As mentioned, constructing the operator map
we face a necessity of selection - to set **K**'=**KA** or **AK**.
Both forms are compatible with the classic map
but are not equivalent due to non-commutativity of the operators: they differ
by factor
a, as
**KA=aAK**.
Let us consider an "intermediate" variant
**K**'=a^{1/2}**KA**=a^{-1/2}**AK**.
For the second equation let us keep the symmetric form:
**A**'=**AKA**.

Substitute expressions for the matrix elements
*K _{mn}*=d

Let us pass now to the Schrodinger representation
and find the matrix of the evolution operator **U**.
As we have **UK**'=**KU**, we obtain
*U*_{m,n-1}a^{n-1/2}=*U*_{m+1,n}.
Analogously, the condition **UA**'=**AU** yields
*U*_{m,n-1}a^{2n-m-1}=*U*_{mn}.
As follows from the first expression, the diagonal elements satisfy
*U*_{m+1,m+1}=*U*_{mm}a^{m+1/2},
so

Next, with a use of the second expression we have

.

It is easy to check that the resulting expression for
*U*_{mn} has the correct
period *N* in respect to both indices.
Let us select the initial element as
*U*_{00}=*N*^{ -1/2}*e*^{-ip/4}**U**|=1, and the trace of the matrix appears to be real positive.
Finally, we obtain

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