- #1
bjnartowt
- 284
- 3
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’,
[tex]\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 } [/tex]
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[tex]{c_{a'}}({t_0}) = \left\langle {a'|\alpha ,{t_0}} \right\rangle [/tex], 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,
[tex]\left| {\alpha ,t} \right\rangle = \sum\nolimits_{a'} {{c_{a'}}(t) \cdot \left| {a'} \right\rangle } [/tex]
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?