Is it possible to describe spontaneous breaking of the local gauge symmetry in the standard Hamiltonian formalism, without any manual redefinition of the gauge field( A(r) -> A(r)+▽Λ(r) )?(adsbygoogle = window.adsbygoogle || []).push({});

Detailed description of my question is given below.(Quite lengthy.. sorry.)

1. By the 'standard Hamiltonian description of spontaneous symmetry breaking' I mean the following.

- There is a Hamiltonian and an associated continuous symmetry, under which the given Hamiltonian is invariant. Yet, the ground states don't possess the same symmetry as the Hamiltonian. They can be transformed to one another by the given symmetry operation.

- For example, in the case of ferromagnetism, the Heisenberg Hamiltonian is invariant under rotation. Yet, its ground states are oriented to certain directions(they aren't rotationally invariant) and one can transform a ground state to another by applying the rotation operator.

2. A problem when we do the same thing for the local gauge symmetry?

- In quantum mechanics, when the local gauge invariance is discussed in textbooks, they say the physics is invariant under the local gauge transformation exp(iΛ(r)/hbar), if the gauge field is transformed as A(r) -> A(r)+▽Λ(r).

- However, while the Hamiltonian is transformed by a well-defined operation exp(iΛ(r)/hbar), the gauge field is manually transformed as A(r)->A(r)+▽Λ(r). (no such quantum operation exists). Furthermore, the Hamiltonian isn't invariant under exp(iΛ(r)/hbar), although it looks invariant once you redefine the gauge field as A(r) -> A(r)+▽Λ(r).

- This is clearly different from what we normally do when discussing other symmetries as in the part 1 of this thread.

3. What if we quantize the gauge field?

- One may expect that if we quantize the gauge field and write the total Hamiltonian( the electron part + the gauge field part + coupling term), we can use the formalism described in the part 1.

- Yet, the situation is not much better, since the gauge field can't be quantized without specifying the gauge(let's say, the Coulomb gauge). Then, the gauge transformation isn't a quantum operation.

- We can still construct an unitary operator made of photon creation and annihilation operators so that it can transform A(r) to something else. Howver, it will never have any longitudinal degree of freedom, whereas the term ▽Λ(r) appearing in the gauge transformation is solely longitudinal when Fourier transformed to the k-space.

4. What do we need? Higgs?

- I heard that in the electroweak theory, a scalar field(the Higgs field) is introduced to describe the spontaneous breaking of the local gauge symmetry, and the gauge bosons(W and Z) acquire the longitudinal degree of freedom and the mass.

- Then, do we need a fictitious scalar field(such as Higgs?) to describe the logitudinal degree of freedom of photons and achieve the full Hamiltonian formalism for the local gauge transformation, or is it simply impossible?

Cf. The global gauge symmetry - a simpler question

- Since the global gauge transformation multiplies 'exp(iθ)' to each particle, it can be written as 'exp(iNθ)' in the 2nd quantization notation, where N is the number operator.

- The Hamiltonian describing a superconductor or a superfluid is invariant under 'exp(iNθ)' but its ground states, namely the BCS ground state for the superconductor or the coherent state for the superfluid, are not invariant under exp(iNθ). Actually, exp(iNθ) transforms a ground state to another one(with a different phase). Therefore a complete analogy with the ferromagnetism is established.

Heisenberg Hamiltonian <-> superconductor(superfluid) Hamiltonian

rotational symmetry <-> global gauge symmetry

ferromagnetic ground state <-> BCS ground state (coherent state)

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# Spontaneous breaking of the local gauge symmetry(such as that in superconductors)?

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