- #1
Kashmir
- 465
- 74
If ##U## is an unitary operator written as the bra ket of two complete basis vectors :##U=\sum_{k}\left|b^{(k)}\right\rangle\left\langle a^{(k)}\right|##
##U^\dagger=\sum_{k}\left|a^{(k)}\right\rangle\left\langle b^{(k)}\right|##
And we've a general vector ##|\alpha\rangle## such that ##|\alpha\rangle=\sum_{a^{\prime}}\left|a^{\prime}\right\rangle\left\langle a^{\prime} \mid \alpha\right\rangle##
Sakurai writes at pg 50 :
"how can we obtain ##\left\langle b^{\prime} \mid \alpha\right\rangle##, the expansion coefficients in the new basis? answer is very simple: Just multiply (1.5.9) by ##\left\langle b^{(k)}\right|##
##
\left\langle b^{(k)} \mid \alpha\right\rangle=\sum_{l}\left\langle b^{(k)} \mid a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle=\sum_{l}\left\langle a^{(k)}\left|U^{\dagger}\right| a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle .
##
##(1.5 .1##
In matrix notation, (1.5.10) states that the column matrix for ##|\alpha\rangle## in the new basis can be obtained just by applying the square matrix ##U^{\dagger}## to the colum matrix in the old basis:
##\quad(\mathrm{New})=\left(U^{\dagger}\right)(##old ##)##"So if the matrix representing ##U^\dagger## is applied on to the matrix representing ##|\alpha\rangle## ,it gives the vectors representation in the new basis. But when I apply ##U^\dagger## onto say an basis vector ##\left|a_{1}\right\rangle## ,it doesn't give me the vectors representation in new basis as shown below :
##\begin{aligned} U^{\dagger}\left|a_{1}\right\rangle &=\sum_{k}\left|a^{k}\right\rangle\left\langle b^{k} \mid a_{1}\right\rangle \\ &=\sum_{k}\left(\left\langle b^{k} \mid a_{1}\right\rangle\right) \cdot\left|a^{k}\right\rangle \end{aligned}##
##U^\dagger=\sum_{k}\left|a^{(k)}\right\rangle\left\langle b^{(k)}\right|##
And we've a general vector ##|\alpha\rangle## such that ##|\alpha\rangle=\sum_{a^{\prime}}\left|a^{\prime}\right\rangle\left\langle a^{\prime} \mid \alpha\right\rangle##
Sakurai writes at pg 50 :
"how can we obtain ##\left\langle b^{\prime} \mid \alpha\right\rangle##, the expansion coefficients in the new basis? answer is very simple: Just multiply (1.5.9) by ##\left\langle b^{(k)}\right|##
##
\left\langle b^{(k)} \mid \alpha\right\rangle=\sum_{l}\left\langle b^{(k)} \mid a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle=\sum_{l}\left\langle a^{(k)}\left|U^{\dagger}\right| a^{(l)}\right\rangle\left\langle a^{(l)} \mid \alpha\right\rangle .
##
##(1.5 .1##
In matrix notation, (1.5.10) states that the column matrix for ##|\alpha\rangle## in the new basis can be obtained just by applying the square matrix ##U^{\dagger}## to the colum matrix in the old basis:
##\quad(\mathrm{New})=\left(U^{\dagger}\right)(##old ##)##"So if the matrix representing ##U^\dagger## is applied on to the matrix representing ##|\alpha\rangle## ,it gives the vectors representation in the new basis. But when I apply ##U^\dagger## onto say an basis vector ##\left|a_{1}\right\rangle## ,it doesn't give me the vectors representation in new basis as shown below :
##\begin{aligned} U^{\dagger}\left|a_{1}\right\rangle &=\sum_{k}\left|a^{k}\right\rangle\left\langle b^{k} \mid a_{1}\right\rangle \\ &=\sum_{k}\left(\left\langle b^{k} \mid a_{1}\right\rangle\right) \cdot\left|a^{k}\right\rangle \end{aligned}##