, for any α.  That means we cannot have a term in the potential proportional to F, because F will change. Nor can we have one proportional to F2, because

  • F2 → F2 exp[2iα] 

will change too.  But if we remember how complex conjugation works, then we realize that the complex conjugate of F, written F*, will change the other way:

  • F* → (F exp[iα])* → F* exp[-iα]

so then

  • F* F = |F|2 → (F* exp[-iα]) (F exp[+iα]) = F* F exp[-iα] exp[+iα] = = F* F exp[-iα+iα] = F* F

is unchanged, and can appear as a term in V.

Not all fields need to change the same way.  For instance, perhaps a field H transforms, for the same value of α, as 

  • H → H exp[+iα/2]

Notice the factor of ½. More generally a field can be transformed by multiplying by exp[+iqα], where q is called “the charge of that field under U(1)“. In what we’ve done so far, F has charge 1 and H has charge ½.  How can we combine these fields to find something that is invariant under U(1) and can appear as a term in V?  Let me give you two examples, one that works and one that doesn’t, so you can see the pattern.  

What about FH?  That won’t work:

  • F H → F exp[+iα] H exp[+iα/2] = F H exp[+iα + iα/2] = F H exp[+i(3/2)α]

Interestingly, you see that F has charge 1, H has charge ½, and the combination acts as though it has charge 3/2.  The charges add! So that tells us that if we want a term that doesn’t transform under the U(1), we need a combination with charge zero!  

We already saw one example: if F has charge 1, then F* has charge -1, and so F* F has charge zero and can appear in V.  What else can we do?  How about F* H2 ?  Aha!  That combination has charge zero, and indeed

  • F* H2 → F* exp[-iα] (H exp[+iα/2])2  = F* H2 exp[-iα+iα/2+iα/2] =  F* H2

For a U(1) symmetry, that trick — finding combinations of fields whose charges add to zero — is all we need to find allowable terms for the potential.

U(1)xU(1)

In the Standard Model we have SU(2)xU(1) gauge symmetries, but let’s start with something simpler: U(1)xU(1).  Suppose I told you that the H field has charge -½ under the first U(1) and charge +½ under the second U(1).  What would that mean?

Since we have two U(1)s, we have two angles, one for each, and let’s call them α and β.  If we do a symmetry transformation from the first U(1), then

  • H → H exp[-iα/2]

the -½ coming from the H field’s charge under the first U(1).  If we do a symmetry transformation from the second U(1), then

  • H → H exp[+iβ/2]

with a positive sign because the charge of H under the second U(1) is +½. But since both of these are symmetries, we can do both at once:

  • H → H exp[-iα/2] exp[+iβ/2] = H exp[-iα/2 + iβ/2]

and under this transformation V must be invariant, for any and all α and β.  Fortunately, |H|2 = H* H is indeed invariant under this transformation, as is |H|4. Indeed all terms appearing in the Standard Model’s potential V are proportional to powers of |H|2.

But if we are in a gauge theory, where there is a gauge boson W that goes with the first U(1) and a second gauge boson X that goes with the second U(1), then (as I told you last time) V must contain terms combining H with W and X.  (Note that whereas H is a complex field, W and X are real, so we don’t need to worry about their complex conjugates.) How do we find these terms?

It turns out that for U(1) symmetries, what we are supposed to do is apply the following rule

  • look at the transformation for each field,
    • H → H exp[-iα/2 + iβ/2]
  • take what’s in the exponential, without the “i”,
    • -α/2 + β/2
  • replace the angle with the corresponding gauge boson times its interaction strength, i.e.
    • replace α with gW
    • replace β with g’X
      • (-½ gW + ½ g’X )
  • multiply by the field
    • H (-½ gW + ½ g’X )
  • complex-square the result (i.e. multiply it by its complex conjugate) times ½
    • ½ | H (-½ gW + ½ g’X ) |

This gives us the potential:

  • ½ | H (-½ gW + ½ g’X ) |= ⅛  |H|2 [-gW +g’X]2 = ⅛  |H|2 [gW – g’X]2

(The minus sign difference in the last two expressions doesn’t matter since it appears in something squared, and (-1)2 = +1.  Notice also the absolute value signs in the first expression; this is H (-½ gW + ½  g’X ) times its complex conjugate, so that |H|2 appears, not H2.  Otherwise this expression wouldn’t be invariant under the two U(1) symmetries.)

