- #1
rubbergnome
- 15
- 0
Hello everyone. I was kind of "working on a project" when I stumbled onto something, which I can't get if it's wrong or just shallow; it must be wrong, because I haven't seen it anywhere and it seems a quite general statement.
Suppose we are given a bosonic Fock space H generated in the usual manner by acting which any number of creation operators, which can be fields [itex]\psi^\dagger (x)[/itex] which create particles at chosen spacetime positions or their modes [itex]a^\dagger (k)[/itex] which create particles at chosen spacetime momenta. The underlying space of positions is not [itex]\mathbf{R}^3[/itex] in my case, but I don't think it changes things much.
Let [itex]|0\rangle[/itex] be the vacuum. Then, as usual, we act with [itex]N = n + m[/itex] operators
[itex]|x_1,x_2,...,x_n ; k_1,k_2,...,k_m\rangle = \prod_{i,j=1}^{n,m} \psi^\dagger(x_i)a^\dagger(k_j)|0\rangle[/itex].
As in my case, every state in the Fock space represents a geometrical object in a sense, i thought of searching for a moduli space in which coordinates could parametrize these states. I thought that if one sets a maximal number [itex]N[/itex] of particles, one needs for every [itex]n \leq N[/itex] a sequence of [itex]n[/itex] coordinates, let them be positions or momenta. Then, one can rearrange these coordinates in a triangular [itex]N \times N[/itex] matrix. As the underlying vector space of positions/momenta is equipped in my case with an abelian product law, giving it in fact an algebra structure, cannot one consider this a mapping to the Heisenberg Algebra, as it is isomorphic to the algebra of triangular matrices? The commutation relations can be carried along as there is a distinction between position and momentum in coordinates.
The identity element would not be the vacuum, I think, but the zero vector. Maybe one could give the matrix algebra another row/column to get coordinates for [itex]N=0[/itex] which corresponds to the vacuum; and then take the quotient of the algebra with a suitable equivalence relation/group such that the resulting algebra gives only one vacuum. Where am I wrong?
I also didn't know whether to post this in this section or the Mathematics section, as the question is related to both. Move this thread as you please, of course. Thanks in advance.
Suppose we are given a bosonic Fock space H generated in the usual manner by acting which any number of creation operators, which can be fields [itex]\psi^\dagger (x)[/itex] which create particles at chosen spacetime positions or their modes [itex]a^\dagger (k)[/itex] which create particles at chosen spacetime momenta. The underlying space of positions is not [itex]\mathbf{R}^3[/itex] in my case, but I don't think it changes things much.
Let [itex]|0\rangle[/itex] be the vacuum. Then, as usual, we act with [itex]N = n + m[/itex] operators
[itex]|x_1,x_2,...,x_n ; k_1,k_2,...,k_m\rangle = \prod_{i,j=1}^{n,m} \psi^\dagger(x_i)a^\dagger(k_j)|0\rangle[/itex].
As in my case, every state in the Fock space represents a geometrical object in a sense, i thought of searching for a moduli space in which coordinates could parametrize these states. I thought that if one sets a maximal number [itex]N[/itex] of particles, one needs for every [itex]n \leq N[/itex] a sequence of [itex]n[/itex] coordinates, let them be positions or momenta. Then, one can rearrange these coordinates in a triangular [itex]N \times N[/itex] matrix. As the underlying vector space of positions/momenta is equipped in my case with an abelian product law, giving it in fact an algebra structure, cannot one consider this a mapping to the Heisenberg Algebra, as it is isomorphic to the algebra of triangular matrices? The commutation relations can be carried along as there is a distinction between position and momentum in coordinates.
The identity element would not be the vacuum, I think, but the zero vector. Maybe one could give the matrix algebra another row/column to get coordinates for [itex]N=0[/itex] which corresponds to the vacuum; and then take the quotient of the algebra with a suitable equivalence relation/group such that the resulting algebra gives only one vacuum. Where am I wrong?
I also didn't know whether to post this in this section or the Mathematics section, as the question is related to both. Move this thread as you please, of course. Thanks in advance.