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- Thread starter ArjSiv
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- #3

Haelfix

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Like are they seperable or not.

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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):

[tex]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 [/tex]

Then for any operator O on the composite space we get

[tex]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[/tex]

[tex]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 [/tex]

Then for any operator O on the composite space we get

[tex]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[/tex]

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- #5

vanesch

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

As others pointed out, the operators of the form [tex]U_A \otimes U_B[/tex] span the algebra of operators, but not all operators in that algebra are of that form themselves.

Exactly like the vectors of [tex]H = H_A \otimes H_B[/tex] which are not all of the form [tex] |\psi_A> \otimes |\psi_B > [/tex], but the entire space is nevertheless spanned by those vectors.

- #6

reilly

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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

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