Higgs and Duality

[QuantumT]

Is the Higgs boson involved in duality?

If we interpret duality as energy gaining mass upon measurement, is that then the Higgs being "inserted" at measurement?

[Strange]

40 minutes ago, QuantumT said:

Is the Higgs boson involved in duality?

What duality are you referring to? Wave-Particle? If so, then all particles, including the Higgs boson, have both wave-like and particle-like properties.

41 minutes ago, QuantumT said:

If we interpret duality as energy gaining mass upon measurement

Energy and mass are equivalent. But why would measurement change that?

[QuantumT]

11 minutes ago, Strange said:

What duality are you referring to? Wave-Particle? If so, then all particles, including the Higgs boson, have both wave-like and particle-like properties.

Energy and mass are equivalent. But why would measurement change that?

So you're saying that measurement does not cause a physical wave-particle change?

[Strange]

27 minutes ago, QuantumT said:

So you're saying that measurement does not cause a physical wave-particle change?

It doesn't cause a change. You can measure a wave-like property (e.g. wavelength) or you can measure a particle-like property (eg. location). 

But even if it did, I don't see the connection to mass or energy changing.

[QuantumT]

5 minutes ago, Strange said:

It doesn't cause a change. You can measure a wave-like property (e.g. wavelength) or you can measure a particle-like property (eg. location). 

But even if it did, I don't see the connection to mass or energy changing.

Thank you for taking the time to answer 

[swansont]

13 hours ago, QuantumT said:

So you're saying that measurement does not cause a physical wave-particle change?

Change implies it was one and became the other.

"Wave-particle duality" is a watered-down description used so that one can still keep hold of classical notions and not be as frightened of and confused by quantum mechanics as if one were exposed to it all at once. It's not really the bedrock upon which one should build.