Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let $A$ be a Hermitian operator corresponding to some observable. If we prepare $N$ identical systems in the state $\psi$ and measure this observable in each system, the average of the measurements (for large $N$) will be $\langle \psi | A | \psi \rangle$.

In the standard model, $\langle 0 | \phi^0(x) | 0 \rangle = v/ \sqrt{2} = 174\ GeV\ \ \forall x$.

Questions:

  1. What observable, if any, corresponds to $\phi^0(x)\ $?

  2. Is $v/ \sqrt{2}$ really the average of a large number of measurements of identical systems in the vacuum state?

(For precision, you could modify question 1 to "what observable corresponds to $\int d^4 x\ f(x) \phi^0(x)$," where $f:\mathbb{R}^4 \to \mathbb{R}$ is a smooth function localized about some $x_0$.)

share|improve this question
add comment

1 Answer 1

In the Standard Model, $\phi(x)$ is an observable; it's the value of a component of the Higgs field. It's analogous to the value of the electric field's z-component at a point. (A slight subtlety: $\phi^0(x)$ isn't actually a gauge-invariant quantity, so it's not actually an observable itself. But it's a perfectly good gauge-fixed representation of $\sqrt{||\phi||^2}$, which is a gauge invariant observable.)

$v$, meanwhile, is one of the parameters of the model; it's supposed to have a specific fixed value. We don't know precisely what this value is, and we infer it by averaging repeated measurements of the observable $\phi^0(x)$. But the Standard Model itself treats $v$ as fixed.

share|improve this answer
    
Thanks for the correction re: gauge invariance; let me continue to use the gauge-fixed version for convenience. I guess I am still not clear on what it means for $\phi^0(x)$ to be an observable in the same sense that $E_z(x)$ is. I can imagine measuring $E_z(x)$ by putting a macroscopic charged object at $x$ and measuring its acceleration. How do I measure $\phi^0(x)$? –  user22037 Apr 6 '13 at 19:16
    
The answer to that question is not going to be satisfying. You find some current that $\phi^0(x)$ couples to, and you vary the current via some experimental procedure. We hopefully are able to think about this procedure in terms of observables you already know, like momenta of Standard Model particles. If you can: great, you have an experimental paper! If not, you have an experimental challenge... –  user1504 Apr 11 '13 at 0:33
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.