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According to electroweak theory, the photon ($\gamma^0$) and weak bosons ($W^+, W^-, Z^0$) are all linear combinations or superpositions of the weak hypercharge boson ($B$) and the weak isospin bosons ($W_1, W_2, W_3$):

\begin{align} W^+&=1/\sqrt2\ (W_1-iW_2\ ) \\ W^-&=1/\sqrt2\ (W_1+iW_2\ ) \\ Z^0&=W_3\cos\theta_W-B\sin\theta_W \qquad\quad\theta_W\approx 28.13\deg \\ \gamma^0&=W_3\sin\theta_W+B\cos\theta_W \end{align} Considering this, why won't wavefunction collapse occur upon observation of the photon? Is this because we must measure a property of the $B$ or $W_3$ bosons that is not shared among them and we do not have a way to do that yet?

To clarify, official question: upon observation of the photon, why doesn't the wavefunction collapse so we observe a $B$ or $W_3$ boson?

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    $\begingroup$ You observe a photon by having it absorbed by a detection apparatus. What does this have to do with its resolution to two unphysical states that are not even propagating mass eigenstates? Why should its wavefunction collapse into either of these states? Where did this misconception come from? $\endgroup$ – Cosmas Zachos Feb 5 at 20:10
  • $\begingroup$ In non-relativistic quantum mechanics, measurement of an observable leaves the system in an eigenstate of the observable ('collapses' the state to an eigenstate). In your post, you stipulate that a photon is observed but you don't stipulate what the observable is. Also, what is the observable that has the two eigenstates $B$ or $W_3$? $\endgroup$ – Hal Hollis Feb 5 at 20:42
  • $\begingroup$ @HalHollis: The second and third sentences of your comment make sense to me, but I'm confused by this: In non-relativistic quantum mechanics, measurement of an observable leaves the system in an eigenstate of the observable ('collapses' the state to an eigenstate). It seems to me that this is true in both relativistic and non-relativistic QM (assuming the Copenhagen interpretation), and in any case this question is clearly about relativistic QM, since we're talking about photons. $\endgroup$ – Ben Crowell Feb 5 at 21:26
  • $\begingroup$ Hi @BenCrowell, I confess the first sentence seems odd. When I started writing the comment, I was heading in a different direction than what I ended up in and, after deleting some stuff in the edit box, I didn't adequately revise the first sentence. $\endgroup$ – Hal Hollis Feb 5 at 23:20
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We are presently living in the world where the symmetry you describe has been broken, i.e. the particles instead of being massless have acquired through the Higgs mechanism a mass, a huge one for the W and Z , and zero for the photon.

Energy conservation is an absolute law , and masses are part of the energy budget of a system under study. Thus the photon and the Ws and Z cannot mix the interactions, they are distinct particles described by different wavefunctions: the photon by a quantized Maxwell equation, the W and Z by Klein Gordon solutions.

Before symmetry breaking your question might have some sense, except as everything is massless I do not see how an experiment to detect them can be set up.

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