# Unitary operators preserve normalization in arbitrary basis

## Homework Statement

To test my knowledge of Sakurai, I asked myself to: "Prove that an operator being unitary is independent of basis."

## The Attempt at a Solution

I want to show the expansion coefficients’ squared magnitudes sum to unity at time “t”, given that they do at time t = t0.
Consider expanding an arbitrary –ket at t0, in the arbitrary basis a’,

$$\left| {\alpha ,{t_0}} \right\rangle = {\bf{I}}\left| {\alpha ,{t_0}} \right\rangle = \left( {\sum\nolimits_{a'} {\left| {a'} \right\rangle \left\langle {a'} \right|} } \right)\left| {\alpha ,{t_0}} \right\rangle = \sum\nolimits_{a'} {\left| {a'} \right\rangle \left\langle {a'|\alpha ,{t_0}} \right\rangle } = \sum\nolimits_{a'} {{c_{a'}}({t_0}) \cdot \left| {a'} \right\rangle }$$

Note that the work directly above is a ket that has not yet time-evolved, so the expansion using the identity-operator and the notation of the expansion coefficients$${c_{a'}}({t_0}) = \left\langle {a'|\alpha ,{t_0}} \right\rangle$$, is still our “business-as-usual” expansion from Chapter 1.

Now, consider a ket, whatever that ket may be, at some later time “t”. The time “t” is just a label, so we can use the same procedure that we used in [I.8] to construct the time-evolved ket,

$$\left| {\alpha ,t} \right\rangle = \sum\nolimits_{a'} {{c_{a'}}(t) \cdot \left| {a'} \right\rangle }$$

So we see that in [I.8] and [I.9], we have two sets of coefficients, and , respectively. I’m not sure how these two sets of coefficients are related to one another....well, I know they sum to unity in both bases. Can't find the expression for how coefficients change in a basis!

What's the expression for the coefficients in an expansion before and after a basis-switch?

## Answers and Replies

I am not sure if that is what you want, but:

Write

$$|\alpha,t\rangle=U(t-t_0)|\alpha,t_0\rangle$$

Then you can calculate $$c_{a''}(t)$$ in terms of $$c_{a'}(t_0)$$

and matrix elements

$$\langle a''|U(t-t_0)|a'\rangle$$