Transformations in vector space

  • #26
Fredrik
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A passive transformation is supposed to take a basis to a basis, right? But the function [itex]A\mapsto U(t)^\dagger AU(t)[/itex] takes an operator to an operator. So it's acting on the wrong type of object.

say now i want to express that rotation by changing the basis and keeping the ket fixed. then will i not have to rotate the basis by same angular velocity as the propagator rotated the ket and IN THE SAME DIRECSION?
The components of a vector x in a basis [itex]\{e_i\}[/itex] are the [itex]x_i[/itex] defined by [itex]x=x_i e_i[/itex]. The i component of Ux in the same basis is [itex]U_{ij}x_j[/itex]. This follows from [itex]Ux=U(x_je_j)=x_jUe_j=x_j(Ue_j)_i e_i=x_jU_{ij}e_i[/itex]. The components of x in the basis [itex]e_i'=Ue_i[/itex] are the [itex]x_i'[/itex] defined by [itex]x=x_i'e_i'[/itex], and we have [itex]x=x_j'e_j'=x_j'Ue_j=x_j'(Ue_j)_i e_i=x_j'U_{ij}e_i[/itex]. So we have [itex]x_i=U_{ij}x_j'[/itex], and therefore [itex]U_{ik}^*x_i=U_{ik}^*U_{ij}x_j'=\delta_{kj}x_j'=x_k'[/itex].

In other words, when we apply the unitary transformation U to the vector x, its components changes according to

[tex]x_i\rightarrow U_{ij}x_j[/tex]

and when we apply it to the basis vectors, the components change according to

[tex]x_i\rightarrow x_i'=U_{ji}^*x_j[/tex]

So yes, you would have to apply [itex]U^\dagger[/itex] to the basis vectors to get the components to change the same way as when you apply U to the vector. (This argument probably needs some modifications to work when we're dealing with an infinite number of dimensions).

See this post for a reminder about the connection between linear operators and matrices. (This is probably the most important detail from linear algebra, so you should make sure that you understand it perfectly).
 
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