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


  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 $x_0$.)

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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.

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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

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