Transformations in vector space

Fredrik
Staff Emeritus
Gold Member
A passive transformation is supposed to take a basis to a basis, right? But the function $A\mapsto U(t)^\dagger AU(t)$ 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 $\{e_i\}$ are the $x_i$ defined by $x=x_i e_i$. The i component of Ux in the same basis is $U_{ij}x_j$. This follows from $Ux=U(x_je_j)=x_jUe_j=x_j(Ue_j)_i e_i=x_jU_{ij}e_i$. The components of x in the basis $e_i'=Ue_i$ are the $x_i'$ defined by $x=x_i'e_i'$, and we have $x=x_j'e_j'=x_j'Ue_j=x_j'(Ue_j)_i e_i=x_j'U_{ij}e_i$. So we have $x_i=U_{ij}x_j'$, and therefore $U_{ik}^*x_i=U_{ik}^*U_{ij}x_j'=\delta_{kj}x_j'=x_k'$.

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

$$x_i\rightarrow U_{ij}x_j$$

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

$$x_i\rightarrow x_i'=U_{ji}^*x_j$$

So yes, you would have to apply $U^\dagger$ 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).