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Suppose $U$ is a unitary operator, $A$ a vector self-adjoint operator and $v$ a fixed vector. Is it true that $U[v\cdot (U^\dagger A U)]^2U^\dagger$ equals to $(v\cdot A)^2$? I am mainly confused because of the dotting with $v$. Thank you.


For example, if it is simply $U[(U^\dagger A U)]^2U^\dagger$ then clearly this equals to $A$. but I don't understand how to deal with the $v\cdot$


Added Context: As Muphrid pointed out, "the linear operators act on functions, but the vector v belongs to a finite vector space (and hence, A is a vector of linear operators)".

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How is $[v\cdot (U^\dagger A U)]^2$ to be understood? –  Hans Engler Nov 20 '12 at 3:27
    
How do you "dot" a vector with an operator? –  wj32 Nov 20 '12 at 3:27
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This sounds like a quantum mechanics problem, where the linear operators act on functions, but the vector $v$ belongs to a finite vector space (and hence, $A$ is a vector of linear operators). Is that similar to the context of the problem as you encountered it? –  Muphrid Nov 20 '12 at 4:47
    
@Muphrid: Precisely! –  Darren Nov 20 '12 at 9:18

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up vote 1 down vote accepted

Given the context,the answer is yes. Let's consider just one "component" of the full vector of operators.

$$U[v_x U^\dagger A_x U][v_x U^\dagger A_x U]U^\dagger = v_x^2 A_x^2$$

since you get a lot of $UU^\dagger$ or $U^\dagger U$ (which are all 1). The logic extends to when the full vector of operators is considered, as the components of $v$ will commute with the operators because the vector is constant.

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Thank you for the answer and for clarifying the context! –  Darren Nov 20 '12 at 16:30

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