Composite Hilbert Spaces and Operators

[ArjSiv]

So, say I have a composite hilbert space H = H_A \otimes H_B, can I write any operator in H as U_A \otimes U_B?

Thanks

 

[dextercioby]

Not really. IT should be something like U_{A}\otimes\hat{1}_{B}+\hat{1}_{A}\otimes U_{B}

 

[Haelfix]

You also need some assumptions on the properties of the particular hilbert spaces in question (Ha and Hb).

Like are they seperable or not.

 

[MaverickMenzies]

You can find some operators that are simple tensor products of local operators, however, in general any operator acting on the composite space can be decomposed as a linear combination of products of dyad operators. To see this use the completeness relation (i.e. that the sum of projection operators equals the identity for both spaces):

I_A \otimes I_B = \sum_j \sum_k \vert e_j \rangle \langle e_j \vert \otimes \vert f_k \rangle \langle f_k \vert

Then for any operator O on the composite space we get

O = I . O . I = \sum_j \sum_k \sum_m \sum_n \langle e_j , f_k \vert O \vert e_m, f_n \rangle \vert e_j \rangle \langle e_m \vert \otimes \vert f_k \rangle \langle f_n \vert

 

[vanesch]

So, say I have a composite hilbert space H = H_A \otimes H_B, can I write any operator in H as U_A \otimes U_B?

As others pointed out, the operators of the form U_A \otimes U_B span the algebra of operators, but not all operators in that algebra are of that form themselves.
Exactly like the vectors of H = H_A \otimes H_B which are not all of the form |\psi_A> \otimes |\psi_B >, but the entire space is nevertheless spanned by those vectors.

 

[reilly]

The answer is a simple no. Take for example the hydrogen atom. The basic idea of Rutherford and later, Bohr, is 1. that protons, nucleii, and electrons are indepedent. Which means the initial proton-electon states occur in different subspaces. However, the potential involves the coordinate of both electron and proton.

Thus the hilbert space in which the hydrogen atom lives cannot be decomposed into separate spaces for the constituants(sp?). This is discussed in most QM texts.

Regards.
Reilly Atkinson