Because of the rule I just outlined, we don’t obtain

  • ⅛  |H|2 [(gW)2 + (g’X)2]

in the Standard Model potential, which would give W a rest mass and X a rest mass (here’s a reminder as to why). We instead get

  • ⅛  |H|2 [gW – g’X]2

which means that only one combination of W and X, namely gW – g’X, has a rest mass. A separate combination, namely g’W + gX, has none.

This is the math that explains why the Standard Model has a Z boson with a large rest mass and a photon with no rest mass!  (In a sense, this was Glashow’s simple but crucial insight.)

Next time I’ll be more technically explicit about this statement, so we can find out how much mass the Z boson actually has.  And then we’ll do SU(2)xU(1) completely.  Then we’ll do the triplet model… and finally, as the most complex step, I’ll tell you how we get the rule described above.

12 responses to “5. The Massless Photon

  1. «x and y are not invariant, but the equation which relates them is invariant! »
    This sounds like the definition of a proportionality, ratios making up the equality of ratios vary but the rule of the proportionality stays same. What is the relationship?

    • It’s not the *definition* of a proportionality. Instead, a proportionality is an *example* of an equation with a symmetry. If x/y = 2, then x –> a x , y –> a y is a symmetry that leaves the equations invariant. (And if you take the logarithm of this equation you will get the case I covered in the text, since log x + log y = log 2 and log x –> log x + log a .)

      An example of an equation with a symmetry which cannot be though of as a proportionality or its log would be the equation

      (cos w)^2 + (sin w)^2 = 1

      which has the symmetry w –> w + a where a is any number. That’s clearly not a proportionality.

  2. kashyap vasavada

    Typo: In the eq. under the figure,
    in eq.1 Im F should have – sign. In eq. 2 Re F should have + sign

  3. kashyap vasavada

    replace α with gW replace β with g’X (-½ gW + ½ g’X )
    Do you mean replace exp(i α) with gW. What is the significance of replacing exp transformation by fields?

    • No, it is correct as stated: replace α with gW and replace β with g’X . Otherwise you would get (g W) (g’ X) multiplied and not (-½ gW + ½ g’X ) .

      Why is it correct as stated? I plan to explain this, but it requires beginning calculus [unless I can think of a way around that] and so I’m currently planning to leave this til the end.

      • kashyap vasavada

        Thanks Matt. I will look forward to that explanation. I wonder if this is similar to exponentiation in Lie Algebra.

        • It is very much so. Essentially, gauge bosons are valued in the algebra, not the group, so they are related to the logarithm of group transformations.

          • kashyap vasavada

            OK! I understand gauge theory is after all algebra. But aren’t particles members of various representations of groups? So what is meant by the word “valued” ?

            • The electron in electromagnetism is rotated by a group element e^{i alpha}. The “alpha” is part of the algebra, and that’s also where the gauge fields appear.

              In the strong interactions, the quarks are triplets under SU(3). That means they are rotated by a group element U = exp[i theta T], where T is an element of the su(3) algebra, and transforms in the 8 of su(3), the adjoint representation. The gauge fields also appear exponentiated in the same way, and that’s why (unlike matter fields) they are always in the adjoint representation.

              Does that help?

  4. kashyap vasavada

    Thanks. I should do some reading. But let me ask a question anyway! If exponentiation (U(1), SU(2), SU(3)) is the same for matter fields and gauge fields depending on internal quantum numbers and multiplets are similar then where does gauge boson ness come in?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s