# A “Linear” Mapping - What am I talking about?

My situation:

I have a fixed initial state $|\psi_i \rangle$ which is a ($1 \times n$) column vector. I apply a linear operator $\hat{A}(\phi_{1,2,3,...,x})$, which has a number of variables, to this initial state to receive another ($1 \times n$) column vector - my final state.

i.e.

$|\psi_f \rangle = \hat{A}(\phi_{1,2,3,...,x}) |\psi_i \rangle$

I'm running experiments for a particular application, where I'm looking into whether taking uniform steps in the space that contains $\phi_{1,2,3,...,x}$ will result in uniform steps in the space that contains $|\psi_f \rangle$. It it does, is it fair to call it a linear mapping? Or is that going against what people would normal think of when they hear that term?

I hope that makes sense! Thank you,

Pete

-
Are you trying to say that the output is directly proportional to the input? – John Douma Apr 24 '13 at 3:39

It's a parameterized family of linear maps. That is, it's a linear map in $| \psi_i \rangle$ but a possibly nonlinear map in whatever other variables are around. (I don't know what you mean by "uniform steps.")
By uniform steps I mean that I choose points that are equally spaced in the space that contains the variables $\phi_{1,2,3,...,x}$. I guess I do mean directly proportional - for the corresponding points in the final state space to also be spaced equally each basis could only be times by a constant. I think my confusion is because I'm not sure how to picture the problem. I'm interested in the behaviour of the final state as I change the variables. Is it okay to think of a point in the space containing the variables mapping to a point in the space containing the final state? – Pete Apr 24 '13 at 8:42