Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Are you trying to say that the output is directly proportional to the input? –  user69810 Apr 24 '13 at 3:39

1 Answer 1

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.")

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.