What does it mean, The Higgs boson is an excitation of the higgs field

[silvercats]

How come Higgs Bosons have mass if Higgs field itself gives other thin

How come Higgs Bosons have mass if Higgs field itself gives other thing their mass?

 

[silvercats]

What does it mean, The Higgs boson is an excitation of the higgs field? Isn't the higgs field simply made of Higgs Bosons?

 

[Bill_K]

What does it mean, The Higgs boson is an excitation of the higgs field?

The electron is an excitation of the electron field, the photon is an excitation of the electromagnetic field, and the Higgs boson is an excitation of the Higgs field. The world is made of quantum fields, and all the particles we see are just excitations of those fields. "Excitation" means an eigenstate with well-defined energy.

Isn't the higgs field simply made of Higgs Bosons?

The Higgs field they talk about is an aspect of the vacuum, and isn't made of anything in the normal sense. It's part of a completely uniform and unchanging background, upon which physics takes place.

 

[silvercats]

Then what is an excitation of an electron field?

 

[silvercats]

isn't made of anything in the normal sense..

Everything has to be made up of something :O. isn't it

 

[Bill_K]

Then what is an excitation of an electron field?

An excitation of THE electron field is an electron. Repeating what I said - particles are not the fundamental objects in the world. There are not 10⁸⁰ (or whatever the number is) independent electrons running around, they are all elementary excitations of a single universe-pervading field. (And this is one good reason why they are identical!)

Same for the other types of elementary particles.

 

[DrDu]

The terminology is not very fortunate: A Higgs boson is comparable to a sound wave in some solid, where the solid is the equivalent of the Higgs field. The Higgs field itself is also composed of bosons, but which are not identical to the excitations having been observed at CERN.

 

[mfb]

The Higgs field gives a mass both to other particles and the Higgs particle. In addition, the Higgs particle is the only particle which can have a mass on its own.

 

[silvercats]

Higgs particle is the only particle which can have a mass on its own.

Some kind of another field is giving the Higgs boson its mass possibly?

 

[silvercats]

Thanks all!

 

[Morgoth]

Well I'm not sure about what I'm going to say, but I am open to corrections ^_^.
it's not Higgs Boson that gives the masses... It is the Higgs Interaction with particles, that breaks the symmetry of electroweak theory and weak interaction bosons (W,Z) appear with mass while electromagnetic interaction boson (photon) appears massless. In that sense the Higgs gives mass to Ws,Zs.
The masses of all particles in QFT appear as interactions of fields with themselves. For example a Klein Gordon field has a term of m^2 multiplying the coupling of the field with itself. Same on Dirac's case, and so on. And that's due to symmetries of your problem. If you put additional interaction, these symmetries can break and that is what the Higgs Field does (spontaneous symmetry breaking). So I am not sure if it gives mass, or if it just differs them...

 

[Bill_K]

It is the Higgs Interaction with particles, that breaks the symmetry of electroweak theory...

If you put additional interaction, these symmetries can break and that is what the Higgs Field does

Almost, but not quite. The interaction is not what breaks the symmetry. The entire Lagrangian, including the interaction terms of the Higgs with the other particles, is gauge invariant. That's a basic requirement!

The breaking of the symmetry is spontaneous, meaning that it is due to the nonzero vacuum expectation value of the Higgs field.

For example, the usual Dirac mass term can be written in terms of right- and left-handed spinors as m(ψ_Lψ_R + ψ_Rψ_L). This would not be gauge invariant, since ψ_L transforms as an SU(2) doublet while ψ_R transforms as an SU(2) singlet. To replace it with something that is gauge invariant we introduce a Higgs field φ, and write G(ψ_Lφψ_R + ψ_Rφψ_L), where G is a coupling constant. This will be gauge invariant, for example, if φ is also an SU(2) doublet (the usual assumption). After the symmetry is spontaneously broken, φ acquires the vacuum expectation value
$$\left(\begin{array}{c}0\\v\end{array}\right)$$
and the Lagrangian becomes Gv(ψ_Lψ_R + ψ_Rψ_L) which acts as a Dirac mass term.