If $U$ is an unitary operator written as the bra ket of two complete basis vectors i.e

$U=\sum_{k}\left|b^{(k)}\right\rangle\left\langle a^{(k)}\right|$


$U^\dagger=\sum_{k}\left|a^{(k)}\right\rangle\left\langle b^{(k)}\right|$

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 :

$\begin{aligned} U^{\dagger}\left|a^{1}\right\rangle &=\sum_{k}\left|a^{k}\right\rangle\left\langle b^{k} \mid a^{1}\right\rangle \\ &=\sum_{k}\left(\left\langle b^{k} \mid a^{1}\right\rangle\right) \cdot\left|a^{k}\right\rangle \end{aligned}$

  • $\begingroup$ Can downvoters please comment what should be improved about the question? $\endgroup$ Aug 17, 2022 at 10:36
  • $\begingroup$ I didn't downvote but $\left < b^k | a_1 \right > \neq b^k \left | a_1 \right >$. $\endgroup$ Aug 17, 2022 at 10:44
  • $\begingroup$ I also didn't downvote, but you are using a multitude of notations which makes the question very difficult to read. For example, Is $\left|a_1\right>=\left|a^{(1)}\right>$? $\left|b^k\right>=\left|b^{(k)}\right>$? Why do you change mid-question? What is $\left|a'\right>$? $\left|b'\right>$? $\endgroup$ Aug 17, 2022 at 11:14
  • $\begingroup$ @ConnorBehan, where did i write that? $\endgroup$
    – Kashmiri
    Aug 17, 2022 at 12:22
  • $\begingroup$ @ɪdɪətstrəʊlə, edited it. $\endgroup$
    – Kashmiri
    Aug 17, 2022 at 12:26

2 Answers 2


$|a^1\rangle =\sum_{j}(\langle b^{j}|a^1\rangle) |b^{j}\rangle=\sum_{j}(\langle a^j|U^{\dagger}|a^{1}\rangle) |b^j\rangle$


I think that at the end you are looking for something different. Following the sakurai page, it states that $U\vec{a} =\vec{b}$ without the transpose symbol. In fact if you apply only U you obtain the correct matrix. the U trasposed is useful to obtain the "old" vectors $\vec{a_{k}}$ from the new basis formed by $\vec{b_{k}}$ vector


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