0
$\begingroup$

This question is an exact duplicate of:

How does operator matrix transform under change of basis?

If $\rvert \beta\rangle$ and $\rvert \alpha \rangle$ are two bases related by transformation

$ \rvert\beta_m\rangle = \sum_n S_{mn} \rvert\alpha_n\rangle$

and $O^\beta$ and $O^\alpha$ are the representations of the operator in the two bases, how are $O^\beta$ and $O^\alpha$ related to each other?

This is my attempt at solution.

We know

$ \rvert\beta_m\rangle = \sum_n S_{mn} \rvert\alpha_n\rangle$

$ \langle\beta_m\rvert = \sum_n \langle\alpha_n\rvert S_{nm}^*$

Then

$O^\beta_{ij}=\langle \beta_i \rvert O \rvert \beta_j\rangle$

$O^\beta_{ij}=\sum_k \sum_l S^*_{ki}\langle \alpha_k \rvert O \rvert \alpha_l\rangle S_{jl}$

$O^\beta_{ij}=\sum_k \sum_l S^*_{ki} O^\alpha_{kl} S_{jl}$

I am not sure how to proceed from here.

$\endgroup$

marked as duplicate by Brian Moths, ACuriousMind, Kyle Kanos, hft, John Rennie Sep 24 '15 at 7:19

This question was marked as an exact duplicate of an existing question.

  • 6
    $\begingroup$ How is this any different from your previous question physics.stackexchange.com/q/208578? If you're not satisfied with the answer, please clarify the question rather than posting a new (almost identical!) one. $\endgroup$ – Norbert Schuch Sep 23 '15 at 16:34
  • $\begingroup$ @NorbertSchuch I need to show that density matrix transforms as an operator matrix. I am unable to see the equivalence of transformation in the two cases. $\endgroup$ – Yogesh Yadav Sep 23 '15 at 17:50
  • $\begingroup$ You are asking exactly the same question: Given an operator $O$ which I expand in two bases, how are the corresponding representations related. $\endgroup$ – Norbert Schuch Sep 23 '15 at 17:51
  • $\begingroup$ In the other question, I want to show that the $\rho^\alpha_{jj^{'}}=\sum_i c_{ij^{'}}^* c_{ij}$ (as given in the other question) form of the density matrix transforms as any operator matrix. $\endgroup$ – Yogesh Yadav Sep 23 '15 at 17:53
  • $\begingroup$ That's not whay your question says. This is only what you do in your attempt to explain what you tried to solve the question. You should edit your first question if what you want to know is in fact different. $\endgroup$ – Norbert Schuch Sep 23 '15 at 17:55
0
$\begingroup$

Actually you're done: you expressed the operator in $\beta$-basis representation w.r.t. the operator in the $\alpha$-basis representation. Isn't that what you would like to achieve?

Another formulation of what you've done is that under a state (unitary) transformation

$$U: | \Psi \rangle \rightarrow |\Psi'\rangle = U |\Psi \rangle$$ the operators transform under a similarity trasnformation

$$U: O \rightarrow U' = U^* O U$$

which is exactly the form of your result.

$\endgroup$

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